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		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810909</id>
		<title>01572192 Rxn Dynamics</title>
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		<updated>2020-05-22T18:17:27Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below in Figure 16 where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased while the vibrational energy is decreased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of the reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 18. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 19, the momentum of the H atom is reduced and the translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810900</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810900"/>
		<updated>2020-05-22T18:15:11Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below in Figure 16 where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased while the vibrational energy is decreased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 18. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 19, the momentum of the H atom is reduced and the translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810866</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810866"/>
		<updated>2020-05-22T18:05:19Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below in Figure 16 where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased while the vibrational energy is decreased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 18. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 19, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810856</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810856"/>
		<updated>2020-05-22T18:02:53Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below in Figure 16 where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 18. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 19, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810828</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810828"/>
		<updated>2020-05-22T17:54:36Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below in Figure 16 where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810804</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810804"/>
		<updated>2020-05-22T17:49:01Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Distance vs Time plot for the estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810715</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810715"/>
		<updated>2020-05-22T17:26:10Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810706</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810706"/>
		<updated>2020-05-22T17:23:42Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Activation Energy */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactant or product state. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810687</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810687"/>
		<updated>2020-05-22T17:20:07Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* PES inspection and location of transition state */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy level, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involves the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810670</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810670"/>
		<updated>2020-05-22T17:17:18Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition State Theory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant as compared to the recrossing of the barrier to revert back to the reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810662</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810662"/>
		<updated>2020-05-22T17:15:37Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reactive and unreactive trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to the reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810650</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810650"/>
		<updated>2020-05-22T17:13:24Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction path and dynamic trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes rather than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momentum values need to change in the reverse process, causing the plot to be reflected about both axes.&lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810648</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810648"/>
		<updated>2020-05-22T17:12:34Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition state */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.&lt;br /&gt;
&lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810643</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810643"/>
		<updated>2020-05-22T17:11:28Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momentum vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momentum vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated and the same final momenta values are used but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but with one being the reverse of the other. In one case, the atom is moving away from the molecule and in the other case the atom is approaching the molecule. However, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810626</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810626"/>
		<updated>2020-05-22T17:03:34Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction path and dynamic trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momentum vs Time&amp;quot; plots remained the same. However, BC and AB distances and momenta are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations which can be observed in both the distance vs time and the momentum vs time plot. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810610</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810610"/>
		<updated>2020-05-22T17:00:09Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction path and dynamic trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the system forms molecule BC and atom A is displaced. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810594</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810594"/>
		<updated>2020-05-22T16:56:32Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction path and dynamic trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs as seen in Figures 5-8. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810575</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810575"/>
		<updated>2020-05-22T16:53:18Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition state */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the simulation, values of approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the forces indicate a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;, one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from a local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and have positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810546</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810546"/>
