https://chemwiki.ch.ic.ac.uk/api.php?action=feedcontributions&feedformat=atom&user=Ng611ChemWiki - User contributions [en]2026-05-16T06:57:29ZUser contributionsMediaWiki 1.43.0https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01502946&diff=812978MRD:015029462020-07-07T17:05:14Z<p>Ng611: </p>
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<div>'''''<u>4/5 - A generally good first section. You provided clear and well reasonsed responses to the questions. Your second section was also generally good but had 1 or 2 areas of misunderstanding that you should try to address. </u>'''''<br />
<br />
== Exercise 1 ==<br />
<br />
=== Question 1) Definition of the TS ===<br />
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''<br />
<br />
The transition state is defined as the maximum located on a minimum energy path. Mathematically, it is a saddle point on the potential energy surface. At TS coordinates, šæ''V''(''x,y'')/šæ''x'' = 0 and šæ''V''(''x,y'')/šæy = 0, where x and y are r<sub>1</sub> and r<sub>2</sub> in this case. Additionally at TS coordinates, the second partial derivative with respect to one of the coordinates is positive and negative with respect to the other coordinate, i.e either šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>''x'' > 0 and šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>''y'' < 0 or šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>''x'' < 0 and šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>y > 0. This fact allows the saddle point to be simply distinguished from a local minimum as at the latter, both šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>''x'' > 0 and šæ<sup>2</sup>''V''(''x,y'')/šæ<sup>2</sup>''y ''> 0.<br />
<br />
'''''<u>A solid definition of the TS and saddle point, well done. </u>'''''<br />
<br />
=== Question 2) H + H<sub>2</sub> TS position ===<br />
'''Report your best estimate of the transition state position''' ('''r<sub>ts</sub>''') '''and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.'''<br />
<br />
If r<sub>1</sub> = r<sub>2</sub> = r<sub>ts</sub> and p<sub>1</sub> = p<sub>2</sub> = 0, the atoms are expected to stay still and not oscillate as the attractive and repulsive forces are perfectly balanced. Therefore, E<sub>kin</sub> of the atoms will be 0, given such initial conditions and E<sub>pot</sub> = const., as illustrated by the figure below. The TS coordinates can also be estimated by the surface plot and the reaction trajectory, which at mentioned initial conditions will converge into a single spot on the E<sub>pot</sub> surface, which is the transition state. Using this method, r<sub>ts</sub> was estimated to be equal to approximately '''90.77''' '''pm''' for the H + H<sub>2</sub> system.<br />
<br />
'''''<u>Sensible estimate for the TS, and well reasoned. Well done!</u>'''''<br />
{|style="margin: 0 auto;"<br />
| [[File:01502946 Ex1 transition state contourplot.png|thumb|320px|center|Location of the TS]]<br />
| [[File:01502946 Ex1 TS distvtime.png|thumb|400px|center|Distances as a function of time]]<br />
|}<br />
<br />
=== Question 3) MEP&dynamic trajectories ===<br />
'''Comment on how theĀ ''mep''Ā and the trajectory you just calculated differ.'''<br />
<br />
For the MEP path, no vibrational motion of the formed H<sub>2</sub> is observed unlike in the case of dynamic mode. Over the course of the reaction, some of the initial potential energy is changed into kinetic energy. When molecular hydrogen forms, some of the kinetic energy is converted into vibrational energy, which is indicated by the oscillations on the PE surface. In the case of MEP, the reaction proceeds via a pathway with smallest possible total energy. In order to minimise the total energy, vibrational energy must be minimised as well. Consequently, no vibrational motion is observed in the MEP mode. As the MEP mode corresponds to infinitely slow motion of the nuclei, no changes in momenta or velocities are observed either. Furthermore, total energy of the system is not conserved in the MEP mode. The total potential energy of products is lower than the potential energy at TS but as the system has no kinetic energy at any point, E<sub>pot</sub> must be equal to E<sub>tot</sub>. This seemingly violates the law of conservation of energy, but that it is just the result of assumptions made in the MEP mode. On the other hand, energy is conserved in the dynamics mode as motion of the atoms is not taken to be infinitely slow.<br />
<br />
'''''<u>good explanation!</u>'''''<br />
<br />
{|style="margin: 0 auto;"<br />
| [[File:01502946 Ex1 valleyfloorDynamics.png|thumb|350px|center|Reaction path by the dynamics mode]]<br />
| [[File:01502946 Ex1 valleyfloorMEP.png|thumb|350px|center|Reaction path by the MEP mode]]<br />
|}<br />
<br />
''Look at the āInternuclear Distances vs Timeā and āMomenta vs Timeā. What would change if we used the initial conditionsĀ '''r<sub>1</sub>'''Ā =Ā '''r<sub>ts</sub>'''Ā andĀ '''r<sub>2</sub>'''Ā =Ā '''r<sub>ts</sub>'''+1Ā pm instead?''<br />
* All three atoms are the same elements and only the definitions of r<sub>1</sub> and r<sub>2</sub> are interchanged, therefore the outcome of the reaction will still be chemically the same. The only difference is that the chemical bond will be formed between the hydrogen in the middle and the proton that previously ended up as a free radical (instead of H<sub>A</sub> + H<sub>B</sub> + H<sub>C</sub> --> H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub>, we get H<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>). On Internuclear Dist. vs Time and Momenta vs Time plots, the lines which previously represented parameters associated with r<sub>A-B</sub> now represent r<sub>B-C</sub> and vice versa (see figures below).<br />
<br />
{|style="margin: 0 auto;"<br />
| [[File:01502946 Ex1 dynamics INdistvstime.png|thumb|350px|center|Internuclear distances vs time]]<br />
| [[File:01502946 Ex1 dynamics momentavstime.png|thumb|350px|center|Momenta vs time]]<br />
|}<br />
<br />
''Take note of the final values of the positionsĀ '''r<sub>1</sub>'''(t)Ā '''r<sub>2</sub>'''(t) andĀ Ā '''p<sub>1</sub>'''(t)Ā '''p<sub>2</sub>'''(t) for your trajectory for large enough t. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?''<br />
* The atoms return from the products to the transition state and stay there as the TS structure described in question 2. Essentially, it is the reverse process of TS structure converting into products.<br />
<br />
=== Question 4) A nice table '''''<u>(I agree, it is a very nice table)</u>''''' ===<br />
'''Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory.'''<br />
<br />
'''''Note:''''' In this case, r<sub>1</sub> and r<sub>2</sub> correspond to r<sub>A-B</sub> and r<sub>B-C</sub>, respectively. The units for p are gĀ·mol<sup>-1</sup>Ā·pmĀ·fs<sup>-1</sup>.<br />
<br />
{| class="wikitable"<br />
!p<sub>1 </sub><br />
!p<sub>2</sub><br />
!E<sub>tot</sub> (kJ/mol)<br />
!Reactive?<br />
!Description of the dynamics<br />
!Illustration of the trajectory<br />
|-<br />
|<nowiki>-2.56</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-414.27</nowiki><br />
|Yes<br />
|H<sub>A</sub>-H<sub>B</sub> has very little vibrational energy. Collision with H<sub>C</sub> occurs with sufficient energy to convert H<sub>A</sub>-H<sub>B</sub> into H<sub>B</sub>-H<sub>C</sub>, which oscillates slightly as the distance between H<sub>A</sub> and the newly formed H<sub>2</sub> increases.<br />
|[[File:01502946 Ex1 table condition1.png|350px|]]<br />
|-<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-4.1</nowiki><br />
|<nowiki>-420.08</nowiki><br />
|No<br />
|The collision of H<sub>C</sub> of H<sub>A</sub>-H<sub>B</sub> is not energetic enough to form H<sub>B</sub>-H<sub>C</sub>. The energy barrier is not crossed and the H<sub>2</sub> molecule continues oscillating as H<sub>C</sub> recedes.<br />
|[[File:01502946 Ex1 table condition2.png|350px|]]<br />
|-<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-413.98</nowiki><br />
|Yes<br />
|H<sub>A</sub>-H<sub>B</sub> has some vibrational energy. The collision occurs with sufficient energy and H<sub>A</sub>-H<sub>B</sub> is converted into H<sub>B</sub>-H<sub>C</sub>. The newly formed dihydrogen oscillates more heavily than the initial molecule. The reaction proceeds via a slightly higher energy pathway than in the first case as the vibrational energy of the starting material is greater.<br />
|[[File:01502946 Ex1 table condition3.png|350px|]]<br />
|-<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.1</nowiki><br />
|<nowiki>-357.27</nowiki><br />
|No<br />
|H<sub>C</sub> collides with H<sub>A-B</sub> with sufficient energy to form H<sub>B-C </sub>but as a result of the product having high vibrational energy, the starting materials are reformed. The vibrational energy is illustrated by the high energy oscillations seen in the figure on the right.<br />
|[[File:01502946 Ex1 table condition4.png|350px|]]<br />
|-<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.6</nowiki><br />
|<nowiki>-349.48</nowiki><br />
|Yes<br />
|H<sub>C</sub> collides with H<sub>A</sub>-H<sub>B</sub> once again with sufficient energy to form H<sub>B</sub>-H<sub>C</sub>, but as r<sub>1</sub> is small enough and H<sub>B</sub>-H<sub>C</sub> undergoes high energy oscillations, H<sub>A</sub>-H<sub>B</sub> is reformed. However, in this case H<sub>C</sub> experiences a high attractive force to H<sub>A</sub>-H<sub>B</sub> and stays relatively close to the molecule. Upon stretching of the H<sub>A</sub>-H<sub>B</sub> bond, H<sub>B</sub>-H<sub>C </sub>is reformed and H<sub>A</sub> departs as a single atom. H<sub>B</sub>-H<sub>C</sub> oscillates heavily, but attractive forces to H<sub>A</sub> decrease continuously due to increasing r<sub>1</sub>. <br />
|[[File:01502946 Ex1 table condition5.png|350px|]]<br />
|}<br />
''What can you conclude from the table?''<br />
* The reactants must be sufficiently high in energy for the reaction to occur, but their energy must be also be suitably distributed between translational and vibrational forms. Otherwise, the reaction can possibly revert back to starting materials even after crossing the TS due to high energy oscillations brought about by the collision, as illustrated by the different initial conditions listed in the table above.'''''<u>Good conclusion. </u>''''' <br />
<br />
=== Question 5) Transition state theory ===<br />
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''<br />
<br />
The differences between TST predictions and experimental rate constant values are the direct result of assumptions made in calculations based on TST.<sup>1</sup><br />
# The motion of nuclei is described only by classical mechanics and any quantum effects are neglected. This has a great influence on simple reactions involving light particles (H, D), as possible quantum tunneling is not taken into account. In reality, light particles can tunnel through narrow low energy barriers and this is especially likely in given simple case. Therefore in the case of reaction H + H<sub>2</sub>, employing TST will result in underestimation of the rate constant. Nevertheless, other influences of TST assumptions should be considered in conjunction with this one and therefore the rate constant might still be overestimated, regardless of quantum mechanical effects. In reactions involving more massive atoms or higher energy barriers, neglecting quantum behaviour of particles and tunneling will likely have no effect on the predicted rate constant as the probability of tunneling decreases exponentially with increasing particle mass. '''''<u>Good! Keep in mind that even for hydrogen and deuterium, tunneling is still a weak effect. This is why-- among other things-- processes like nuclear fusion are so difficult. </u>''''' <br />
# Every collision with sufficient kinetic energy will definitely cross the TS and will go on to form the products. Furthermore, conversion to products is considered to be irreversible: reactants that have reached the transition state will not revert back to starting materials. In reality, the TS can revert back to starting materials as well. As illustrated by the entries in the table above, products are often able to recross the TS and convert back into starting materials, given a suitable distribution of translational and vibrational energies. Additionally, reactions leading back to starting materials can still occur, although this will have a small effect if the forward equilibrium constant is very large. As a consequence, these assumptions of TST would overestimate the rate constant for reforming starting materials decrease the total reaction rate. '''''<u>Good!</u>''''' <br />
# The transition state structure is in quasi-equilibrium with the reactants and the particles follow a Boltzmann distribution of energies. In addition, TS structures are sufficiently long-lived to reach a Boltzmann distribution of energies as well before converting to products. Although the thermal equilibrium is not very important in H + H<sub>2</sub> system, in very fast reactions the nuclei might not reach a Boltzmann distribution of energies before colliding. Furthermore at very high temperatures, higher energy levels are populated, thus breaking the assumption that only low vibrational states are populated. If the average energies of particles are different from what is predicted, the fraction of sufficiently energetic collisions leading to TS cannot be accurately estimated either. This could result in poor estimates for the rate constant as crossing the energy barrier to form TS is rate limiting.<br />
'''''<u>Good explanation of the deficiencies of TST. On balance, taking all of the above into consideration, will TST overestimate or underestimate?</u>'''''<br />
<br />
== Exercise 2 ==<br />
<br />
=== Question 1) Energetics of F + H<sub>2</sub> and H + HF ===<br />
'''By inspecting the potential energy surfaces, classify the F + H<sub>2</sub>Ā and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''<br />
<br />
The reaction between '''F + H<sub>2</sub>''' is exothermic, as the energy of the products (HF + H) is lower than that of the reactants. Conversely, the reaction '''H + HF''' is endothermic, as the products are higher in energy than the reactants. The bond energies of H-F and H-H bonds are 565 kJ/mol and 436 kJ/mol, respectively.<sup>2</sup> As the H-F bond is stronger than the H-H bond, energy is released as heat upon formation of HF from F and H<sub>2</sub>. In the reverse reaction, more energy is needed to put into breaking of the H-F bond, which results in the reaction being endothermic.<br />
<br />
=== Question 2) F + H<sub>2</sub> TS position ===<br />
'''Locate the approximate position of the transition state.'''<br />
<br />
In this case, '''A''' = F, '''B''' = '''C''' = H and therefore, r<sub>1</sub> = r<sub>A-B</sub> and r<sub>2</sub> = r<sub>B-C</sub><br />
<br />
At the transition states, all repulsive and attractive forces are balanced. Consequently, E<sub>tot</sub> = E<sub>pot</sub> = const.<br />
<br />
Such conditions are satisfied at: '''r<sub>1</sub> = 181 pm''' and '''r<sub>2</sub> = 74.5 pm''', given p<sub>1</sub> = p<sub>2</sub> = 0 gĀ·mol<sup>-1</sup>Ā·pmĀ·fs<sup>-1</sup>. This result was obtained through trial and error, but the TS could also be located through decreasing the spacing between isoenergetic contour lines and searching for the saddle point (although in this case even the smallest spacing does not give a very precise answer). The equilibrium bond lengths of H<sub>2</sub> and HF are r<sub>H2</sub> = 74 pm and r<sub>HF</sub> = 91.7 pm, respectively.<sup>2</sup> The large difference between r<sub>1</sub> and r<sub>HF</sub> and small difference between r<sub>2</sub> and r<sub>H2</sub> indicates an early transition state as the TS structure is highly similar to the starting materials. The reaction is exothermic and an early TS is also predicted by the Hammond postulate: the TS structure (and its energy) must be similar to starting materials (and their energy) as the reaction cannot proceed before the energy barrier leading to the TS has been crossed.<br />
<br />
'''''<u>Sensible answer, well done. </u>'''''[[File:01502946 Ex2 FplusH2 TSloc.png|thumb|centre|Location of the TS]]<br />
<br />
=== Question 3) Activation energies ===<br />
'''Report the activation energy for both reactions.'''<br />
<br />
Reaction: F + H<sub>2</sub> ('''E<sub>a</sub> =''' '''0.995 kJ/mol''')<br />
<br />
Reaction: H + HF ('''E<sub>a</sub> =''' '''126.57 kJ/mol''')<br />
<br />
'''''<u>Sensible estimates for Ea in both directions. I'd say that your Ea for F+H2 is ever so slightly too low, but only a little. </u>'''''<br />
<br />
The activation energies were determined with the help of energy vs time plots by the following procedure.<br />
# Firstly, the total energy at an arbitrarily chosen position (r<sub>1</sub> = 74 pm, r<sub>2</sub> = 300 pm) on the "starting materials valley" floor was determined. This corresponded to the total energy of starting materials before the reaction (E<sub>SM</sub>). The r values correspond to a H<sub>2</sub> molecule at equilibrium bond length and a fluorine atom far away from H<sub>2</sub>. Note that p<sub>1</sub> = p<sub>2</sub> = 0 (same in further steps) and that using a larger r<sub>2</sub> value can give a slightly smaller total energy value, but this is a sufficient approximation.<br />
# Secondly, the total energy at TS (E<sub>TS</sub>) was determined (see the previous question).<br />
# Finally the total energy at an arbitrarily chosen position (r<sub>1</sub> = 300 pm, r<sub>2</sub> = 91.7 pm) on the "products valley" floor was determined. This corresponded to the total energy of products after the reaction (E<sub>PR</sub>). The r values correspond to a HF molecule at equilibrium bond length and an hydrogen atom far away from HF.<br />
The difference between E<sub>TS</sub> and E<sub>SM</sub> was then calculated to yield the activation energy of reaction F + H<sub>2</sub>. The activation energy of the reverse reaction was found similarly for E<sub>TS</sub> and E<sub>PR</sub>.<br />
<br />
=== Question 4) A reactive F + H<sub>2</sub> trajectory ===<br />
'''Identify a set of initial conditions that results in a reactive trajectory for the F + H<sub>2</sub>, and look at the āAnimationā and āMomenta vs Timeā.'''Ā '''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''<br />
<br />
Similarly to 2), '''A''' = F, '''B''' = '''C''' = H and therefore r<sub>1</sub> = r<sub>A-B</sub> and r<sub>2</sub> = r<sub>B-C</sub>. The figure below illustrates the outcome of reaction F + H<sub>2</sub> at the following initial conditions: <br />
<br />
'''r<sub>1</sub>''' = 180 pm, '''p<sub>1</sub>''' = -1.4 gĀ·mol<sup>-1</sup>Ā·pmĀ·fs<sup>-1</sup>;<br />
<br />
'''r<sub>2</sub>''' = 74 pm, '''p<sub>2</sub>''' = -0.4 gĀ·mol<sup>-1</sup>Ā·pmĀ·fs<sup>-1</sup>.<br />
<br />
NB: the step number must be set higher than 500 at size 0.1 fs to see the outcome.<br />
<br />
{|style="margin: 0 auto;"<br />
| [[File:01502946 Ex2 reactivepathFplusH2.png|thumb|300px|center|The reactive path]]<br />
| [[File:01502946 Ex2 reactivepathFplusH2 energyvstime.png|thumb|350px|center|Energy as a function of time]]<br />
| [[File:01502946 Ex2 reactivepathFplusH2 momentavstime.png|thumb|300px|center|Momenta of the species as a function of time]]<br />
|}<br />
As the initial momenta are relatively small, the potential energy is approximately equal to the total energy. Initially, there is a small component of kinetic energy, but as the initial momenta are small, its contribution is negligible. As H<sub>2</sub> approaches F, the kinetic energy component increases quickly. When HF is formed and atomic hydrogen departs, a portion of the potential energy is converted into vibrational energy. The potential and kinetic energy components change accordingly as the formed diatomic molecule oscillates. The H-F oscillations are considerably stronger than the initial H-H oscillations. Note that the potential energy does not reach the total energy value as the atomic hydrogen will have kinetic energy as well. As the single proton cannot undergo vibrational motion on its own and no force acts upon it (due to increasing r<sub>2</sub> value), it will move with constant velocity and therefore constant kinetic energy.<br />
<br />
Experimentally, the reaction could be confirmed by femtosecond absorption spectroscopy, in which molecules are excited electronically by a short laser pulse (lasting ca 1 ps) to a dissociative state. After a short interval, the excited molecule is exposed to another pulse at the absorption wavelength of one of the products. From the presence of dissociation product absorption bands and their relative intensities with respect to the excited molecule absorption bands (HF in this case), composition of gas-phase reaction mixtures can be probed.<sup>3 </sup>In addition, vibrational or rotational spectroscopy could be used too as H<sub>2</sub> and HF give distinct bands in the spectra. This is due to the excitation frequencies of H<sub>2</sub> being higher than those of HF as fluorine is ca 19x more massive than hydrogen. It is known from spectroscopy that v<sub>vib</sub> ā ā(k/Ī¼) and v<sub>rot</sub> ā 1/Ī¼, where v<sub>vib</sub> and v<sub>rot</sub> are vibrational and rotational frequencies (Hz), respectively, k is the bond force constant and Ī¼ is the reduced mass of the diatomic species. As Ī¼(H<sub>2</sub>)<Ī¼(HF), both the values of v<sub>vib </sub>and v<sub>rot</sub> are expected to be greater for H<sub>2</sub>. Although k(H<sub>2</sub>)<k(HF), the difference in reduced masses outweighs the difference in force constants and v<sub>vib</sub> will nevertheless be greater for H<sub>2</sub>. '''''<u>Remember that H2 is symmetric though and therefore the transition is likely to be IR inactive. If you were to run this reaction, any IR signatures would be from HF. I think you're a bit confused here about what you're looking for. fs-TA would not really be suitable for this experiment as you've described it (as a photodissociation followed by a probe). Your vibrational spectroscopy idea is better but you've not explained how this could be used to monitor energy release. Think about what happens if a molecule is vibrationally excited and what changes to the IR spectrum would occur.</u>''''' <br />
<br />
NB: The relationship v<sub>rot</sub> ā 1/Ī¼ assumes a rigid rotor model, but experimentally any centrifugal distortions must be taken into account.<br />
<br />
''Setup a calculation starting on the side of the reactants of F + H<sub>2</sub>, at the bottom of the well r<sub>HH</sub>Ā =Ā 74Ā pm, with a momentum p<sub>FH</sub>Ā =Ā -1.0Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, and explore several values of p<sub>HH</sub>Ā in the range -6.1 to 6.1Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>Ā (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.''<br />
* Conversion to products seems to be more likely at low p values (in the range -2 to 2 ''g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>''). Although conversion is likely at values around -6.1 and 6.1 ''g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> ''as well, small changes in p value seem to have a great influence on the outcome of the reaction. On the other hand, at small initial momenta values, small changes (+-0.05 ''g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'') seem to have little effect on the outcome.<br />
<br />
''For the same initial position, increase slightly the momentum p<sub>FH</sub>Ā =Ā -1.6Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, and considerably reduce the overall energy of the system by reducing the momentum p<sub>HH</sub>Ā =Ā 0.2Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. What do you observe now?''<br />
*The products are formed. Increasing p<sub>HH</sub> has little effect on the outcome until values of ca 2 - 1.6''Ā ''g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>.<br />
<br />
=== Question 5) Distribution of energy ===<br />
'''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''<br />
<br />
In endothermic reactions, the TS occurs early in the reactant channel. This means that the TS structure of the starting materials best resembles the TS structure. On the other hand, in exothermic reactions, the TS occurs late in the products channel. This means that the TS structure of the products is more close to the TS structure. According to Polyani's empirical rules, the translational component of reactant energy is more important in crossing an early TS and vibrational energy is more important in overcoming a late TS.<sup>4</sup> <br />
<br />
As illustrated by the exothermic reaction between F and H<sub>2</sub>, products are unlikely to form if the vibrational component of H<sub>2</sub> is high and translational component is low. If H<sub>2</sub> is just busy oscillating and the colliding species do not have sufficient kinetic energy, the early TS is unlikely to be crossed and the H-H bond is not broken. Although formation of the stronger H-F bond is energetically favoured, it cannot happen if the collision itself is not energetic enough. <br />
<br />
Conversely, in the endothermic reaction between H and HF, the vibrational energy of HF has a key role in the success of the reaction. Even if the kinetic energy of H is high but the vibrational energy of HF is low, products are again unlikely to be formed. In order to cross the late TS, the strong HF bond has to have sufficient vibrational energy in order to be broken upon collision with the free H. As shown previously, the fluorine atom attracts the leaving hydrogen quite strongly so unless the HF bond oscillates heavily before the collision, H<sub>2</sub> cannot form as the strong attractive forces of fluorine cannot be overcome even when collisions occur with high kinetic energies.<br />
<br />
As a result, reactions might not proceed even if the system has enough energy to overcome the TS barrier or to irreversibly convert to the products having crossed the TS. The key to successful reactions lies among other factors in the distribution of energy between different modes of molecular motion.<br />
<br />
'''''<u>Good answer. I'd have liked to have seen an example for the late TS case though. </u>'''''<br />
<br />
== References ==<br />
1. Steinfeld J. I., Francisco J. S., Hase W. L. Chemical Kinetics and Dynamics 2nd Edition, pg 315-318. Prentice-Hall.<br />
<br />
2. Atkins P., de Paula J. (2014) Physical Chemistry 10th Edition, pg 985. Oxford University Press.<br />
<br />
3. Atkins P., de Paula J. (2014) Physical Chemistry 10th Edition, pg 898. Oxford University Press.<br />
<br />
4. Steinfeld J. I., Francisco J. S., Hase W. L. Chemical Kinetics and Dynamics 2nd Edition, pg 272-274. Prentice-Hall.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01502997&diff=812913MRD:015029972020-06-25T22:48:32Z<p>Ng611: </p>
<hr />
<div><blockquote>'''''<u>4/5 - Generally a strong report. There are one or two minor points that are commented, but otherwise an excellent piece of work!!</u>'''''</blockquote><br />
<br />
== H + H<sub>2</sub> system ==<br />
===Dynamics from the transition state region ===<br />
====Identification of Transition state====<br />
A transition state is a local maximum on a potential energy surface diagram (see figure1 and figure2 on the right) and can be defined mathematically by Ī“V(ri)/Ī“ri=0. Both the transition state (local maximum on potential energy surface) and the reactants/products (local minimum on potential energy surface) can be represented by when the first derivative of potential energy equals to 0 (Ī“V(ri)/Ī“ri=0). However, the transition state can be distinguished from the reactants or products using the second derivative of potential energy. When the second derivative is less than 0, this represents a local maximum and so the transition state. On the other hand, a second derivative greater than 0 corresponds to a local minimum, therefore the reactants or products. <br />
<br />
'''''<u>Good, what is this kind of mathematical object called?</u>''''' <br />
[[File:Nwc18_Surface_Plot_at_TS.png|thumb|Figure1-Potential Energy Surface Diagram of this reaction. Transition state represented by the black dot.]]<br />
[[File:Nwc18 contour plot at TS.png|thumb|Figure2-Counter plot showing Transition state at the red cross.]]<br />
<br />
====Estimation of transition state position====<br />
An estimated transition state position (r<sub>ts</sub>) for the reaction between H and H<sub>2</sub> will be when all three <br />
hydrogen atoms are 90.775 pm apart '''''<u>Well the farthest hydrogen atoms are 2x90.775 pm apart, but you have the right idea</u>'''''. At this position the force acting in either directions to form or dissociate a hydrogen <br />
hydrogen bond is close to 0. Hence, the distance between the atoms does not change with time and can be seen from figure3 on the <br />
right. <br />
<br />
'''''<u>Good TS estimate</u>''''' <br />
[[File:Nwc18 Internuclear distance agaisnt time plot.png|thumb|Figure3-A Internuclear distance against time plot at transition state postion]]<br />
====MEP and Dynamic calculations of trajectories====<br />
The minimum energy path (MEP) also known as the reaction path is, in this case, the reaction between H<sub>A</sub>-H<sub>B</sub> <br />
and H<sub>C</sub> to form H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> that requires the minimal amount of energy. Calculations <br />
show that this is the collinear approach of the H<sub>C</sub> atom along the H<sub>A</sub>-H<sub>B</sub> bond.<ref name="Atkins" /> <br />
Figure4 and Figure5 on the right shows the change in internuclear distance with time of the MEP after transition state and when <br />
masses of the atoms with there movement in gaseous phase (dynamics) are taken into account respectively. When calculated using MEP <br />
the graph shows smooth lines that corresponds to no vibration in the atoms whilst the dynamic calculation shows an oscillatory <br />
behaviour due to the vibration of the atoms. This atomic vibration is greatest within the atoms forming the new bonds as free atoms <br />
exhibit more translational than vibrational movement.<br />
<br />
The oscillating behaviour is only seen when masses of the atoms are taken into account, such as in the dynamic calculation, as this corresponds to the present of both kinetic and potential energy. Hence, as the atoms in H<sub>B</sub>-H<sub>C</sub> vibrates the energy is constantly inter-converting between kinetic and potential energy. When considering MEP calculations however, the masses of the atoms and so potential energy are not taken into account. Hence, the smooth curves.<br />
<br />
Furthermore, the plateau of the curves at around 194 pm representing H<sub>A</sub> and H<sub>B</sub> (r<sub>1</sub>) distance as well as H<sub>A</sub> and H<sub>C</sub> distance suggests that the atom H<sub>A</sub> and molecule H<sub>B</sub>-H<sub>C</sub> formed after the reaction stoped moving away from each other at 195 pm. This is different to the dynamic calculation where the newly formed products moved away from each other infinitely due to the potential energy being converted into kinetic energy of the products and can be shown by the exponential curves representing r<sub>1</sub>.<br />
<br />
'''''<u>Good!</u>'''''<br />
<br />
====The reverse reaction of reactants to transition state==== <br />
Below shows the internuclear distance against time(left) and momenta against time(right) plot. The top figures represents the reformation of H<sub>A</sub>-H<sub>B</sub> after the transition state was reached and the atom H<sub>C</sub> and molecule H<sub>A</sub>-H<sub>B</sub> moving further apart with time. The bottom figures represents the molecule H<sub>A</sub>-H<sub>B</sub> being approached by atom H<sub>C</sub> and eventually forming the transition state. The two distance against time graph are mirror images of each other and the momenta against time graphs are 180<sup>o</sup> rotations (C<sub>2</sub> symmetry) of each other. <br />
<br />
By putting the bottom and top reaction animations together. A trajectory of H<sub>C</sub> approaching H<sub>A</sub>-H<sub>b</sub>, forming a transition state then reforming the initial reactants which rebound from each other can be seen. This suggests that no reaction has taken place as the transition state exerts momentum towards the reactant. <br />
<br />
Furthermore, as the transition state is a local maximum on the potential energy surface the situations shown represent trajectories going down the valley from the transition state. '''''<u> unsure about your last bit, but everything else is broadly correct. Well done.</u>''''' <br />
<br />
[[File:Nwc18_r2+1_distant_plot.png|300px]] [[File:Nwc18_r2+1_momenta_plot.png|300px]] <br />
[[File:Nwc18_reverse_distance_plot.png|300px]] [[File:Nwc18_reverse_momenta_plot.png|300px]] <br />
[[File:Nwc18_MEP.png|thumb|Figure4-Internuclear distance against time plot calculated using MEP]]<br />
[[File:Nwc18_dynamics.png|thumb|Figure5-Internuclear distance against time plot calculated using Dyamics]]<br />
<br />
===Reactive and unreactive trajectories ===<br />
The reaction trajectory (represented by the black lines on the contour plots below) is a path that can be taken by the atoms involved in a reaction. Different to a reaction path, it does not have to be either of the reactants, products or transition state configuration and is used to explore the potential energy surface of a reaction. <br />
<br />
The reaction path, on the other hand, is a path that connects the reactants, products and transition state. <br />
====Energy and reactivity====<br />
Summary of reactive and unreactive trajectories<br />
{| class="wikitable" border=1<br />
! p<sub>1</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 ||414.280 ||Reactive ||H<sub>C</sub> approached H<sub>A</sub>-H<sub>B</sub> and forms the stable products H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> via the transition state. The products moved away from each other after the reaction. An oscillatory behaviour can be seen as AB distance increases from 75 pm suggesting that H<sub>A</sub>- H<sub>B</sub> bind has broken and the vibration of the release H<sub>A</sub> is causing this oscillation.||[[File:Nwc18_table1.png|200px]]<br />
|-<br />
| -3.1 || -4.1 ||420.077 ||Unreactive ||H<sub>C</sub> approached H<sub>A</sub>-H<sub>B</sub> without enough energy to reach the transition state and so does not result in a reaction but stays as H<sub>C</sub> and H<sub>A</sub>-H<sub>B</sub> which moved away from each other after the approach.||[[File:Nwc18_table2.png|200px]]<br />
|-<br />
| -3.1 || -5.1 ||413.977 ||Reactive ||This is the same as the first case where H<sub>C</sub> approached H<sub>A</sub>-H<sub>B</sub> and forms the stable products H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> via the transition state.The products moved away from each other after the reaction. An oscillatory behaviour can be seen as AB distance increases from 75 pm suggesting that H<sub>A</sub>- H<sub>B</sub> bind has broken and the vibration of the release H<sub>A</sub> is causing this oscillation. However, as the molecule H<sub>A</sub>- H<sub>B</sub> started with a larger momentum compared to case 1 the atoms within the molecule is oscillation more. Hence, the presence of an oscillating behaviour before approaching the transition state. ||[[File:Nwc18_table3.png|200px]]<br />
|-<br />
| -5.1 || -10.1 ||357.277 ||Unreactive ||Barrier Recrossing. H<sub>C</sub> approached H<sub>A</sub>-H<sub>B</sub> with enough energy to form the products H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> via the transition state. However, the products formed with excess energy then reacted to reform the reactants which moved away from each other after the reformation. ||[[File:Nwc18_table4.png|200px]]<br />
|-<br />
| -5.1 || -10.6 ||349.477 ||Reactive ||Barrier Recrossing. H<sub>C</sub> approached H<sub>A</sub>-H<sub>B</sub> with enough energy to form the products H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> via the transition state. However, the products formed with excess energy then reacted to reform the reactants which reacted again, with enough energy, to form the products that finally moved away from each other after the reformation. ||[[File:Nwc18_table5.png|200px]]<br />
|}<br />
Conclusion: Not all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive. As shown in the table above, trajectories with a momenta (kinetic energy) too high may result in the reformation of the reactants in order to dissipate the excess energy. This may not react again to form the products if the reactants no longer have enough energy to react. Hence, it is crucial to have the right amount of kinetic energy for a reaction to proceed successfully. <br />
Also, having a greater kinetic energy and hence total energy does not guarantee the succession of the reaction if the same starting position is not kept as H<sub>C</sub> may not be approaching H<sub>A</sub>- H<sub>B</sub> in the direction (along the bond) which results in the MEP. Hence, more energy may be required to overcome the activation barrier and form stable products.<br />
====Transition state Theory====<br />
The transition state theory (TST) separates the reaction into two regions, the reactant region and the product region, by the transition state.<ref name="TST" /> When applying TST to potential energy surfaces a few assumptions apply:<ref name="TST" /><br />
<br />
1. The Born-Oppenheimer approximation applies<br />
<br />
2. Quantum-tunnelling negligible <br />
<br />
3. The energies of atoms in the reactant region are Boltzmann distributed<br />
<br />
4. System that has reached transition state with momentum towards the product configuration will not reenter the reactant region<br />
<br />
However, as discussed in section 1.2.1, products formed with enough energy can react to reform reactants by barrier recrossing and so reenter the reactant region. Hence, the TST overestimates reaction rates and calculations will result in rate values greater than the experimental rate values. '''''<u>Good! Keep in mind that QM effects to play a role (tunneling will increase the rxn rate relative to TST) but this is totally outweighed by TS recrossing. </u>'''''<br />
<br />
==F-H-H system ==<br />
===PES inspection===<br />
====Energetics and transition states of the reaction between F + H<sub>2</sub> and HF + H ====<br />
The reaction between F and H<sub>2</sub> forming a new F-H bond is exothermic whilst the reaction between H-F and H forming a new H-H bond is endothermic. This is due to the H-F bond (565 kJ/mol) <ref name="Bond_Strength" /> being stronger and so requiring more energy to break compared to the H-H bond( 432 kJ/mol)<ref name="Bond_Strength" />. More energy is also released when the stronger H-F bond is formed compared to the H-H bond. Hence, an overall energy is given out during the reaction between F and H<sub>2</sub> and an overall energy is required for the reaction between H-F and H. This can be seen in the Figures below (F + H<sub>2</sub> on the left and F-H + H on the right) where A represents the F atom and B/C represents the H atoms.<br />
[[File:Nwc18_F+H2_surface_plot3.png|300px]] [[File:Nwc18_FH+H_surface_plot3.png|300px]] <br />
Both plot may seem to be the same but the figure on the left representing F + H<sub>2</sub> happened in the direction where AB decreases (right to left) whilst the figure on the right representing F-H + H happened in the direction where AB distance increases. Hence, the reactant energy (represented by the local minimum on the right of the graph) for the reaction F + H<sub>2</sub> is higher than the product energy (represented by the local minimum on the left of the graph). This indicates an exothermic reaction where the excess energy from the reactant is given off to the surrounding. On the other hand, the reactant energy (represented by the local minimum on the left of the graph) for the reaction F-H + H is lower than the product energy (represented by the local minimum on the right of the graph). This indicates an endothermic reaction where extra energy is required from the surrounding to allow the occurrence/initiation of the reaction. <br />
<br />
The transition state of the reaction between F and H<sub>2</sub> is located at AB (where F=A, H=B and H=C) distance of approximately 181.104 pm and BC distance of approximately 74.488 pm.(shown by the red cross in figure6 below) <br />