		<updated>2020-05-22T16:47:17Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition state */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0 for a saddle point, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810534</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810534"/>
		<updated>2020-05-22T16:40:42Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|300px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|300px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810532</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810532"/>
		<updated>2020-05-22T16:38:58Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 19. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which state that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810526</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810526"/>
		<updated>2020-05-22T16:32:42Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810519</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810519"/>
		<updated>2020-05-22T16:30:42Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy do not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810511</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810511"/>
		<updated>2020-05-22T16:27:48Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the exit channel, showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption peak due to the transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the vibrational ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810506</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810506"/>
		<updated>2020-05-22T16:23:42Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Activation Energy */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore there are errors associated with the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810487</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810487"/>
		<updated>2020-05-22T16:18:50Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Introduction */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810484</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810484"/>
		<updated>2020-05-22T16:18:01Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;br /&gt;
&lt;br /&gt;
4. J. C. Polanyi, &#039;&#039;Science, &#039;&#039;1987, &#039;&#039;&#039;236&#039;&#039;&#039;, 680-690.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810476</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810476"/>
		<updated>2020-05-22T16:15:02Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, in &#039;&#039;Chemical Kinetics and Dynamics, &#039;&#039;Prentice-Hall, 1998, ch. 10, pp. 289-290. &lt;br /&gt;
&lt;br /&gt;
2. K. J. Laidler, in &#039;&#039;Chemical Kinetics&#039;&#039;, Harper-Collins, 3&amp;lt;sup&amp;gt;rd&amp;lt;/sup&amp;gt; ed., 1987, ch. 12, pp. 460-466. &lt;br /&gt;
&lt;br /&gt;
3. I. N. Levine, &#039;&#039;Physical Chemistry&#039;&#039;, McGraw-Hill, 6&amp;lt;sup&amp;gt;th&amp;lt;/sup&amp;gt; ed., 2009, ch. 22. pp. 880-892.&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810407</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810407"/>
		<updated>2020-05-22T15:50:27Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules which says that in an exothermic reaction, translational energy is more effective in crossing the barrier than vibrational energy, and vice versa for an endothermic reaction. The reason for this observation is that in an exothermic reaction with early transition state, high vibrational energy only allows the molecule to vibrate from side to side but does not provide enough energy to allow the molecule to cross over the barrier. On the other hand, in an endothermic reaction with late transition state, vibrationally excited molecules can cross the barrier while molecules with high translational energy will simply hit the repulsive inner wall of the potential surface and bounce back to the entrance channel.   &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810305</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810305"/>
		<updated>2020-05-22T15:25:18Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* H + H2 system */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. By obtaining 2 values for ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; , one positive and one negative, it indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test discussed above and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.90 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second vibrational excited state. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule is measured as it decays back to the ground state.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810162</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810162"/>
		<updated>2020-05-22T14:29:31Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition State Theory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur leading to the formation of products and the lack of account for such phenomenon in the classical model underestimates the reaction rate. Nonetheless, the effect of quantum tunnelling is less significant compared to recrossing of the barrier to revert back to reactants. Therefore, the transition state theory is more likely to overestimate the reaction rate constant.      &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state, also known as overtones. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule as it decays back to the ground state is measured.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810144</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810144"/>
		<updated>2020-05-22T14:23:36Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reactive and unreactive trajectory */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state, also known as overtones. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule as it decays back to the ground state is measured.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810138</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810138"/>
		<updated>2020-05-22T14:22:05Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Transition state */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state, also known as overtones. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule as it decays back to the ground state is measured.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810111</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=810111"/>