[[File:Nwc18_TS_F+H2_2.png|thumb|center|Figure6 Position of transition state of F + H<sub>2</sub>.]]<br />
The transition state of the reaction between F-H and H is located at AB distance (where A/B=H and C=F) of approximately 74.488 pm and BC distance of approximately 181.104 pm. (shown by the red cross in figure7 below) <br />
[[File:Nwc18_TS_FH+H.png|thumb|center|Figure7 position of transition state of HF + H.]]<br />
<br />
====Locating activation energy for both reactions====<br />
The activation energy of an reaction is the energy difference between transitions state and the reactant. Hence, the activation energy can be calculated for both system from the transition state energy that was determined to be -433.981 kJ/mol and their reactant energy which can be determined from a energy against time plot. <br />
<br />
The transition energy for both reactions are the same as the transition state are the same for both reaction.<br />
<br />
For the exothermic reaction between F and H<sub>2</sub> the reactant energy was found from the figures below to be -434.364 kJ/mol. The activation energy is therefore (-433.981+434.364=)0.383 kJ/mol. '''''<u>This is a little on the low side (the true value is >1), likely because you didn't let the simulation run for long enough. Your TS estimate appears sensible. </u>'''''<br />
[[File:Nwc18_Energy_plot_exothermic2.png|300px]] [[File:Nwc18_contour_plot_exothermic.png|300px]] <br />
The figure on the left represents the energy vs time plot and the figure on the right represents the contour plot which shows <br />
the reaction going in the direction that forms the reactants from the transition state. <br />
A=F B=H C=H<br />
For the endothermic reaction between HF and H the reactant energy was found from the figures below to be -560.435 kJ/mol. The activation energy is therefore (-433.981+560.435=) 126.454 kJ/mol.<br />
[[File:Nwc18_Energy_plot_endothermic.png|300px]] [[File:Nwc18_contour_plot_endothermic2.png|300px]] <br />
The figure on the left represents the energy vs time plot and the figure on the right represents the contour plot which shows <br />
the reaction going in the direction that forms the reactants from the transition state.<br />
A=H B=H C=F<br />
The calculated activation energies confirm that F + H<sub>2</sub> is an exothermic reaction and HF + H is an endothermic reaction due to the reasons explained in 2.1.1.<br />
<br />
===Reaction dynamics===<br />
====F+H<sub>2</sub> system-Energy conservation====<br />
[[File:Nwc18_initial_conditions.png|thumb|Figure8 a reactive trajectory for F + H<sub>2</sub>.]]<br />
[[File:Nwc18_ex2_momenta.png|thumb|Figure9 momenta vs time graph for F + H<sub>2</sub>.]]<br />
[[File:Nwc18_ex2_animation.png|thumb|Figure10 Animation of F + H<sub>2</sub>.]] <br />
[[File:Nwc18_ex2_animation2.png|thumb|Figure10 Animation of F + H<sub>2</sub>.]]<br />
Figure8 below shows a reactive trajectory for the reaction F + H<sub>2</sub> where the initial conditions were set to AB=181 pm and BC=74 pm. (where A=F B=H C=H) The initial momentum were set to 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> for both reactant. This initial condition was used as it is close to the conditions of the transition state which resembles the reactant but not exactly the same as the conditions of the transition state so forces required to initiate the reaction is not 0. As the momentum are set to 0 the kinetic energies of the reactants are also 0.<br />
Figure8 shows the reaction started with little oscillatory behaviour which effect increased as the reaction progress to form the products. This behaviour can be explained by Figure9 where the initial vibrational (potential) energy of molecule H<sub>2</sub> was converted to translational (kinetic) energy constantly (due to conservation of energy) causing one of the H atom to approach F and to exert van der waals forces on each other. This attractive force between the atoms caused further conversion of more of the H<sub>2</sub> vibrational energy to its kinetic energy. Eventually, the attractive forces between the H and F atom became large enough to overcome the H-H bond strength and so the H-H bond break and H-F bond form. Some of the original vibrational energy of H<sub>2</sub> converted to kinetic energy were then converted to the vibrational energy of HF. The rest of the original vibrational energy were converted to the kinetic energy of the H atom released and so the H atom moves away from the HF molecule formed. <br />
The increase in amplitude of the momentum of both A-B (HF) and B-C (H<sub>2</sub>) suggests that vibrational energy are converted to kinetic energy due to the conservation of energy as only vibrational energy was present before the start of the reaction. The effect of the conversion, and so amount of energy converted, became significant before 200 fs when the reaction has started. <br />
<br />
The animation (figure9 and figure10) of the reaction is also a good demonstration for this. <br />
<br />
The release of the reaction energy could be confirmed experimentally using bomb calorimetry. The calorimeter used is an isochronic system that measures the change in internal energy in the form of heat (temperature change) using the equation below.<ref name="Calorimetry" /><br />
<br />
q<sub>cal</sub>=C<sub>cal</sub>ĪT=āq<sub>v,system</sub><br />
<br />
C<sub>cal</sub> is the heat capacity of the calorimeter.<br />
<br />
'''''<u>Bomb calorimetry is generally not suitable as it doesn't distinguish between translational and vibrational KE. </u>'''''<br />
<br />
====F+H<sub>2</sub> system-Energy and reactivity====<br />
When introducing an initial momentum to the reactants and hence, putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration the reaction does not always end up forming the products (depending on the amount of extra energy put in). However, it may form products as an intermediate with high enough energy to react and reform the reactants. This effect is demonstrated in the figures below. Therefore, a right amount of energy and momentum in the correct direction are required for the reaction to proceed and end up on the product side. <br />
<br />
The initial distance between AB and BC are kept the same as in 2.2.1 but p<sub>HF</sub> = -1.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>HH</sub> varied from -6.1 to 6.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>.<br />
p<sub>HH</sub>=-6.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> Overall=unreactive<br />
[[File:Nwc18_contour_-6.1.png|300px]] [[File:Nwc18_momenta_-6.1.png|300px]] <br />
p<sub>HH</sub>=-3.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> Overall=unreactive<br />
[[File:Nwc18_contour_-3.0.png|300px]] [[File:Nwc18_momenta_-3.0.png|300px]]<br />
p<sub>HH</sub>=0.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> Overall=reactive<br />
[[File:Nwc18_contour_0.0.png|300px]] [[File:Nwc18_momenta_0.0.png|300px]]<br />
p<sub>HH</sub>=3.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> Overall=reactive<br />
[[File:Nwc18_contour_3.0.png|300px]] [[File:Nwc18_momenta_3.0.png|300px]]<br />
p<sub>HH</sub>=6.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> Overall=unreactive<br />
[[File:Nwc18_contour_6.1.png|300px]] [[File:Nwc18_momenta_6.1.png|300px]]<br />
<br />
The above finding can be further proved by increasing p<sub>HF</sub> by a little in the same direction but greatly reducing the magnitude of p<sub>HH</sub> and hence reducing the overall energy of the system. <br />
p<sub>HF</sub>=-1.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> <br />
Overall=reactive<br />
[[File:Nwc18_contour_0.2.png|300px]] [[File:Nwc18_momenta_0.2.png|300px]]<br />
The reaction is reactive despite the decrease in energy of the system.<br />
In conclusion, the increase in energy of the system may increase the oscillation strength and frequency of molecules as well as increase kinetic energy of the reactants and products (proportional to magnitude of momentum) but it does not necessarily increase the chance of succession of the reaction. <br />
====H+HF system====<br />
The initial condition for a reactive trajectory of H + HF was set to AB=200 pm, BC=92 pm, p<sub>HF</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>HH</sub>=-17 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. (A=H B=H C=F) This is shown in the figure below. <br />
<br />
p<sub>HH</sub> was set to be above the activation energy to ensure that the H atom approaches HF with high enough kinetic energy to initiate the reaction and form stable products.<br />
<br />
[[File:Nwc18_HHF_contour_-17.png|300px]] [[File:Nwc18_HHF_momenta_-17.png|300px]]<br />
<br />
Further determination of reactive trajectories by decreasing the value of p<sub>HH</sub> and increase the value of p<sub>HF</sub> shows that a small displacement of p<sub>HH</sub> from the initial condition requires a even smaller increase in p<sub>HF</sub> to result in a reactive trajectory. Hence, translational energy has a bigger effect on the system than the vibrational energy. However, as the displacement of p<sub>HH</sub> from the initial condition becomes larger, the effect of vibrational energy on the system increases and overcomes the effect of translation energy. Hence, a greater increase in p<sub>HF</sub> ( compared to the increase in p<sub>HH</sub>) is required to result in a reactive trajectory. This is illustrated in the figures below. <br />
<br />
p<sub>HF</sub>=0.40 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=-16 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
[[File:Nwc18_HHF_contour_-16_0.40.png|300px]] [[File:Nwc18_HHF_momenta_-16_0.40.png|300px]]<br />
<br />
p<sub>HF</sub>=2.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=-15 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
[[File:Nwc18_HHF_contour_-15_2.0.png|300px]] [[File:Nwc18_HHF_momenta_-15_2.0.png|300px]]<br />
<br />
p<sub>HF</sub>=4.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=-14 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
[[File:Nwc18_HHF_contour_-14_4.png|300px]] [[File:Nwc18_HHF_momenta_-14_4.png|300px]]<br />
<br />
p<sub>HF</sub>=7.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=-12 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
[[File:Nwc18_HHF_contour_-12_7.png|300px]] [[File:Nwc18_HHF_momenta_-12_7.png|300px]]<br />
<br />
p<sub>HF</sub>=11.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>HH</sub>=-9 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
[[File:Nwc18_HHF_contour_-9_11.png|300px]] [[File:Nwc18_HHF_momenta_-9_11.png|300px]]<br />
<br />
The two momentum are set to be acting in opposite directions so a smaller (decreasing) magnitude of p<sub>HH</sub> is required each time to result in a reactive trajectory.<br />
<br />
====Polanyi's empirical rules====<br />
Polanyi's rule states that vibrational energy is more efficient in promoting a late transition state reaction than translational energy.<ref name="Polanyi"/><br />
This explains the previous experiments where the vibrational energy becomes more efficient in promoting the endothermic, late transition state, reaction of H + HF than the translational energy.<br />
<br />
Further experiments on the exothermic, early transition state, F + H<sub>2</sub> reaction can be conducted to see if translational energy is more efficient in promoting a early transition state reaction than vibrational energy.<br />
<br />
'''''<u>Good examples! Absolutely correct. </u>'''''<br />
<br />
==References==<br />
<references><br />
<br />
<ref name="Atkins"> Atkins, P. W. (1940) Chapter 18 Reaction dynamics. ''Atkins' Physical chemistry''. Eleventh edition. Oxford, United Kingdom : Oxford University Press 2018. ISBN 9780191092183 ; ISBN 9780198769866.</ref> <br />
<br />
<ref name="TST"> Bligaard, T. and NĆørskov, J. K. (2008) āHeterogeneous catalysisā, in Chemical Bonding at Surfaces and Interfaces. Elsevier, pp. 255ā321. doi: 10.1016/B978-044452837-7.50005-8.</ref><br />
<br />
<ref name="Bond_Strength">Covalent Bond Strengths | Grandinetti Group (no date). Available at: https://www.grandinetti.org/covalent-bond-strengths (Accessed: 21 May 2020).</ref> <br />
<br />
<ref name="Calorimetry"> Bomb Calorimetry (no date). Available at: https://ch301.cm.utexas.edu/section2.php?target=thermo%2Fthermochemistry%2Fbomb-calorim.html (Accessed: 21 May 2020).</ref> <br />
<br />
<ref name="Polanyi">Zhang, Z. et al. (2012) āTheoretical study of the validity of the polanyi rules for the late-barrier Cl + CHD3 reactionā, Journal of Physical Chemistry Letters. American Chemical Society, 3(23), pp. 3416ā3419. doi: 10.1021/jz301649w.</ref> <br />
<br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01508310&diff=812912MRD:015083102020-06-25T22:36:37Z<p>Ng611: </p>
<hr />
<div><br />
'''''<u>3/5 - Some very very good aspects to this report, unfortunately you let yourself down with some fairly hard-to-ignore gaps in several basic areas.</u>'''''<br />
<br />
==H-H-H system==<br />
===Dynamics from the transition state region===<br />
The transition state is the point on the potential energy surface that has āV(ri)/āri=0, where ri is the distance between two atoms.The transition state is a local maximum and its second derivative ā<sup>2</sup>V(ri)/āri<sup>2</sup> is < 0, whereas a local minimum will have a positive second derivative instead. <br />
<br />
'''''<u>NO! The TS is not a local maximum. Look again at your TS and carefully consider the curvature in ALL directions. </u>''''' <br />
[[File:Hhh_surfaceplot_01508.png|400px|thumb|centre|a potential energy surface plot of H and H<sub>2</sub> with respect to interatomic distances.]]<br />
[[File:HHHSURFACE_015.png|400px|thumb|centre|the potential energy surface plot at a different angle.]]<br />
<br />
===Trajectories from r<sub>1</sub>=r<sub>2</sub>: locating the transition state===<br />
At the transition state, r<sub>1</sub>=r<sub>2</sub>=90.8 pm by estimation as this transition state is symmetric. At this separation, the forces acting along AB and BC are both -0.004 kJ.mol<sup>-1</sup>.pm<sup>-1</sup>, which is close to zero and the eigenvalues for the Hessian matrix are -0.028 and +0.167 kJ.mol<sup>-1</sup>.pm<sup>-2</sup>, indicating the presence of transition state. The Internuclear Distance vs Time plot suggests the distance between H<sub>A</sub>, H<sub>B</Sub> and H<sub>C</sub> is equal, about 90 pm and the distance is constant over time suggests the atoms are static at the transition state.<br />
<br />
'''''<u>Good TS estimate. </u>'''''<br />
[[File:Animation101508310.png|400px|centre|thumb|picture showing H and H<sub>2</sub> at the symmetric transition state.]]<br />
[[File:Optimisation1_01508310.png|400px|centre|thumb|values for separations and eigenvalues for Hessian Matrix at the transition state.]]<br />
[[File:Internucleardistance_01508310.png|400px|centre|thumb|Internuclear Distance vs Time plot between the H atoms at the transition state.]]<br />
<br />
===Calculating the reaction path===<br />
When r<sub>A-B</sub> is set to r<sub>ts</sub>+1 (91.8 pm) and r<sub>B-C</sub> equals r<sub>ts</sub> (90.8 pm),the transition state rolls towards H<sub>B</sub> and H<sub>C</sub>, H<sub>A</sub> remains as an atom. MEP and Dynamic reaction paths are different as MEP set the momenta to zero in every step of the reaction path, products are therefore static and hence the reaction path is relatively straight. The dynamic reaction path is more realistic. It takes the momenta of gaseous molecules into account,i.e. The reactants and the products have vibrational energies, that is illustrated by the wavy line along the reaction path. <br />
[[File:Surface_Plot1mep_015083.png|400px|centre|thumb|the minimum energy path (MEP) of the H-H-H system]]<br />
[[File:01569348_MEP_AB1_dynamic.png|400px|centre|thumb|the dynamic reaction path of the H-H-H system]]'''''<u>Good!</u>'''''<br />
<br />
====MEP and Dynamics calculations comparisons====<br />
'''r<sub>1</sub>=91.8 pm, r<sub>2</sub>=90.8 pm, momenta set to 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
<br />
H<sub>B</sub> and H<sub>C</sub> form a H<sub>2</sub> molecule whereas H<sub>A</sub> moves away from the molecule. The distance between B and C decreased to 74 pm, which is the hydrogen molecule bond length <ref name="reference 1"/>.In MEP calculation, the momenta are set to zero at each time step.<br />
<br />
[[File:MepH2bondlength_01508.png|centre|400px|thumb|Internuclear Distances vs Time plot for H-H-H system using MEP calculation.]]<br />
<br />
[[File:MepH2momenta015083.png|centre|400px|thumb|Momenta vs Time plot for H-H-H system using MEP calculation.]]<br />
<br />
Using Dynamics calculation, the outcome of the reaction is the same. The transition state rolls towards H<sub>B</sub> and H<sub>C</sub> and a hydrogen molecule is formed. One noticeable difference is that the hydrogen molecule now has some vibational energy as indicated by the wavy line along B-C in the Internuclear Distance vs Time plot.As for the Momenta vs Time plot, the system is linear, therefore the angle theta is zero. H<sub>A</sub> and H<sub>B</sub> are moving away from each other at an increasing velocity until the momentum reaches 5 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>.The oscillation along the B-C bond corresponds to the vibration of the bond. <br />
[[File:DynamicsH2_015083.png|centre|400px|thumb|Internuclear Distances vs Time plot for H-H-H system using Dynamics calculation.]]<br />
[[File:DynamicsH2monenta_01508.png|centre|400px|thumb|Momenta vs Time plot for H-H-H system using Dynamics calculation.]]<br />
<br />
'''r<sub>1</sub>=90.8 pm, r<sub>2</sub>=91.8 pm, momenta set to 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
<br />
Once the length of r<sub>1</sub> and r<sub>2</sub> switches, the shapes of the graphs are identical but instead of H<sub>B</sub> and H<sub>C</sub> forming a molecule, H<sub>A</sub> and H<sub>B</sub> now form the hydrogen molecule in the products.<br />
[[File:Mepinter2_01508.png|centre|400px|thumb|Internuclear Distances vs Time plot for H-H-H system using MEP calculation.]]<br />
[[File:Momentamep2_015.png|centre|400px|thumb|Momenta vs Time plot for H-H-H system using MEP calculation.]]<br />
[[File:InterDynamic_015083.png|centre|400px|thumb|Internuclear Distances vs Time plot for H-H-H system using Dynamics calculation.]]<br />
[[File:Momentadynamics_015083.png|centre|400px|thumb|Momenta vs Time plot for H-H-H system using Dynamics calculation.]]<br />
<br />
'''Trajectory starting with the final position and reversed momentum to the reaction path above'''<br />
<br />
This trajectory starts at the end of the reaction path above and finishes at where the above reaction started. it is essentially a reverse of what happened in the reaction above. '''''<u> Good. </u>'''''<br />
[[File:Reverse_015083.png|centre|400px|thumb|Internuclear Distances vs Time plot for H-H-H system using Dynamics calculation (r<sub>1</sub>=70.0 pm,r<sub>2</sub>=353 pm,p1=-3.21 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p2=-5.07 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>) .]]<br />
<br />
===Recative and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>!! E<sub>tot</sub>/kJmol<sup>-1</sup>!! Recative?!! Description of the dynamics!! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1||-414 || Yes||The translational energy of the reactants can surmount the potential energy barrier and form the products. The ramaining energy is converted into vibrational energy of the products. || [[File:Table1_01508310.png|350px ]]<br />
|-<br />
| -3.1 || -4.1||-420 || No||The reactant H<sub>2</sub> has vibrational energy and its r<sub>AB</sub> motion is rapid. Before it could cross the energy barrier, the molecule slams into the repulsive inner wall of the potential surface and bounces back into the reactant channel. || [[File:Table2_0150.png|350px]]<br />
|-<br />
| -3.1 || -5.1||-414|| Yes||The reactants have both the translational and vibrational energies. The energy is used to surmount the potential energy barrier and converted into the vibrational energy of the products. || [[File:Table3_01508.png|350px]]<br />
|-<br />
| -5.1 || -10.1||-357 || No||The reactants have sufficient energy to cross the transition state and the products did form for a certain amount of time. However, before the the system can exit the product channel, it recrosses the transition state and form the reactants again, hence no reaction occurs. ||[[File:Table4_01508.png|350px]]<br />
|-<br />
| -5.1 || -10.6||-394 || Yes||The system has sufficient energy to cross the transition state but it recrosses the transition state shortly after the reaction path hits the potential energy surface. However, the system is able to cross the transition state for the third time and finally exits the product channel and forms the products. || [[File:-10.6_01508.png|350px]]<br />
|-<br />
|}<br />
In order to form products, a system must have sufficient energy to cross the transition state. Only molecules with the correct phase <ref name="reference 2"/> can exit the product channel and form stable products. Recrossing the transition state can lead to either product formation or no reaction. It is possible to cross the transition state multiple times.<br />
<br />
===Transition State Theory===<br />
<br />
Transition State Theory is classical, it doesn't take quantum tunnelling into account.Quantum tunnelling is probable for this reaction as it involves the transfer of a hydrogen atom, particularly at lower temperatures <ref name="reference 3"/>.At low temperatures, the H atom can tunnel through the potential energy barrier instead of passing over the barrier, hence the activation energy is lowered and the reaction rate increases. However, recrossing the energy barrier is possible as seen with the simulations, this can have a negative effect on the reaction rate. At high temperatures, the system is thermally activated and the behavious is approximately classical<ref name="reference 4"/>. Overall, the situation can be complicated as the theory assumes quasi-equilibrium between reactants and transition state (activated complexes). The activated complex can then either form products or revert back to the reactants with or without quantum tunnelling. The experimental reaction rate might still be greater than that predicted by the theory. '''''<u>It is possible but generally this is not the case. </u>'''''<br />
<br />
==F-H-H system==<br />
===PES inspection===<br />
'''F+H<sub>2</sub> (F<sub>1</sub>-H<sub>1</sub>-H<sub>2</sub>)'''<br />
<br />
F+H<sub>2</sub> is exothermic as the reactants are higher in energy than the products therefore during the reaction, energy is released to the surroundings. H-F bond is stronger than H-H bond as the products are lower in energy and therefore more stable.The transition state is located approximately at r<sub>F<sub>1</sub>-H<sub>1</sub></sub>=181.10 pm and r<sub>H<sub>1</sub>-H<sub>2</sub></sub>=74.49 pm. This is verified by the forces (close to zero) and eigenvalues for hessian matrix (they are of opposite signs). The activation energy is estimated by tilting the transition state towards the reactants and is shown to be 424 kJ.mol<sup>-1</sup>.<br />
[[File:Fh2_015083.png |400px|centre|thumb|Surface plot for F<sub>1</sub>-H<sub>1</sub>-H<sub>2</sub> system using Dynamic Calculation, the transition state is identified as the black dot.]]<br />
[[File:OptimisationFhh1_01508.png|400px|centre|thumb|values for separations and eigenvalues for Hessian Matrix at the transition state.]]<br />
[[File:F1H1H2_EA_0150.png|400px|centre|thumb|Counter plot for estimating the activation energy using MEP calculation.]]<br />
[[File:EA1_01508.png|400px|centre|thumb|parameters for the activation energy.]]<br />
'''H+HF (H<sub>1</sub>-H<sub>2</sub>-F<sub>1</sub>)'''<br />
<br />
This reaction is endothermis as the reactants are lower in energy than the products therefore energy is taken in from the surroundings to form the products. H-F bond is still stronger than H-H as the reactans are lower in energy and more stable. The transition state is located at approximately r<sub>H<sub>1</sub>-H<sub>2</sub></sub>=280 pm and r<sub>H<sub>2</sub>-F<sub>1</sub></sub>=92 pm. It is located with the same reasons stated above.With the same method, the activation energy is estimated to be 434 kJ.mol<sup>-1</sup>. The two activation energies should be identical as these two reactions are essentially the reverse of one another.<br />
<br />
'''''<u>NO!! Look at your PES-- if you go in one direction, you spend most of your time going downhill, the opposite case is seen if you go in the opposite direction. The Ea values are not at all the same. Think about this as well: do you think fluorine will react more easily that H + HF?</u>'''''<br />
<br />
[[File:H+HF_01508310.png|400px|centre|thumb|Surface plot for H<sub>1</sub>-H<sub>2</sub>-F<sub>1</sub> system using Dynamic Calculationthe, the transition state is identified as the black dot.]]<br />
[[File:HHF_015083.png |400px|centre|thumb|values for separations and eigenvalues for Hessian Matrix at the transition state.]]<br />
[[File:EA2_01508.png|400px|centre|thumb|Counter plot for estimating the activation energy using MEP calculation.]]<br />
[[File:EA2_PARAMETERS.png|400px|centre|thumb|parameters for the activation energy.]]<br />
<br />
===Reaction dynamics===<br />
===F+H<sub>2</sub> system===<br />
<br />
Initially, the reactants have kinetic energy in the form of vibrational and translational energies, and as it crosses the potential energy barrier, the kinetic energy is converted into potential energy. Once the system is exiting the product channel, the potential energy is converted into the vibrational energy of the product.One good experimental technique to prob the electronic structure of the reacting species will be IR spectroscopy.In the anharmonic oscillator model, exciting molecules in the relaxed state promotes excitation from the ground state to the first energy level mainly. This will give rise to a single peak in the IR spectrum. Howver, when exciting a sample of excited molecules, many transitions are possible. For example, from 0-->1, the fundamental or 1-->2, the first hot band. They will show up in the spectrum as 2 peaks, with the hot band being at a lower wavenumber as the energy gap decreases up the potential well. The intensity of the hot band decreases over time as the electrons can fall back to lower energy levels and energy is emitted as radiation. As this system is linear, rotational is not important electronic energy is not considered here.Photoelectron spectroscopy is also a viable method. <ref name="reference 5"/> '''''<u> Interesting choice!</u>'''''<br />
<br />
<br />
[[File:Reactiondynamicsfh2_0150.png||400px|centre|thumb|contour plot for F+H<sub>2</sub>.]]<br />
[[File:Reactiondynamicsparameterāā015.png |400px|centre|thumb|parameters for the reactive trajectory.]]<br />
====Varying the momentum of HH in the F+H<sub>2</sub> system.====<br />
Conditions:r<sub>FH</sub>=182 pm, r<sub>HH</sub>=74 pm, P<sub>FH</sub>=-1.0g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
{| class="wikitable" border="1"<br />
|+ caption<br />
|-<br />
|P<sub>HH</sub>/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>|| Contour Plot<br />
|-<br />
|0 || [[File:0_015083.png |400 px|centre]] <br />
|-<br />
|2 || [[File:2_0508.png |400 px|centre]] <br />
|-<br />
|4 || [[File:4_01508.png ā|400 px|centre]] <br />
|-<br />
|6.1|| [[File:6_01508.png|400 px|centre]] <br />
|-<br />
| -2|| [[File:-2_01508.png|400 px|centre]] <br />
|-<br />
| -4|| [[File:-4.1_01508.pngā|400 px|centre]]<br />
|-<br />
| -6.1|| [[File:Fit1_01508.png|400 px|centre]] <br />
|-<br />
|}<br />
Although the amount of energy that is put into H-H vibration is much more than the activation energy of the reaction, the reaction does not always occur. The large amount of energy is demonstrated by the high frequency of those wavy lines. Recrossing the transition states multiple times can lead to no reaction or product formation depending on how the tragectory collides with the potential well.<br />
<br />
[[File:Reactiondynamic_last01588.png|400px|centre|thumb|contour plot for r<sub>FH</sub>=182 pm, f<sub>HH</sub>=74 pm, P<sub>FH</sub>=-1.6g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>,P<sub>HH</sub>=0.2g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
The system now has much less vibrational energy into the reactants than before. The reaction is still successful and the trajectory recrosses the transition states fewer times than before.This can be explained by the increased momentum of FH so that F can approach the closest H with more vibrational energy.<br />
<br />
===H+HF system===<br />
{| class="wikitable" border="1"<br />
!Graph Number!! p<sub>1</sub>(HH)/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>(HF)/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>!! E<sub>tot</sub>/kJmol<sup>-1</sup>!! Recative?!! Illustration of the trajectory<br />
|-<br />
|1|| -2 || -10||-397|| No|| [[File:Final1_01508.png|350px ]]<br />
|-<br />
|2|| 0|| -8||-400 || No|| [[File:Final2_015.png|350px]]<br />
|-<br />
|3||5 || -6||-360|| No|| [[File:Final3_015.png|350px]]<br />
|-<br />
|4|| 5 || -3||-389 || Yes||[[File:Final4_015.png|350px]]<br />
|-<br />
|}<br />
<br />
By referening to Polanyl's empirical rules, an exothermic reaction has an early barrier, which happens in the reactant channel. It tends to favour vibrational excitation of the products. On the other hand, an endothermic reaction has a late barrier, which occurs in the product channel. It tends to favour low product vibrational excitation. The location of the energy barrier is also crucial for a successful reaction. Translational energy of the reactants is the most effective at passing the early barrier whereas vibrational energy of the reactants is the most effective at surmounting the late barrier. This is undrstandable because in an endothermic reaction, if the reactants have mainly translational energy, it is more likely that the reactants will hit the potential well wall and results in no reaction.This reaction is endothermic, and hence it has a late barrier. In all the graphs, the reactants all have vibrational energy as indicated by the wavy line in the reactant channel.<ref name="reference 2"/><br />
<br />
'''''<u>An excellent systematic study, well done!</u>'''''<br />
<br />
==References==<br />
<br />
<references><br />
<ref name="reference 1">R. DeKock and H. Gray, Chemical structure and bonding, Univeristy Science Books, 1989.</ref><br />
<ref name="reference 2">S. Jeffrey I., Experimental Chemical Dynamics, Prentice-Hall, Upper Saddle River, 2nd edn., 1989.</ref><br />
<ref name="reference 3">L. Keith J., Theories of reaction rates, Harper & Row, New York;London, 3rd edn., 1987.</ref><br />
<ref name="reference 4">D. Brougham, A. Horsewill and R. Jenkinson, Chemical Physics Letters, 1997, 272, 69-74.</ref><br />
<ref name="reference 5">S. Bradforth, D. Arnold, D. Neumark and D. Manolopoulos, The Journal of Chemical Physics, 1993, 99, 6345-6359.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:SMYM&diff=812911MRD:SMYM2020-06-25T22:24:48Z<p>Ng611: </p>
<hr />
<div>'''5/5 - An excellent report overall. Well done!'''<br />
<br />
== Exercise 1: H + H<sub>2</sub> system ==<br />
<br />
=== Transition state on an energy potential diagram ===<br />
<br />
( "On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?" )<br />
<br />
On a potential energy surface plot, the transition state is the maximum on the minimum energy pathway which links the reactant and product.<ref>Steinfeld, J.; Fransicso, J.; Hase, W.; ''Chemical kinetics and dynamics; '' p289-290; Pearson 1998</ref> Such a point is also referred to as a saddle point. The minimum energy pathway is portrayed in Figure 1, with the saddle point marked with 'X'. <br />
<br />
With the transition state being a stationary point, the derivative with respect to each coordinate must equal to zero.<br />
<br />
<math>\frac{\partial V(r_1)}{\partial r_1}\ = \frac{\partial V(r_2)}{\partial r_2}\ = 0 </math> , where <math>V</math> is the potential energy, <math>r_1</math> is the AB distance, and <math>r_2</math> is the BC distance.<br />
<br />
Mathematically, a saddle point on a surface plot is a local minimum in one direction but a local maximum in the orthogonal direction. Diagonal vectors relative to <math>r_1</math> and <math>r_2</math> can be set up (see Figure 2) so that the transition state is a maximum along <math>j</math> and a minimum along <math>i</math>:<br />
<br />
<math>\frac {\partial^2 V}{\partial j^2}<0 </math> , <math>\frac {\partial^2 V}{\partial i^2}>0 </math><br />
<br />
[[File:SMYM Screenshot 2020-05-08 at 17.58.49.png|thumb|Figure 1. A potential energy surface plot showing the minimum energy path. The transition state is marked with 'X'.|574x574px|left]]<br />
[[File:SMYM Contour eg.png|thumb|center|Figure 2. Potential energy contour plot with coordinates i and j defined.|297x297px]]<br />
<br />
<br />
<br />
'''''<u>EXCELLENT!</u>'''''<br />
=== Locating the transition state ===<br />
<br />
( "Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory." )<br />
<br />
The following two statements of the transition state must be true:<br />
<br />
# The transition state is a maximum on the minimum energy pathway, which means that moving away from the transition state will result in a steep decrease in potential energy. If the reactants begin at the transition state and are given zero momentum, they will remain there, giving no trajectories. <br />
# Since the potential energy surface plot is symmetrical for the H + H<sub>2</sub> system, r<sub>1</sub> must equal r<sub>2</sub> at the transition state. <br />
<br />
Utilizing the above statements, the transition state can be found by setting r<sub>1</sub> = r<sub>2</sub>, and p<sub>1</sub> = p<sub>2</sub> = 0. The transition state would be successfully located if no trajectory is shown, and forces along AB and BC direction equal zero. After fine-tuning the decimal places to F = 0 , the transition state is found to be at 90.7742 pm. The contour plot and the energy and forces outputs are shown in Figure 3. <br />
<br />
'''''<u>Good!</u>''''' <br />
<br />
The finding of the transition state at 90.7742 pm is also supported by the internuclear distances versus time plot (Figure 4)as two straight lines are shown, showing a constant internuclear distance at all times.<br />
When the reactants are not at the transition state, the internuclear distances show as two oscillating lines due to the asymmetrical vibrations of the two reactants. When the transition state is approached, the molecules achieve symmetrical vibrations.<br />
<br />
[[File:SMYM Screenshot 2020-05-08 at 17.14.01.png|thumb|Figure 3. Contour plot of the transition state with the energy and forces outputs|553x553px|left]]<br />
[[File:SMYM Dis vs time.png|centre|thumb|Figure 4. Internuclear distances vs. time plot of the transition state|306x306px]]<br />
<br />
<br />
'''''<u> Good!</u>'''''<br />
<br />
<br />
=== Reaction path calculation with MEP and Dynamics ===<br />
<br />
( "Comment on how the mep and the trajectory you just calculated differ." )<br />
<br />
To calculate the reaction path, ''r<sub>1</sub>'' was set as 91.7742 pm, and ''r<sub>2</sub>'' was set as 90.7742 pm, both momentums were equal to zero. The contour plot of the trajectories for both MEP and dynamic simulations are displayed in Figure 5.<br />
[[File:SMYM Screenshot 2020-05-11 at 11.31.48.png|centre|thumb|683x683px|Figure 6. The dynamic calculation (left) and the MEP calculation (right) of the path]]<br />
<br />
<br />
The minimum energy path (MEP) is the path on the potential energy surface with the lowest energy, hence the blackline traces the valley floor in the MEP graph (right of Fig. 6). The reaction path is straight, indicating that there is no vibration of the atoms. This is the case because for each step of the MEP, the momentum is reset to zero, resulting in an infinitesimally low-speed trajectory. There is no remaining inertia for the atoms to 'roll up' the potential energy surface, leading to no oscillation. Although useful for characterizing the reaction, the MEP does not reflect motion of the atoms in reality as their masses have been ignored. On the other hand, vibrations along the minimum energy pathway are seen in the dynamics calculation (left of Fig. 6). The inertia of the atoms are retained in this calculation, allowing enough momentum for the atoms to reach a higher potential on the valley surface.<br />
<br />
'''''<u>Excellent!</u>'''''<br />
<br />
=== Reactive and unreactive trajectories ===<br />
<br />
( "Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?" )<br />
<br />
The following trajectories were carried out with the initial positions of r<sub>1</sub> = 74 pm and r<sub>2</sub> = 200 pm<br />
<br />
{| class="wikitable" border=1<br />
! p<sub>1</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Yes ||Initially, the trajectory shows no H <sub>a</sub> - H<sub>b</sub> bond vibration. H<sub>a</sub> - H<sub>b</sub> and H<sub>c</sub> then overcome the transition state with enough momentum such that the newly formed H <sub>b</sub> - H<sub>c</sub> vibrates.||[[File:SMYM Traj1.png|thumb|266x266px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || No || H <sub>a</sub> - H<sub>b</sub> bond vibration could be seen at the start of the trajectory due to increased p<sub>1</sub>, but this trajectory does not arrive at the transition state due to the decreased p<sub>2</sub>. The decreased momentum between H<sub>b</sub> and H<sub>c</sub> resulted in lower kinetic energy when the two collided, and thus H<sub>c</sub> is repelled by H<sub>b</sub> and not enough energy was present to overcome the potential energy barrier. ||[[File:SMYM Traj2.png|thumb|269x269px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Yes || This trajectory is similar to the first example in the sense that it successfully overcomes the energy barrier at the transition state, and the H <sub>b</sub> - H<sub>c</sub> bond formed vibrates in a similar fashion as example one. The only difference is that now there is H<sub>a</sub> - H<sub>b</sub> bond vibration at the start of the reaction due to the increase of p<sub>1</sub>. Unlike example two, H<sub>c</sub> has enough kinetic energy to collide with H <sub>a</sub> - H<sub>b</sub> due to the increased p<sub>2</sub>. ||[[File:SMYM Traj3.png|thumb|273x273px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || No || This reaction has approximately twice the momentum of the first example. H<sub>a</sub> - H<sub>b</sub> and H<sub>c</sub> collide with lots of kinetic energy but the reaction is unsuccessful. Straight after the collision, H<sub>a</sub> - H<sub>b</sub> bond stretches and H<sub>b</sub> - H<sub>c</sub> forms. However, the very strong vibration of H<sub>b</sub> - H<sub>c</sub> causes the bond to break. H<sub>a</sub> - H<sub>b</sub> is reformed again. ||[[File:SMYM Traj4.png|thumb|276x276px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Yes || Here, the reactants again have high momentum and collide with lots of kinetic energy. Similarly, H<sub>b</sub> - H<sub>c</sub> momentarily breaks due to the strong bond vibration and H<sub>a</sub> - H<sub>b</sub> reform. However, the vibration between H<sub>a</sub> - H<sub>b</sub> is so strong that H<sub>b</sub> is ejected to H<sub>c</sub>, once again forming H<sub>b</sub> - H<sub>c</sub>. This step pushes the molecules over the transition state and the products are formed. Due to the high momentum, there are strong vibrations between the newly formed H<sub>b</sub> - H<sub>c</sub>. ||[[File:SMYM Traj5.png|thumb|276x276px]]<br />
|}<br />
<br />
In conclusion, having enough kinetic energy during the collision is not the main factor for a successful collision, as the high kinetic energy can cause excessive vibration which breaks the newly formed bond. The molecules must have the right combination of momentum and kinetic energy that leads them onto a successful trajectory.<br />
<br />
=== Transition state theory ===<br />
<br />
( "Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?" )<br />
<br />
Transition state makes two crucial assumptions that influence the difference between the observed reaction rate and the observed one:<br />
<br />
# All molecules which possess the necessary kinetic energy to overcome the activation energy barrier will react and become products.<br />
# No quantum effects are observed. <br />
<br />