		<updated>2020-05-22T14:10:30Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into vibrational kinetic energy of the product molecule HF. This can be seen from Figure 14 whereby at small AB distance, there is large change in AB distance over time as the system moves down the product channel, thus showing a high vibrational energy in the product molecule HF. Experimentally, the formation of higher vibrational energy HF molecule can be probed using IR spectroscopy. Two IR absorptions can be observed in vibrationally excited molecules of HF as a result of the transition from the ground state to the first vibrational excited state, as well as the transition from the first excited state to the second excited state, also known as overtones. In contrast, ground state HF molecules will only produce 1 absorption due to transition from the ground state to the first excited state. Another possible way of detecting the vibrationally excited HF is with the use of Infrared Chemiluminescence. In this technique, infrared emission from the excited HF molecule as it decays back to the ground state is measured.       &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809874</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809874"/>
		<updated>2020-05-22T11:58:59Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 19: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809872</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809872"/>
		<updated>2020-05-22T11:58:19Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equally to the efficiency of reaction. Furthermore, it is worth noting that even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809860</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809860"/>
		<updated>2020-05-22T11:54:08Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
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=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. This suggests that the translational energy and vibrational energy does not contribute equal to the efficiency of reaction. Furthermore, it is worth noting even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809512</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809512"/>
		<updated>2020-05-22T09:03:22Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction. Furthermore, it is worth noting even at the limits of the range, not all values close to the limit give rise to a reactive trajectory as barrier recrossing might occur. For example, the case of p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 6.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;is shown below where an unreactive trajectory is obtained due to barrier recrossing.        &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809501</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809501"/>
		<updated>2020-05-22T08:50:59Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; despite the system having more energy than the activation energy needed for the reaction.       &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809495</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809495"/>
		<updated>2020-05-22T08:41:49Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]] &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.       &lt;br /&gt;
&lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809490</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809490"/>
		<updated>2020-05-22T08:40:31Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
The initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1. &amp;lt;/sup&amp;gt;When p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is only reactive near the limits of the range around -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) and 5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.       &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809470</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809470"/>
		<updated>2020-05-22T08:21:38Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F + H2 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.      &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;      &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809454</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809454"/>
		<updated>2020-05-22T08:10:13Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state and vibrational energy has a larger influence on the efficiency of reaction. The initial conditions for FH + H is set to be r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 110 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 240 pm, p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = 4.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = -8.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. As p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is decreased from -8.0 to -2.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;while p&amp;lt;sub&amp;gt;HF &amp;lt;/sub&amp;gt;is increased from 4.0 to 11.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, the momentum of the incoming H atom is reduced and translational energy of the system is lowered, while the vibrational energy of H-F is increased. As a result, a reactive trajectory is obtained as shown in Figure 18. Hence, for a late transition state, the vibrational energy has a larger contribution to the efficiency of the reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809439</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809439"/>
		<updated>2020-05-22T07:59:07Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]&lt;br /&gt;
At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
[[File:110,4_240,-8.png|thumb|none|400px|Figure 18: Unreactive trajectory for F-H + H with high translational energy and low vibrational energy ]]&lt;br /&gt;
[[File:110,11_240,-2.png|thumb|none|400px|Figure 18: Reactive trajectory for F-H + H with high vibrational energy and low translational energy ]]&lt;br /&gt;
&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state. In this case, as the momentum of the incoming H atom (p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt;) is decreased, lowering the translational energy of the system, while the vibrational energy of H-F is increased, a reactive trajectory is obtained. Hence, for a late transition state, the vibrational energy has a larger influence on the efficiency of reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:110,4_240,-8.png&amp;diff=809438</id>
		<title>File:110,4 240,-8.png</title>
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		<updated>2020-05-22T07:55:32Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
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	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:110,11_240,-2.png&amp;diff=809436</id>
		<title>File:110,11 240,-2.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:110,11_240,-2.png&amp;diff=809436"/>
		<updated>2020-05-22T07:55:16Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: &lt;/p&gt;
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	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809416</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809416"/>