The first assumption leads to a significant overestimation of the reaction rate as in reality molecules that possess the necessary energy to overcome the activation energy barrier often do in fact not form products. This can be seen above in the reaction in which although the molecules possess enough energy to overcome the barrier, the internal bond vibrations are so strong that no products are formed. <br />
<br />
The second assumption causes a slight underestimation of the reaction rate. If quantum effects are considered, tunneling becomes significant. This is especially true for protons as they are light and have a tendency to tunnel. The tunneling effect is observed when a reactant molecule does not have enough kinetic energy to overcome the activation energy barrier but still forms the product. This can be explained by the molecule not going over the activation energy barrier but travels through the potential energy surface.<ref>Steinfeld, J.; Fransicso, J.; Hase, W.; ''Chemical kinetics and dynamics; '' p316-318; Pearson 1998</ref><br />
<br />
To summarise, the reaction rates predicted by the transition state theory causes significant overestimations compared to the observed ones. '''''<u>You're correct, but why, given you have two effects acting in opposite directions. </u>'''''<br />
<br />
== EXERCISE 2: F - H - H system ==<br />
<br />
=== PES inspection ===<br />
<br />
( "By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?" )<br />
[[File:SMYM Screenshot 2020-05-19 at 14.58.24.png|centre|thumb|724x724px|Figure 7. Potential energy surface plots for (left) HF + H --> H2 + F and (right) H2 + F --> HF + F]]<br />
<br />
The valley floor of Figure 7 indicates that H<sub>2</sub> and F are higher in energy than H and HF. This means that when H and HF are converted to H<sub>2</sub> and F, more energy is put in to break the H-F than the energy released from the formation of H-H, making the reaction endothermic. Vice versa, the conversion from H<sub>2</sub> and F to H and HF would be exothermic. This information indicates that H-F bond is stronger than H-H bond. This can be proven by their bond dissociation energy. H-F has a higher bond dissociation energy of 568.6 kJ/mol compared to H-H (436.002 kJ/mol).<ref>T. L. Cottrell, ''The Strengths of Chemical Bonds, 2d ed''., Butterworth, 1958</ref><br />
<br />
<br />
<br />
[[File:SMYM Screenshot 2020-05-21 at 12.30.21.png|thumb|480x480px|Figure 8. The transition state location of the F-H-H system]]<br />
<br />
=== Locating the transition state ===<br />
<br />
The approximate transition state position was found by varying r<sub>1</sub> and r<sub>2</sub> with p<sub>1</sub> = p<sub>2</sub> = 0, until the forces along AB and BC equal to zero. Unlike the H + H<sub>2</sub> system, the potential energy surface is not symmetrical so r<sub>1</sub> will not be equal to r<sub>2</sub> at the transition state. <br />
<br />
After fine-tuning, the transition state was found to be at r<sub>1</sub> = 180.855 pm and r<sub>2</sub> = 74.493 pm, where the kinetic energy and the forces acting on the atoms equal to zero (Figure 8).<br />
<br />
<br />
<br />
<br />
<br />
<br />
=== The activation energy for both reactions ===<br />
<br />
The activation energy could be obtained by looking at the energy difference between the reactants and the transition state. According to Figure 8, the energy of the transition state was -433.981 kJ/mol. To find the energy of the reactants, r<sub>1</sub> is set as the local minimum of the reactant which is the bond length of either HF or H<sub>2</sub>. r<sub>2</sub> is then given a large value such that reactant 2 is far away from reactant 1 where there is no interaction between the two and the momentums were set to zero.<br />
<br />
H<sub>2</sub> + F --> HF + H: <br />
<br />
Atom A = H, Atom B = H, Atom C = F, r<sub>1</sub> = 74 pm , r<sub>2</sub> = 1000 pm<br />
The energy calculated for the reactants were -435.100 kJ/mol, which results in the activation energy of 1.119 kJ/mol.<br />
<br />
HF + H --> H<sub>2</sub> + F:<br />
<br />
Atom A = H, Atom B = H, Atom C = F, r<sub>1</sub> = 91 pm , r<sub>2</sub> = 1000 pm<br />
The energy calculated for the reactants were -560.404 kJ/mol, which results in the activation energy of 126.423 kJ/mol.<br />
<br />
'''''<u>Good TS estimate and good Ea values. Well done!</u>'''''<br />
<br />
=== Reaction dynamics ===<br />
[[File:SMYM Animation.png|thumb|Figure 9. Momentum vs. time plot of the reaction H2 + F --> HF + H]]<br />
( "In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally." )<br />
<br />
A momentum versus time graph of a successful reaction trajectory is shown in Figure 9. For such a trajectory, r<sub>1</sub> was set to 179 pm and r<sub>2</sub> was set to 74 pm, both momentum were equal to zero. The reaction between H<sub>2</sub> and F is exothermic and since the energy is conserved, the products must show an elevation in energy. This can be seen in the strong oscillation of the momentum A-B (the H-F bond), which demonstrates that a large part of the energy is transformed into the vibrational energy of the H-F bond. Apart from that, constant momentum is observed between B-C. This suggests that the H atom is traveling in a constant velocity away from the H-F molecule and so there is a gain of translational energy. The energy released from the reaction as translational and vibrational energy can be measured in the form of heat. One of the measurement technique is by using a bomb calorimeter. The bomb calorimeter is an isolated reaction vessel surrounded by a jacket of water. The change in temperature of the water is measured and with the known heat capacity of the bomb calorimeter material and water, the heat released by the reaction, and therefore the energy can be calculated. However, the bomb calorimeter cannot distinguish between translational and vibrational energies. <br />
<br />
A better alternative would be to use infrared chemiluminescence.<ref>Atkins, P; de Paule, J; ''Physical Chemistry 11th edition; ''p. 889-890; Oxford University (2018)</ref> If during the reaction the energy is mostly converted to vibrational energy, the molecule is going to be in a high energy vibrational state. The molecule will then decay into lower vibrational energy states and emit photons while doing so. These photons can be measured, and if a high amount of photons is detected, most of the energy released during the reaction is in the form of vibrational energy. On the other hand the absence indicates that most of the energy is converted into translation energy.<br />
<br />
'''''<u>Good!</u>'''''<br />
<br />
=== Polanyi's empirical rule === <br />
<br />
( "Discuss how the distribution of energy between different modes (translation and vibration) affects the efficiency of the reaction, and how this is influenced by the position of the transition state." )<br />
<br />
The Polanyi's empirical rule states that vibrational energy is more effective at promoting reactions with a late transition state (endothermic reactions), whereas translational energy is more effective at promoting reactions with an early transition state (exothermic reactions).<br />
<br />
'''H2 + F'''<br />
<br />
A series of simulations were carried out to determine the empirical rules. The initial conditions were set as r<sub>HH</sub> = 74 pm and r<sub>HF</sub> = 220 pm, with p<sub>HF</sub> kept at -1.0 g mol<sup>-1</sup> pm fs<sup>-1</sup> and p<sub>HH</sub> varied between -6.1 to 6.1 g mol<sup>-1</sup> pm fs<sup>-1</sup>.<br />
<br />
{| class="wikitable" border=1<br />
!Reaction No.<br />
! p<sub>HH</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>HF</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? <br />
|-<br />
|1<br />
| -6.0 || -1.0|| -403.893 || No<br />
|-<br />
|2<br />
| -4.0 || -1.0|| -421.893 || Yes<br />
|<br />
|-<br />
|3<br />
|<nowiki>-2.0</nowiki><br />
|<nowiki>-1.0</nowiki><br />
|<nowiki>-431.893</nowiki><br />
|No<br />
|-<br />
|4<br />
| 0 || -1.0 || -433.893 || No<br />
|-<br />
|5<br />
|2.0<br />
|<nowiki>-1.0</nowiki><br />
|<nowiki>-427.893</nowiki><br />
|No<br />
|-<br />
|6<br />
| 4.0 || -1.0 ||-413.893||Yes<br />
|-<br />
|7<br />
| 6.0 || -1.0 || -391.893 ||No<br />
|-<br />
|8<br />
|<nowiki>-1.6</nowiki><br />
|0.2<br />
|<nowiki>-432.774</nowiki><br />
|Yes<br />
|}<br />
<br />
In these reactions, p<sub>HH </sub>signifies the vibrational momentum of the H-H bond and p<sub>HF </sub>signifies the translational momentum of F towards H<sub>2</sub>. For the reactions ran, only 2, 6and 8 were reactive and the p<sub>HH</sub> of reaction 2 and 6 have a higher absolute value than reaction 8, showing that more vibrational energy is put in than translational.<br />
<br />
The total energy of reaction 2 and 6 is higher than reaction 8, indicating that reaction 2 and 6 are not as efficient as reaction 8. This can be explained by the exothermic nature of the reaction, which according to Polanyi's rule, is better promoted by translational energy. Since more vibrational energy was present in reaction 2 and 6, the reaction was less efficient. <br />
<br />
'''HF + H'''<br />
<br />
For this reaction, the initial condition of r<sub>HF</sub> = 91 pm and r<sub>HH</sub> = 220 pm is used, with both p<sub>HF</sub> and p<sub>HH </sub>being varied According to Polanyi's rules the reverse of the above would be true, since this reaction is endothermic. Therefore, it is expected that high vibrational momentum and low translational momentum would produce the most efficient reaction (I.e. lower total energy).<br />
<br />
<br />
{| class="wikitable" border=1<br />
!Reaction No.<br />
! p<sub>HH</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>HF</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? <br />
|-<br />
|1<br />
|<nowiki>-22</nowiki>||-0.1||-77.798||Yes<br />
|-<br />
|2<br />
|<nowiki>-10</nowiki><br />
|<nowiki>-10</nowiki><br />
|<nowiki>-556.972</nowiki><br />
|No<br />
|-<br />
|3<br />
|<nowiki>-0.1</nowiki>||-26||-206.404|| Yes<br />
|}<br />
Reaction 1 was given with high vibrational energy and low translational energy. Although reactive, the total energy was relatively high, showing an inefficient reaction. When the conditions are reversed in reaction 3, the total energy lowers by almost three-fold, showing a large increase in efficiency. Reaction 2 was given the same vibrational and translational momentum and resulted in an unsuccessful reaction. All the above observations support the Polanyi's empirical rule for endothermic reactions. <br />
<br />
'''''<u>Good systematic analysis! Well done!</u>''''' <br />
<br />
== References ==<br />
<references><br />
<br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01566447&diff=812910MRD:015664472020-06-25T22:18:50Z<p>Ng611: </p>
<hr />
<div>'''''<u>5/5 - An excellent report, well done. Some minor comments throughout (see the below)</u>'''''<br />
<br />
== H + H<sub>2</sub> system ==<br />
===Dynamics from Transition State region ===<br />
H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub> -> H<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub><br />
<br />
<br />
The progress of the reaction above can be mapped on a potential energy surface (PES). The trajectory shown on the potential energy surface represents the change in potential energy, r<sub>AB</sub> and r<sub>BC</sub> as reaction proceeds given an initial set of parameters (r<sub>AB</sub>, r<sub>BC</sub>,p<sub>AB</sub>, p<sub>BC</sub> at time=0, where r is distance and p is momenta). <br />
<br />
<br />
[[File:01566447HH2Surface_Plot1_1.PNG|900px|thumb|center|<i>Potential energy surface for reaction of H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub></i>.]]<br />
<br />
<br />
<br />
On the PES, the transition state (TS) is a <b>first order saddle point</b>. It is the maxima on the minimum energy path, and the minima in the directions orthogonal to this path.<br />
<br />
Assuming a set of initial parameters that result in the successful formation of the products H<sub>A</sub>-H<sub>B</sub> and H<sub>C</sub>, the trajectory shown passes though a saddle point which represents the TS of the reaction. It is mathematically defined as <b>āV(r<sub>AB</sub>)/ār<sub>AB</sub>=āV(r<sub>BC</sub>)/ār<sub>BC</sub>=0</b>. Since Force=āV(r)/ār, zero gradient means zero force along AB and BC. To distinguish the TS from a local minima on the PES, we can use the Hessian matrix to compute the Hessian eigenvalues. For local energy minima, both Hessian eigenvalues will be positive. <b>For a TS, one Hessian eigenvalue is positive while the other is negative.</b><br />
<br />
'''''<u>Good definition!</u>'''''<br />
<br />
====Locating TS position====<br />
If initial r<sub>AB</sub> and r<sub>BC</sub> distances are set to the distances corresponding to the TS, with no initial momentum, the force acting on the atoms is 0 thus would remain there indefinitely. Since the H + H<sub>2</sub> system is symmetrical, r<sub>AB</sub>=r<sub>BC</sub> at the TS. <br />
<br />
[[File:01566447r_ts_2.png|500px|thumb|center|<i>Internuclear distance-time graph of r<sub>AB</sub>=r<sub>BC</sub>=90.8pm at zero momenta, internuclear distances at TS</i>]]<br />
<br />
<br />
The TS position <b>r<sub>ts</sub></b> is approximately <b>90.8 pm</b>, where r<sub>ts</sub>=r<sub>AB</sub>=r<sub>BC</sub>. The parameters set for the plotted trajectory r<sub>AB</sub>=r<sub>BC</sub>=90.8 pm and p<sub>AB</sub>=p<sub>BC</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. Oscillation is at a minimum as shown by the horizontal lines on the Internuclear distance-time graph above, which means the system is at the midpoint on the ridge, the atoms are stationary and system is at the transition state: r<sub>ts</sub>=(r<sub>AC</sub>-r<sub>AB</sub>)=(r<sub>AC</sub>-r<sub>BC</sub>)=90.8 pm as illustrated in the graph.<br />
<br />
'''''<u>Good TS estimate. </u>'''''<br />
<br />
===Trajectories from r<sub>AB</sub> = r<sub>ts</sub>, r<sub>BC</sub> = r<sub>ts</sub>+1pm===<br />
There are two ways to calculate the trajectory, 'Dynamics' and 'Minimum energy path(mep)', the latter of which resets the momenta and velocity to 0 at every time step. Results obtained from using the 2 different methods are shown below. Products formed are H<sub>A</sub>-H<sub>B</sub> and H<sub>C</sub>.<br />
<br />
====Trajectory from 'Dynamics':====<br />
{| class="wikitable" border="1"<br />
| [[File:01566447dynamics_contour.png|thumb|<i>Contour plot</i>]] || [[File:01566447dynamics_energy.png|thumb|<i>Energy-Time plot</i>]]<br />
|}<br />
<br />
The trajectory calculated using the 'Dynamic' option shows greater oscillation, which means the product molecules vibrate and possess kinetic energy. Potential energy is converted to kinetic energy as shown in the Energy-Time graph where the kinetic energy increases whilst potential energy decreases.<br />
<br />
====Trajectory from 'mep':==== <br />
{| class="wikitable" border="1"<br />
| [[File:01566447mep_contour.png|thumb|<i>Contour plot</i>]] || [[File:01566447mep_energy.png|thumb|<i>Energy-Time plot</i>]]<br />
|}<br />
<br />
Whereas in the 'mep' contour plot, the trajectory is smooth and follows the valley floor of the PES, kinetic energy throughout the reaction path is constantly 0 as shown in the Energy-Time graph. Atoms possess no kinetic energy to vibrate as a result of removing the inertial effect, thus no oscillation is observed from the trajectory on the contour plot. This is not representative of reality because atoms have mass and thus have inertia.<br />
<br />
The mep connects the 2 energy minima on the PES and the energy maxima along the MES corresponds to the saddle point that represents the transition state of the reaction.<br />
<br />
'''''<u>Good explanation!</u>'''''<br />
<br />
====Changing initial conditions====<br />
If we used the initial conditions r<sub>AB</sub> = r<sub>ts</sub>+1pm, r<sub>BC</sub> = r<sub>ts</sub>, reaction would proceed in the opposite direction instead, products formed are H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub>.<br />
<br />
After setting the initial parameters with the final positions and reversing the signs of the momenta, trajectories from the Dynamics calculation differs from that of mep. For the dynamics calculation, reaction proceeds back towards the initial position. For the mep calculation, the pathway proceeds in the opposite direction and continues to follow the valley floor of the PES.<br />
<br />
'''''<u>Why not back this up with an example?</u>'''''<br />
<br />
===Reactive and Unreactive Trajectories===<br />
{| class="wikitable" border="1" align=ācenterā<br />
|+ Table of calculated trajectories from different combinations of momenta, using 'Dynamics' calculation type:<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ mol<sup>-1</sup> !! Reactive? !! Description of Dynamics !! Illustration of trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Yes || As H<sub>A</sub> approaches H<sub>B</sub>-H<sub>C</sub>, AB distance decreases with very small vibrations due to the low momenta, eventually collides with H<sub>B</sub>-H<sub>C</sub> with enough energy to cross the transition state(TS) which breaks the H<sub>B</sub>-H<sub>C</sub> bond and forms the H<sub>A</sub>-H<sub>B</sub> bond, thus BC distance increases. The trajectory after the TS oscillates more due to the greater momenta along BC. || [[File:01566447Surface_Plot_1.png|300px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || No || As H<sub>A</sub> approaches H<sub>B</sub>-H<sub>C</sub>, AB distance decreases. The trajectory is wavy due to the momenta along AB. However, system does not have enough energy to cross reach the TS before H<sub>A</sub> starts moving away from H<sub>B</sub>-H<sub>C</sub> without any H-H bond formation or cleavage. || [[File:01566447Surface_Plot_2.png|300px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Yes || Dynamics of this system is very similar to the 1st system. The difference is the momenta along AB is larger than that of the 1st system, resulting in a more wavy trajectory before the TS which means more vibrations along AB. Reaction crosses the TS once and forms the products H<sub>A</sub>-H<sub>B</sub> and H<sub>C</sub>. || [[File:01566447Surface_Plot_3.png|300px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || No || As H<sub>A</sub> approaches H<sub>B</sub>-H<sub>C</sub>, due to the larger momenta, there are larger vibrations along AB, enough energy to reach the TS and form H<sub>A</sub>-H<sub>B</sub> and H<sub>C</sub>. However, momenta along BC is too large, resulting in very large vibrations, which is sufficient for system to cross the TS again and break the H<sub>A</sub>-H<sub>B</sub> bond to reform the reactants, H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub>. || [[File:01566447Surface_Plot_4.png|300px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Yes || H<sub>A</sub> and H<sub>B</sub>-H<sub>C</sub> has sufficient energy to overcome the energy barrier to form the TS and form the products H<sub>A</sub>-H<sub>B</sub> and H<sub>C</sub>. The products possess a lot of vibrational energy as shown by the large oscillations after TS is crossed.|| [[File:01566447Surface_Plot_5.png|300px]]<br />
|}<br />
<br />
From the table, we can conclude that the hypothesis that trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive is false. Comparing the 1st and the 4th graph, we can see the 4th has greater momenta thus more kinetic energy, but the trajectory is unreactive, while the 1st graph shows a reactive trajectory. This means that whether a trajectory is reactive or not is not dependent only on the size of its values, because the TS can be recrossed. The momentas must be of the right combination such that the resulting products are formed after crossing the TS(once or more than once). Atoms must possess the right amount of vibrational energy and possess the correct orientation for a reaction to proceed successfully.<br />
<br />
===Transition State Theory (TST)===<br />
TST has certain assumptions which may lead to deviations in reaction rate values from experimental values.<br />
<br />
Assumptions<ref name="TSTassumptions" />:<br />
1)Born-Oppenheimer Approximation where electronic and nuclear motions can be separated.<br />
<br />
2)Reactant molecules are distributed among their states according to the Maxwell-Boltzmann distribution.<br />
<br />
3)Molecular systems that have crossed the TS in the direction of the products cannot recross the TS again to reform reactants<br />
<br />
4)In the TS, motion along the reaction coordinate may be separated from the other motions and treated classically as translation, ignoring any quantum tunneling effects.<br />
<br />
5)Even in the absence of equilibrum between the reactant and product molecules, the TS that are becoming products are distributed among their states according to Maxwell-Boltzmann laws.<br />
<br />
Assumption 4 which states that quantum effects are ignored in the TST does not hold in reality, because quantum tunneling effects do occur, which means reactant molecules do not need to overcome the activation energy and cross the TS to form products. Hence in reality, experimental reaction rates would be larger than those predicted using the TST.<br />
<br />
Assumption 3 is also another assumption that does not represent what happens in reality. As seen in the table of reactive and unreactive trajectories, the 4th reaction(p<sub>1</sub>=-5.1, p<sub>2</sub>=-10.1) is an example of how the TS can be crossed again even after formation of products. Leading to lower experimental reaction rates compared to the predicted values.<br />
<br />
Effect of recrossing of TS is likely to be larger than that of quantum tunneling. Hence the overall effect is that the <b>predicted rate constants using TST is an overestimation of the experimental rate constants.</b><br />
<br />
'''''<u>Absolutely correct, well done!</u>'''''<br />
<br />
== F-H-H system ==<br />
===PES inspection===<br />
The reaction of <b>F<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub></b> -> F<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub> is as shown in the PES below(left to right). As F<sub>A</sub>H<sub>B</sub> distance decreases, the potential energy decreases, indicating reaction in this direction is <b>exothermic</b>.<br />
<br />
[[File:01566447_endoexo.png|600px|thumb|center|PES of F + H<sub>2</sub> and FH + H system]]<br />
<br />
Whereas the reaction in the opposite direction(right to left): <b>F<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub></b> -> F<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub>, shows an increase in energy, indicating the reaction is <b>endothermic</b>.<br />
<br />
====How bond strength affects energetics of reaction====<br />
{| class="wikitable" border="1"<br />
|+ Bond strengths<br />
! Bond !! Bond energy (kJ mol<sup>-1</sup>)<ref name="BondStrength" /><br />
|-<br />
| H-H || 436<br />
|-<br />
| F-H || 569<br />
|}<br />
<br />
As shown in the table above, the bond strength of H-H is weaker than F-H. This means that for the reaction of F + H<sub>2</sub>, energy released from forming the F-H bond is greater than the energy required to break the H-H bond. The reaction releases energy thus it is exothermic.<br />
<br />
On the other hand, the reaction of FH + H is endothermic because energy required to break the F-H bond is greater than the energy released from the cleavage of the H-H bond. Overall, energy is required for the reaction to proceed successfully.<br />
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====Transition state====<br />
The approximate transition state was located with the help of Hammond's postulate. Below are the parameters, force, energy and Hessian eigenvalues for the transition state. The forces are close to or at zero, the Hessian eigenvalues have opposing signs which indicates this is a saddle point and the transition state. <b>r<sub>AB</sub>=180.6pm and r<sub>BC</sub>=74.5pm.</b><br />
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[[File: 01566447TS_parameters.PNG|500px|thumb|center|Parameters and calculated values for the TS of F-H-H system]]<br />
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A summary of the reaction showing the structure of the reactants, products and transition state is shown below:<br />
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[[File:01566447FHH_Reaction.jpg|700px|center]]'''''<u>Excellent!</u>'''''<br />
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====Activation energy====<br />
The energies of the TS, reactants and products were calculated using an mep trajectory and shown in the figure below:<br />
[[File: 01566447_activationE.png|600px|thumb|center|Energy against reaction coordinate graph and calculation of activation energies]]<br />
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'''''<u>Good estimates and nice figure. </u>'''''<br />
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The approximate activation energy for the <b>exothermic reaction(F + H<sub>2</sub>) is 1.118 kJ mol<sup>-1</sup></b>.<br />
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The approximate activation energy for the <b>endothermic reaction(FH + H) is 126.564 kJ mol<sup>-1</sup></b>.<br />
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===Reaction Dynamics===<br />
====Mechanism of Energy release====<br />
[[File:01566447_Plot2_momenta.png|500px|thumb|center|Momenta-Time graph of a reactive trajectory for the reaction of F + H<sub>2</sub>]]<br />
From the momenta graph above, we can see the oscillations of the product molecule is much larger than that of the reactant molecules. This implies that during the reaction, the decrease in potential energy releases energy in the form of kinetic energy, more specifically vibrational energy as shown by the large oscillations of product molecule after TS is crossed. This gain in vibrational energy can be experimentally measured using Infrared Chemiluminescence. The intensities of the IR absorption peaks can be used to calculate the relative populations of the product at each vibrational energy level. Products with more vibrational energy will populate higher energy levels, giving rise to overtones which has a smaller wavenumber compared to the fundamental peak. Over time, the overtones will become smaller while the fundamental peak becomes larger. This is because product molecules in the excited state release kinetic energy as heat and fall to the ground state, less product is at higher vibrational states. <br />
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This provides an alternative way to experimentally confirm the mechanism of energy release. The heat released, converted from kinetic energy can be measured using calorimetry. The increase in temperature due to kinetic energy is measured and used to calculate the amount of kinetic energy released. However, we are unable to differentiate between vibrational kinetic energy and translational kinetic energy using this method. Therefor, infrared chemiluminescence is preferred over calorimetry.<br />
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'''''<u> Absolutely correct. </u>'''''<br />
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====Polanyi's Empirical Rules====<br />
For exothermic reactions with early TS, more translational energy over vibrational energy tends to lead to more effective reactions. This can be illustrated by the cases studied using the reaction of F<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub>. Whereas for endothermic reactions with late TS, more vibrational energy over translational energy usually leads to more effective reactions and is shown by the cases studied for reaction of F<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>.<ref name="Polanyi" /> For both cases, there must be enough kinetic energy(vibrational and translational) to overcome the activation barrier for reaction to proceed.<br />
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<b>Exothermic F<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub>:</b><br />
For positions r<sub>AB</sub>=190pm and r<sub>BC</sub>=74pm,<br />
[[File: 01566447_Plot1.png|400px|thumb|center|Plot 1: Unreactive trajectory, p<sub>AB</sub>=-1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>BC</sub>=-6.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
[[File: 01566447_Plot2.png|400px|thumb|center|Plot 2: Reactive trajectory, p<sub>AB</sub>=-1.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>BC</sub>=0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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The momenta along AB represents translational energy of F, whereas momenta along BC represents vibrational energy of H<sub>2</sub>. In Plot 1, momenta along BC is larger, which means more vibrational energy over translational energy was used to drive the reaction, but it did not succeed. On the other hand, in Plot 2, momenta of AB is larger which means more translational energy instead of vibrational energy was used to drive the reaction. The reaction was successful as seen from the reactive trajectory. This is in line with Polanyi's Empirical rules, where an exothermic reaction with early TS is more effectively driven by translational energy.<br />
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<b>Endothermic F<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>:</b><br />
For positions r<sub>AB</sub>=92pm and r<sub>BC</sub>=225pm,<br />
[[File: 01566447_Plot3.png|400px|thumb|center|Plot 3: Unreactive trajectory, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>BC</sub>=-10 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
[[File: 01566447_Plot4.png|400px|thumb|center|Plot 4: Unreactive trajectory, p<sub>AB</sub>=-20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>BC</sub>=-1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
[[File: 01566447_Plot5.png|400px|thumb|center|Plot 5: Reactive trajectory, p<sub>AB</sub>=-29 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and p<sub>BC</sub>=-1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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''<u>Good examples for the most part. '''Good attempt with the final figure, but what you've really done is just 'blasted through' the barrier rather than crossing it as your trajectory doesnt go anywhere near the TS! It is a tricky trajecotyr to get right though. '''</u>''<br />
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The momenta along AB represents vibrational energy of F-H, while the momenta along BC represents the translational energy of the lone H atom. For Plot 3, momenta along BC is quite high, which means the system has a lot of translational energy which is ineffective in driving the reaction, resulting in an unreactive trajectory. For Plot 4, momenta along BC is reduced whereas momenta along AB increases, which raises the amount of vibrational energy compared to translational energy. However, there is still not enough vibrational energy to yield a reactive trajectory. Lastly, in Plot 5, momenta along AB is large enough to effectively drive the reaction and produce a reactive trajectory. This is consistent with Polanyi's empirical rules which states endothermic reactions with late TS are more effectively driven by vibrational energy.<br />
<br />
==References==<br />
<references><br />
<ref name="TSTassumptions">J. I. Steinfeld, J. S. Francisco, W. L. Hase, in <i>Chemical Kinetic and Dynamics</i>, Prentice-Hall, 2<sup>nd</sup> ed., 1998, ch. 10, pp. 287-321.</ref><br />
<ref name="BondStrength">Chemistry LibreTexts, https://chem.libretexts.org/Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_202_-_General_Chemistry_II/Unit_5%3A_Molecular_Shape/5.1%3A_Bond_Strength, (accessed May 2020).</ref><br />
<ref name="Polanyi">K. J. Laidler, in <i>Chemical Kinetics</i>, Harper-Collins, 3<sup>rd</sup> ed, 1951, ch.12, pp. 460-471.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD01506162&diff=812909MRD015061622020-06-25T22:07:15Z<p>Ng611: </p>
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<div>'''''<u>3/5 - A good report overall. There are several gaps in your knowledge that you should fill before moving on (see comments). I would have liked to have seen more detail in places (highlighted throughout). Good use of figures!</u>'''''<br />
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= Exercise 1: H + H<sub>2</sub> system =<br />
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==Dynamics from the transition state region==<br />
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1. The transition state is mathematically defined as having āV(r<sub>1</sub>)/ār<sub>1</sub> = āV(r<sub>2</sub>)/ār<sub>2</sub>=0 and āV<sup>2</sup>(r<sub>i</sub>)/ār<sub>i</sub> <0 '''''<u> This describes a potential energy maximum, rather than a TS. What mathematical structure does a TS resemble and how would you define it mathematically?</u>''''' The energy goes down most deeply at transition state along the mininum energy path linking reactants and products. A transition state can be identified by having constant internuclear distances over time.A local minimum has āV<sup>2</sup>(r<sub>i</sub>)/ār<sub>i</sub> >0.<br />
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==Trajectories from r<sub>1</sub> =r<sub>2</sub>: locating the transition state==<br />
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[[File:Q2_90_pm_01506162.png| 300 px| left |thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=90 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Internuclear_Distances_vs_Time_for_transition_state_01506162.png| 300 px| center | thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Q2_92_pm_01506162.png| 300 px| left | thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=92 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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2. My best estimate of the transition state position is r<sub>1</sub> = r<sub>2</sub>= 90.775 pm. The trajectory with this estimate and zero initial momentum has constant internuclear distances throughout the time. It doesn't "roll" towards the reactants or products proving that the trajectory is exactly at transition state.<br />
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'''''<u>Correct TS estimate, well done</u>'''''<br />
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==Trajectories from r<sub>1</sub> = r<sub>ts</sub> + Ī“, r<sub>2</sub> = r<sub>ts</sub> ==<br />
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[[File: Q3_dynamics_plot_01506162.png|300 px | left | thumb | Contour plot with initial conditions set as r<sub>1</sub>= 91.775 pm, r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p2<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and calculation type set as Dynamics ]]<br />
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[[File: Q3_MEP_plot_01506162.png | 300 px | right | thumb | Contour plot with initial conditions set as r<sub>1</sub>= 91.775 pm r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p2<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and calculation type set as MEP]]<br />
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3. As shown in the pictures above, Mep doesn't have oscillatory behaviour corresponding to vibrations since momenta are always reset to zero in each time step. Meanwhile the trajectory calculated from calculation type being Dynamics has oscillations .<br />
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'''''<u> Good. A little more detail would be useful. How is the direction od descent in the PES calculation determined?</u>'''''<br />
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== Reactive and unreactive trajectories ==<br />
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4.<br />
{| class="wikitable" border="1"<br />
|+ <br />
! p1/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p2/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Yes || A approaches BC which has no oscillatory behaviour. A bond between A and B forms and the bond between B and C breaks at transition state. C dissociates from new molecule AB while AB oscillates. <br />
|| [[File: Contour_plot_for_first_set.png| 200px]]<br />
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| -3.1 || -4.1 || -420.077 || No || A approches BC which possesses oscillatory behaviour. No bond forms between A and B and the bond between B and C doesn't break when A gets close to B. Then A moves further away from BC while BC continues to oscillate. || [[File: Contour_plot_for_second_set_yqc01506162.png | 200px]]<br />
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| -3.1 || -5.1 || -413.977 || Yes || A approaches BC which oscillates in a low amplitude. A bond between A and B forms and the bond between B and C breaks at transition state. C dissociates from new molecule AB while AB oscillates. <br />
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|| [[File: Contour_plot_for_third_set.png | 200px]]<br />
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| -5.1 || -10.1 || -357.277 || No || A approaches BC which doesn't oscillate. A bond between A and B forms and the bond between B and C breaks at transition state. AB oscillates and C moves away from AB to an extent. Barrier recrossing occurs in which the bond between AB breaks and BC reforms while A moves away from BC. || [[File: Contour_plot_for_fourth_set.png | 200px]]<br />
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| -5.1 || -10.6 || -349.477 || Yes || A approaches BC with no oscillation. A bond between A and B forms then breaks followed by BC reforming. The bond between B and C breaks then the bond between A and B reforms. C dissociates from AB while AB oscillates. || [[File: Contour_plot_for_fifth_set.png | 200px]]<br />
|}<br />
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One can conclude from the table that in order to have a reactive trajectory, translational energy of incoming atom needs to pair up with adequate vibrational energy of the molecule e.g. high, high and low, low.<br />
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'''''<u>I'm not quite sure what you're getting at here. In general, it is hard to predict which barriers will be recrossed simply from looking at trajectories.</u>'''''<br />
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== Transition State Theory ==<br />
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[[File:Contour_plot_for_fourth_set.png | thumb| center | Contour plot with initial conditions set as r<sub>1</sub>= 74 pm, r<sub>2</sub>= 200 pm, p<sub>1</sub>= -5.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>2</sub>= -10.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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5. Transition State Theory relies on three key assumptions and they have different effects in causing the differences between the estimated reaction rates and the experimental values. In this case, we are using the simulation as a much simpler model of real-life experiments. The first assumption which states all trajectories with KE>E<sub>a</sub> are reactive overestimates the reaction rates since in reality barrier recrossing could have occurred. Above is an illustration of barrier recrossing. The second assumption which states the transition states are in quasi equilibrium with the reactants varies the estimated value from the real one to a limited extent. The third assumption which states quantum tunneling is not to be considered underestimates the reaction rate since in reality quantum tunneling does occur. Unfortunately, no illustration from the simulation can be done since our simulation doesn't take quantum tunneling into account either. Overall, Transition State Theory overestimates reaction rates since there are much more barrier recrossing than quantum tunneling in real life experiments.<br />
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'''''<u>Good!</u>'''''<br />
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= Exercise 2: F-H-H system =<br />
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== PES Inspection ==<br />
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[[File: Q6_exothermic_surface_plot_01506162.png| thumb | left | Surface plot with A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Q6_endothermic_surface_plot_01506162.png | thumb | right | Surface plot with A=H, B=H, C=F, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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6. By inspecting the surface plots of reactive trajectories, F + H<sub>2</sub> is exothermic and H + HF is endothermic. This is because H-F has higher bond enthalpy which means it is stronger than H-H. In F+ H<sub>2</sub>, more energy is released to form H-F than required to break H-H. In H + HF, more energy is required to break H-F than released to form H-H.<br />
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'''''<u>Good!</u>'''''<br />
[[File: Q8_TS_zero_energy_01506162.png | thumb | center| 600 px | LEPS GUI screenshot of approximated transition state]]<br />
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[[File: Ex._2_transition_state_distance_vs_time_01506162.png | thumb |center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 181 pm, r<sub>BC</sub>= 74.49 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Ex._2_AB=180_pm,_BC=74.49_pm_distance_vs_time.png | thumb | center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 180 pm, r<sub>BC</sub>= 74.49 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Ex._2_AB=181_pm,_B=75_pm_distance_vs_time_01506162.png | thumb | center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 181 pm, r<sub>BC</sub>= 75 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
7. Transition state was located by using Hammond postulate. I understand that since F + H<sub>2</sub> is exothermic, it has an early transition state and the transition state is reactant-like. Therefore I started by using A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and varied AB and BC distances in order to make forces along AB and BC equal to zero. The approximate position of transition state found is r<sub>HF</sub>=181 pm, r<sub>HH</sub>=74.49 pm which has zero force along the bonds as shown in the screenshot above. Three internuclear distances vs time plots are also shown above to further demonstrate the validity of the position of transition state.<br />
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[[File: SM_pre_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: SM_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: SM_pro_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: TS_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of transition state of F + H<sub>2</sub>]]<br />
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[[File: P_pre_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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[[File: P_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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[[File: P_pro_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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8. Activation energy of F + H<sub>2</sub> was found by measuring potential energies of reactants, transition state and products with kinetic energy=0, momenta=0 and calculation type being MEP. Potential energy of reactants was found by placing the incoming F atom very far away from H<sub>2</sub> molecule at equilibrium bond length =74 pm and kept increasing the distance between F and H<sub>2</sub> until potential energy stopped changing. This resulted in F having no effect on H<SUB>2</SUB> 's potential energy which was found out to be -435.100 kJ mol<sup>-1</sup>. Above pictures are an illustration of the approach. The potential energy of transition state was recorded by setting initial conditions in transition state position approximated above and it was found out to be -433.981 kJ mol<sup>-1</sup>. Potential energy of the products was found by using the same method to find the potential energy of reactants and was found out to be -560.404 kJ mol<sup>-1</sup>. Above pictures are illustrations of the process.H + HF is a reverse reaction of F + H<sub>2</sub> and its activation energy can be calculated from these values as well. E<sub>a</sub> for F + H<sub>2</sub> was therefore found out to be -433.981-(-435.100)=1.119 kJ mol<sup>-1</sup>. E<sub>a</sub> for H + HF was found out to be -433.981-(-560.404)=126.423 kJ mol<sup>-1</sup>. <br />
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'''''<u>Good estimates for Ea in both directions, well done. </u>''''' <br />
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== Reaction dynamics ==<br />
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[[File: Momentum_vs_time_as_proof_of_release_of_reaction_energy_01506162.png | thumb | 600 px | center | Momenta vs time for F + H<sub>2</sub>, A=F , B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=-20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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9. The potential energy released in the reaction of F + H<sub>2</sub> is first converted to kinetic energy which is used to promote the products in vibrational ground state to vibrational first state. The vibrational activation would cause an increase in overtone intensity and decrease in fundamental intensity in IR absorbance spectrum. Overtime, the vibrational first state relaxes back to the ground state emitting IR. The vibrational relaxation would cause overtone intensity to decrease and fundamental peak intensity to increase in IR absorbance spectrum. The mechanism of the release of the reaction energy can be thus experimentally confirmed by measuring the emission of IR during vibrational relaxation by FTIR '''''<u>Not FTIR, as it doesn't have adequate time resolution, but yes, you're largely correct.</u>'''''.<br />
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[[File:Q10_high_translation,_low_vibration_contour_01506162.png | 300 px | thumb | left | Surface plot for H + HF, A=H, B=H, C=F, with initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=-20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Q10_low_translation,_high_vibration_surface_plot_01506162.png | 300 px | thumb | right | Surface plot for H + HF, A=H, B=H, C=F, with initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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10. H + HF is an endothermic reaction thus has a late transition state which according to Hammond's postulate, adopts the product structure. As shown above are the surface plots for H + HF with H having high translational energy, HF having low vibrational energy and H having low translational energy, HF having high vibrational energy respectively. In the case of high translational energy of incoming H atom and high vibrational energy of HF molecule, the incoming H atom flings down the reaction channel, hits the potential wall with high translational energy and therefore bounces back. This leads to an unreactive trajectory. In the case of low translational energy of incoming H atom and high vibrational energy of HF molecule, incoming H atom approaches HF molecule whose direction of vibrational motion aligns with the direction that goes to the products. Since H has low energy, it doesn't hit the wall and bounce back. Also, HF's vibrational motion is with sufficient amplitude to cross the energy barrier and form products. This results in a reactive trajectory. According to Polanyi's empirical rules, vibrational energy is more efficient in promoting a reaction with late transition state meanwhile translational energy is more efficient in promoting a reaction with early transition state.<br />
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'''''<u>Good examples. Did you take any examples for the F+H2 reaction?</u>'''''</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD01506162&diff=812908MRD015061622020-06-25T21:26:49Z<p>Ng611: </p>
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<div><br />
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= Exercise 1: H + H<sub>2</sub> system =<br />
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==Dynamics from the transition state region==<br />
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1. The transition state is mathematically defined as having āV(r<sub>1</sub>)/ār<sub>1</sub> = āV(r<sub>2</sub>)/ār<sub>2</sub>=0 and āV<sup>2</sup>(r<sub>i</sub>)/ār<sub>i</sub> <0 '''''<u> This describes a potential energy maximum, rather than a TS. What mathematical structure does a TS resemble and how would you define it mathematically?</u>''''' The energy goes down most deeply at transition state along the mininum energy path linking reactants and products. A transition state can be identified by having constant internuclear distances over time.A local minimum has āV<sup>2</sup>(r<sub>i</sub>)/ār<sub>i</sub> >0.<br />
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==Trajectories from r<sub>1</sub> =r<sub>2</sub>: locating the transition state==<br />
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[[File:Q2_90_pm_01506162.png| 300 px| left |thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=90 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Internuclear_Distances_vs_Time_for_transition_state_01506162.png| 300 px| center | thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Q2_92_pm_01506162.png| 300 px| left | thumb| Internuclear distances vs time for trajectory with initial conditions r<sub>1</sub>=r<sub>2</sub>=92 pm, p<sub>1</sub>=p<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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2. My best estimate of the transition state position is r<sub>1</sub> = r<sub>2</sub>= 90.775 pm. The trajectory with this estimate and zero initial momentum has constant internuclear distances throughout the time. It doesn't "roll" towards the reactants or products proving that the trajectory is exactly at transition state.<br />
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'''''<u>Correct TS estimate, well done</u>'''''<br />
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==Trajectories from r<sub>1</sub> = r<sub>ts</sub> + Ī“, r<sub>2</sub> = r<sub>ts</sub> ==<br />
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[[File: Q3_dynamics_plot_01506162.png|300 px | left | thumb | Contour plot with initial conditions set as r<sub>1</sub>= 91.775 pm, r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p2<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and calculation type set as Dynamics ]]<br />
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[[File: Q3_MEP_plot_01506162.png | 300 px | right | thumb | Contour plot with initial conditions set as r<sub>1</sub>= 91.775 pm r<sub>2</sub>=90.775 pm, p<sub>1</sub>=p2<sub>2</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and calculation type set as MEP]]<br />
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3. As shown in the pictures above, Mep doesn't have oscillatory behaviour corresponding to vibrations since momenta are always reset to zero in each time step. Meanwhile the trajectory calculated from calculation type being Dynamics has oscillations .<br />
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== Reactive and unreactive trajectories ==<br />
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4.<br />
{| class="wikitable" border="1"<br />
|+ <br />
! p1/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p2/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
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| -2.56 || -5.1 || -414.280 || Yes || A approaches BC which has no oscillatory behaviour. A bond between A and B forms and the bond between B and C breaks at transition state. C dissociates from new molecule AB while AB oscillates. <br />
|| [[File: Contour_plot_for_first_set.png| 200px]]<br />
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| -3.1 || -4.1 || -420.077 || No || A approches BC which possesses oscillatory behaviour. No bond forms between A and B and the bond between B and C doesn't break when A gets close to B. Then A moves further away from BC while BC continues to oscillate. || [[File: Contour_plot_for_second_set_yqc01506162.png | 200px]]<br />
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| -3.1 || -5.1 || -413.977 || Yes || A approaches BC which oscillates in a low amplitude. A bond between A and B forms and the bond between B and C breaks at transition state. C dissociates from new molecule AB while AB oscillates. <br />
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|| [[File: Contour_plot_for_third_set.png | 200px]]<br />
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| -5.1 || -10.1 || -357.277 || No || A approaches BC which doesn't oscillate. A bond between A and B forms and the bond between B and C breaks at transition state. AB oscillates and C moves away from AB to an extent. Barrier recrossing occurs in which the bond between AB breaks and BC reforms while A moves away from BC. || [[File: Contour_plot_for_fourth_set.png | 200px]]<br />
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| -5.1 || -10.6 || -349.477 || Yes || A approaches BC with no oscillation. A bond between A and B forms then breaks followed by BC reforming. The bond between B and C breaks then the bond between A and B reforms. C dissociates from AB while AB oscillates. || [[File: Contour_plot_for_fifth_set.png | 200px]]<br />
|}<br />
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One can conclude from the table that in order to have a reactive trajectory, translational energy of incoming atom needs to pair up with adequate vibrational energy of the molecule e.g. high, high and low, low.<br />
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== Transition State Theory ==<br />
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[[File:Contour_plot_for_fourth_set.png | thumb| center | Contour plot with initial conditions set as r<sub>1</sub>= 74 pm, r<sub>2</sub>= 200 pm, p<sub>1</sub>= -5.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>2</sub>= -10.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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5. Transition State Theory relies on three key assumptions and they have different effects in causing the differences between the estimated reaction rates and the experimental values. In this case, we are using the simulation as a much simpler model of real-life experiments. The first assumption which states all trajectories with KE>E<sub>a</sub> are reactive overestimates the reaction rates since in reality barrier recrossing could have occurred. Above is an illustration of barrier recrossing. The second assumption which states the transition states are in quasi equilibrium with the reactants varies the estimated value from the real one to a limited extent. The third assumption which states quantum tunneling is not to be considered underestimates the reaction rate since in reality quantum tunneling does occur. Unfortunately, no illustration from the simulation can be done since our simulation doesn't take quantum tunneling into account either. Overall, Transition State Theory overestimates reaction rates since there are much more barrier recrossing than quantum tunneling in real life experiments.<br />
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= Exercise 2: F-H-H system =<br />
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== PES Inspection ==<br />
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[[File: Q6_exothermic_surface_plot_01506162.png| thumb | left | Surface plot with A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Q6_endothermic_surface_plot_01506162.png | thumb | right | Surface plot with A=H, B=H, C=F, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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6. By inspecting the surface plots of reactive trajectories, F + H<sub>2</sub> is exothermic and H + HF is endothermic. This is because H-F has higher bond enthalpy which means it is stronger than H-H. In F+ H<sub>2</sub>, more energy is released to form H-F than required to break H-H. In H + HF, more energy is required to break H-F than released to form H-H.<br />
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[[File: Q8_TS_zero_energy_01506162.png | thumb | center| 600 px | LEPS GUI screenshot of approximated transition state]]<br />
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[[File: Ex._2_transition_state_distance_vs_time_01506162.png | thumb |center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 181 pm, r<sub>BC</sub>= 74.49 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Ex._2_AB=180_pm,_BC=74.49_pm_distance_vs_time.png | thumb | center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 180 pm, r<sub>BC</sub>= 74.49 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File: Ex._2_AB=181_pm,_B=75_pm_distance_vs_time_01506162.png | thumb | center | Internuclear distances vs time for A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 181 pm, r<sub>BC</sub>= 75 pm, p<sub>AB</sub>=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
7. Transition state was located by using Hammond postulate. I understand that since F + H<sub>2</sub> is exothermic, it has an early transition state and the transition state is reactant-like. Therefore I started by using A=F, B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and varied AB and BC distances in order to make forces along AB and BC equal to zero. The approximate position of transition state found is r<sub>HF</sub>=181 pm, r<sub>HH</sub>=74.49 pm which has zero force along the bonds as shown in the screenshot above. Three internuclear distances vs time plots are also shown above to further demonstrate the validity of the position of transition state.<br />
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[[File: SM_pre_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: SM_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: SM_pro_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of reactants of F + H<sub>2</sub>]]<br />
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[[File: TS_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of transition state of F + H<sub>2</sub>]]<br />
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[[File: P_pre_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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[[File: P_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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[[File: P_pro_max_energy_01506162.png | 600 px | thumb | center | Illustration of finding the potential energy of products of F + H<sub>2</sub>]]<br />
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8. Activation energy of F + H<sub>2</sub> was found by measuring potential energies of reactants, transition state and products with kinetic energy=0, momenta=0 and calculation type being MEP. Potential energy of reactants was found by placing the incoming F atom very far away from H<sub>2</sub> molecule at equilibrium bond length =74 pm and kept increasing the distance between F and H<sub>2</sub> until potential energy stopped changing. This resulted in F having no effect on H<SUB>2</SUB> 's potential energy which was found out to be -435.100 kJ mol<sup>-1</sup>. Above pictures are an illustration of the approach. The potential energy of transition state was recorded by setting initial conditions in transition state position approximated above and it was found out to be -433.981 kJ mol<sup>-1</sup>. Potential energy of the products was found by using the same method to find the potential energy of reactants and was found out to be -560.404 kJ mol<sup>-1</sup>. Above pictures are illustrations of the process.H + HF is a reverse reaction of F + H<sub>2</sub> and its activation energy can be calculated from these values as well. E<sub>a</sub> for F + H<sub>2</sub> was therefore found out to be -433.981-(-435.100)=1.119 kJ mol<sup>-1</sup>. E<sub>a</sub> for H + HF was found out to be -433.981-(-560.404)=126.423 kJ mol<sup>-1</sup>. <br />
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== Reaction dynamics ==<br />
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[[File: Momentum_vs_time_as_proof_of_release_of_reaction_energy_01506162.png | thumb | 600 px | center | Momenta vs time for F + H<sub>2</sub>, A=F , B=H, C=H, initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=-20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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9. The potential energy released in the reaction of F + H<sub>2</sub> is first converted to kinetic energy which is used to promote the products in vibrational ground state to vibrational first state. The vibrational activation would cause an increase in overtone intensity and decrease in fundamental intensity in IR absorbance spectrum. Overtime, the vibrational first state relaxes back to the ground state emitting IR. The vibrational relaxation would cause overtone intensity to decrease and fundamental peak intensity to increase in IR absorbance spectrum. The mechanism of the release of the reaction energy can be thus experimentally confirmed by measuring the emission of IR during vibrational relaxation by FTIR.<br />
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[[File:Q10_high_translation,_low_vibration_contour_01506162.png | 300 px | thumb | left | Surface plot for H + HF, A=H, B=H, C=F, with initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=-20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= 1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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[[File:Q10_low_translation,_high_vibration_surface_plot_01506162.png | 300 px | thumb | right | Surface plot for H + HF, A=H, B=H, C=F, with initial conditions set as r<sub>AB</sub>= 200 pm, r<sub>BC</sub>= 74 pm, p<sub>AB</sub>=1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>= -20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]]<br />
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10. H + HF is an endothermic reaction thus has a late transition state which according to Hammond's postulate, adopts the product structure. As shown above are the surface plots for H + HF with H having high translational energy, HF having low vibrational energy and H having low translational energy, HF having high vibrational energy respectively. In the case of high translational energy of incoming H atom and high vibrational energy of HF molecule, the incoming H atom flings down the reaction channel, hits the potential wall with high translational energy and therefore bounces back. This leads to an unreactive trajectory. In the case of low translational energy of incoming H atom and high vibrational energy of HF molecule, incoming H atom approaches HF molecule whose direction of vibrational motion aligns with the direction that goes to the products. Since H has low energy, it doesn't hit the wall and bounce back. Also, HF's vibrational motion is with sufficient amplitude to cross the energy barrier and form products. This results in a reactive trajectory. According to Polanyi's empirical rules, vibrational energy is more efficient in promoting a reaction with late transition state meanwhile translational energy is more efficient in promoting a reaction with early transition state.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:aec_18&diff=812907MRD:aec 182020-06-25T21:23:53Z<p>Ng611: </p>
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<div>'''''<u>4/5 - A nearly perfect report, well done-- the only major issue is a lack of examples for the endothermic reaction in the final section. Otherwise, everything here is excellent, well done. </u>'''''<br />
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== Introduction ==<br />
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=== Transition State ===<br />
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==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====<br />
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On a potential energy surface, the transition state is mathematically defined by; āV(ri)/āri=0 whereby the gradient of the curve is zero. The transition state is a local maximum between two minima which the lowest energy reaction path passes through. The potential energy of a given reaction rises an amount equal to the activation energy and here the transition state point is located. After this activated state is formed the energy will decrease to the products energy. The relative energies of reactants in terms of the relative product energies determine the endothermic or exothermic nature of the reaction.<sup>1</sup><br />
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The transition state is found at the 'saddle point' <sup>2</sup> where the derivatives of the slopes in orthogonal directions are zero. The a maximum point can be distinguished from a minimum by by looking at a second differential and finding if its less than or more than zero. For a positive second differential, the point will be a minimum and for negative a maximum. <br />
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[[File:Screen Shot 2020-05-22 at 15.27.33.png|300px|thumb|Figure 1: Transition state location for a symmetrical system]]Good answer! '''''<u>Think also about the directions you need to take the second derivatives along. Where will these directions lie on your PES?</u>'''''[[File:Screen Shot 2020-05-22 at 21.58.37.png|300px|thumb|Figure 2: Potential energy profile for an exothermic (left) and endothermic (right) system.]]<br />
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A potential energy profile for both exothermic and endothermic reactions are shown below: the activation energy is indicated on the diagram. The general course of the reaction in this experiment is B and C are initially joined by a chemical bond which A distorts when the in contact at the right conditions (distance, r values and momentum, p values), causing an exchange in atoms as new bond forms and existing bonds break. In this report r<sub>1</sub> = BC bond distance and r<sub>2</sub> = AB. The reaction co-ordinate on the x axis represents the progress of the reaction and thus the changes in atomic distortion as described above. In figure 2 the gradient at the TS shown is zero and can be confirmed to be a maximum with the second differential giving a negative value.<br />
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At the transition state two reactant molecules have come into such close proximity that a small change will send them in the direction of the products (or back to reactants). <br />
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For a symmetrical system such as H-H-H described in section 2, the transition state lies at the point shown in figure 1. The reactant and product energies are equal and the TS therefore lies along the y=x line (r<sub>1</sub>=r<sub>2</sub> at the TS). <br />
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Transition state theory is used with statistical thermodynamics to provide a more detailed calculation of rate constants by identifying the contributing factors in the reaction model. <sup>3</sup><br />
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===Locating the transition state for the H-H-H system ===<br />
====Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory====<br />
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The transition state shown in fig.1 can be estimated by finding distances AB=r<sub>1</sub> and BC=r<sub>2</sub> whereby the force /kJ.mol-1.pm-1 is equal to zero. As the H + H<sub>2</sub> surface is symmetric, r<sub>1</sub> and r<sub>2</sub> are equal at this point (TS lies on y=x line). The momentum is set to zero at the transition state point. <br />
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It was found that a distance of r<sub>1</sub>=r<sub>2</sub>=90.775 pm (90.8 pm to 1.d.p) gave forces of 0.00 kJ.mol-1.pm-1 along AB and BC; this is where the transition state position can be estimated to. It can also be noted that at this point one eigenvalue (Ļ<sub>2</sub>) is positive and the other is negative, in accordance to Hessian matrix, more on this in section 4.<br />
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'''''<u>Correct answer, well done!</u>'''''<br />
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At the transition state point <br />
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{| class="wikitable" border="1"<br />
|+ Table to show locating transition state trial and error runs<br />
! r<sub>1</sub>=r<sub>2</sub> distance /pm !! Force along AB and BC /kJ.mol<sup>-1</sup>.pm<sup>-1</sup> !! Ļ<sub>2</sub> kJ.mol<sup>-1</sup>.pm<sup>-2</sup> !! Etot !! Illustration <br />
|-<br />
| 100 || -1.66 || -0.077 and +0.091 || -403.554 || [[File:Screen Shot 2020-05-22 at 22.03.10.png|300px]]<br />
|-<br />
| 80 || +2.505 || +0.081 and +0.301 || -391.146 || [[File:Screen Shot 2020-05-22 at 22.03.10.png|300px]]<br />
|-<br />
| 90 || +0.132 || +0.022 and +0.175 || -415.276 || [[File:Screen Shot 2020-05-22 at 22.05.18.png|300px]]<br />
|-<br />
| 90.8 || -0.004 || +0.028 and +0.167 || -415.378 || [[File:Screen Shot 2020-05-20 at 15.58.28.png|300px]]<br />
|}<br />
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As the point reaches the transition state for this reaction, the bond distance lines on the internuclear distance against time plot becomes horizontal. At this point the distances are not fluctuating as the bonds do not oscillate and so appear as constant over time on the graph.<br />
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'''''<u>Nice!</u>'''''<br />
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==Trajectory investigation==<br />
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===Trajectories from r1 = rts+Ī“, r2 = rts===<br />
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====Comment on how the mep and the dynamic trajectory differ====<br />
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The minimum energy reaction path is a path in which the velocities are re-set to zero after every time step. <br />
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We can set the conditions of the reaction to where it is 'nugged' '''''<u>lol </u>'''''slightly from the trastition state point, still at zero momenta. The conditions below were used:<br />
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r<sub>1</sub> = rts + 1 and r<sub>2</sub> = rts pm : r<sub>1</sub> = 91.8 and r<sub>2</sub> = 90.8<br />
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On the left we can see the MEP trajectory shows the path of the reacting molecules does not oscillate; and on the right the dynamic trajectory which shows the molecule is oscillating. This is expected for a mep as the particle has infintely slow motion as its velocity is reset to zero in every step. This trajectory shows the reaction conditions to be unreactive as the reactants are reformed at the end (energy does not reach activation energy and TS is not passed). <br />
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'''''<u>Absolutely correct, well done!</u>''''' <br />
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{| class="wikitable" border="1"<br />
|+ MEP and dynamic trajectories<br />
! MEP !! Dynamic<br />
|-<br />
| [[File:Screen Shot 2020-05-22 at 16.28.09.png|300px]] || [[File:Screen Shot 2020-05-20 at 15.26.50.png|250px]]<br />
|}<br />
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If we were to change the initial conditions r<sub>1</sub> = rts and r<sub>2</sub> = rts+1 pm : r<sub>1</sub> = 90.8 and r<sub>2</sub> = 91.8, we can observe the effect this will have on the āInternuclear Distances vs Timeā plot and the āMomenta vs Timeā plot. For these examples (fig.3) the reaction does occur as the internuclear distance plot shows BC distance decreasing and AB (reactant) increasing. The momenta plots show that the BC bond is distorted back and forth (oscillates) whilst the AB bond gradually breaks. This is reversed in fig.4 when the transition state in 'nugged' in the opposite direction.<br />
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{| class="wikitable" border="1"<br />
|+ Dynamic trajectories for 'nugged' TS<br />
! internuclear distance vs time plot !! momenta vs time plot<br />
|-<br />
| [[File:Screen Shot 2020-05-20 at 15.57.07.png|300px|thumb|left|Figure 3a: Distance time plot for r1 = 90.8 pm and r2 = 91.8 pm.]] || [[File:Screen Shot 2020-05-20 at 15.57.27.png|300px|thumb|centre|Figure 3b: Momentum time plot for r1 = 90.8 pm and r2 = 91.8 pm.]] <br />
|-<br />
| [[File:Screen Shot 2020-05-20 at 16.14.23.png|300px|thumb|left|Figure 4a: Distance time plot for r1 = 91.8 pm and r2 = 90.8 pm.]] || [[File:Screen Shot 2020-05-20 at 16.14.39.png|300px|thumb|centre|Figure 4b: Momentum time plot for r1 = 91.8 pm and r2 = 90.8 pm.]] <br />
|}<br />
<br />
The final distance and momentum values can be found from these plots (using the 'get last geometry' tool) and a calculation can be set up where the initial positions correspond to the final positions of the trajectory above. And when momenta values signs are reversed. By reversing the momenta the reaction now occurs as the molecules collide with enough energy to reach activation. When both signs are positive (first row) no reaction occurs as the molecules move in opposite directions. <br />
<br />
{| class="wikitable" border="1"<br />
|+ Table to show distance and momenta time graphs with momenta values sign reversed<br />
! r1 distance /pm !! r2 distance /pm !! p1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Internuclear distance vs time || Momenta vs time<br />
|-<br />
| 74.02 || 352.63 || 3.21 || 5.07 || [[File:Screen Shot 2020-05-20 at 16.33.46.png|300px]] || [[File:Screen Shot 2020-05-20 at 16.33.15.png|300px]]<br />
|-<br />
| 74.02 || 352.63 || -3.21 || -5.07 || [[File:Screen Shot 2020-05-20 at 16.45.34.png|300px]] || [[File:Screen Shot 2020-05-20 at 16.46.04.png|300px]]<br />
|}<br />
<br />
'''''<u>Did you round your momenta here? You should see the reaction happen in reverse, starting at the products and ending at the TS, where it stays. You seem to have overshot a bit as your momenta dont converge to 0. </u>'''''<br />
<br />
=== Reactive and unreactive trajectories ===<br />
<br />
====Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?====<br />
<br />
In the following table the values r1 and r2 were set to 74 pm and 200 pm respectively. <br />
<br />
{| class="wikitable" border="1"<br />
|+ caption<br />
! p1 /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p2 /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Etot kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Yes || The reaction proceeds to completion shown by the trajectory from reactants across the transition state towards the product channel with increased BC bond distance. || [[File:Screen Shot 2020-05-20 at 20.29.05.png|300px]]<br />
|-<br />
| -3.1 || -4.1|| -420.077 || No || The trajectory initially moves down the reactant channel towards the products but then changes direction and passes back to reactants (increasing AB value), without reaching the transition state. The increased amount of oscillation indicates higher vibrational energy than in example one. || [[File:Screen Shot 2020-05-20 at 20.38.29.png|300px]]<br />
|-<br />
| -3.1 || -5.1|| -413.977 || Yes || The reaction proceeds to completion shown by the trajectory from reactants to pass the transition state and down the product channel. The oscillations are similar to example one as energy is relatively similar. || [[File:Screen Shot 2020-05-20 at 20.48.54.png|300px]]<br />
|-<br />
| -5.1 || -10.1|| -357.277 || No || Despite having a large amount of energy the reaction does not go to completion. Unlike example two, this trajectory does pass the transition state but then reforms the reactants. This is contrary to the assumptions made in the transition state theory. || [[File:Screen Shot 2020-05-20 at 20.56.22.png|300px]]<br />
|-<br />
| -5.1 || -10.6|| -349.477 || Yes || The reaction goes to completion. the trajectory travels over the transition state once and then unlike in example one, crosses back over the transition state towards the reactants, only to then recross for the third time a leading to a trajectory down the product channel. || [[File:Screen Shot 2020-05-20 at 21.03.13.png|300px]]<br />
|}<br />
<br />
From the table there seems to be no correlation between the total energy of each system and wether the reaction proceeds or not. However there does seem to be increased correlation between the increase in energy (less negative) and the ability for the trajectory course to pass over the transition state multiple times. There is also no correlation between increased total momenta and reactivity.<br />
<br />
== Transition State Theory ==<br />
<br />
====Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?====<br />
<br />
There are three main assumptions transition theory makes when predicting values<sup>3</sup>:<br />
<br />
1. Molecular systems which pass the transition state will continue to products and do not turn back and form reactant molecules again. For example there can be fluctuations between reactants and transition state but once transition state is reached all is converted to product R <=> TS -> P. Once the energy barrier between reactants and transition state is overcome (activation energy) the reaction will proceed to form products. <br />
<br />
2. The energy distribution among the reactant molecules is in accordance with the Maxwell-Boltzmann distribution. It is assumed the concentration of TS can be calculated using quasi-equilibrium theory, as the system only deviates from equilibrium by an infinitesimal amount. <br />
<br />
3. Quantum effects can be ignored and a reaction can be fully assumed as classical motion. Effects such as quantum tunnelling are ignored, leading to an underestimate of rate.<br />
<br />
Example four in the table does not obey assumption 1 above as it displays a trajectory that passes over the transition state and then passes back over to reform reactants. Example 5 also crosses the transition state and then back to reactants (and then crosses for a third time), not obeying this assumption. The rate determining step is the TS-> P and so, as a result the TS theory would overestimate the actual observed rate constant for these trajectories. This mean that overall the TS theory overestimates our experimental values, despite the fact effects such as quantum tunnelling are ignored which would lead to an underestimate (smaller contribution compared to the explained above).<br />
<br />
'''''<u>This is absolutely the case, well done!</u>'''''<br />
<br />
== F - H - H System ==<br />
<br />
===PES inspection===<br />
<br />
====By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?====<br />
<br />
The F + H<sub>2</sub> system was analysed using the following conditions:<br />
<br />
A = F : B = H : C = H<br />
<br />
r1=BC=74 pm : r2=AB=230pm <br />
<br />
For an unsymmetrical system, the potential energy surface shown in fig.5 can be analysed to determine the exothermic or endothermic nature of the reaction. From this plot we can see that the energy of the reactants is larger than the energy of the products (AB product is a smaller energy value than BC, the reactant) and so by using this in conjunction with fig.2 we can see that the overall process would be exothermic, as energy is released between the BC bond break and AB bond formation. <br />
<br />
[[File:Screen Shot 2020-05-20 at 22.07.03.png|300px|thumb|Figure 5: Potential energy surface plot for an exothermic system.]]<br />
<br />
This knowledge can be related to the strength of the chemical species involved, as the reactant bond strength H-H is less than the product bond strength H-F and so more energy is given out when bonds form compared to the energy required to break bonds. <br />
<br />
In the second experiment, the H H-F system, the following conditions were used:<br />
<br />
A = H : B = H : C = F<br />
<br />
r1=BC=74 pm : r2=AB=230pm <br />
<br />
Here, in fig.6 we can see the surface plot for this reaction, which shows it to be endothermic. This can be seen from the fact the products are higher in energy than the reactants (AB larger value than BC) and so the energy required to form bonds is larger than the energy released when bonds are broken. This evidence supports the theory that the H-F bond is stronger than the H-H bond, of which the exact bond strength values can be calculated as seen in the following section. '''''<u>Good!</u>''''' <br />
<br />
[[File:Screen Shot 2020-05-21 at 16.50.25.png|300px|thumb|Figure 6: Potential energy surface plot for an endothermic system.]]<br />
<br />
====Locate the approximate position of the transition state====<br />