		<updated>2020-05-22T07:31:56Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics F-H + H */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
For an early transition state in the case of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, increase in energy of the translational motion increases the efficiency of reaction. Between Figure 16 and 17, the momentum p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;is significantly reduced, indicating a reduction in the vibrational energy of the H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule whereas the magnitude of p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;is increased, showing an increase in translational energy when F approaches the H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule. A reactive trajectory is obtained in Figure 17 when the translational energy is increased, illustrating that the translational energy is more important for an early transition state reaction. On the other hand, in the reaction of F-H with H, it involves a late transition state. In this case, as the momentum of the incoming H atom (p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt;) is decreased, lowering the translational energy of the system, while the vibrational energy of H-F is increased, a reactive trajectory is obtained. Hence, for a late transition state, the vibrational energy has a larger influence on the efficiency of reaction. This observation aligns with Polanyi&#039;s empirical rules.  &lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809364</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809364"/>
		<updated>2020-05-22T06:34:24Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics F-H + H===&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809361</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809361"/>
		<updated>2020-05-22T06:33:01Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics ===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1, -6.0 and -5.8 but not -5.9 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15). On the other end of the range, the trajectory is reactive at 5.9 but not 6.0 and 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809358</id>
		<title>01572192 Rxn Dynamics</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=01572192_Rxn_Dynamics&amp;diff=809358"/>
		<updated>2020-05-22T06:20:49Z</updated>

		<summary type="html">&lt;p&gt;Jz14418: /* Reaction Dynamics */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Introduction ==&lt;br /&gt;
&lt;br /&gt;
In this simulation, the reaction between a H atom and a H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule is being studied. The H atom approaches H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;in a collinear manner and result in the formation of a new H-H bond while displacing a H atom from the initial molecule.   &lt;br /&gt;
&lt;br /&gt;
== H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==            &lt;br /&gt;
&lt;br /&gt;
=== Transition state ===&lt;br /&gt;
[[File:Screenshot 2020-05-20 at 6.57.44 PM.png|thumb|none|400px|Figure 1: Identification of transition state]]&lt;br /&gt;
&lt;br /&gt;
The transition state is mathematically defined as the saddle point on the potential energy surface diagram as the transition state lies at an energy maximum along the minimum energy pathway from the reactant to the product. The saddle point is a stationary point and its first partial derivatives are 0. Also, in the second partial derivative test, H is smaller than 0, where H = f&amp;lt;sub&amp;gt;xx&amp;lt;/sub&amp;gt;(x,y) f&amp;lt;sub&amp;gt;yy&amp;lt;/sub&amp;gt;(x,y) - [f&amp;lt;sub&amp;gt;xy&amp;lt;/sub&amp;gt;(x,y)]&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt;. In other words, the second partial derivative is positive with respect to one direction and negative with respect to another. &lt;br /&gt;
&lt;br /&gt;
Using the programme, value of the forces at approximately 0 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; pm&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;indicates a stationary point and by having a positive and a negative value for ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;as part of the Hessian eigenvalue, the transition state can be identified. ω&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; is a force constant and obtaining 2 values, one positive and one negative, indicates that a particular point is a maximum with respect to one direction and minimum with respect to another, therefore confirming that it is a saddle point, and thus corresponds to the transition state. One the other hand, in a more simplistic 2D picture, the transition state is a maximum point and can be identified using its second derivative which is smaller than 0. For a local minimum, the ω&amp;lt;sup&amp;gt;2 &amp;lt;/sup&amp;gt;values will show 2 positive values as the curvature from the local minimum is positive in all directions. Mathematically, a local minimum has H &amp;gt; 0 in the second partial derivative test and has positive second partial derivative values. &lt;br /&gt;
&lt;br /&gt;
[[File:Dist_vs_time2.png|thumb|none|400px|Figure 2: Estimation of transition state position]]&lt;br /&gt;
&lt;br /&gt;
At the transition state, the 3 atoms are equally apart and AB=BC. The best estimate of transition state position r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; is found to be 90.8 pm. As shown in Figure 2, the internuclear distance AB and BC are equal and remains constant over 50 fs. This shows that the trajectory is on the ridge and thus a good estimate of the transition state is obtained. On the other hand, if the system is not at the transition state, the &amp;quot;Distance vs Time&amp;quot; plot will show that the system undergoes an oscillatory motion with the AB and BC distances changing over time as the system oscillates about the position of the transition state along the diagonal of the contour plot shown in Figure 1. If the system is displaced from the transition state, it will roll downhill to an energy minimum given by either the reactants or the products. In that case, there will be an increase in the AC distance over time as the atom that is formed moves away from the molecule.        &lt;br /&gt;
                                                                                                                &lt;br /&gt;
=== Reaction path and dynamic trajectory ===&lt;br /&gt;
[[File:MEP_traj_new.png|thumb|none|400px|Figure 3: MEP trajectory]]&lt;br /&gt;
[[File:dynamic_traj_new.png|thumb|none|400px|Figure 4: Dynamic trajectory]]&lt;br /&gt;
&lt;br /&gt;