<br />
The transition state of both reactions can be approximated by trial and error method. We know from part one that the transition state can be found by using a calculation that results in the forces along AB and BC to be zero. We also know that the Hessian matrix can be used to identify that one eigenvalue is positive and one negative at the transition state. Furthermore, Hammond's postulate can be used to indicate the location of the TS. Hammond's postulate states that the transition state will resemble either the reactants or the products, depending on which it is closest in energy to. <br />
The TS for an exothermic reaction (fig.2 and fig.5) will be closer in energy to the reactants, whereas for an endothermic reaction it will be closer to the products.<br />
<br />
Using this above information 'trials' were conducted to locate the transition state. This was found to be at the position indicated in the table below.<br />
<br />
{| class="wikitable" border="1"<br />
|+ Transition state location of F + H2 system<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1=p2 momentum /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Etot /kJ.mol<sup>-1</sup> !! Forces along BC /kJ.mol<sup>-1</sup>.pm<sup>-1</sup> !! Forces along AB /kJ.mol<sup>-1</sup>.pm<sup>-1</sup> !! Ļ<sup>2</sup> /kJ.mol<sup>-1</sup>pm<sup>-2</sup><br />
|-<br />
| 74.5 || 181.5 || 0 || -433.981 || -0.007 || +0.000 || -0.002 and +0.332<br />
|}<br />
<br />
[[File:Screen Shot 2020-05-21 at 16.35.48.png|300px|thumb|left|Figure 7: internuclear distance vs time plot at the transition state of F +H2 system.]]<br />
<br />
<br />
At this point the internuclear distance vs time plot shows two horizontal straight lines as the product and reactant bonds are not changing (in oscillation). <br />
<br />
The transition state for the H H-F system is the same as in the previous example but with the values reversed, where A=H B=H C=F;<br />
<br />
<br />
[[File:Screen Shot 2020-05-21 at 16.52.58.png|300px|thumb|Figure 8: internuclear distance vs time plot at the transition state of H +HF system.]]<br />
<br />
<br />
{| class="wikitable" border="1"<br />
|+ Transition state location of H + HF system<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1=p2 momentum /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Etot /kJ.mol<sup>-1</sup> !! Forces along BC /kJ.mol<sup>-1</sup>.pm<sup>-1</sup> !! Forces along AB /kJ.mol<sup>-1</sup>.pm<sup>-2</sup> !! Ļ2 /kJ.mol<sup>-1</sup>pm<sup>-2</sup><br />
|-<br />
| 181.5 || 74.5 || 0 || -433.981 || +0.000 || -0.007 || +0.332 and-0.002 <br />
|}<br />
<br />
====Report the activation energy for both reactions.====<br />
<br />
The activation energy can be found by calculating the initial energies of the reactants and finding the difference between this and the total energy. There are two way of doing this which have been explored.<br />
<br />
The first is to set the bond distance of the F atom to so far away from the H<sub>2</sub> that it can be assumed to be not interacting with the H<sub>2</sub>. From this we can work out the energy of the H-H bond alone, and use this value in the activation energy. For the F + H<sub>2</sub> example, the total energy plateaus at approximately AB=700pm where total energy = -435.057.<br />
<br />
{| class="wikitable" border="1"<br />
|+ Transition state location of F + H<sub>2</sub> system where A=F B=H C=H<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1=p2 momentum /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Etot /kJ.mol<sup>-1</sup><br />
|-<br />
| 74.5 || 700 || 0 || -435.057<br />
|}<br />
<br />
The total energy of the system at the transition state is taken as -433.981 kJ.mol<sup>-1</sup>. Therefore these values give us an overall activation energy of:<br />
<br />
Activation energy:<br />
-433.981--435.057 = 1.076 kJ.mol<sup>-1</sup><br />
<br />
The other method is to perform a series of mep plots and find the value the 'jump' in energy tends to; this is the activation energy, fig.9.<br />
<br />
[[File:Screen Shot 2020-05-21 at 18.05.16.png|300px|thumb|Figure 9: MEP plot for calculating activation energy.]]<br />
<br />
For the H HF system the H-F bond distance r1 is set to 91 pm and again r2 is increased until no interactions occur.<br />
<br />
{| class="wikitable" border="1"<br />
|+ Transition state location of H + HF system where A=H B=H C=F<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1=p2 momentum /g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! Etot /kJ.mol<sup>-1</sup><br />
|-<br />
| 91 || 700 || 0 || -560.404<br />
|}<br />
<br />
Activation energy:<br />
-433.981--560.404 = 126.432 kJ.mol<sup>-1</sup><br />
<br />
'''''<u>Good assumptions for Ea in both directions, well done!</u>'''''<br />
<br />
== Reactions dynamics ==<br />
<br />
====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.====<br />
<br />
The table below results in a reactive trajectory for the F + H<sub>2</sub> system:<br />
<br />
{| class="wikitable" border="1"<br />
|+ F + H<sub>2</sub> system where A=F B=H C=H reactive trajectory<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1(pHH) momentum /g.mol-1.pm.fs-1 !! p2(pHF) momentum /g.mol-1.pm.fs-1 !! Etot /kJ.mol-1<br />
|-<br />
| 74.5 || 150 || -9 || -1 || -363.066<br />
|}<br />
<br />
[[File:Screen Shot 2020-05-22 at 20.45.42.png|300px|thumb|Figure 10: Momenta vs time for the reactive trajectory described in the table.]]<br />
<br />
In this reaction, potential energy is converted to kinetic energy. From the momenta time plot above, we can see the potential energy in B-C as the bond is broken converts to kinetic from the oscillating motion of the A-B bond (increase and decrease in momentum on graph). The two possibilities for the release in bond energy from the reactants to the products is through translational or vibrational kinetic energy (rotational and electronic energy are ignored). <br />
<br />
Whether the reaction releases energy via translational or rotational energy can be found by IR spectroscopy.<br />
<br />
In both reactions the H-H and H-F will initially be at the ground state. As the energy of the systems increases the electrons can be promoted to the first energy level. From this excited state one of two things may occur: the electron might decay back to the ground state or gain enough energy to be further promoted to the second excited state. In IR spectroscopy the number of peaks can determine which of these two processes occurred. If the electron was promoted to the first state only then one peak will be shown on the absorption spectra. This is the fundamental peak and will be the largest as most transitions occurring will be this one. If some electrons are promoted from 1st to 2nd state then two peaks will be shown on the spectra. The second peak will be at a lower wavenumber as the energy difference between the 1st and 2nd state is smaller than the energy difference between the 1st and ground state (anharmonic oscillator).<br />
<br />
If the process was via vibrational energy then the two peaks will show on the spectrum as described above. If it was by by translational then only the fundamental peak will be present as there is no promotion to higher energy levels.<br />
<br />
'''<u>''Good discussion!''</u>'''<br />
<br />
====Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.====<br />
<br />
Polanyi's rules<sup>4</sup> state that vibrational energy is more efficient than translational energy in activating a late-barrier reaction, with the reverse true for an early barrier reaction. Going by this theory, the exothermic F + H<sub>2</sub> reaction should be activated by mainly translational energy; and the endothermic H + HF reaction mainly vibrational energy. <br />
<br />
{| class="wikitable" border="1"<br />
|+ F + H<sub>2</sub> system where A=F B=H C=H reactive trajectory<br />
! BC r1 bond distance /pm !! AB r2 bond distance /pm !! p1(pHH) momentum /g.mol-1.pm.fs-1 !! p2(pHF) momentum /g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! illustration !! description of dynamics<br />
|-<br />
| 74 || 200 || -6.1 || -1 || -402.504 || [[File:Screen Shot 2020-05-22 at 21.43.56.png|300px]] || Does not react. The reactants have greater vibrational energy (and therefore less translational energy) and the trajectory 'bounces' back to the reactants and does not follow the path to the products.<br />
|-<br />
| 74 || 200 || 0 || -1 || -433.614 || [[File:Screen Shot 2020-05-22 at 21.49.09.png|300px]] || Reactive. The reactants now have lower vibrational energy (and therefore higher translational energy) than the products - as the momenta is set to 0 for p1 (BC). The trajectory does not 'bounce' back to the reactants and instead follows the path to form the products.<br />
|}<br />
<br />
Vibrational energy is better for overcoming the endothermic energy barrier as the direction of vibrational motion can align with the direction that goes to the products. If translational energy was solely used the trajectory would not curve around in the correct direction to form products.<br />
<br />
Exothermic reactions have a much smaller energy barrier and so do not require the energy to be in the correct direction as with endothermic reactions. They work better with higher translational energy.<br />
<br />
'''''<u>Excellent, although I'd have liked to have seen some examples for endothermic reactions as well. </u>'''''<br />
<br />
==References==<br />
<br />
1. Atkins, P. and De Paula, J., 2013. Atkins' Physical Chemistry. 11th ed. Oxford: Oxford University Press, pp.792-794.<br />
<br />
2. Saddle points. 2020. [video] Directed by K. Academy. Available at: <https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions-videos/v/saddle-points> [Accessed 22 May 2020].<br />
<br />
3. Steinfeld, J.and Francisco, J. and Hase, W. Chemical kinetics and dynamics. 2nd ed. Upper Saddle River: Prentice-Hall, pp.287-321.<br />
<br />
4. Chemistryviews.org. 2020. New Rules For Reaction Dynamics :: News :: Chemistryviews. [online] Available at: <https://www.chemistryviews.org/details/news/1378289/New_Rules_for_Reaction_Dynamics.html> [Accessed 22 May 2020].<br />
<br />
==Bibliography==<br />
<br />
1. Atkins, P. and De Paula, J., 2013. Atkins' Physical Chemistry. 11th ed. Oxford: Oxford University Press.<br />
<br />
3. Steinfeld, J.and Francisco, J. and Hase, W. Chemical kinetics and dynamics. 2nd ed. Upper Saddle River: Prentice-Hall.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cb5418&diff=812906MRD:cb54182020-06-25T21:12:05Z<p>Ng611: /* EXERCISE 1: H + H2 system */</p>
<hr />
<div><br />
'''''<u>4/5 - A good report overall, with only one or two minor errors. Well done. </u>'''''<br />
<br />
== Molecular Reaction Dynamics: Applications to Triatomic systems ==<br />
<br />
== EXERCISE 1: H + H<sub>2</sub> system ==<br />
=== Question 1 ===<br />
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''<br />
<br />
{| class="wikitable" border="1"<br />
|-<br />
| [[File:Q1.1Surface Plot.png|thumb|A potential energy surface plot of a H + H<sub>2</sub> system]]|| <br />
[[File:Q1.1.2Surface Plot.png|thumb|An internuclear distance vs. time plot of a H + H<sub>2</sub> system]]<br />
|}<br />
<br />
The transition state is mathematically defined on the potential energy surface plot as the point where the first partial derivative of the potential with respect to the internuclear distance r<sub>1</sub> (distance A-B) and r<sub>2</sub> (distance B-C) energy is equal to 0, shown mathematically as, (dV(r<sub>i</sub>)/dr<sub>i</sub>)=0. In other words, it is the maximum on the minimum energy path of a reactant-product pathway. The transition state is a saddle point where it is either a local maxima or a local minima, this is shown mathematically via the second partial derivative where the value is either positive or negative <ref>Moore, J. W., Moss, S. J. & Coady, C. J. Potential-Energy Surfaces and Transition-State Theory. vol. 60 (1983).</ref>. <br />
<br />
It is identified through the 'rolling ball' analogy. The 'rolling ball' analogy portrays a 'ball' moving by applying momentum from an area near or at the transition state on the potential energy surface plot towards either the products or reactants, the two local minima. By applying the opposite momentum, the 'ball' will roll towards the other local minimum.<br />
<br />
'''''<u>Good response. What direction are the 1st/2nd partial derivatives taken along?</u>'''''<br />
<br />
=== Question 2 ===<br />
'''Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.'''<br />
<br />
{| class="wikitable" border="1"<br />
|-<br />
| [[File:Cb5418E1Q2.1.png|thumb|A contour plot of a H + H<sub>2</sub> system at r<sub>ts</sub>=90.78 pm]]|| <br />
[[File:Cb5418E1Q2.2.png|thumb|An internuclear distance vs. time plot of a H + H<sub>2</sub> system at r<sub>ts</sub>=90.78 pm]]<br />
|}<br />
<br />
The transition state position (r<sub>ts</sub>) is estimated to be 90.78 pm. The contour plot shows the position of r<sub>ts</sub> for the symmetric H + H<sub>2</sub> system. Due to the symmetry of the H + H<sub>2</sub> system, the transition state condition is r<sub>1</sub>=r<sub>2</sub>. If for instance, the trajectory is started on the ridge at r<sub>1</sub>=r<sub>2</sub> then the gradient is equal to 0 in right angled directions with respect to the ridge. This means that the trajectory will oscillate solely on the ridge and will not fall off.<br />
<br />
The internuclear distance vs. time plot at r<sub>ts</sub>=90.78 shows no fluctuations in the distance of A-C and B-C. This confirms that the system undergoes a periodic symmetric vibration.<br />
<br />
'''''<u>Sensible estimate, well done. </u>'''''<br />
<br />
=== Question 3 ===<br />
'''Comment on how theĀ ''mep''Ā and the trajectory you just calculated differ.'''<br />
<br />
{| class="wikitable" border="1"<br />
|-)<br />
| [[File:Cb5418E1Q3.1.png|thumb|A contour plot of a H + H<sub>2</sub> showing the reaction path via a minimum energy pathway (''mep'')]]|| <br />
[[File:Cb5418E1Q3.2.png|thumb|A contour plot of a H + H<sub>2</sub> showing the reaction path via dynamics calculation]]<br />
|}<br />
<br />
The trajectories for r<sub>1</sub> = r<sub>ts</sub>+Ī“, r<sub>2</sub> = r<sub>ts</sub> where Ī“=1 pm and r<sub>ts</sub>=90.78 pm are shown on the two contour plots. The minimum energy path (''mep'') does not show a realistic representation of the reaction path, since gaseous atoms have mass so their motion will be inertial. As a result, little to no oscillation is observed via the minimum energy path, whereas some oscillation is observed via the reaction path calculated by dynamics. The reaction path calculated by dynamics is a better representation of the motion of atoms and its pathway.<br />
<br />
'''''<u>Correct - the MEP is inaccurate as far as realism goes. However, it does have uses. Try to think of what they would be.</u>'''''<br />
<br />
=== Question 4 ===<br />
'''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''<br />
<br />
The table shows five different trajectories with different values of momenta (p<sub>1</sub> and p<sub>2</sub>) for the parameters r<sub>1</sub>=75 pm and r<sub>2</sub>=200 pm. <br />
<br />
{| class="wikitable" border=1<br />
! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactive? !! Illustration of the trajectory !! Description of the dynamics<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Yes || [[File:Cb5418E1Q4.1.png|200px]] || The reaction path starts from the diatomic B-C+A and transition towards A-B+C. As seen on the contour plot, the reaction path oscillates slightly as it approaches the transition state. Once the reaction takes place the path becomes much more smooth. This is due to the higher kinetic energy in B-C needed to overcome the activation barrier which means a higher internal vibrational energy. Therefore, greater vibration is observed during the first half of the reaction.<br />
|-<br />
| -3.1 || -4.1 || -420.007 || No|| [[File:Cb5418E1Q4.2.png|200px]] || This is an example of an unreactive reaction where there is not enough kinetic energy to overcome the activation barrier. The molecule A does not have enough kinetic energy to overcome the transition energy, thus it is unable to approach the diatomic B-C. Instead, the molecule A uses the remaining kinetic energy to bounce back. Similarly, significant oscillation is seen here due to the high internal vibrational energy.<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Yes|| [[File:Cb5418E1Q4.3.png|200px]] || This is very similar to the first scenario where there is enough kinetic energy to overcome the activation barrier. Both the diatomic B-C and the molecule A have enough kinetic energy to approach each other causing the reaction to take place. More oscillation in the reaction path for B-C and this gradually smoothens out as it approaches molecule A forming the diatomic A-B.<br />
|-<br />
| -5.1 || -10.1 || -357.277 || No || [[File:Cb5418E1Q4.4.png|200px]] || In this scenario, a process called barrier recrossing occurs. Initially, molecule A oscillates greatly and overcomes the transitions state. At this point, molecule A reacts with the diatomic B-C and the bond A-B is formed. Despite that, the reaction reverses back to its initial states as the reactants molecule A and diatomic B-C reforms<ref>Komatsuzaki, T. & Nagaoka, M. Study on āregularityā of barrier recrossing motion. J. Chem. Phys. 105, 10838ā10848 (1996).</ref>.<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Yes || [[File:Cb5418E1Q4.5.png|200px]] || The reaction proceeds in this scenario, however there is much greater oscillation due to greater kinetic energy in the system. There is plenty of kinetic energy to overcome the activation barrier for this reaction. The sporadic reaction path is due to very high internal vibrational energy, especially around the area of the transition state. However, once the B-C bond is broken and the A-B bond is formed the reaction path smooths out as energy is used up.''' <br />
|}<br />
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Conclusively, the hypothesis: "It would seem fair to assume that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier.", is partially accepted. This is because there are many different factors which affect the reactiveness of the process and not just solely on the kinetic energy. An example of this is barrier recrossing.<br />
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=== Question 5 ===<br />
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br><br />
'''<br />
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The Transition State Theory predictions for reaction rate values are not as accurate when compared to the experimental values. The main reason behind this is due to the numerous assumptions made to fit the theory. The theory itself was made to provide a better understanding of the reaction rates of elementary reactions. A pivotal assumption is the estimation of R<sub>cl</sub>, the average transmission rate/frequency, where trajectories which have a greater kinetic energy than the activation barrier are reactive. However, this is not always the case as seen with the barrier recrossing effect. Another key assumption is that the kinetic energy is modelled via the Boltzmann distribution along the reaction coordinate. This means that the intermediates are long-lived, but for most experimental cases the intermediates are actually short-lived. Another assumption is that the Transition State Theory is considered classically. Simply saying that the trajectory is described solely by the velocity and not the wavefunction. When compared to the experimental cases this theory ignores the effect of quantum tunnelling, where reactants with not enough kinetic energy to overcome the barrier are able to tunnel through the it instead. It is also mentioned that in Transition State Theory, the pre-exponential factor depends on temperature. However, at high temperatures new transition states are available due to the population of vibrational, rotational and translational energy gaps which contradicts the theory<ref>Pilling, M. J., and Seakins, Paul W. Reaction Kinetics. Oxford: Oxford UP, 1995. Print.</ref>. As mentioned before, all of these assumptions are made to fulfil the theory, but these discrepancies mentioned in the extract will lead different reaction rate values.<br />
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'''''<u>A very detailed and entirely correct answer, well done. In these simple, limiting cases though will TS theory overestimate or underestimate the rate?</u>'''''<br />
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== EXERCISE 2: F - H - H system ==<br />
=== PES inspection ===<br />
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==== Question 1 ====<br />
'''By inspecting the potential energy surfaces, classify the F+H<sub>2</sub> and H+HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''<br />
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{| class="wikitable" border="1"<br />
|-<br />
| [[File:Cb5418E2Q1.png|thumb|A potential energy surface plot of a F - H - H system]]<br />
|}<br />
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As shown on the potential energy surface plot, at lower bond distances the potential energy well of H-F (AB) is deeper in energy when compared to H-H (BC). For this reason, F+H<sub>2</sub> is exothermic and H+HF is endothermic. Therefore, the F-H bond is a stronger bond than H-H.<br />
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'''''<u>Good!</u>'''''<br />
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==== Question 2 ====<br />
'''Locate the approximate position of the transition state.'''<br />
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{| class="wikitable" border="1"<br />
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| [[File:Cb5418E2Q2.1.png|thumb|A contour plot of a H - H - F system at transition state]]|| <br />
[[File:Cb5418E2Q2.2.png|thumb|An internuclear distance vs. time plot of a H - H - F system at transition state]]<br />
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The approximate position of the transition state is determined as follows: F-H (AB)=181.15 pm, H-H (BC)=74.49 pm. This is the position where for both AB and BC, the momenta equate to zero. The position of the transition state in this system was determined using a Hessian matrix. The Hessian matrix proposes one positive eigenvalue which is close to 1 and a negative eigenvalue which is close to 0.<br />
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==== Question 3 ====<br />
'''Report the activation energy for both reactions.'''<br />
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{| class="wikitable" border="1"<br />
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| [[File:Cb5418E2Q3.1.png|thumb|A contour plot of a H - H - F system for F + H<sub>2</sub>]]|| <br />
[[File:Cb5418E2Q3.2.png|thumb|An energy vs. time plot of a H - H - F system for F + H<sub>2</sub>]]<br />
|}<br />
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| [[File:Cb5418E2Q3.3.png|thumb|A contour plot of a H - H - F system for H + HF]]|| <br />
[[File:Cb5418E2Q3.4.png|thumb|An energy vs. time plot of a H - H - F system for H + HF]]<br />
|}<br />
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The activation energies for both reactions are calculated by subtracting E<sub>TOT</sub> of each reaction by the E<sub>TOT</sub> of the transition state, this is due to the manipulation of Hammond's postulate. These values are stated in the table below. <br />
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{|class="wikitable" border="1"<br />
|+'''Activation energies for F + H<sub>2</sub> and H + HF reactions'''<br />
! '''Reactants''' !! '''Endothermic/exothermic''' !! '''E<sub>TOT</sub>/(kJ.mol<sup>-1</sup>)''' !! Activation Energy/(kJ.mol<sup>-1</sup>)<br />
|-<br />
| TS || - || -433.981 || -<br />
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| F-H + H || Endothermic || -560.617 || 126.636<br />
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| F + H-H || Exothermic || -435.059 || 1.078<br />
|}<br />
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'''''<u>A sensible set of Ea values, well done!</u>'''''<br />
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=== Reaction dynamics ===<br />
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==== Question 4 ====<br />
'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the āAnimationā and āMomenta vs Timeā. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''<br />
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{| class="wikitable" border="1"<br />
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| [[File:Cb5418E2Q5.1.png|thumb|A contour plot of a H - H - F system for F + H<sub>2</sub> for a specific condition]]|| <br />
[[File:Cb5418E2Q5.2.png|thumb|A momenta vs. time plot of a H - H - F system for F + H<sub>2</sub> for a specific condition]]<br />
|}<br />
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The set of initial conditions that results in a reactive trajectory for the F + H<sub>2</sub> is as follows: F-H (AB)=200 pm, H-H (BC)=75 pm, p<sub>AB</sub>= -1.0 g.mol<sup>-1</sup>, p<sub>BC</sub>= -1.0 g.mol<sup>-1</sup>. From the momenta vs. time plot shown, the B-C plot which corresponds to the H-H bond oscillates at a much smaller extent early in the reaction, this is due to the vibrational bond energy. This is modelled quantum mechanically by a simple harmonic oscillator. As the diatomic H-H approaches the fluorine atom, F, the the B-C bond starts to oscillate more irregularly and to a larger extent. Eventually, the H-H bond breaks and instead the H-F bond is formed. This reaction is shown on the plot at the point where the A-B bond oscillates aggressively. The exothermic process leads to a release in energy to the surroundings as heat energy via internal vibrational relaxation. This massive release in heat energy will cause a sharp rise in temperature which could be monitored by a thermometer to confirm experimentally that the reaction is occurring or has occurred.<br />
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'''''<u>No, because translational KE will also be detected as heat. Another method has to be used. Can you think of what it is?</u>'''''<br />
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==== Question 5 ====<br />
'''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''<br />
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The Polanyi's empirical rules simply state that vibrational energy is more efficient in the promotion of a late-barrier reaction than translational energy and vice versa for an early-barrier reaction<ref>Zhang, Z., Zhou, Y., Zhang, D. H., CzakĆ³, G. & Bowman, J. M. Theoretical study of the validity of the Polanyi rules for the late-barrier Cl + CHD3 reaction. J. Phys. Chem. Lett. 3, 3416ā3419 (2012).</ref>. This is basically saying that the early transition state is favoured by translational energy and the late transition state is favoured by vibrational energy . <br />
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'''''<u>Correct definition, well done. </u>''''' <br />
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{| class="wikitable" border="1"<br />
|-<br />
| [[File:Cb5418E2Q4.png|thumb|A contour plot of a F - H - H system for H + HF (low vibrational energy)]]<br />
|}<br />
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The initial conditions as follows are applied to the contour plot shown: F-H (AB)=85 pm, H-H (BC)=100 pm, p<sub>AB</sub>= -0.25 g.mol<sup>-1</sup>, p<sub>BC</sub>= -5.0 g.mol<sup>-1</sup>.<br />
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An example of a H + HF endothermic reaction is shown on the contour plot. This plot is a good example of Polanyi's empirical rules. The reaction path is for low vibrational but high translational energies. Since this is an endothermic reaction, there is a late transition state. According to Polanyi's empirical rules, vibrational energy is favoured over translational energy for this path, but since vibrational energy is low, the result of the reaction is unsuccessful. The reaction path stopped at a certain point and reversed back to its original state, regardless of the fact that the translational energy is greater than the activation barrier.<br />
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|-<br />
| [[File:Cb5418E2Q4.1.png|thumb|A contour plot of a F - H - H system for H + HF (high vibrational energy)]]<br />
|}<br />
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'''''<u>This plot doesn't appear to be correct -- it seems that you've included a plot for the F+H2 reaction by accident, or else you started in the products channel.</u>''''' <br />
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This contour plot shows the result of the trajectory upon increasing the vibrational energy of the system. The trajectory is now reactive resulting in the formation of its products. This is because the gaseous atom motion aligns more towards the products allowing the reaction to complete successfully as it is now orthogonal to the minimum energy path.<br />
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== References ==</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:kb4817&diff=812905MRD:kb48172020-06-25T21:01:20Z<p>Ng611: </p>
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<div><br />
'''''<u>3/5 - Some very good aspects to this report, but there were one or two large gaps in your understanding which needed to be corrected. See comments below. </u>'''''<br />
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== Molecular Reaction Dynamics ==<br />
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== <u>H<sub>2</sub> + H</u> ==<br />
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=== <u>H<sub>2</sub> + H Transition State</u> ===<br />
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The transition state on a potential energy surface diagram is defined as the point where the derivative (gradient) of the both reactant's and product's potential is 0 with respect to their bond distances. This can be further defined by taking the second derivative to establish whether it is a maxima or minima where a result less than one is minima and vice versa. The peak should be a maxima on the minimal energy pathway (MEP). The associated forces acting on both bonds, AB and BC, at this point are 0, which allows you to distinguish between other local minima where one of the bonds will not have zero force energy.<br />
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For a homo-nuclear reaction, the transition state is identified by the intersection of A-B and B-C as seen in ''Figure 1'' where the bond distances are equal causing the forces acting on them to equal 0. <br />
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'''''<u>You're on the right track but you're still missing some important points. What mathematical structure does the TS resemble and how does this affect the gradient?</u>''''' <br />
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[[File:exercise_one_d_t2.PNG|thumb|center|''Figure 1: Inter-nucleic distance vs time'']]<br />
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This intersection is at approximately 90.78 pm as shown by the single dot of the function representing 'gradient = 0' in ''Figure 2''.<br />
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'''''<u> Correct estimate for the TS, well done</u>'''''<br />
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[[File:exercise_one_V_r.PNG|thumb|center|''Figure 2: Potential energy surface for Hydrogen reaction at p=0 and bond distance AB=BC=90.78 pm'']]<br />
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To identify other transition states, the atoms involved in the reaction will be stationary therefore change in momentum is zero and therefore so is the derivative. So by calculating the bond distances for which these are zero you can calculate the transition state. Hammond's postulate also states that if a reaction is exothermic the TS will resemble reactants and vice versa.<br />
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=== Trajectories ===<br />
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When calculating the MEP compared to the dynamic pathway of a reaction the main variable is the lack of oscillations of the bond lengths (i.e the vibrations between atoms). As we see in the AB bond, which is one pm longer than the BC bond, the vibrations of the bond length affect the inertia of atoms, however in the MEP the time steps always reset momenta to zero regardless of vibration and so vibrations are not observed. Both trajectories go towards forming the BC bond. MEP corresponds to purely translation energy whereas dynamic plotting corresponds to both translation and symmetric vibration.<br />
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'''''<u>Absolutely correct, well done!</u>'''''<br />
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[[File:birds_eye_MEP.PNG|thumb|left|''Figure 3: MEP where AB=91.78 pm, BC=90.78 pm'']] [[File:birds_eye_dynamic.PNG|thumb|center|''Figure 4: Dynamic pathway where AB=91.78 pm, BC=90.78 pm'']]<br />
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This can be further highlighted by the ''Figure 5'' and ''Figure 6'' where distance and momenta oscillate for dynamic and not MEP. For AB=BC + 1 pm at time t=40 for ''Figure 5'' '''AB=280.22 pm, BC=73.76 pm''', and ''Figure 6'' '''p<sub>AB</sub>=5.02 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>=2.34 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''.<br />
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[[File:d_t_1.PNG|thumb|left|''Figure 5: Dynamic inter-nucleic distances at initial AB=91.78 pm, BC=90.78 pm'']] [[File:p_t_1.PNG|thumb|center|''Figure 6: Dynamic momenta at initial AB=91.78 pm, BC=90.78 pm'']]<br />
[[File:d_t_2.PNG|thumb|left|''Figure 7: MEP inter-nucleic distances at initial AB=91.78 pm, BC=90.78 pm'']] [[File:p_t_2.PNG|thumb|center|''Figure 8: MEP momenta at initial AB=91.78 pm, BC=90.78 pm'']]<br />
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For '''BC=AB + 1 pm''' at last geometric point, the exact opposite occurs '''AB=74.03 pm, BC=352.60 pm''', and '''p<sub>AB</sub>=3.20 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>BC</sub>=5.06 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''.<br />
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If we then plotted these final values as initial values and reversed the sign of momentum we obtain ''Figure 9'', ''Figure 10'' and ''Figure 11''. We see that atom C approaches AB but never reacts and ends up at the same starting position as the initial reaction conditions.<br />
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'''''<u>Actually it looks like they've basically reached the TS all over again! This is exactly what should happen, assuming that the positions/momenta aren't rounded off.</u>'''''<br />
[[File:amendment3.PNG|300px|thumb|left|''Figure 9: Dynamic contour plot'']] [[File:amendment1.PNG|thumb|300px|center|''Figure 10: Dynamic Internucleic distances'']] <br />
[[File:amendment2.PNG|300px|thumb|left|''Figure 11: Dynamic momenta'']]<br />
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{|class="wikitable" border="1"<br />
|+'''Table of reactivity of different momenta but values AB=74 pm, BC=200 pm'''<br />
! '''p<sub>AB</sub>/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>''' !! '''p<sub>BC</sub>/g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>''' !! '''E<sub>TOT</sub>/(kJ.mol<sup>-1</sup>)''' !! Reactive? !! Diagram !! Description of Dynamics<br />
|-<br />
| -2.56 || -5.1 || -414.280 || YES || [[File:kb14817.PNG|300px]] || AB + C -> A + BC where purely translation energy in initial approach of C develops oscillations due to the bond having formed and vibrating.<br />
|-<br />
| -3.1 || -4.1 || -420.077 || NO || [[File:kb24817.PNG|300px]] || C approaches AB, however not enough energy to overcome activation energy therefore no reaction<br />
|-<br />
| -3.1 || -5.1 || -413.977 || YES || [[File:kb34817.PNG|300px]] || Same outcome as first reaction with AB having slightly higher momenta and C having same momenta resulting on overall higher energy used in reaction<br />
|-<br />
| -5.1 || -10.1 || -357.277 || NO || [[File:kb444817.PNG|300px]] || Reaction AB + C -> A + BC does occur, however it quickly dissolves back into the original reactants. This is called barrier recrossing. Recrossing is defined as passage over the potential barrier separating the two species followed by return to the original side<sup>1</sup>.<br />
|-<br />
| -5.1 || -10.6 || -349.477 || YES || [[File:kb554817.PNG|300px]] || Reaction occurs but also undergoes barrier recrossing twice. This therefore yields the products A + BC.<br />
|}<br />
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=== Transition State Theory ===<br />
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Overall the theoretical values of Transition State Theory (TST) underestimate the experimental values. This is due to assumptions made in TST that are not true when analysing in quantum mechanically. <br />
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'''''<u>This is incorrect. TS theory overestimates the rate of reaction. This is because the no-recrossing assumption (assumption 1 in your list) is a much stronger assumption than any other. QM tunneling is a weak effect, even for light atoms, and has an inverse dependence on mass, meaning that it becomes even less significant moving away from hydrogen. Your other assumptions are also not as relevant as (1). Think about why assumption (1) leads to an overestimation. </u>''''' <br />
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Assumptions made by TST:<br />
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1) R ā TS -> P : Once there is enough energy to form the transition state (activation energy) the TS will form the products and no back reaction occurs<br />
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2) There is no quantum tunneling effect which essentially allows a slight bypass of the activation energy maxima<br />
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3) There is a quasi-equilibrium between reactants and transition state meaning there is a Boltzmann distribution between the two at any given time<sup>2</sup><br />
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As TST treats the rate classically no quantum tunneling is factored in. The rate is therefore an underestimation as tunneling would provide a lower entry barrier for the formation of products and therefore more can be formed at a given time. It also assumes that TS molecules live long enough to form a Boltzmann distribution, which isn't true for many reactions so once the TS is formed it forms the products. This also assumes that we are dealing with a bulk amount reactants we are dealing with however the program being used only considers one molecule. It also assumes that at higher temperatures it still only passes through the energy maxima in the minimum energy pathway, however due to the increased energy of the system access to higher transition states are now available.<br />
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== <u>F + H<sub>2</sub> and FH + H systems</u> ==<br />
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F + H<sub>2</sub> is exothermic and FH + H is endothermic. This tells us that the F-H bond is stronger than the homo-diatomic H-H bond as energy is released when forming F-H and energy must be put in to break this bond and form H-H instead. We observe this through the surface plot ''Figure 12'' where the potential well of HF (AB) at lower bond distances is much deeper than that of H-H (BC) at lower bond distances.<br />
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[[File:kbe2_1.PNG|300px|thumb|center|''Figure 12: F-H-H Surface Plot'']]<br />
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=== Transition State ===<br />
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The approximate transition state position will be where both momenta equal zero and the forces acting on AB and BC are zero at given bond distances. The distances for this system are '''F-H=AB=181.15 pm''' and '''H-H=BC=74.49 pm''' as seen in ''Figure 13'' and ''Figure 14''. This was located using the Hessian matrix where one eigenvalue had to be positive and close to one and the other eigenvalue negative and close to 0.<br />
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[[File:kbe2_2.PNG|300px|thumb|left|''Figure 13: F-H-H Transition state'']] [[File:kbe2_3.PNG|300px|thumb|center|''Figure 14: F-H-H Transition state inter-nuclear distance'']]<br />
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=== Activation energy === <br />
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To calculate activation energy, AB was plotted on a transition state MEP contour graph as normal and then BC was given a very large value to completely separate the two and vice versa for BC calculation. The last geometry of this was then plotted to establish the energy of reactants. These were then subtracted from the transition state energy to give the activation energy. In the case of '''F + H<sub>2</sub> (1)''' and '''FH + H (2)''' , the total energy of '''(1)''' is calculated and subtracted from the transition state energy to give the exothermic activation energy, and total energy of '''(2)''' subtracted from the transition state energy to give the endothermic activation energy. An alternative method to work out the energy of reactants is to -1 pm from AB whilst not changing BC and observe the dip into the well and vice versa as shown in ''Figures 15,16,17, and 18''.<br />
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[[File:AAkb4817.PNG|300px|thumb|left|''Figure 15: Contour plot for F + H<sub>2</sub> Reactants'']] [[File:AAAkb4817.PNG|300px|thumb|center|''Figure 16: Energy v Time plot for F + H<sub>2</sub> Reactants'']]<br />
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[[File:FH2kb4817.PNG|300px|thumb|left|''Figure 17: Contour plot for FH + H Reactants'']] [[File:FHkb4817.PNG|300px|thumb|center|''Figure 18: Energy v Time plot for FH + H Reactants'']]<br />
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{|class="wikitable" border="1"<br />
|+'''Activation energies'''<br />
! '''Reactants (R)''' !! '''Endo/exothermic''' !! '''E<sub>TOT</sub>/(kJ.mol<sup>-1</sup>)''' !! Activation Energy/(kJ.mol<sup>-1</sup>)<br />
|-<br />
| TS || N/A || -433.981 || N/A<br />
|-<br />
| FH + H || Endo || -560.617 || 126.636<br />
|-<br />
| F + HH || Exo || -435.059 || 1.078<br />
|}<br />
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'''''<u>Good estimates for Ea in both directions, well done!</u>'''''<br />
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=== Reaction Dynamics ===<br />
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The reaction F + H<sub>2</sub> is exothermic and releases radiative heat as a result. This temperature increase is due to the increased vibrations of the newly formed F-H, which must compensate for the excess energy as it is itself buried deeper in energy (conservation of energy principle). This can be seen in ''Figure 15'' where the momenta of the newly formed AB(=F-H) oscillates. To test this you could take the IR spectrum of the sample before and after it has reacted to see an initial peak in ground state of the molecule and once it reacts an overtone appears which is smaller and at lower wavenumbers (due to asymmetric overlap of anharmonic levels). This overtone decreases to 0 over time as all the molcules return to the ground state. <br />
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A set of reaction conditions where '''AB=FH=175 pm, BC=HH=74 pm, and p<sub>AB</sub>= -1.6 g.mol<sup>-1</sup>, p<sub>BC</sub>= +0.2 g.mol<sup>-1</sup>''' allows a reaction to take place where the oscillations of the FH molecule are observed afterwards<br />
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'''''<u>Good answer. What other techniques can you think of?</u>'''''<br />
[[File:reactionkb4817.PNG|thumb|left|''Figure 19: F-H-H Reaction system Contour graph '']] [[File:reaction2kb4817.PNG|thumb|center|''Figure 20: F-H-H Reaction system Momenta Time graph '']]<br />
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Polanyi's empirical rules state: vibrational energy is more efficient in promoting a late-barrier reaction (transition state resembling products) than translational, whereas the reverse is true for an early barrier reaction<sup>3</sup>. The reaction FH + H is endothermic and has a late transition state. Therefore, vibrational energy should encourage the formation of products and form them at a greater rate.<br />
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'''''<u>Correct. </u>'''''<br />
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If we take an example for endothermic reaction H + HF, where there is a late transition stage, and plot AB=FH=91 pm and BC=HH= 100 pm, and p<sub>AB</sub>= -0.1 g.mol<sup>-1</sup>, p<sub>BC</sub>= -5 g.mol<sup>-1</sup> we see there is no reaction (''Figure 20''). When the H collides with FH, it hits with too much force and in a straight line and so bounces further back then the original position which means it is too far for it form a HH bond, even if the translational energy is greater than the activation energy. When we have higher vibrational energy, its motion aligns more in the direction of the products as its orthogonal to the MEP. For an exothermic reaction, translational energy means the trajectory can fall to the lower energy region. Vibrations complicate this trajectory as it is in a different direction to the MEP.<br />
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[[File:deeznuts.PNG|thumb|center|''Figure 21: Contour graph for low vibration, high translation energies'']]<br />
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'''''<u>I also need examples for the inverse case here as well. </u>'''''<br />
=='''Bibliography'''==<br />
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<u>1) Recrossings and Transition-State Theory; Huw O. Pritchard; The Journal of Physical Chemistry A 2005 109 (7), 1400-1404; DOI: 10.1021/jp045262s</u><br />
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<u>2)Current Status of Transition-State Theory; Donald G. Truhlar, Bruce C. Garrett, and Stephen J. Klippenstein; The Journal of Physical Chemistry 1996 100 (31), 12771-12800; DOI: 10.1021/jp953748q</u><br />
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<u>3)Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction; Zhaojun Zhang, Yong Zhou, Dong H. Zhang, GĆ”bor CzakĆ³, and Joel M. Bowman; The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419; DOI: 10.1021/jz301649w</u></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01532430mrd&diff=812904MRD:01532430mrd2020-06-25T20:44:57Z<p>Ng611: /* MRD:01532430 */</p>
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<div>= MRD:01532430 =<br />
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'''''<u> 4/5 - Excepting a couple of small points (see the comments) this is a good report, well done. </u>'''''<br />
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== EXERCISE 1: H + H<sub>2</sub>Ā system ==<br />
<blockquote>'''''Q. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''</blockquote><blockquote>A.</blockquote>i) '''How is the transition state mathematically defined?'''<br />
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At the transition state the kinetic energy is 0 as it has all been converted to potential energy and the gradient of the potential is 0. Therefore it can mathematically defined using the first partial derivative: '''āV(r<sub>i</sub>)/ār<sub>i</sub>=0.''' <br />
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ii) '''How can the transition state be identified?''' <br />
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The transition state is also the minimum energy point reactants must reach in order for products to form, this naturally forms a maximum point when graphed. In this case the transition point is represented at the plane on the saddle point which gives the maximum point. <br />
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iii) '''How can it be distinguished from a local minimum of the potential energy surface?''' <br />
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As mentioned above the Transition state can be identified using the first partial derivative of potential energy. Also mentioned above is that the transition state is a maximum, so using the second derivative we can distiniguish it from the minimum point. The maximum point would give a negative value, while the minimum point will give a positive value. <br />
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'''''<u>Almost. Remember that with a saddle point, it is a minimum along one set of co-ordinates and a maximum along another set. A rigorous definition of the TS requires this behaviour to be accounted for.</u>'''''<br />
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<blockquote>'''''Q. Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.'''''</blockquote>A. Knowing that the surface of H + H<sub>2, </sub>the transition state should be at a point where r1 = r2. As well as this the momentum is 0 as the atoms are also stationary at this point, therefore p1=p2=0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>.<br />
[[File:1.2 01532430.png|centre|thumb|''Figure 1: Internuclear Distances vs Time for exercise 1.'']]<br />
Trial and error in a region between 0.74 pm and 1.00 pm was used initially. This was decided as the standard equilibrium bond distance for H2 is 0.74 so it can be assumed the value for the transition state is larger. This was continued until a horizontal line was obtained in the Internuclear Distances vs Time, figure 1. This indicated no change in distance and therefore stationary atoms and therefore the transition state. This was obtained at 90.750 pm.<br />
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'''''<u>Correct estimate for the TS, well done.</u>'''''<blockquote>'''Q. Comment on how theĀ ''mep''Ā and the trajectory you just calculated differ.'''</blockquote>A.'''Setting up the trajectories. '''For this section the trajectory was changed so the positionsĀ are '''r<sub>1</sub>'''Ā =Ā '''r<sub>ts</sub>'''+1Ā pm,Ā Ā '''r<sub>2</sub>'''Ā =Ā '''r<sub>ts. </sub>''' [[File:1.3 MEP2 01532430.png|thumb|Figure 3.1 (Dynamic) Contour Plot]]'''So with R1 = 91.750 and R0 = 90.750''', the reaction has come through the other side of the transition state and now the reaction is proceeding downhill from the maximum point. <nowiki> </nowiki>Momenta was kept at 0. Plots had their calculation types differed to MEP (steps = 3000) (figures 2) and dynamic (figures 3), these were then compared. [[File:1.3 MEP 01532430.png|left|thumb|Figure 2.1: (MEP) Contour Plot]] <br />
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'''Comparing figures 2.2 and 3.2''' . The dynamic figure 3.2 shows minor<br />
oscillations while the MEP figure 2.2 shows a flat line and no oscillations for<br />
B-C. This indicates in the dynamic figure there is vibrational motion, while in MEP there is only<br />
transitional motion and only vibration motion, this canĀ <br />
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be explained using<br />
evidence from figures 2.3 and 3.3.Ā <br />
[[File:1.3 INT 01532430.png|left|thumb|Figure 2.2: (MEP) Internuclear Distance vs Time Plot]]<br />
[[File:1.3 INT2 01532430.png|thumb|Figure 3.2: (Dynamic) Internuclear Distance vs Time Plot]]<br />
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'''Comparing 2.3 and 3.3''', the MEP<br />
plot (figure 2.3) the horizontal line shows there is no momentum. Yet in the<br />
dynamic plot (figure 3.3) there is an oscillation of momentum. Despite the atoms<br />
not being given initial momenta the conversion of potential energy to kinetic<br />
energy for the dynamic plot results in the movement of the atoms and therefore,<br />
momentum. This occurs due to the atoms starting at a downhill position. As<br />
indicated earlier MEP doesn't involve any conversion to KE and so with every<br />
step momenta is still 0, resulting in the horizontal line<br />
[[File:1.3 MOM 01532430.png|left|thumb|Figure 2.3: (MEP) Momenta vs Time]]<br />
[[File:1.3 MOM2 01532430.png|thumb|Figure 3.3: (Dynamic) Momenta vs Time]]<br />
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'''''<u>A sensible discussion of the differences, well done. </u>'''''<br />
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<blockquote>'''''Q. How does Internuclear Distances vs Time and''<br />
Momenta vs Time change with r<sub>1</sub>Ā =Ā r<sub>ts</sub>Ā andĀ r<sub>2</sub>Ā =Ā r<sub>ts</sub>+1'''''Ā <nowiki/>'''''</blockquote>A. <br />
[[File:1.4 MOM 01532430.png|thumb|296x296px|Figure 4.3: Momenta vs Time]]<br />
Reversing the distances, so R1 = 90.750, R2 = 91.750 produced these plots. Plot 4.1 gave a contour plot almost identical to plot 3.1 but reflected in the x=y line, this shows the direction of attack is symmetrical and now reversed under these conditions. Plot 4.2 is also very similar to plot 3.2 with the exception that the variables, A-B and B-C have been swapped, this occurs due to the symmetric and opposite direction of attack for the reaction. The same thing occurs when comparing figures 4.3 and 3.3. <br />
[[File:1.4 01532430.png|left|thumb|Figure 4.1: Contour Plot ]]<br />
[[File:1.4 INT 01532430.png|centre|thumb|Figure 4.2: Internuclear Distance vs Time]]<br />
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=== Reactive and unreactive trajectories ===<br />
<blockquote>'''''Q. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''''</blockquote>A.<br />
{| class="wikitable"<br />
!p<sub>1</sub>/Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
!p<sub>2</sub>/Ā g.mol<sup>-1</sup>.pm.fs<sup>-1</sup><br />
!E<sub>tot</sub><br />
!Reactive?<br />
!Description of the dynamics<br />
!Illustration of the trajectory<br />
|-<br />
|<nowiki>-2.56</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-414.280</nowiki><br />
|Yes<br />
|During the reactant stage Hb-Hc has minimal vibrational energy and Ha moves towards Hb-Hc. During the transition state the atoms move very slowly. After the TS the Ha-Hb and Hc move away fast and have more vibrational energy then Hb-Hc did. This is observed from the increased frequency and amplitude following the transition state.<br />
|[[File:2.56 01532430.png|thumb|Figure 5]]<br />
|-<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-4.1</nowiki><br />
|<nowiki>-420.077</nowiki><br />
|No<br />
|The contour diagram shows the reactants never reach the transition state. This is due to the kinetic energy not reaching the required activation energy. The diagram also shows the reactant molecules moving back on each other, but with less vibrational energy. This is seen with the smaller amplitude when compared to the initial attack. <br />
|[[File:3.1 01532430.png|thumb|Figure 6]]<br />
|-<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-413.977</nowiki><br />
|Yes<br />
|The reactants, Hb-Hc move towards together relatively quickly compared to plot one. This high momentum results in the reaction be successful. As well as this the vibrational energy is higher for the reactants and products, this is due to the higher initial kinetic energy and higher momentum. <br />
|[[File:3.1 2 01532430.png|thumb|Figure 7<br />
]]<br />
|-<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.1</nowiki><br />
|<nowiki>-357.277</nowiki><br />
|No<br />
|This reaction is very interesting. The high momentum causes the reactants to initially surpass the activation barrier and the products form. However, the overly high momentum and KE results in the products to have a very high vibrational energy and this results in the reactants reforming. Therefore the reaction is not successful.<br />
|[[File:10.1 01532430.png|thumb|Figure 8<br />
]]<br />
|-<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.6</nowiki><br />
|<nowiki>-349.477</nowiki><br />
|Yes<br />
|This reaction is also very interesting. The kinetic energy of the reactants is sufficient to form products however momentarily due to the high vibrational energy, the reactant reforms after the initial product, but then products also reform after that.<br />
|[[File:10.6 01532430.png|thumb|Figure 9<br />
]]<br />
|}<br />
From this table, the conclusions reached are that there must be sufficient kinetic energy to overcome the activation energy for the reaction to be successful, plot 2 shows this. On the other hand, plot 4 shows that even with high momentum and sufficient KE the reaction will potentially not be successful due to the reactants reforming. Therefore, the kinetic energy must be higher then the activation energy to pass the transition state but not too high of a KE, as this can result in a vibrational energy too high resulting in reactants reforming. <br />
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==== Transition State Theory ====<br />
<blockquote>'''''Q. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''</blockquote>A. A key factor in the transition state theory is that once the reactants have passed the transition state and activation barrier the reaction is irreversible. This doesn't coincide with the results from the table above, plots 4 and 5, indicating the transition state theory isn't accurate. This repeated crossover of the transition state indicates the prediction of theory are higher then experimental values. <ref>Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L.Ā ''Chemical Kinetics and Dynamics''. 2nd ed. Upper Saddle River: Prentice-Hall, 1989. Print.</ref><br />
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'''''<u>Good description of the limitations of TS theory. What other assumptions does TS theory make that don't hold up?</u>'''''<br />
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== EXERCISE 2: F - H - H system ==<br />
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=== PES inspection ===<br />
<blockquote>'''''Q. By inspecting the potential energy surfaces, classify the F + H<sub>2</sub>Ā and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''</blockquote>A surface plot was used to investigate the energetics of the F + H2 --> HF + H reaction. Figure 10 shows this. A=F, B=H, C=H.<br />
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The energy of the reactants is on the left and is clearly higher then the energy of the products on the right. This indicates a release of energy, therefore, making the reaction exothermic. This evidence relates to the bond strength by showing the product H-F is more stable and therefore its bond strength is higher then the reactant, H2. This coincides with literature values. HH has a lower bond enthalpy then HF, ( 436 KJmol-1 and 564 KJmol-1 respectively. <ref>Blanksby S. J., Ellison G. B. (April 2003). "Bond dissociation energies of organic molecules". Accounts of Chemical Research. 36 (4): 255ā63.</ref><br />
[[File:Surface Plot F + H2 01532430.png|left|thumb|Figure 10: Surface Plot of F-H-H System]]<br />
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<blockquote>'''''Q. Locate the approximate position of the transition state.'''''</blockquote>A. Locating the position of the transition state requires a similar strategy to exercise 1. Momentum was set to p1=p2=0 and R<sub>F-Hb</sub> = 181.1 pm and R<sub>Hb-Hc</sub> = 74.5 pm. The bond distance of the hydrogen atom was determined using the knowledge gained in exercise 1 that the equilibrium bond length is 0.74. HF bond distance was determined using estimation and trial and error. The transition state of this reaction was satisfied when the Internuclear Bond Distance vs Time plot produced horizontal lines, figure 11. As mentioned in exercise 1, this result in this plot indicates the atoms having no vibrational motion and of course, no change in intermolecular distance.#<br />
[[File:Trans 01532430.png|centre|thumb|Figure 11: Intermolecular Distances vs Time at Transition State for F-H-H ]]<br />
<blockquote>'''''Q. Report the activation energy for both reactions.'''''</blockquote>A. The table illustrates the trajectories chosen to investigate the activation energy of both reactions and the results obtained. <br />
{| class="wikitable"<br />
!Reaction<br />
!Set Trajectories (pm)<br />
!Energy of Reactants<br />
!Energy of Transition State<br />
!Activation Energy<br />
!Energy vs Time Plot of Reaction<br />
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|F + H<sub>2</sub> -> HF + H<br />
|A-B = 182.1<br />
B-C = 74.5<br />
|<nowiki>-435.02</nowiki><br />
|<nowiki>-433.98</nowiki><br />
|1.04 KJmol<sup>-1</sup><br />
|[[File:182.1 01532430.png|thumb|Figure 12]]<br />
|-<br />
|H + HF -> H<sub>2</sub> + F<br />
|A-B = 180.1<br />
B-C = 74.5<br />
|<nowiki>-559.23</nowiki><br />
|<nowiki>-433.98</nowiki><br />
|125.25 KJmol<sup>-1</sup><br />
|[[File:Energ v time 01532430.png|thumb|Figure 13]]<br />
|}<br />
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'''''<u>Sensible estimates for Ea in both directions, well done. </u>'''''<br />
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=== Reaction Dynamics ===<br />
<blockquote>'''''Q. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. '''''</blockquote>A. The trajectory of the following plots are AB - 180.1 and BC - 74.5, and were viewed dynamically at 3000 steps and 0.1 fs. <br />
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Figure 14.1 shows the reaction going beyond the transition state and the reaction being successful. Figure 14.2 shows that energy is conserved, by the green horizontal constant energy line. Figure 14.3 shows the oscillation of the reactant, products and their high vibrational energy. An animation would also show the reaction being complete.<br />
[[File:7a 01532430.png|left|thumb|Figure 14.1: Contour Plot ]]<br />
[[File:7c 01532430.png|thumb|Figure 14.3]]<br />
[[File:7b 01532430.png|centre|thumb|Figure 14.2]]<blockquote>'''''Q. Explain how this could be confirmed experimentally.'''''</blockquote>Infrared chemiluminescence can be used to determine the mechanism by measuring the populations of vibrational states of products. <ref>Atkins P. and de Paula J.Ā ''Physical Chemistry''Ā (8th ed., W.H.Freeman 2006) p.886Ā ISBNĀ 0-7167-8759-8</ref><br />
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'''''<u>How though? Give some detail. </u>'''''<br />
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<blockquote>Q. '''''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''''</blockquote>A. Hammonds Postulate states that exothermic reactions have an early transition state that resembles the reactants more then the products, while endothermic reactions have a late transition state resembling the products.<ref>George S. Hammond<br />
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1955Ā <em>77</em>Ā (2), 334-338<br />
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DOI: 10.1021/ja01607a027<br />
</ref> While Polanyi adds to this, stating for exothermic reactions with these late transition states, high transitonal, low vibrational energies produce the most efficient reactions.<ref>Zhang Z, Zhou Y, Zhang DH, CzakĆ³ G, Bowman JM. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction.Ā ''J Phys Chem Lett''. 2012;3(23):3416ā3419. doi:10.1021/jz301649w</ref> Figure 15 shows the F + H2 -> HF + H reaction being unsuccessful, and this can be attributed to the high momentum, and therefore high vibrational energy. Figure 16 however shows a reaction with lower momentum and yet the reaction is successful unlike figure 15. This coincides with the assumption of Hammonds Postulate and Polanyi.<br />
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{| class="wikitable"<br />
!P<sub>AB</sub><br />
!P<sub>BC</sub><br />
!Reaction Successful?<br />
!Contour Plot of Reaction<br />
|-<br />
|<nowiki>-1</nowiki><br />
|<nowiki>-2</nowiki><br />
|No<br />
|[[File:Last contour 1 01532430.png|thumb|Figure 15]]<br />
|-<br />
|<nowiki>-1</nowiki><br />
|<nowiki>-0.5</nowiki><br />
|Yes<br />
|[[File:Last contour 2 01532430.png|thumb|Figure 16]]<br />
|}<br />
On the other hand the reaction H + HF -> H<sub>2</sub> + F is endothermic. So according to hammond postulate and polanyi, there will be a late transition state and this favours the reactants having a high vibrational energy. This is indicated in figures 17 where the reaciton is unsuccessful with a low momentum and yet is successful with a high momentum in figure 18. <br />
{| class="wikitable"<br />
!PAB<br />
!PBC<br />
!Reaction Successful<br />
!Contour Plot of Reaction<br />
|-<br />
|<nowiki>-10</nowiki><br />
|<nowiki>-1</nowiki><br />
|No<br />
|[[File:Last contour 3 01532430.png|thumb|Figure 17]]<br />
|-<br />
|<nowiki>-20</nowiki><br />
|<nowiki>-1</nowiki><br />
|Yes<br />
|[[File:Last contour 4 01532430.png|thumb|Figure 18]]<br />
|}<br />
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<blockquote>'''''<u>Good!</u>'''''</blockquote></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yr2comptb1217&diff=812903MRD:yr2comptb12172020-06-25T20:31:19Z<p>Ng611: /* EXERCISE 1: H + H2 system */</p>
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== '''''<u>Overall - 5/5. An absolutely exemplary report, well done!</u>''''' ==<br />
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== EXERCISE 1: H + H2 system ==<br />
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==== Dynamics from the transition state region ====<br />
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''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''<br />
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The reaction path is the minimum energy pathway that links the minima between the reactants and products. The transition state is the first order saddle point in the potential energy surface (PES), the maximum on the minimum energy path that links the reactants and products. At this point, there is a maximum potential, and the potential for the coordinate orthogonal to the reaction has a minimum. Mathematically, the transition state can be found at the point where the gradient of the potential is equal to zero; āV('''r<sub>i</sub>''')/ā'''r<sub>i</sub>'''=0. From this point, the energy goes down most steeply along the minimum energy path that links the reactants and products. <br />
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To identify the transition state, begin trajectories near to the point, with no initial momentum, and observe whether the trajectory follows towards the reactants or products. If the trajectory is started at the transition state exactly without any momentum it will remain there forever as the gradient of the potential is zero and the trajectory cannot 'roll' towards anywhere unless there is a change in structure.<br />
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For a saddle point, the function at the point must also obey the following rule; āV<sup>2</sup>('''r<sub>i</sub>''')/ā'''r<sub>i</sub><sup>2</sup>'''> 0 for all coordinates bar the reaction coordinate. Along the reaction coordinate, āV<sup>2</sup>('''r<sub>i</sub>''')/ā'''r<sub>i</sub><sup>2</sup>'''< 0. This is because the transition state point is a maximum along only one direction, along the reaction coordinate, and is a minimum along all other coordinates. For a local minima; āV<sup>2</sup>('''r<sub>i</sub>''')/ā'''r<sub>i</sub><sup>2</sup>'''> 0 for all coordinates, including that of the reaction coordinate.<br />
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'''''<u>Good description of the PES!</u>'''''<br />
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==== Trajectories from r<sub>1</sub> = r<sub>2</sub>: locating the transition state ====<br />
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''Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.''<br />
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The position of the transition state, r<sub>ts</sub>,is found to be when r<sub>1</sub>=r<sub>2</sub> as the potential energy surface is symmetric. Similarly, it is found that the reaction is concerted, so the value of r<sub>ts</sub> will be greater than the bond distance of 74pm. Looking at the plot of Internuclear Distances vs Time using the initial conditions from before, ''Figure 1'', the transition state can be seen to occur when the r<sub>1</sub> and r<sub>2</sub> < 100pm. The best estimate for r<sub>ts</sub> was found to be at 91pm. As seen in the contour plot below, ''Figure 2'', the trajectory follows neither towards the reactants or products, but remains at this potential. This further supports that this is the transition state position. In the plot of Internuclear Distances vs Time, ''Figure 3'', there is also minimal oscillation due to being a transition state and no change in the bond distances.<br />
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{|style="margin: 0 auto;"<br />
| [[File:rTS_INvsT_under100.png|200px|thumb|upright|''Figure 1: Internuclear Distances vs Time with initial conditions'']]<br />
| [[File:Surface_Plot_rTS.png|200px|thumb|upright|''Figure 2: Contour plot at r<sub>ts</sub>'']]<br />
| [[File:rTS_tb1217.png|200px|thumb|upright|''Figure 3: Internuclear Distances vs Time plot at r<sub>ts</sub>'']]<br />
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'''''<u>Your estimate for the PES is correct and well justified, well done. </u>'''''<br />
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==== Trajectories from r<sub>1</sub>Ā = r<sub>ts</sub>+Ī“, r<sub>2</sub>Ā = r<sub>ts</sub> ====<br />
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''Comment on how the mep and the trajectory you just calculated differ''<br />
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For the mep, ''Figure 4,'' the trajectory follows the valley floor to form H<sub>1</sub> and H<sub>2</sub>-H<sub>3</sub>, whereas using dynamics, ''Figure 5'', the trajectory does not follow the valley floor in a straight line, but follows a wavy line along the valley floor. This is because in the dynamics calculation, the velocity of the species are not reset to zero, hence have kinetic energy and vibrate. The vibrations of the H<sub>2</sub>-H<sub>3</sub> is what is depicted as the wavy line that the trajectory follows in ''Figure 5''. In the mep, the species has no kinetic energy as the velocity is being reset to 0, hence the bond does not vibrate and the trajectory simply follows the valley floor.<br />
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'''''<u>Good!</u>'''''<br />
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{|style="margin: 0 auto;"<br />
| [[File:Mep_tb1217_surface_plot.png|200px|thumb|upright|''Figure 4: Surface plot using mep'']]<br />
| [[File:Dynamic_tb1217_surfaceplot.png|200px|thumb|upright|''Figure 5: Surface plot using dynamics'']]<br />
|}<br />
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''Look at the āInternuclear Distances vs Timeā and āMomenta vs Timeā. What would change if we used the initial conditionsĀ '''r<sub>1</sub>'''Ā =Ā '''r<sub>ts</sub>'''Ā andĀ '''r<sub>2</sub>'''Ā =Ā '''r<sub>ts</sub>'''+1Ā pm instead?''<br />
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''Take note of the final values of the positionsĀ '''r<sub>1</sub>'''(t)Ā '''r<sub>2</sub>'''(t) andĀ Ā '''p<sub>1</sub>'''(t)Ā '''p<sub>2</sub>'''(t) for your trajectory for large enough t.''<br />
* ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.''<br />
''What do you observe?''<br />
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When using the final positions, 74pm and 240pm, and final momenta values reversed, -3.2 gmol<sup>-1</sup>pmfs<sup>-1</sup> and -5 gmol<sup>-1</sup>pmfs<sup>-1</sup>, the trajectory began at the reactants phase, and then followed the reverse trajectory from before, ''Figure 6 and Figure 7''. When the conditions are set to r<sub>1</sub>Ā =Ā r<sub>ts</sub>Ā andĀ r<sub>2</sub>Ā =Ā r<sub>ts</sub>+1Ā pm, the trajectory followed to form the reactants, this is because when the structure is changed even slightly from the transition state structure, the trajectory follows the minima path towards either the reactants or products, depending on which is closer in energy to. The initial conditions set with r<sub>1</sub>Ā =Ā r<sub>ts</sub>Ā andĀ r<sub>2</sub>Ā =Ā r<sub>ts</sub>+1Ā pm cause the trajectory to follow a path towards the reactants as the conditions are set near to the transition state and are closer to the reactant's structure than the product's. When using the final positions and reverse of the final momenta, the trajectory begins where the previous trajectory ended and is only able to nearly reach the transition state, but not overcome it to form the products, hence falls back to form the reactants.<br />
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'''''<u>Good!</u>'''''<br />
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{|style="margin: 0 auto;"<br />
| [[File:Inital_beforechangig_tb1217.png|200px|thumb|upright|''Figure 6: Contour plot using r<sub>1</sub>= r<sub>ts</sub> and r<sub>2</sub>= r<sub>ts</sub>+1pm'']]<br />
| [[File:Surfaceplot_backtoreactants_tb1217.png|200px|thumb|upright|''Figure 7: Contour plot using final positions and reverse of final momenta'']]<br />
|}<br />
<br />
==== Reactive and unreactive trajectories ====<br />
''For the initial positions r<sub>1</sub> = 74 pm and r<sub>2</sub> = 200 pm, run trajectories with the following momenta combination and complete the table''<br />
<br />
{| class="wikitable" border=1<br />
! p<sub>1</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.8<br />
|| Yes <br />
|| The momenta is sufficient to reach and overcome the TS energy barrier and form the products. Intramolecular vibrations are also observed.<br />
|| [[File:Plot1_table_tb1217.png|230px]]<br />
|-<br />
| -3.1 || -4.1 || -420.7<br />
|| No<br />
|| The transition state energy and position is never reached and overcome, hence the reaction does not occur and the initial H1-H2 bond remains intact. Intramolecular vibrations are also observed.<br />
||[[File:Plot2_table_tb1217.png|230px]]<br />
|-<br />
| -3.1 || -5.1 || -414.0<br />
|| Yes <br />
|| The trajectory reaches the transition state and follows through to form the products.<br />
||[[File:Plot3_table_tb1217.png|230px]]<br />
|-<br />
| -5.1 || -10.1 || -357.3<br />
|| No<br />
|| The trajectory reaches the transition state region and crosses it. However, even though the bond in the product forms, the reactants are then reformed. This is due to barrier recrossing.<br />
||[[File:Plot4_table_tb1217.png|230px]]<br />
|-<br />
| -5.1 || -10.6 || -349.5<br />
|| Yes<br />
|| The trajectory follows a pathway where the product is made, before reverting back to the reactants, and finally returning to and forming the products.<br />
||[[File:Plotfinal_table_tb1217.png|230px]]<br />
|}<br />
<br />
''What can you conclude from the table?''<br />
<br />
Some of the above trajectories have a greater energy than a successful trajectory, but the reaction still does not occur. This is because the total energy of the system cannot always determine whether the reaction occurs or not. The trajectory that the reaction follows needs to be determined and calculated in order to determine whether a reaction occurs or not.<br />
<br />
====Transition State Theory====<br />
<br />
''Given the results you have obtained, how will Transition State Theory (TST) predictions for reaction rate values compare with experimental values?''<br />
<br />
[[File:Plot4_table_tb1217.png|256px|thumb|left|''Figure 8: Barrier Recrossing'']]<br />
<br />
In TST, it is assumed that there is an equilibrium between the activated complex and the reactants. TST also assumes that once the energy barrier has been crossed, the reaction will be successful; once the transition state region, and activation energy, is passed- the trajectory will follow to the products and cannot go back to the reactants <ref name="TST" />. However, as shown in the table above, this is not always the case. <br />
<br />
For example, in ''Figure 8'', barrier recrossing is observed. These barrier recrossings are not observed by TST due to the above assumption, and leads to the reaction rate constant being overestimated. This is because, in TST, these reactions would have been classified as reaching completion, but due to barrier recrossing, they do not. This leads to an overestimation in the reaction rate constant.<br />
<br />
TST is based on the assumption that atomic nuclei behave according to classical mechanics and also ignores phenoma such as quantum tunnelling through the energetic barrier<ref name="Quantum" />. In quantum tunnelling, particles are able to pass through barriers even when their energy is less than the energy needed to go over the barrier. This leads to an underestimation of the reaction rate constant. However, the effect from ignoring quantum tunnelling is much smaller than the effect from ignoring barrier recrossing. The rate of quantum tunnelling also decreases as the size of the species increases, so as the atom size gets larger; the effect of quantum tunnelling diminishes. <br />
<br />
Overall, TST leads to an overestimation in the reaction rate constant compared with experimental values.<br />
<br />
'''''<u>An exemplary answer-- absolutely correct, well done!</u>'''''<br />
<br />
== EXERCISE 2: F - H - H system ==<br />
<br />
===PES inspection===<br />
<br />
''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.''<br />
<br />
[[File:Ts_f2h_tb1217.png|200px|thumb|right|''Figure 9: Contour plot at r<sub>ts</sub>]]<br />
{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"<br />
|+ Atom Assignment<br />
! A!! B !! C<br />
|-<br />
| F|| H ||H<br />
|}<br />
<br />
Reaction 1: F + H<sub>2</sub><br />
<br />
The r<sub>ts</sub> was found to be at r<sub>1</sub>= 182pm and r<sub>2</sub>= 74pm. The transition state is early and close to the reactants, and according to Hammond's postulate, this indicates that the reaction is exothermic. This is because the bond strength of the bonds made, H-F, is greater than the bond strength of the bonds broken, H-H. The activation energy is 1.5 kjmol<sup>-1</sup>.<br />
<br />
Reaction 2: H + HF<br />
<br />
The r<sub>ts</sub> was found to be the same, at r<sub>1</sub>= 182pm and r<sub>2</sub>= 74pm. The transition state for this reaction is late, indicating the reaction is endothermic. This is because the energy needed to break the bond, H-F, is greater than the energy released from the bond made, H-H. The activation energy is 126.5 kjmol<sup>-1</sup>.<br />
<br />
'''''<u>Sensible estimates for Ea in both directions. Well done. </u>'''''<br />
<br />
=== Reaction dynamics ===<br />
* ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''<br />
<br />
In the reaction, total energy must be conserved. This reaction is exothermic, as the bond energy of the bond made is greater than the bond energy of the bond broken. This means there is a release of energy. In this reaction, this energy is released as both translational and vibrational energy, and this release of energy can be measured experimentally in two ways. <br />
The first method is using calorimetry. This method will record the temperature changes caused due to the release of translational and vibrational energy combined. <br />
<br />
The energy released that is deposited into the vibrational energy of the new bond formed can also be observed individually. This can be observed using Infrared Chemiluminescence (IRCL) <ref name="IRCL" />. This methods allows us to record the release of vibrational energy separately from the release of translational energy. Once vibrationally excited products form, they emit infrared photons due to the vibrational relaxation that occurs. This method records intensities of the infrared emission lines from the vibrationally excited molecules and measure the populations of the vibrational states in the product, confirming that reaction energy is released into vibrational energy of the bond <ref name="FollowReaction" />. <br />
<br />
'''''<u>Excellent answer, well done!</u>''''' <br />
<br />
* Setup a calculation starting on the side of the reactants of F + H<sub>2</sub>, at the bottom of the well r<sub>HH</sub>&nbsp;=&nbsp;74&nbsp;pm, with a momentum p<sub>FH</sub>&nbsp;=&nbsp;-1.0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, and explore several values of p<sub>HH</sub> in the range -6.1 to 6.1&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.<br />
<br />
The reaction varies between proceeding and not proceeding towards the products when the values are changed for p<sub>HH</sub> in the range -6.1 to 6.1&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. The varying results have been displayed below, with descriptions as to what is happening in each trajectory. This data and trajectories show how the exothermic reaction varies with a range of vibrational energy inputs.<br />