Figure 3 and 4 above illustrate the reaction path (minimum energy path, mep) and the trajectory from dynamic calculations respectively when the system is slightly displaced from the transition state. BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;is used. In the mep, the trajectory follows the valley floor and no oscillation of the molecule AB is observed. This is because the momenta in each step is being set to 0. For the dynamic calculation, the trajectory shows an oscillatory behaviour as the distance AB varies along the trajectory. &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rt+1_dist_time.png|thumb|none|400px|Figure 5: Distance vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts+1_momenta.png|thumb|none|400px|Figure 6: Momenta vs Time plot when BC= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]                                                                                                                    &lt;br /&gt;
&lt;br /&gt;
[[File:r1=rts_dist_time.png|thumb|none|400px|Figure 7: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm]]&lt;br /&gt;
[[File:r1=rts_momenta.png|thumb|none|400px|Figure 8: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm ]]   &lt;br /&gt;
&lt;br /&gt;
When the system is displaced in the opposite direction, where BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; and AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, the shape of both the &amp;quot;Internuclear distance vs Time&amp;quot; and the &amp;quot;Momenta vs Time&amp;quot; plots remained the same. However, BC and AB distances are now represented by different lines on each of the graphs. Initially, when BC is displaced from r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; (BC= r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; +1 pm), the system forms molecule AB and atom C as indicated by the increase in BC distance over time while that of AB shows an oscillatory motion due to molecular vibrations. As the calculation changes to AB = r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm, AB is now being displaced and the system forms molecule BC and atom A. As a result, distance AB increases with time while distance BC shows oscillatory motion. These 2 situations represent the reactant and the product state of the system and illustrates the move towards an energy minimum when the system is displaced from its transition state.                                                                                                                                                                                                                                                                                                                                                                                &lt;br /&gt;
&lt;br /&gt;
[[File:last_position_first.png|thumb|none|400px|Figure 9: Distance vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]&lt;br /&gt;
[[File:Last_momenta_first.png|thumb|none|400px|Figure 10: Momenta vs Time plot when AB= r&amp;lt;sub&amp;gt;ts &amp;lt;/sub&amp;gt;+ 1 pm for the approach to transition state]]  &lt;br /&gt;
&lt;br /&gt;
When initial positions of a calculation correspond to the final positions of a trajectory previously calculated, using the same final momenta values but with their signs reversed, it was found that the new &amp;quot;Internuclear distance vs Time&amp;quot; plot generated is a mirror image of the previous plot about the internuclear distance axis. This is because both calculations are fundamentally based on the same process but one being the reverse of the other. In one case, the atom is moving away from the molecule and the other being the approach of the atom towards the molecule. On the other hand, the new &amp;quot;Momenta vs Time&amp;quot; plot is a reflection in both the momentum and time axes just than just a mirror image. This is because, unlike distance, momentum is a vector quantity with direction, thus the signs of the momenta values need to change in the reverse process, causing the plot to be reflected about both axes.       &lt;br /&gt;
&lt;br /&gt;
=== Reactive and unreactive trajectory ===&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; / kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;  !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280&lt;br /&gt;
|| Yes&lt;br /&gt;
|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table_1_traj_new.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.007&lt;br /&gt;
||No||Atom A approaches the molecule BC but does not have sufficient energy to overcome the activation barrier and returns back to the reactant state. ||[[File:Table 2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977&lt;br /&gt;
||Yes|| Atom A approaches the molecule BC, passing through the transition state and forms the product AB, displacing atom C. &lt;br /&gt;
||[[File:Table3_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.227&lt;br /&gt;
|| No&lt;br /&gt;
|| Atom A approaches the molecule BC, crossing over the barrier and forms the molecule AB. However, after the formation of AB, the system reverts back to reactants made up of molecule BC. This is because AB has high vibrational energy which destabilises the product molecule. As a result, the system recrosses the barrier, leading to the reformation of BC. Overall, there is no reaction. &lt;br /&gt;
||[[File:Table4_2_traj.png|thumb|none|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 ||-349.477|| Yes&lt;br /&gt;
|| Atom A approaches molecule BC, crossing over the barrier and forms molecule AB. In this case, AB formed has lower vibrational energy, thus the system does not have sufficient energy to recross the barrier. Hence, BC molecule remains intact and C is displaced, leading to the product state.  &lt;br /&gt;
||[[File:Table5_traj.png|thumb|none|200px]]&lt;br /&gt;
|}&lt;br /&gt;
From the table, it can be concluded that whether a reaction occurs or not does not depend on the total energy of the system but how the energy of the system is being distributed. For example, comparing the last 2 examples in the table, despite the last example having a greater total energy, the energy is distributed differently than that of the penultimate example, hence the system was not able to recross the barrier, therefore leading to the formation of products.    &lt;br /&gt;
&lt;br /&gt;
=== Transition State Theory ===&lt;br /&gt;
As the transition state theory (TST) is based on a number of assumptions, the assumptions will influence the prediction of the reaction rate values, causing them to differ from experimental values. One assumption of the Transition State Theory is that systems that have crossed the transition state in the direction of products cannot reform reactants. However, in the table above, it was observed that barrier recrossing is possible. Thus, based on this assumption, the transition state theory overestimates the reaction rate. On the other hand, the transition state theory is based on classical mechanics and does not take into account quantum effects. For example, quantum tunnelling could occur and the lack of account for such phenomenon in the classical model underestimates the reaction rate.    &lt;br /&gt;