<br />
{| class="wikitable" border=1<br />
! p<sub>HH</sub> !! E<sub>tot</sub> !! style="text-align: center;" |Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -6.0|| -404.0<br />
|style="text-align: center;" | No<br />
|| The trajectory follows to react and form the products but then recrosses the transition state barrier to form the reactants. Barrier recrossing occurs.<br />
|| [[File:-6.0_tb1217.png|230px]]<br />
|-<br />
| -5.8|| -406.1<br />
|style="text-align: center;" | Yes<br />
|| The trajectory follows from the reactants to form the products, but takes a long time to do so. The reaction has to be observed for a much longer time in order to observe the reaction proceeding to form the reactants. When observed for a shorter time, only barrier recrossing is observed. Overall the trajectory of the reaction is not smooth and the structure changes between the reactants, products and transition state.<br />
|| [[File:-5.8_tb1217.png|230px]]<br />
|-<br />
| -3|| -428.0<br />
|style="text-align: center;"| No<br />
|| The reaction does not proceed at all and remains within the entry channel as the reactants. <br />
|| [[File:-3_tb1217.png|230px]]<br />
|-<br />
| 4|| -414.0<br />
|style="text-align: center;"| Yes<br />
|| The reaction proceeds and forms the products but also changes between the products, reactants and transition state structure. The reaction also has to be observed for a long time to observe a successful reaction. When observed for a shorter length of time, only barrier recrossing is observed and the reaction does not appear to proceed.<br />
|| [[File:4_tb1217.png|230px]]<br />
|-<br />
| 6.0|| -392.0<br />
|style="text-align: center;"| Yes<br />
|| Similar to the trajectory dynamics above, the reaction follows barrier recrossing when observed for a short timespan but eventually the reaction does proceed to form the products.<br />
|| [[File:6.0_tb1217.png|230px]]<br />
|-<br />
|}<br />
<br />
From the above data, the same concept applies where the total energy does not determine whether or not a trajectory is reactive. Overall, the reaction would have been seen to proceed according to Transition State Theory as an energy greater than the activation energy was put in, however due to barrier recrossing and inefficient energy input (vibrational), not all of the reactions did proceed to form products. The reactions were also observed to not proceed at first, but for some trajectories, once the time the reaction was observed for was increased; the reaction does proceed. This is because lots of barrier recrossing occurs where the species change between products, reactants and the transition state positions when given lots of vibrational energy.<br />
<br />
* For the same initial position, increase slightly the momentum p<sub>FH</sub>&nbsp;=&nbsp;-1.6&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, and considerably reduce the overall energy of the system by reducing the momentum p<sub>HH</sub>&nbsp;=&nbsp;0.2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. What do you observe now?<br />
<br />
This is a reactive trajectory, following through from the entry channel with the reactants towards the exit channel and forming the products, ''Figure 10''. The trajectory is smooth unlike the ones from above and there is no recrossing of the transition state barrier or reforming of the reactants. In this trajectory, there is overall less energy in the system yet the reaction proceeds as there is greater translational energy than before. This is an effect supported by Polanyi's Empirical rules for early barrier, exothermic reactions.<br />
<br />
[[File:Prcoedsdin_tb1217.png|230px|thumb|centre|''Figure 10: Contour plot of the successful trajectory'']]<br />
<br />
* Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of p<sub>HH</sub> above the activation energy (an H atom colliding with a high kinetic energy).<br />
<br />
The initial conditions were set as below to give a contour plot shown in ''Figure 11''. This reaction trajectory is not reactive as there is lots of translational energy but low vibrational energy. This is also an observation from Polanyi's Empirical rules, as this is a late barrier, endothermic reaction, it is not efficient to input translational energy, and are more likely to obtain a successful trajectory by inputting more vibrational energy.<br />
<br />
{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"<br />
|+ Initial conditions<br />
! Distance (pm) !! Momentum(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
|-<br />
|style="text-align: center;"|91<br />
|style="text-align: center;"|-0.2<br />
|-<br />
|style="text-align: center;"|220<br />
|style="text-align: center;"|-10<br />
|}<br />
<br />
[[File:Animation_tb1217.png|230px|thumb|centre|''Figure 11: Contour plot of unreactive trajectory'']]<br />
<br />
* Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. <br />
<br />
A reactive trajectory was found using the conditions below.<br />
<br />
{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"<br />
|+ Initial conditions<br />
! Distance (pm) !! Momentum (g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
|-<br />
|style="text-align: center;"|91<br />
|style="text-align: center;"|17<br />
|-<br />
|style="text-align: center;"|220<br />
|style="text-align: center;"|-1<br />
|}<br />
<br />
This produced the following contour plot, ''Figure 13''. In this trajectory, there is greater vibrational energy and hence the reaction is successful, following Polanyi's Empirical rules.<br />
<br />
[[File:Succesfful_tb1217.png|230px|thumb|centre|''Figure 12: Contour plot'']]<br />
<br />
=== Polanyi's empirical rules and Hammond's Postulate ===<br />
<br />
[[File:Polanyi_tb1217.jpg|200px|thumb|left|''Figure 13: Efficiency of translational vs vibrational energy inputs for endothermic and exothermic reactions<ref name="Polanyi_Images" />'']]<br />
Hammond's Postulate is as follows, the structure of the transition state for an exothermic reaction is reached relatively early in the reaction, and so it resembles the reactants. The transition state structure for an endothermic reaction is reached relatively late and resembles the products more than the reactants<ref name="H.Postulate" />.<br />
<br />
Polanyi's Emperical rules, ''Figure 13'', shows that vibrational energy is more efficient for a late barrier reaction to proceed than translational energy, and translational energy is more efficient for a early barrier reaction to proceed successfully.<br />
<br />
====Exothermic and early barrier reaction: H<sub>2</sub> + F ====<br />
<br />
Following Polanyi's Emperical rules, this reaction is more likely to proceed with translational energy than vibrational energy as it is exothermic, hence translational energy is more efficient<ref name="Polanyi" />. This is further supported by the trajectories of the reactions above. Even when a range of large vibrational energies were put into the reaction, the reaction still did not always proceed and resulted in barrier recrossing. Even if the reaction did proceed, the trajectory was not smooth, and the structure would convert between products, reactants, and the transition state. However, when the energy was put in as translational energy, even a small increase led to a successful trajectory; with little vibrational energy and reduced energy in the system overall compared to the range tested as seen in the table. This supports Polanyi's Empirical rules that for exothermic reactions, translational energy is an efficient energy input for a successful reaction to occur.<br />
<br />
To further support this, the following initial conditions were set, reactions with around the same total energy but one with more vibrational energy input than translational energy, and another with more translational energy input than vibrational energy, ''Figure 14 and Figure 15''. The reaction is successful and follows a smooth trajectory to form the products for the reaction with a greater translational energy input, supporting Polanyi's Empirical rules.<br />
<br />
{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"<br />
|+ Initial Conditions<br />
! Distance(pm)!! Momentum(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)!! !! Distance(pm) !! Momentum(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
|-<br />
|style="text-align: center;"|225<br />
|style="text-align: center;"|-1.0<br />
|| <br />
|style="text-align: center;"|225<br />
|style="text-align: center;"|-1.6<br />
|-<br />
|style="text-align: center;"|74<br />
|style="text-align: center;"|-1.2 <br />
|| <br />
|style="text-align: center;"|74<br />
|style="text-align: center;"|0.2<br />
|}<br />
<br />
{|style="margin: 0 auto;"<br />
|[[File:-2_tb1217.png|200px|thumb|upright|''Figure 14: Unsuccessful trajectory with more vibrational energy'']]<br />
|[[File:Prcoedsdin_tb1217.png|200px|thumb|upright|''Figure 15: Successful trajectory with more translational energy'']]<br />
|}<br />
<br />
====Endothermic and late barrier reaction: HF + H ====<br />
<br />
For an endothermic reaction, Polanyi's Empirical rules state that inputting vibrational energy is more efficient in getting a successful reaction than translational energy. This rule is supported by the trajectories carried out below. When more energy is input as translational energy than vibrational energy, the reaction does not proceed, ''Figure 16''. The successful trajectory has a large amount of vibrational energy as p<sub>HF</sub> is large, and little translational energy, ''Figure 17''. <br />
<br />
{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"<br />
|+ Initial Conditions<br />
! Distance(pm)!! Momentum(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)!! !! Distance(pm) !! Momentum(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
|-<br />
|style="text-align: center;"|91<br />
|style="text-align: center;"|0.5<br />
|| <br />
|style="text-align: center;"|91<br />
|style="text-align: center;"|17<br />
|-<br />
|style="text-align: center;"|225<br />
|style="text-align: center;"|-9 <br />
|| <br />
|style="text-align: center;"|225<br />
|style="text-align: center;"|-1<br />
|}<br />
<br />
{|style="margin: 0 auto;"<br />
|[[File:Notworking_endo_tb1217.png|200px|thumb|upright|''Figure 16: Unsuccessful trajectory with more translational energy'']]<br />
|[[File:Succesfful_tb1217.png|200px|thumb|upright|''Figure 17: Successful trajectory with more vibrational energy'']]<br />
|}<br />
<br />
Overall, the data and trajectories recorded support Polanyi's Empirical rules and Hammond's Postulate.<br />
<br />
'''''<u>Absolutely correct. Excellent and detailed explanation of Polanyi's rules</u>'''''<br />
<br />
=References=<br />
<br />
<br />
<references><br />
<ref name="TST">K.J. Laidler, Chemical Kinetics, 1987, 88-98</ref><br />
<ref name="Quantum"> D. Sholl and J. A. Steckel, Density Functional Theory: A Practical Introduction, Wiley, 2011.</ref><br />
<ref name="FollowReaction">J. L. Schreiber and J. . Polanyi, Faraday Discuss. Chem. Soc, 1977, 62, 267ā290</ref><br />
<ref name="IRCL">J. R. Ferraro and L. J. Basile, Fourier Transform Infrared Spectra: Applications to Chemical Systems, Elsevier Science, 2012.</ref><br />
<ref name="H.Postulate">W. Brown, C. Foote, B. Iverson and E. Anslyn, Organic Chemistry, Cengage Learning, 2008.</ref><br />
<ref name="Polanyi">J. C. Polanyi, Angew. Chemie Int. Ed. English, 1987, 26, 952ā971</ref><br />
<ref name="Polanyi_Images">J. C. POLANYI, Science (80-. )., 1987, 236, 680ā690.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:RGF&diff=812570MRD:RGF2020-06-01T16:25:47Z<p>Ng611: /* Potential Energy Surface Plots and Transition States */</p>
<hr />
<div>==Molecular Reaction Dynamics for Triatomic Systems==<br />
<br />
==H + H<sub>2</sub> systems==<br />
<br />
OVERALL: 4/5. '''''<u>A very detailed and thorough report in most places. One or two missing bits held your report back from full marks. See the comments throughout for detailed feedback. </u>'''''<br />
<br />
===Potential Energy Surface Plots and Transition States===<br />
<br />
A potential energy surface maps the progress of a reaction as a function of r<sub>AB</sub> and r<sub>BC</sub>. The trajectory of the reaction is the relative positions of the atoms at each instant in time, and shows how these relative positions lead to a change in potential energy of the system. The trajectory is displayed as a black line on the plots. The transition state is displayed as a saddle point on a potential energy surface, and is defined as the maximum on the minimum energy path.<ref>J. S. Francisco and W. L. Hase, <i>Chemical kinetics and dynamics</i>, Prentice-Hall, Upper Saddle River, 1989.<br />
</ref> The reactants and products form minima on the minimum energy path due to there being no unfavorable interactions between H and H<sub>2</sub>, however when the atom approaches H<sub>2</sub> they begin to repel each other, increasing the potential energy. The maximum repulsion is reached at the transition state. The transition state is mathematically defined as:āV(r<sub>AB</sub>)/ār<sub>AB</sub>=āV(r<sub>BC</sub>)/ār<sub>BC</sub>=0. It can be distinguished from a local minimum of the potential energy surface as āV<sup>2</sup>(r<sub>AB</sub>)/ār<sub>AB</sub><sup>2</sup> > 0, since it is a minimum point, and āV<sup>2</sup>(r<sub>BC</sub>)/ār<sub>BC</sub><sup>2</sup> < 0, since it is a maximum point.<br />
<br />
'''''<u>A sensible description, but where do these directions lie on the TS?</u>'''''<br />
<br />
<br />
[[File:TS_surface3_RGF.png|thumb|500px|center|Figure 1: A surface plot showing the transition state as the maximum of the minimum energy path.]]<br />
<br />
===Estimating the Transition State Position===<br />
<br />
<br />
[[File:TS_IDvT1_RGF.png|thumb|500px|center|Figure 2: An internuclear distance against time graph for H + H<sub>2</sub>.]]<br />
<br />
Since the potential energy surface for the H + H<sub>2</sub> system is symmetric, the transition state is when the distances between AB and BC are equal.<ref>N. E. Henriksen and F. Y. Hansen, <i>Theories of molecular reaction dynamics : the microscopic foundation of chemical kinetics</i>, Oxford University Press, Oxford, 2018.<br />
</ref> <i>Figure 2</i> allows this distance to be estimated at 85 - 95 pm. The initial conditions were set to p<sub>1</sub> = p<sub>2</sub> = 0.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and r<sub>AB</sub> = r<sub>BC</sub>. Different distance values were tested until the internuclear distance against time graph had a gradient of zero, and the animation showed the system undergoing a periodic symmetric vibration. This gave the estimate of the transition state position, r<sub>ts</sub>, to be equal to 90.8 pm. Giving the energy at the transition state to be -415.4 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>A sensible estimate.</u>'''''<br />
<br />
[[File:TS_INvT2_RGF.png|thumb|500px|center|Figure 3: An internuclear distance against time graph showing the transition state position to be equal to 90.8 pm.]]<br />
<br />
===Reaction Path===<br />
<br />
The MEP (minimum energy path) calculates the reaction path by using a trajectory that has the particles moving infinitely slowly. It does this by resetting the momenta to zero in each time step, this causes the MEP calculation to follow the valley floor throughout the whole reaction. In the dynamic calculations, the particles have a momentum that causes an oscillating nature, where the energy is continually switching from potential to kinetic energy. This can be seen in the wavy nature of the trajectory as it continually goes through peaks and troughs of potential energy. The dynamic calculation is more realistic as atoms have a mass and their motion will be inertial.<br />
<br />
'''''<u> A sensible explanation, well done</u>'''''<br />
<br />
The initial conditions were set to slightly displace the transition state towards the products and with an initial momenta of zero.<br />
<br />
[[File:MEP_RGF.png|thumb|500px|center|Figure 4: MEP calculation for trajectory.]]<br />
[[File:dynamic_RGF.png|thumb|500px|center|Figure 5: Dynamic calculation for trajectory.]]<br />
<br />
===Reactive and Unreactive Trajectories===<br />
<br />
Setting the initial conditions to r<sub>AB</sub>=74 pm and r<sub>BC</sub>= 200 pm, different values for momenta were tested to see if higher energy guaranteed a reactive trajectory. <br />
<br />
{| class="wikitable" border="1"<br />
|-<br />
!p<sub>1</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>2</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! E<sub>tot</sub> / kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56<br />
| -5.1<br />
| -414.3<br />
| Yes<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> (which doesn't oscillate due to the relatively low momentum for p<sub>1</sub>) and has sufficient energy in the correct modes to overcome the activation energy, forming H<sub>b</sub>-H<sub>c</sub>. The two products move away from each other and H<sub>b</sub>-H<sub>c</sub> oscillates due to p<sub>2</sub> having sufficient momentum. <br />
|[[File:-2.56_RGF.png|450px|center]]<br />
|-<br />
| -3.1<br />
| -4.1<br />
| -420.1<br />
| No<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> (which oscillates), but H<sub>c</sub> has insufficient energy in the correct modes, causing an unreactive trajectory.<br />
|[[File:3.1_RGF.png|450px|center]]<br />
|-<br />
| -3.1<br />
| -5.1<br />
| -414.0<br />
| Yes<br />
| This reaction follows a similar trajectory to the first reaction. However, due to the increased momentum of p<sub>1</sub>, H<sub>a</sub>-H<sub>b</sub> oscillates while it approaches H<sub>c</sub>.<br />
| [[File:3.1(2)_RGF.png|450px|center]]<br />
|-<br />
| -5.1<br />
| -10.1<br />
| -357.3<br />
| No<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> and has sufficient energy to overcome the activation energy, and cross the transition state. Due to the high momentum of p<sub>2</sub>, H<sub>b</sub>-H<sub>c</sub> is oscillating with high kinetic energy, so as the two particles begin to translate away from each other, a second transition state is crossed and H<sub>a</sub>-H<sub>b</sub> reforms. Resulting in no reaction being observed.<br />
| [[File:5.1_RGF.png|450px|center]]<br />
|-<br />
|-<br />
| -5.1<br />
| -10.6<br />
| -349.5<br />
| Yes<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> and has sufficient energy to overcome the activation energy, and cross the transition state. Due to the high momentum of p<sub>2</sub>, H<sub>b</sub>-H<sub>c</sub> is oscillating with high kinetic energy, so as the two particles begin to translate away from each other, a second transition state is crossed and H<sub>a</sub>-H<sub>b</sub> reforms. A third transition state is crossed due to the high energy oscillations, meaning H<sub>b</sub>-H<sub>c</sub> forms as the product. So this reaction passes the transition state three times, resulting in a reactive trajectory. <br />
| [[File:5.1(2)_RGF.png|450px|center]]<br />
|}<br />
<br />
===Transition State Theory===<br />
<br />
Transition state theory (TST) rationalises a reaction as the minimum energy path from the reactants to the products, passing through an energy maxima, which represents the transition state. TST makes many assumptions which affects its reaction rate when compared to experimental rates.<ref>K. J. Laidler, <i>Chemical kinetics</i>, Harper & Row, New York, 1987.<br />
</ref><br />
<br />
ā¢ TST considers the system classically, it is described by a trajectory with a velocity and not a wavefunction. In our model for our potential energy surface, our trajectory is described by a momentum (velocity x mass), so is treated classically.<br />
<br />
ā¢ Since the system is classical, quantum tunneling is ignored. This will lead to TST giving an underestimation of the rate, as a classical view won't allow molecules with insufficient energy to overcome the activation energy by tunneling through the potential barrier.<br />
<br />
ā¢ The kinetic energy along the reaction coordinate follows the Boltzmann distribution. <br />
<br />
ā¢ At the transition state, you can't separate the motion of the system at the lowest point of the saddle point.<br />
<br />
ā¢ TST states that all trajectories with a kinetic energy larger than the activation energy will be reactive, and that every time the transition state is crossed products are formed. This assumption is proved wrong by <i>Figure 5</i>, as it shows that reactants can cross the transition state forming products, and then these products can pass back through the transition state reforming the reactants. This will lead to an overestimation in the TST reaction rate.<br />
<br />
[[File:5.1_RGF.png|500px|center|thumb|Figure 6: Contour plot showing that not all transition state crossings result in a reactive trajectory]]<br />
<br />
Quantum tunneling contributes a relatively low amount to the rate of reaction when compared to the fact that not all transition state crossings form products. Therefore, TST leads to an overestimation of the rate when compared to experimental rates.<br />
<br />
'''''<u>A very detailed and thorough explanation, well done!</u>'''''<br />
<br />
==F - H - H system==<br />
<br />
===Potential Energy Surface===<br />
<br />
The initial conditions were set up so atoms A=F, B=H and C=H. This produced a potential energy surface for which at a large BC distance HF + H would form, and at large AB distance F + H<sub>2</sub> would form. From <i>Figure 7</i>, you can see that the potential energy at large AB distance is more positive than at large BC distance. Therefore, HF + H is at a lower energy than F + H<sub>2</sub>.<br />
<br />
HF + H -> F + H<sub>2</sub> <i>Endothermic</i><br />
<br />
F + H<sub>2</sub> -> HF + H <i>Exothermic</i><br />
<br />
[[File:FH2_1_RGF.png|500px|center|thumb|Figure 7: Surface plot of F - H - H system]]<br />
<br />
===Transition State===<br />
<br />
This energy surface isn't symmetrical, so the two r values can't just be set to the same value to predict the transition state. The transition state must be estimated through its definition, the maxima on the minimum energy curve, via the identification of a saddle point. The Hammond postulate states that for an endothermic reaction (HF + H -> F + H<sub>2</sub>) the transition state will resemble the products, due to it being a late transition state. Therefore, the H<sub>2</sub> distance was set to the bond distance of 74.5 pm, and different distances between F and H (the AB distance) were tested until the internuclear distance-time graph had a gradient of zero. The transition state point can be seen on <i>Figure 7</i> as the black dot at r<sub>HH</sub> = 74.5 pm and r<sub>HF</sub> = 182 pm. Giving the energy at the transition state to be -434.0 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>Well done for backing up your estimate with some evidence. However, ow did you confirm that the system was at a transition state and not a potential energy minimum??</u>'''''<br />
<br />
[[File:FH2_2_RGF.png|500px|center|thumb|Figure 8: Internuclear distance-time graph showing the transition state. ]]<br />
<br />
===Activation Energy===<br />
<br />
Activation energy is the potential energy of the transition state minus the potential energy of the products, E<sub>a</sub> = V<sub>TS</sub> - V<sub>R</sub>. The MEP calculation was used to find the potential energy of the reactants, V<sub>R</sub> = -558.7 kJ.mol<sup>-1</sup>. From the calculation for the transition state, the transition state potential energy was extracted, V<sub>TS</sub> = -434.0 kJ.mol<sup>-1</sup>. From this the activation energy was calculated, E<sub>a</sub> = 124.7 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>What about in the other direction (i.e.: for the F+h2 reaction?)</u>'''''<br />
<br />
===Reaction Dynamics===<br />
<br />
<i>Figure 9</i> shows the momentum-time graph for a reactive trajectory for the exothermic reaction (H<sub>2</sub> + F -> HF + H). Where the initial conditions set the atoms A=H, B=H and C=F, which gave A-B to represent H-H and B-C to represent H-F. Since energy is conserved, the extra energy released from the reaction due to its exothermic nature is converted into vibrational energy, which causes the product molecule, HF, to have increased oscillation (seen in the increase in momentum). <br />
<br />
[[File:IR_1_RGF.png|500px|center|thumb|Figure 9: A momentum-time graph for the exothermic reaction. ]]<br />
<br />
This can be measured experimentally by Raman scattering.<ref>J. A. Koningstein, <i>Introduction to the theory of the Raman effect</i>, Reidel, Dordrecht, 1972.</ref> Electrons that become excited transition into higher vibrational states, this leads to overtone bands being present on the IR spectra. These overtone bands will increase in intensity as the higher vibrational states become more populated, allowing the increased vibrational energy of the system to be monitored.<br />
<br />
'''''<u>Good! What other methods could be used?</u>'''''<br />
<br />
===How the Distribution of Energy Between Different Modes Affect the Efficiency of the Reaction===<br />
<br />
Polanyi's rules state that the position of the transition state determines what mode of energy will be more efficient for the reaction.<ref>J. C. Polanyi, <i>Acc. Chem. Res.</i> 1972, <b>5</b>, 161-168.</ref> They state that for an early barrier transition state (closer to the reactants) translational energy is more efficient, and for a late barrier transition state vibrational energy is more efficient for a reactive trajectory.<br />
<br />
The exothermic reaction (F + H<sub>2</sub> -> HF + H) has an early barrier transition state, so translational energy, p<sub>HF</sub>, should be more efficient for a reactive trajectory than vibrational, p<sub>HH</sub>. <i>Table 1</i> that when p<sub>HF</sub> is constant at -1 and p<sub>HH</sub> is increased, the trajectory stayed unreactive. However, when p<sub>HH</sub> is constant at -1 and p<sub>HF</sub> is increased, the trajectory became reactive. This proves that translational energy is more efficient than vibrational for the exothermic reaction. <br />
{| class="wikitable" border="1"<br />
|+ Table 1: Data for exothermic reaction<br />
|-<br />
!p<sub>HH</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>HF</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! Reactive?<br />
|-<br />
| -1<br />
| -1<br />
| No<br />
|-<br />
| -2<br />
| -1<br />
| No<br />
|-<br />
| -3<br />
| -1<br />
| No<br />
|-<br />
| -1<br />
| -2<br />
| Yes<br />
|-<br />
| -1<br />
| -3<br />
| Yes<br />
|}<br />
<br />
The endothermic reaction (HF + H -> F + H<sub>2</sub>) has a late barrier transition state, so vibrational energy, p<sub>HF</sub>, should be more efficient for a reactive trajectory than translational, p<sub>HH</sub>. <i>Table 2</i> shows that when p<sub>HF</sub> was increased and p<sub>HH</sub> was decreased the trajectory stayed reactive. This proves that vibrational energy is more efficient than translational for the endothermic reaction. <br />
<br />
{| class="wikitable" border="1"<br />
|+ Table 2: Data for endothermic reaction<br />
|-<br />
!p<sub>HH</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>HF</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! Reactive?<br />
|-<br />
| -18<br />
| 0<br />
| Yes<br />
|-<br />
| -10<br />
| 11<br />
| Yes<br />
|-<br />
| 0<br />
| 21<br />
| Yes<br />
|}<br />
'''''<u>A very thorough and systematic analysis, well done. However, we need to see the trajectories as well. </u>'''''<br />
<br />
==References==</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:RGF&diff=812569MRD:RGF2020-06-01T16:25:12Z<p>Ng611: /* Molecular Reaction Dynamics for Triatomic Systems */</p>
<hr />
<div>==Molecular Reaction Dynamics for Triatomic Systems==<br />
<br />
==H + H<sub>2</sub> systems==<br />
<br />
OVERALL: 4/5. '''''<u>A very detailed and thorough report in most places. One or two missing bits held your report back from full marks. See the comments throughout for detailed feedback. </u>'''''<br />
<br />
===Potential Energy Surface Plots and Transition States===<br />
<br />
A potential energy surface maps the progress of a reaction as a function of r<sub>AB</sub> and r<sub>BC</sub>. The trajectory of the reaction is the relative positions of the atoms at each instant in time, and shows how these relative positions lead to a change in potential energy of the system. The trajectory is displayed as a black line on the plots. The transition state is displayed as a saddle point on a potential energy surface, and is defined as the maximum on the minimum energy path.<ref>J. S. Francisco and W. L. Hase, <i>Chemical kinetics and dynamics</i>, Prentice-Hall, Upper Saddle River, 1989.<br />
</ref> The reactants and products form minima on the minimum energy path due to there being no unfavorable interactions between H and H<sub>2</sub>, however when the atom approaches H<sub>2</sub> they begin to repel each other, increasing the potential energy. The maximum repulsion is reached at the transition state. The transition state is mathematically defined as:āV(r<sub>AB</sub>)/ār<sub>AB</sub>=āV(r<sub>BC</sub>)/ār<sub>BC</sub>=0. It can be distinguished from a local minimum of the potential energy surface as āV<sup>2</sup>(r<sub>AB</sub>)/ār<sub>AB</sub><sup>2</sup> > 0, since it is a minimum point, and āV<sup>2</sup>(r<sub>BC</sub>)/ār<sub>BC</sub><sup>2</sup> < 0, since it is a maximum point.<br />
<br />
'''''<u>A sensible description, but where do these points lie on the TS?</u>'''''<br />
<br />
<br />
[[File:TS_surface3_RGF.png|thumb|500px|center|Figure 1: A surface plot showing the transition state as the maximum of the minimum energy path.]]<br />
<br />
===Estimating the Transition State Position===<br />
<br />
<br />
[[File:TS_IDvT1_RGF.png|thumb|500px|center|Figure 2: An internuclear distance against time graph for H + H<sub>2</sub>.]]<br />
<br />
Since the potential energy surface for the H + H<sub>2</sub> system is symmetric, the transition state is when the distances between AB and BC are equal.<ref>N. E. Henriksen and F. Y. Hansen, <i>Theories of molecular reaction dynamics : the microscopic foundation of chemical kinetics</i>, Oxford University Press, Oxford, 2018.<br />
</ref> <i>Figure 2</i> allows this distance to be estimated at 85 - 95 pm. The initial conditions were set to p<sub>1</sub> = p<sub>2</sub> = 0.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> and r<sub>AB</sub> = r<sub>BC</sub>. Different distance values were tested until the internuclear distance against time graph had a gradient of zero, and the animation showed the system undergoing a periodic symmetric vibration. This gave the estimate of the transition state position, r<sub>ts</sub>, to be equal to 90.8 pm. Giving the energy at the transition state to be -415.4 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>A sensible estimate.</u>'''''<br />
<br />
[[File:TS_INvT2_RGF.png|thumb|500px|center|Figure 3: An internuclear distance against time graph showing the transition state position to be equal to 90.8 pm.]]<br />
<br />
===Reaction Path===<br />
<br />
The MEP (minimum energy path) calculates the reaction path by using a trajectory that has the particles moving infinitely slowly. It does this by resetting the momenta to zero in each time step, this causes the MEP calculation to follow the valley floor throughout the whole reaction. In the dynamic calculations, the particles have a momentum that causes an oscillating nature, where the energy is continually switching from potential to kinetic energy. This can be seen in the wavy nature of the trajectory as it continually goes through peaks and troughs of potential energy. The dynamic calculation is more realistic as atoms have a mass and their motion will be inertial.<br />
<br />
'''''<u> A sensible explanation, well done</u>'''''<br />
<br />
The initial conditions were set to slightly displace the transition state towards the products and with an initial momenta of zero.<br />
<br />
[[File:MEP_RGF.png|thumb|500px|center|Figure 4: MEP calculation for trajectory.]]<br />
[[File:dynamic_RGF.png|thumb|500px|center|Figure 5: Dynamic calculation for trajectory.]]<br />
<br />
===Reactive and Unreactive Trajectories===<br />
<br />
Setting the initial conditions to r<sub>AB</sub>=74 pm and r<sub>BC</sub>= 200 pm, different values for momenta were tested to see if higher energy guaranteed a reactive trajectory. <br />
<br />
{| class="wikitable" border="1"<br />
|-<br />
!p<sub>1</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>2</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! E<sub>tot</sub> / kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56<br />
| -5.1<br />
| -414.3<br />
| Yes<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> (which doesn't oscillate due to the relatively low momentum for p<sub>1</sub>) and has sufficient energy in the correct modes to overcome the activation energy, forming H<sub>b</sub>-H<sub>c</sub>. The two products move away from each other and H<sub>b</sub>-H<sub>c</sub> oscillates due to p<sub>2</sub> having sufficient momentum. <br />
|[[File:-2.56_RGF.png|450px|center]]<br />
|-<br />
| -3.1<br />
| -4.1<br />
| -420.1<br />
| No<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> (which oscillates), but H<sub>c</sub> has insufficient energy in the correct modes, causing an unreactive trajectory.<br />
|[[File:3.1_RGF.png|450px|center]]<br />
|-<br />
| -3.1<br />
| -5.1<br />
| -414.0<br />
| Yes<br />
| This reaction follows a similar trajectory to the first reaction. However, due to the increased momentum of p<sub>1</sub>, H<sub>a</sub>-H<sub>b</sub> oscillates while it approaches H<sub>c</sub>.<br />
| [[File:3.1(2)_RGF.png|450px|center]]<br />
|-<br />
| -5.1<br />
| -10.1<br />
| -357.3<br />
| No<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> and has sufficient energy to overcome the activation energy, and cross the transition state. Due to the high momentum of p<sub>2</sub>, H<sub>b</sub>-H<sub>c</sub> is oscillating with high kinetic energy, so as the two particles begin to translate away from each other, a second transition state is crossed and H<sub>a</sub>-H<sub>b</sub> reforms. Resulting in no reaction being observed.<br />
| [[File:5.1_RGF.png|450px|center]]<br />
|-<br />
|-<br />
| -5.1<br />
| -10.6<br />
| -349.5<br />
| Yes<br />
| H<sub>c</sub> approaches H<sub>a</sub>-H<sub>b</sub> and has sufficient energy to overcome the activation energy, and cross the transition state. Due to the high momentum of p<sub>2</sub>, H<sub>b</sub>-H<sub>c</sub> is oscillating with high kinetic energy, so as the two particles begin to translate away from each other, a second transition state is crossed and H<sub>a</sub>-H<sub>b</sub> reforms. A third transition state is crossed due to the high energy oscillations, meaning H<sub>b</sub>-H<sub>c</sub> forms as the product. So this reaction passes the transition state three times, resulting in a reactive trajectory. <br />
| [[File:5.1(2)_RGF.png|450px|center]]<br />
|}<br />
<br />
===Transition State Theory===<br />
<br />
Transition state theory (TST) rationalises a reaction as the minimum energy path from the reactants to the products, passing through an energy maxima, which represents the transition state. TST makes many assumptions which affects its reaction rate when compared to experimental rates.<ref>K. J. Laidler, <i>Chemical kinetics</i>, Harper & Row, New York, 1987.<br />
</ref><br />
<br />
ā¢ TST considers the system classically, it is described by a trajectory with a velocity and not a wavefunction. In our model for our potential energy surface, our trajectory is described by a momentum (velocity x mass), so is treated classically.<br />
<br />
ā¢ Since the system is classical, quantum tunneling is ignored. This will lead to TST giving an underestimation of the rate, as a classical view won't allow molecules with insufficient energy to overcome the activation energy by tunneling through the potential barrier.<br />
<br />
ā¢ The kinetic energy along the reaction coordinate follows the Boltzmann distribution. <br />
<br />
ā¢ At the transition state, you can't separate the motion of the system at the lowest point of the saddle point.<br />
<br />
ā¢ TST states that all trajectories with a kinetic energy larger than the activation energy will be reactive, and that every time the transition state is crossed products are formed. This assumption is proved wrong by <i>Figure 5</i>, as it shows that reactants can cross the transition state forming products, and then these products can pass back through the transition state reforming the reactants. This will lead to an overestimation in the TST reaction rate.<br />
<br />
[[File:5.1_RGF.png|500px|center|thumb|Figure 6: Contour plot showing that not all transition state crossings result in a reactive trajectory]]<br />
<br />
Quantum tunneling contributes a relatively low amount to the rate of reaction when compared to the fact that not all transition state crossings form products. Therefore, TST leads to an overestimation of the rate when compared to experimental rates.<br />
<br />
'''''<u>A very detailed and thorough explanation, well done!</u>'''''<br />
<br />
==F - H - H system==<br />
<br />
===Potential Energy Surface===<br />
<br />
The initial conditions were set up so atoms A=F, B=H and C=H. This produced a potential energy surface for which at a large BC distance HF + H would form, and at large AB distance F + H<sub>2</sub> would form. From <i>Figure 7</i>, you can see that the potential energy at large AB distance is more positive than at large BC distance. Therefore, HF + H is at a lower energy than F + H<sub>2</sub>.<br />
<br />
HF + H -> F + H<sub>2</sub> <i>Endothermic</i><br />
<br />
F + H<sub>2</sub> -> HF + H <i>Exothermic</i><br />
<br />
[[File:FH2_1_RGF.png|500px|center|thumb|Figure 7: Surface plot of F - H - H system]]<br />
<br />
===Transition State===<br />
<br />
This energy surface isn't symmetrical, so the two r values can't just be set to the same value to predict the transition state. The transition state must be estimated through its definition, the maxima on the minimum energy curve, via the identification of a saddle point. The Hammond postulate states that for an endothermic reaction (HF + H -> F + H<sub>2</sub>) the transition state will resemble the products, due to it being a late transition state. Therefore, the H<sub>2</sub> distance was set to the bond distance of 74.5 pm, and different distances between F and H (the AB distance) were tested until the internuclear distance-time graph had a gradient of zero. The transition state point can be seen on <i>Figure 7</i> as the black dot at r<sub>HH</sub> = 74.5 pm and r<sub>HF</sub> = 182 pm. Giving the energy at the transition state to be -434.0 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>Well done for backing up your estimate with some evidence. However, ow did you confirm that the system was at a transition state and not a potential energy minimum??</u>'''''<br />
<br />
[[File:FH2_2_RGF.png|500px|center|thumb|Figure 8: Internuclear distance-time graph showing the transition state. ]]<br />
<br />
===Activation Energy===<br />
<br />
Activation energy is the potential energy of the transition state minus the potential energy of the products, E<sub>a</sub> = V<sub>TS</sub> - V<sub>R</sub>. The MEP calculation was used to find the potential energy of the reactants, V<sub>R</sub> = -558.7 kJ.mol<sup>-1</sup>. From the calculation for the transition state, the transition state potential energy was extracted, V<sub>TS</sub> = -434.0 kJ.mol<sup>-1</sup>. From this the activation energy was calculated, E<sub>a</sub> = 124.7 kJ.mol<sup>-1</sup>.<br />
<br />
'''''<u>What about in the other direction (i.e.: for the F+h2 reaction?)</u>'''''<br />
<br />
===Reaction Dynamics===<br />
<br />
<i>Figure 9</i> shows the momentum-time graph for a reactive trajectory for the exothermic reaction (H<sub>2</sub> + F -> HF + H). Where the initial conditions set the atoms A=H, B=H and C=F, which gave A-B to represent H-H and B-C to represent H-F. Since energy is conserved, the extra energy released from the reaction due to its exothermic nature is converted into vibrational energy, which causes the product molecule, HF, to have increased oscillation (seen in the increase in momentum). <br />