&lt;br /&gt;
== F-H-H system ==&lt;br /&gt;
=== PES inspection and location of transition state===&lt;br /&gt;
[[File:F-H-H_TS.png|thumb|none|400px|Figure 11: Potential Energy Surface Plot]]  &lt;br /&gt;
In this F-H-H system, A is F while B and C are H. From Figure 11, F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; reaction is exothermic while H + FH is endothermic. At large distance of AB, it represents the state of a F atom and a H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;molecule, while at a large distance of BC, it represents FH and H atom. As the F +  H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;state is at a higher energy, a reaction between F and H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;will lead to a lower energy product and therefore the reaction is exothermic. On the other hand, the reaction of H + FH involving the formation of higher energy products, hence the reaction would be endothermic.  &lt;br /&gt;
&lt;br /&gt;
H-F bond is stronger than H-H bond. When F reacts with H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, more energy is released from the formation of the H-F bond as compared to the energy required to break the H-H bond. There is a net release of energy from the reaction and hence the reaction is exothermic. For H + FH, the opposite process involves a net absorption of energy to break the strong H-F, thus the process is endothermic.   &lt;br /&gt;
&lt;br /&gt;
Using the simulation, it was found that the approximate position of the transition state involves a F-H (AB) distance of 180.6 pm and H-H (BC) distance of 74.5 pm. This was aided by the use of Hammond&#039;s postulate. According to Hammond&#039;s postulate, the transition state resembles the structure of either the reactant or the product depending on whichever is closer in energy to the transition state. In the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;, since the process is exothermic, the reaction involves an early transition state. As such, it can be deduced that the F-H distance in the transition state is larger than that of H-H.     &lt;br /&gt;
&lt;br /&gt;
=== Activation Energy ===&lt;br /&gt;
&lt;br /&gt;
[[File:Ea_F-H+H.png|thumb|none|400px|Figure 12: Energy vs Time plot for the formation of FH + H from the transition state]]  &lt;br /&gt;
[[File:Ea_FH+H.png|thumb|none|400px|Figure 13: Energy vs Time plot for the formation of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; from the transition state]]  &lt;br /&gt;
&lt;br /&gt;
When the system is slightly displaced from the transition state, it will move towards either the reactants or products which are lower in energy. The activation energy can be found by taking the difference between the potential energy of the transition state and the reactants or products. Using the mep, the activation energy is found to be 126.5 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; for the H + FH reaction and 0.85 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;for the F + H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;reaction. As the initial state is set such that the system is slightly displaced from the transition state, the values obtained would differ slightly from the true value of the activation energy. Furthermore, the number of steps used in the simulation affects the potential energy of the reactants and products obtained even though a sufficiently large number of steps is used, therefore leading to errors in the activation energy obtained using this method. &lt;br /&gt;
&lt;br /&gt;
=== Reaction Dynamics ===&lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_trag_FH2.png|thumb|none|400px|Figure 14: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;]]  &lt;br /&gt;
&lt;br /&gt;
An example of a reactive trajectory is shown in Figure 14 for the reaction of F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;. The initial conditions are r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;(AB) = 240 pm, p&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= -3 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, r&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;(BC) = 74 pm, p&amp;lt;sub&amp;gt;HH &amp;lt;/sub&amp;gt;= -1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;. When F reacts with H&amp;lt;sub&amp;gt;2 &amp;lt;/sub&amp;gt;to form F-H and H atom, the potential energy of the system decreases. Given the conservation of energy, the loss in potential energy is converted into kinetic energy. The products now have greater kinetic energy which is expressed in terms of a faster rate of translation. Experimentally, there will be an increase in the temperature of the reacting mixture as temperature is proportional to the average translational energy of the molecules and a calorimeter can be used to measure the heat released during the reaction.  &lt;br /&gt;
&lt;br /&gt;
[[File:Reactive_(-6.1).png|thumb|none|400px|Figure 15: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
[[File:6.1.png|thumb|none|400px|Figure 16: Unreactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; and overall energy is -390.8 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]  &lt;br /&gt;
&lt;br /&gt;
When the initial condition is set at r&amp;lt;sub&amp;gt;FH &amp;lt;/sub&amp;gt;= 240 pm, r&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; = 74 pm, with a momentum p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.0 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, when p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; is varied from -6.1 to 6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, it was found that the trajectory is reactive at -6.1 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; (Figure 15) but not at other values in the range. &lt;br /&gt;
&lt;br /&gt;
[[File:lower_energy.png|thumb|none|400px|Figure 17: Reactive trajectory for F + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; with overall energy -432.9 kJ mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; ]]At the same initial position as previously but with a change in momentum where p&amp;lt;sub&amp;gt;FH&amp;lt;/sub&amp;gt; = -1.6 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;and p&amp;lt;sub&amp;gt;HH&amp;lt;/sub&amp;gt; 0.2 g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;, a reactive trajectory is obtained in Figure 17 even though the overall energy of the system has decreased significantly from -390.8 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 16 to -432.9 kJ mol&amp;lt;sup&amp;gt;-1 &amp;lt;/sup&amp;gt;in Figure 17.&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;/div&gt;</summary>
		<author><name>Jz14418</name></author>
	</entry>
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