<br />
[[File:IR_1_RGF.png|500px|center|thumb|Figure 9: A momentum-time graph for the exothermic reaction. ]]<br />
<br />
This can be measured experimentally by Raman scattering.<ref>J. A. Koningstein, <i>Introduction to the theory of the Raman effect</i>, Reidel, Dordrecht, 1972.</ref> Electrons that become excited transition into higher vibrational states, this leads to overtone bands being present on the IR spectra. These overtone bands will increase in intensity as the higher vibrational states become more populated, allowing the increased vibrational energy of the system to be monitored.<br />
<br />
'''''<u>Good! What other methods could be used?</u>'''''<br />
<br />
===How the Distribution of Energy Between Different Modes Affect the Efficiency of the Reaction===<br />
<br />
Polanyi's rules state that the position of the transition state determines what mode of energy will be more efficient for the reaction.<ref>J. C. Polanyi, <i>Acc. Chem. Res.</i> 1972, <b>5</b>, 161-168.</ref> They state that for an early barrier transition state (closer to the reactants) translational energy is more efficient, and for a late barrier transition state vibrational energy is more efficient for a reactive trajectory.<br />
<br />
The exothermic reaction (F + H<sub>2</sub> -> HF + H) has an early barrier transition state, so translational energy, p<sub>HF</sub>, should be more efficient for a reactive trajectory than vibrational, p<sub>HH</sub>. <i>Table 1</i> that when p<sub>HF</sub> is constant at -1 and p<sub>HH</sub> is increased, the trajectory stayed unreactive. However, when p<sub>HH</sub> is constant at -1 and p<sub>HF</sub> is increased, the trajectory became reactive. This proves that translational energy is more efficient than vibrational for the exothermic reaction. <br />
{| class="wikitable" border="1"<br />
|+ Table 1: Data for exothermic reaction<br />
|-<br />
!p<sub>HH</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>HF</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! Reactive?<br />
|-<br />
| -1<br />
| -1<br />
| No<br />
|-<br />
| -2<br />
| -1<br />
| No<br />
|-<br />
| -3<br />
| -1<br />
| No<br />
|-<br />
| -1<br />
| -2<br />
| Yes<br />
|-<br />
| -1<br />
| -3<br />
| Yes<br />
|}<br />
<br />
The endothermic reaction (HF + H -> F + H<sub>2</sub>) has a late barrier transition state, so vibrational energy, p<sub>HF</sub>, should be more efficient for a reactive trajectory than translational, p<sub>HH</sub>. <i>Table 2</i> shows that when p<sub>HF</sub> was increased and p<sub>HH</sub> was decreased the trajectory stayed reactive. This proves that vibrational energy is more efficient than translational for the endothermic reaction. <br />
<br />
{| class="wikitable" border="1"<br />
|+ Table 2: Data for endothermic reaction<br />
|-<br />
!p<sub>HH</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! p<sub>HF</sub> / g.mol<sup>-1</sup>.pm.fm<sup>-1</sup> !! Reactive?<br />
|-<br />
| -18<br />
| 0<br />
| Yes<br />
|-<br />
| -10<br />
| 11<br />
| Yes<br />
|-<br />
| 0<br />
| 21<br />
| Yes<br />
|}<br />
'''''<u>A very thorough and systematic analysis, well done. However, we need to see the trajectories as well. </u>'''''<br />
<br />
==References==</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:me1218&diff=812564MRD:me12182020-06-01T16:11:47Z<p>Ng611: /* Exercise 1 - (H + H2 System) */</p>
<hr />
<div>== Molecular Reaction Dynamics Lab 2020 ==<br />
<br />
'''''<u>OVERALL:4/5 - A good report overall, although you were missing some details in one or two places. One or two minor points are highlighted. </u>'''''<br />
<br />
=== Exercise 1 - (H + H<sub>2</sub> System) ===<br />
The transition state on a potential energy surface plot is defined as the maxima of the minimum energy path where the gradient of the potential (derivative of potential over distances = āV('''r<sub>i</sub>''')/ā'''r<sub>i</sub>''') is equal to zero. For a symmetrical system like H + H<sub>2</sub>, the transition state lies where r<sub>AB</sub>=r<sub>BC</sub>. Therefore, one way the transition state can be found is by generating a plot of internuclear distance against time for the reaction. The intersection between the lines of AB and BC distance over time gives the transition state with intermediate and equal bond lengths. <br />
<br />
The transition state is a saddle point on the potential energy surface unlike any other local minima. It is a maximum of the minimum energy path of the reaction but the potential energy is at a minimum for the the dividing surface (plot orthogonal to the minimum energy path). This has been illustrated in figures 1, 2 and 3. Mathematically, using the second partial derivative test, the saddle point has a negative Hessian determinant.<ref>Khan Academy. Second Partial DerivativeĀ Test. https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/second-partial-derivative-test (accessed: May 15, 2020). </ref><br />
<br />
'''''<u>A sensible description of the TS, well done!</u>''''' <br />
[[File:Dividing surface me1218.png|thumb|Figure 1: Contour plot showing the dividing surface (initial distances r<sub>AB</sub> and r<sub>BC</sub> set to 230 pm and no initial momenta) running through the transition state. ]]<br />
[[File:Surface pot max me1218 2.png|thumb|Figure 2: 3D surface plot showing the maximum of the transition state over the minimum energy path (along the bottom edge of the plot). ]]<br />
[[File:Surface pot min me1218 2.png|thumb|Figure 3: 3D surface plot showing the potential energy minimum of the transition state along the dividing surface.]]<br />
''Table 1: Initial conditions tested in the software.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|230<br />
|<nowiki>-5.2</nowiki><br />
|-<br />
|BC<br />
|74<br />
|0<br />
|}<br />
By initially using the intial conditions in table 1, a plot of internuclear distances against time was generated (figure 4). The intersection between the AB plot and BC plot (at the transition state bond lengths) was found to be at an internuclear distance value of around 91 pm (can be seen from zoomed figure 4 in figure 5). <br />
<br />
'''''<u>A good estimate, although very slightly off from the true value. You can see some small residual vibrations in the plot. </u>''''' <br />
[[File:Internuclear Distances vs time me1218 1.png|thumb|400x400px|Figure 4: Plot of internuclear distance against time of H + H2 system when using conditions in table 1.|none]]<br />
<br />
[[File:Internuclear Distances vs time me1218 1 zoom.png|thumb|399x399px|Figure 5: Zoomed in plot of internuclear distance against time for H + H2 system using conditions in table 1. The intersection shows the transition state of the system. |none]]<br />
<br />
The software was then set up using the conditions in table 2 to confirm whether the transition state was at around r<sub>AB</sub>=r<sub>BC</sub>=91 pm.<br />
<br />
''Table 2: Conditions for transition state in the software.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|91<br />
|0<br />
|-<br />
|BC<br />
|91<br />
|0<br />
|}<br />
As shown from the contour plot generated from these conditions (figure 6), these conditions resulted in no overall trajectory towards products nor reactants. This is because there is no gradient directly at right angles to the transition state. There is only some oscillation of the system on the point as shown by the zoom of the contour plot of figure 6 in figure 7. Figure 8 is a plot of internuclear distance against time at this transition state. The oscillation in bond distances can be seen but there is no overall increase/ decrease in bond lengths between hydrogen's A, B and C. Looking at the initial geometry information in the control panel, the initial forces along AB and BC are around 0 kJ mol<sup>-1</sup> (-0.037 kJ mol<sup>-1</sup>) showing that this is around the transition state. <br />
<br />
The distance r at this transition state between A and B and between B and C can be called r<sub>ts.</sub><br />
[[File:Contour plot at 91 - transition state me1218.png|thumb|Figure 6: Contour plot showing the transition state of the system. |none]]<br />
[[File:Oscillate me1218.png|none|thumb|Figure 7: Zoomed contour plot of figure 6 showing the slight oscillations about the transition state of the system.]]<br />
[[File:Internuclear Distances vs time me1218 2 ts.png|none|thumb|Figure 8: Plot of internuclear distance against time of H + H2 system when using conditions in table 2 (at transition state).]]<br />
A minimum energy path (MEP) calculation was compared against a dynamics calculation (which, unlike an MEP, takes into account the vibrational motions of the system) using conditions where r<sub>AB</sub> = r<sub>ts</sub> and r<sub>BC =</sub><sub> </sub>(r<sub>ts</sub>+1 pm). In the MEP calculation, the system follows exactly the trough (minimum points) of the potential energy surface with a direct trajectory down to the products - see figure 9. <br />
<br />
In the dynamics calculation model, the system oscillates up and down the sides of the valley travelling towards the products - the trajectory is much less direct than in the MEP model - see figure 10.<br />
<br />
The MEP is calculated assuming infinitely slow movement of the system on the potential energy surface (where momentum = 0 at each step calculated). Therefore, the only thing that determines the trajectory is the potential energy given by the potential energy surface and not any vibrational motion. <br />
<br />
'''''<u>A sensible description, well done. </u>''''' <br />
{| class="wikitable"<br />
!MEP<br />
!Dynamics<br />
|-<br />
|[[File:Mep 1 me1218.png|thumb|Figure 9: Minimum energy path contour diagram of H + H<sub>2</sub> system proceeding to products from the transition state.]]<br />
|[[File:Mep dynamics 1 me1218.png|thumb|Figure 10: Dynamics model contour diagram of H + H<sub>2</sub> system proceeding to products from the transition state. ]]<br />
|}<br />
In figure 10, the vibrations (oscillations up and down the potential energy surface 'valleys') can be observed unlike in the MEP in figure 9. The MEP trajectory follows exactly the minimum points of the potential energy surface down from near the transition state to the products. <br />
<br />
''Table 3: Comparing dynamics of system given an initial position of r<sub>AB</sub>=200 pm and r<sub>BC</sub>=74 pm and the following values of p<sub>1</sub> and p<sub>2</sub>''.<br />
{| class="wikitable"<br />
!Eg.<br />
!p<sub>BC </sub>(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
!p<sub>AB</sub>Ā (g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
!E<sub>tot </sub>(kJ mol<sup>-1</sup>)<br />
!Reactive?<br />
!Description of the dynamics<br />
!Illustration of the trajectory<br />
|-<br />
|(1)<br />
|<nowiki>-2.56</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-414.280</nowiki><br />
|Yes<br />
|Small amount of intramolecular vibration in the products.<br />
<br />
Trajectory of the system goes over the transition state ('saddle point') and down towards the products. Reaction proceeds. <br />
|[[File:Contour plot table 1 me1218.png|thumb|Figure 11: Example (1) ]]<br />
|-<br />
|(2)<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-4.1</nowiki><br />
|<nowiki>-420.077</nowiki><br />
|No<br />
|More intramolecular vibration in reactants than in example (1).<br />
<br />
Trajectory of the system doesn't reach the transition state energy maximum and so the reaction doesn't proceed to products. Instead, the system returns to reactants. <br />
|[[File:Contour plot table 2 me1218.png|thumb|Figure 12: Example (2)]]<br />
|-<br />
|(3)<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-413.977</nowiki><br />
|Yes<br />
|Some vibrational motion in the reactants - results in more vibrational motion in the products. <br />
Trajectory of the system does surmount the transition state maximum and reaction proceeds towards the products with large amount of intramolecular vibrational energy. <br />
|[[File:Contour plot table 3 me1218.png|thumb|Figure 13: Example (3)]]<br />
|-<br />
|(4)<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.1</nowiki><br />
|<nowiki>-357.277</nowiki><br />
|No<br />
|System initially has a lot of energy - high p<sub>AB</sub> and p<sub>BC</sub> (momenta). Trajectory runs over transition state region heading towards the products but instead 'rolls' back over the transition state region back towards the reactants (reaction doesn't proceed). This is called 'barrier recrossing'.<br />
This is an example where assumptions made in 'Transition state theory' fall down. <br />
|[[File:Contour plot table 4 me1218.png|thumb|Figure 14: Example (4)]]<br />
|-<br />
|(5)<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.6</nowiki><br />
|<nowiki>-349.477</nowiki><br />
|Yes<br />
|Similarly, the system initially has a lot of energy (momenta) and trajectory runs over the transition state. The system runs back towards reactants for a bit but then surmounts the transition state once again and heads towards the products. Reaction does proceed. <br />
|[[File:Contour plot table 5 me1218.png|thumb|Figure 15: Example (5)]]<br />
|}<br />
From table 3, it can be said that it is not, in fact, the E<sub>tot</sub> nor the combined magnitude of the initial momenta that determines if the reaction will proceed or not but rather whether the energy supplied is in the correct 'mode' for reaction to proceed. For example (4), even with total energy (E<sub>tot</sub>) of -357.277 kJ mol<sup>-1</sup> (quite high relative to others in the table) and an initial momenta of AB roughly double that of example (3), the reaction does not proceed. Here barrier recrossing can be observed. This is when the system crosses the transition state region again - the bonds in the product form but the system doesn't go fully to product before returning back to the reactants (via the transition state again).<br />
<br />
In 'Transition State Theory', one assumption made is that when kinetic energy is greater than the activation energy (and the transition state is reached), the system always goes on to form products. Once the transition state has been reached, the system cannot go back to form reactants. There are, however, exceptions to this assumption as shown in example (4) in table 3. There is 'barrier recrossing' whereby the transition state is reached (and in this example surpassed - going on to form product bonds) but the system ultimately returns to the reactants instead of forming products. <br />
<br />
In terms of the rate of reaction, 'Transition State Theory' will overestimate the rate of reaction. This is because, hypothetically, if 1 out of 10 possible trajectories proceeds as in example (4) in table 3 and undergoes 'barrier recrossing', there would be 1 out of 10 trajectories which don't proceed to products as predicted by 'Transition State Theory'. This means the rate measured experimentally is always equal to or less than the rate which can be predicted using this theory.<ref>Steinfield, J. I.; Francisco, J. S.; Hase, W. L. Statistical approach to reaction dynamics transition state theory. In Chemical Kinetics and Dynamics. 2<sup>nd</sup> Ed.; Upper Saddle River: Prentice-Hall, 1989; pp. 287-321.</ref><br />
<br />
'''<u>''A good description of the limitations of TST. What other features does TST neglect though?''</u>''' <br />
<br />
=== Exercise 2 - (F + H + H System) ===<br />
{| class="wikitable"<br />
!'F + H<sub>2</sub> system'<br />
!'H + HF system'<br />
|-<br />
|[[File:F H2 system me1218 2.png|thumb|Figure 16: Diagram of the atoms involved in the F + H<sub>2</sub> system and the labelling which is used hereafter.]]<br />
|[[File:H HF system me1218.png|thumb|Figure 17: Diagram of the atoms involved in the H + HF system and the labelling which is used hereafter.]]<br />
|}<br />
<br />
The transition state point was found by looking for a 'saddle-point' in the surface plot of the system. The value for r<sub>BC</sub> was set in the centre of the minimum potential energy surface in the reactants. Then, with values of momenta set equal to 0, the value of r<sub>AB</sub> was varied between 135 pm and 200 pm. Values of r<sub>AB</sub> below the transition state value result in an overall trajectory towards the products whilst values about the transition state result in an overall trajectory back to the reactants. By varying r<sub>AB</sub> and observing the overall trajectory, the transition state was found. This location was confirmed by taking note of the forces values in the initial geometry information panel at this transition state. These values were around 0 kJ mol<sup>-1</sup> pm<sup>-1</sup> along both AB and BC. <br />
<br />
For the reaction of F with H<sub>2</sub>, the transition state is located at around the following: r<sub>AB</sub> = 181.15 pm, r<sub>BC</sub> = 74.5 pm. <br />
<br />
This transition state is closer to the reactants side of the potential energy surface and so the transition state can be said (by Hammond's postulate) to be 'reactant like'. Therefore, the transition state is early and the reaction is exothermic. <br />
<br />
By subracting the potential energy of the reactants (found at the minimum potential energy at the extreme of r<sub>AB</sub> bond length) from that of the transition state found, the activation energy of the F + H<sub>2</sub> reaction was found to be 696 J mol<sup>-1</sup> <br />
<br />
{| class="wikitable"<br />
! colspan="3" |System F + H<sub>2</sub>. Early ('reactant-like') transition state determined to be at r<sub>AB</sub> = 181.15 pm and r<sub>BC</sub> = 74.5 pm.<br />
|-<br />
|[[File:F H2 ts contour me1218.png|thumb|Figure 18: Contour diagram of the F + H<sub>2</sub> reaction - showing the early (reactant-like) transition state.]]<br />
|[[File:F H2 ts surface 1 me1218.png|thumb|Figure 19: Surface plot of the F + H<sub>2</sub> reaction showing the transition state at the 'saddle point'.]]<br />
|[[File:F H2 ts surface 2 me1218.png|thumb|Figure 20: Rotated surface plot of the F + H<sub>2</sub> reaction showing the transition state at the 'saddle point'.]]<br />
|} <br />
<br />
For the reaction of H with HF, the transition state is located at around the following: r<sub>AB</sub> = 74.5 pm, r<sub>BC</sub> = 181.15 pm. <br />
<br />
This transition state is closer to the products side of the potential energy surface and so the transition state can be said (by Hammond's postule) to be 'product like'. Therefore, the transition state is late and the reaction is endothermic.<br />
<br />
From the initial potential energies of the transition state and the reactants, the activation energy of the H + HF reaction was found to be 126.327 KJ mol<sup>-1 </sup>(126,327 J mol<sup>-1</sup>).<br />
<br />
'''''<u>Sensible estimates for the activation energies of both reactions. I'd say your F+H2 barrier is slightly too small, but given the size of the barrier, this is expected.</u>''''' <br />
{| class="wikitable"<br />
! colspan="3" |System H + HF. Late ('reactant-like') transition state determined to be at r<sub>AB</sub> = 74.5 pm and r<sub>BC</sub> = 181.15 pm.<br />
|-<br />
|[[File:H HF ts contour me1218.png|thumb|Figure 21: Contour diagram of the H + HF reaction - showing the late (product-like) transition state. ]]<br />
|[[File:H HF ts surface 1 me1218.png|thumb|Figure 22: Surface plot of the H + HF reaction showing the transition state at the 'saddle point'.]]<br />
|[[File:H HF ts surface 2 me1218.png|thumb|Figure 23: Rotated surface plot of the H + HF reaction showing the transition state at the 'saddle point'.]]<br />
|}<br />
<br />
The activation energy for the reaction of H + HF is much higher than that of the reaction of F and H<sub>2</sub>.<br />
<br />
Since the F + H<sub>2</sub> reaction is exothermic, there is more energy released to form the H-F bond than energy required to break the H-H bond. Also, as the H + HF reaction is endothermic, there is more energy required to break the HF bond than energy given back upon formation of the H-H bond. Therefore, it can be deduced that the H-F bond is stronger than the H-H bond. <br />
<br />
An reactive trajectory of the F + H<sub>2</sub> system was set up using the conditions in table 4.<br />
<br />
''Table 4: Conditions used for reactive F + H<sub>2</sub> trajectory.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|225<br />
|<nowiki>-2.0</nowiki><br />
|-<br />
|BC<br />
|74.5<br />
|<nowiki>-0.5</nowiki><br />
|}<br />
A contour diagram of this reactive trajectory is shown in figure 24.<br />
[[File:Reactive contour F H2 me1218.png|none|thumb|Figure 24: Contour plot showing a reactive trajectory of F + H<sub>2</sub> system with conditions in table 4. ]]<br />
[[File:Reactive momenta F H2 me1218.png|none|thumb|Figure 25: Plot of momentum against time for a reactive trajectory of the F + H<sub>2</sub> system with conditions in table 4.. ]]From these plots using the reactive trajectory conditions, the mechanism of release of energy can be illustrated. <br />
<br />
From the contour plot in figure 24, it can be seen that the reactants start off with little vibrational energy. After surmounting the transition state, the products have much more vibrational energy (shown by the oscillations). Also, in the momentum against time plot for this process (figure 25), the oscillations (vibrational energy) in A-B and B-C are small before around 75 s (in the reactants) whereas in the products, the blue line (for A-B momentum) is high - the products have more vibrational energy. Therefore, in this example, the mechanism of release of energy is mostly via vibrational energy. A reactive trajectory for the reverse process (H + HF reaction) would have mostly translational energy released. <br />
<br />
The mechanism for release of a particular reaction can be measured experimentally. This can be done by using calorimetry and IR spectroscopy (to measure vibrational activity) in tandem. <br />
<br />
Taking the F+ H<sub>2</sub> reaction as an example, calorimetry will allow for the total energy released during the reaction to be measured. In the reactants, vibrational modes of H<sub>2</sub> molecule (being symmetrical and hence non-polar) are not IR active whereas those of the HF molecule formed in the products (being asymmetric and polar) are IR active. Therefore, the wavenumbers of the peaks in the IR spectrum of the products can be used to resolve the vibrational energy component to the total energy released in the reaction. <br />
<br />
'''''<u>How though? What features would you expect to see? More detail in your explanation is needed...</u>''''' <br />
<br />
''Table 5: Plots, conditions and descriptions to illustrate Polanyi's rules using the F + H<sub>2</sub> and the H + HF systems.''<br />
{| class="wikitable"<br />
!<br />
! colspan="2" |F + H<sub>2</sub> system<br />
! colspan="2" |H + HF system<br />
|-<br />
|Contour Plot <br />
showing <br />
<br />
trajectory.<br />
|[[File:Reactive contour F H2 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 1 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 2 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 3 me1218.png|thumb]]<br />
|-<br />
|Distances<br />
[pm]<br />
|r<sub>AB</sub>=225, r<sub>BC</sub>=74.5<br />
|r<sub>AB</sub>=225, r<sub>BC</sub>=74.5<br />
| colspan="2" |r<sub>AB</sub>=248, r<sub>BC</sub>=92<br />
|-<br />
|Momenta<br />
[g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|p<sub>AB</sub>= -2.0, p<sub>BC</sub>= -0.5<br />
|p<sub>AB</sub>= -0.5, p<sub>BC</sub>= -10<br />
|p<sub>AB</sub>= -20, p<sub>BC</sub>= -0.1<br />
|p<sub>AB</sub>= -10, p<sub>BC</sub>= -2.0<br />
|-<br />
|Description<br />
|Reactants (F + H<sub>2</sub>) have a lot of translational energy but little vibrational energy (little oscillation up/ down the 'valleys' of the potential energy surface). <br />
<br />
After surmounting the early transition state, the reaction proceeds to products. <br />
<br />
In the products (HF + H), there is a lot of vibrational energy (increased oscillations up/ down the potential energy surface).<br />
|For the same reaction with much more vibrational energy in the reactants but less translational energy, the early transition state is reached but 'barrier recrossing' occurs and system returns to the reactants.<br />
|Reactants (H + HF) have a lot of vibrational energy (large oscillation up/ down the sides of the potential energy surface). <br />
<br />
The system reaches the late transition state and proceeds to products. <br />
<br />
In the products, some of the vibrational energy is converted into translational energy. <br />
<br />
The vibrational energy in the products is lower than in the reactants.<br />
|For the same reaction with much less vibrational energy in the reactants, the late transition state does is not reached and no reaction proceeds.<br />
|}<br />
<br />
'''''<u>Most of your examples are good, but I'd say that one or two dont illustrate what you hope they do. For example, your demonstration of vibrational energy overcoming the barrier for the H+HF reaction also has a huge amount of translational energy.</u>''''' <br />
<br />
Using these plots in table 5, Polanyi's empirical rules can be illustrated. <br />
<br />
Total energy of the system is conserved in the process of moving from reactants to products. The distribution of this energy in different modes (vibrational and translational), however, can change. <br />
<br />
In terms of Polanyi's rules, the table 5 shows that, depending on the position of the transition state, the distribution of supplied energy in different modes affects whether the reaction will proceed or not. It shows that, for an exothermic reaction (with an early transition state) it is more efficient (in terms of reactants proceeding to products) to start with little vibrational energy and high translational energy in the reactants. For an endothermic reaction (with a late transition state), it is more efficient to start with high vibrational energy in the reactants so that the 'longer transition state bond length can be achieved'. It also shows that for an exothermic reaction, high translational energy in the reactants is converted to high vibrational energy in the products. Again, for an endothermic reaction, the converse is true. <br />
<br />
This rule can be demonstrated in real life observations - an early transition state gives greater vibrational energy out in the products - this increase is observed as a temperature increase upon formation of products. Hence exothermic reactions 'give out heat'.<ref>Polanyi, J. C. Some Concepts in Reaction Dyanmics. ''Science. ''[Online] 1987 https://science.sciencemag.org/content/236/4802/680/tab-pdf (accessed: May 15, 2020).</ref><br />
<br />
=== References ===<br />
<references /></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:me1218&diff=812563MRD:me12182020-06-01T16:09:41Z<p>Ng611: /* Molecular Reaction Dynamics Lab 2020 */</p>
<hr />
<div>== Molecular Reaction Dynamics Lab 2020 ==<br />
<br />
=== Exercise 1 - (H + H<sub>2</sub> System) ===<br />
The transition state on a potential energy surface plot is defined as the maxima of the minimum energy path where the gradient of the potential (derivative of potential over distances = āV('''r<sub>i</sub>''')/ā'''r<sub>i</sub>''') is equal to zero. For a symmetrical system like H + H<sub>2</sub>, the transition state lies where r<sub>AB</sub>=r<sub>BC</sub>. Therefore, one way the transition state can be found is by generating a plot of internuclear distance against time for the reaction. The intersection between the lines of AB and BC distance over time gives the transition state with intermediate and equal bond lengths. <br />
<br />
The transition state is a saddle point on the potential energy surface unlike any other local minima. It is a maximum of the minimum energy path of the reaction but the potential energy is at a minimum for the the dividing surface (plot orthogonal to the minimum energy path). This has been illustrated in figures 1, 2 and 3. Mathematically, using the second partial derivative test, the saddle point has a negative Hessian determinant.<ref>Khan Academy. Second Partial DerivativeĀ Test. https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/second-partial-derivative-test (accessed: May 15, 2020). </ref><br />
<br />
'''''<u>A sensible description of the TS, well done!</u>''''' <br />
[[File:Dividing surface me1218.png|thumb|Figure 1: Contour plot showing the dividing surface (initial distances r<sub>AB</sub> and r<sub>BC</sub> set to 230 pm and no initial momenta) running through the transition state. ]]<br />
[[File:Surface pot max me1218 2.png|thumb|Figure 2: 3D surface plot showing the maximum of the transition state over the minimum energy path (along the bottom edge of the plot). ]]<br />
[[File:Surface pot min me1218 2.png|thumb|Figure 3: 3D surface plot showing the potential energy minimum of the transition state along the dividing surface.]]<br />
''Table 1: Initial conditions tested in the software.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|230<br />
|<nowiki>-5.2</nowiki><br />
|-<br />
|BC<br />
|74<br />
|0<br />
|}<br />
By initially using the intial conditions in table 1, a plot of internuclear distances against time was generated (figure 4). The intersection between the AB plot and BC plot (at the transition state bond lengths) was found to be at an internuclear distance value of around 91 pm (can be seen from zoomed figure 4 in figure 5). <br />
<br />
'''''<u>A good estimate, although very slightly off from the true value. You can see some small residual vibrations in the plot. </u>''''' <br />
[[File:Internuclear Distances vs time me1218 1.png|thumb|400x400px|Figure 4: Plot of internuclear distance against time of H + H2 system when using conditions in table 1.|none]]<br />
<br />
[[File:Internuclear Distances vs time me1218 1 zoom.png|thumb|399x399px|Figure 5: Zoomed in plot of internuclear distance against time for H + H2 system using conditions in table 1. The intersection shows the transition state of the system. |none]]<br />
<br />
The software was then set up using the conditions in table 2 to confirm whether the transition state was at around r<sub>AB</sub>=r<sub>BC</sub>=91 pm.<br />
<br />
''Table 2: Conditions for transition state in the software.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|91<br />
|0<br />
|-<br />
|BC<br />
|91<br />
|0<br />
|}<br />
As shown from the contour plot generated from these conditions (figure 6), these conditions resulted in no overall trajectory towards products nor reactants. This is because there is no gradient directly at right angles to the transition state. There is only some oscillation of the system on the point as shown by the zoom of the contour plot of figure 6 in figure 7. Figure 8 is a plot of internuclear distance against time at this transition state. The oscillation in bond distances can be seen but there is no overall increase/ decrease in bond lengths between hydrogen's A, B and C. Looking at the initial geometry information in the control panel, the initial forces along AB and BC are around 0 kJ mol<sup>-1</sup> (-0.037 kJ mol<sup>-1</sup>) showing that this is around the transition state. <br />
<br />
The distance r at this transition state between A and B and between B and C can be called r<sub>ts.</sub><br />
[[File:Contour plot at 91 - transition state me1218.png|thumb|Figure 6: Contour plot showing the transition state of the system. |none]]<br />
[[File:Oscillate me1218.png|none|thumb|Figure 7: Zoomed contour plot of figure 6 showing the slight oscillations about the transition state of the system.]]<br />
[[File:Internuclear Distances vs time me1218 2 ts.png|none|thumb|Figure 8: Plot of internuclear distance against time of H + H2 system when using conditions in table 2 (at transition state).]]<br />
A minimum energy path (MEP) calculation was compared against a dynamics calculation (which, unlike an MEP, takes into account the vibrational motions of the system) using conditions where r<sub>AB</sub> = r<sub>ts</sub> and r<sub>BC =</sub><sub> </sub>(r<sub>ts</sub>+1 pm). In the MEP calculation, the system follows exactly the trough (minimum points) of the potential energy surface with a direct trajectory down to the products - see figure 9. <br />
<br />
In the dynamics calculation model, the system oscillates up and down the sides of the valley travelling towards the products - the trajectory is much less direct than in the MEP model - see figure 10.<br />
<br />
The MEP is calculated assuming infinitely slow movement of the system on the potential energy surface (where momentum = 0 at each step calculated). Therefore, the only thing that determines the trajectory is the potential energy given by the potential energy surface and not any vibrational motion. <br />
<br />
'''''<u>A sensible description, well done. </u>''''' <br />
{| class="wikitable"<br />
!MEP<br />
!Dynamics<br />
|-<br />
|[[File:Mep 1 me1218.png|thumb|Figure 9: Minimum energy path contour diagram of H + H<sub>2</sub> system proceeding to products from the transition state.]]<br />
|[[File:Mep dynamics 1 me1218.png|thumb|Figure 10: Dynamics model contour diagram of H + H<sub>2</sub> system proceeding to products from the transition state. ]]<br />
|}<br />
In figure 10, the vibrations (oscillations up and down the potential energy surface 'valleys') can be observed unlike in the MEP in figure 9. The MEP trajectory follows exactly the minimum points of the potential energy surface down from near the transition state to the products. <br />
<br />
''Table 3: Comparing dynamics of system given an initial position of r<sub>AB</sub>=200 pm and r<sub>BC</sub>=74 pm and the following values of p<sub>1</sub> and p<sub>2</sub>''.<br />
{| class="wikitable"<br />
!Eg.<br />
!p<sub>BC </sub>(g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
!p<sub>AB</sub>Ā (g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>)<br />
!E<sub>tot </sub>(kJ mol<sup>-1</sup>)<br />
!Reactive?<br />
!Description of the dynamics<br />
!Illustration of the trajectory<br />
|-<br />
|(1)<br />
|<nowiki>-2.56</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-414.280</nowiki><br />
|Yes<br />
|Small amount of intramolecular vibration in the products.<br />
<br />
Trajectory of the system goes over the transition state ('saddle point') and down towards the products. Reaction proceeds. <br />
|[[File:Contour plot table 1 me1218.png|thumb|Figure 11: Example (1) ]]<br />
|-<br />
|(2)<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-4.1</nowiki><br />
|<nowiki>-420.077</nowiki><br />
|No<br />
|More intramolecular vibration in reactants than in example (1).<br />
<br />
Trajectory of the system doesn't reach the transition state energy maximum and so the reaction doesn't proceed to products. Instead, the system returns to reactants. <br />
|[[File:Contour plot table 2 me1218.png|thumb|Figure 12: Example (2)]]<br />
|-<br />
|(3)<br />
|<nowiki>-3.1</nowiki><br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-413.977</nowiki><br />
|Yes<br />
|Some vibrational motion in the reactants - results in more vibrational motion in the products. <br />
Trajectory of the system does surmount the transition state maximum and reaction proceeds towards the products with large amount of intramolecular vibrational energy. <br />
|[[File:Contour plot table 3 me1218.png|thumb|Figure 13: Example (3)]]<br />
|-<br />
|(4)<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.1</nowiki><br />
|<nowiki>-357.277</nowiki><br />
|No<br />
|System initially has a lot of energy - high p<sub>AB</sub> and p<sub>BC</sub> (momenta). Trajectory runs over transition state region heading towards the products but instead 'rolls' back over the transition state region back towards the reactants (reaction doesn't proceed). This is called 'barrier recrossing'.<br />
This is an example where assumptions made in 'Transition state theory' fall down. <br />
|[[File:Contour plot table 4 me1218.png|thumb|Figure 14: Example (4)]]<br />
|-<br />
|(5)<br />
|<nowiki>-5.1</nowiki><br />
|<nowiki>-10.6</nowiki><br />
|<nowiki>-349.477</nowiki><br />
|Yes<br />
|Similarly, the system initially has a lot of energy (momenta) and trajectory runs over the transition state. The system runs back towards reactants for a bit but then surmounts the transition state once again and heads towards the products. Reaction does proceed. <br />
|[[File:Contour plot table 5 me1218.png|thumb|Figure 15: Example (5)]]<br />
|}<br />
From table 3, it can be said that it is not, in fact, the E<sub>tot</sub> nor the combined magnitude of the initial momenta that determines if the reaction will proceed or not but rather whether the energy supplied is in the correct 'mode' for reaction to proceed. For example (4), even with total energy (E<sub>tot</sub>) of -357.277 kJ mol<sup>-1</sup> (quite high relative to others in the table) and an initial momenta of AB roughly double that of example (3), the reaction does not proceed. Here barrier recrossing can be observed. This is when the system crosses the transition state region again - the bonds in the product form but the system doesn't go fully to product before returning back to the reactants (via the transition state again).<br />
<br />
In 'Transition State Theory', one assumption made is that when kinetic energy is greater than the activation energy (and the transition state is reached), the system always goes on to form products. Once the transition state has been reached, the system cannot go back to form reactants. There are, however, exceptions to this assumption as shown in example (4) in table 3. There is 'barrier recrossing' whereby the transition state is reached (and in this example surpassed - going on to form product bonds) but the system ultimately returns to the reactants instead of forming products. <br />
<br />
In terms of the rate of reaction, 'Transition State Theory' will overestimate the rate of reaction. This is because, hypothetically, if 1 out of 10 possible trajectories proceeds as in example (4) in table 3 and undergoes 'barrier recrossing', there would be 1 out of 10 trajectories which don't proceed to products as predicted by 'Transition State Theory'. This means the rate measured experimentally is always equal to or less than the rate which can be predicted using this theory.<ref>Steinfield, J. I.; Francisco, J. S.; Hase, W. L. Statistical approach to reaction dynamics transition state theory. In Chemical Kinetics and Dynamics. 2<sup>nd</sup> Ed.; Upper Saddle River: Prentice-Hall, 1989; pp. 287-321.</ref><br />
<br />
'''<u>''A good description of the limitations of TST. What other features does TST neglect though?''</u>''' <br />
<br />
=== Exercise 2 - (F + H + H System) ===<br />
{| class="wikitable"<br />
!'F + H<sub>2</sub> system'<br />
!'H + HF system'<br />
|-<br />
|[[File:F H2 system me1218 2.png|thumb|Figure 16: Diagram of the atoms involved in the F + H<sub>2</sub> system and the labelling which is used hereafter.]]<br />
|[[File:H HF system me1218.png|thumb|Figure 17: Diagram of the atoms involved in the H + HF system and the labelling which is used hereafter.]]<br />
|}<br />
<br />
The transition state point was found by looking for a 'saddle-point' in the surface plot of the system. The value for r<sub>BC</sub> was set in the centre of the minimum potential energy surface in the reactants. Then, with values of momenta set equal to 0, the value of r<sub>AB</sub> was varied between 135 pm and 200 pm. Values of r<sub>AB</sub> below the transition state value result in an overall trajectory towards the products whilst values about the transition state result in an overall trajectory back to the reactants. By varying r<sub>AB</sub> and observing the overall trajectory, the transition state was found. This location was confirmed by taking note of the forces values in the initial geometry information panel at this transition state. These values were around 0 kJ mol<sup>-1</sup> pm<sup>-1</sup> along both AB and BC. <br />
<br />
For the reaction of F with H<sub>2</sub>, the transition state is located at around the following: r<sub>AB</sub> = 181.15 pm, r<sub>BC</sub> = 74.5 pm. <br />
<br />
This transition state is closer to the reactants side of the potential energy surface and so the transition state can be said (by Hammond's postulate) to be 'reactant like'. Therefore, the transition state is early and the reaction is exothermic. <br />
<br />
By subracting the potential energy of the reactants (found at the minimum potential energy at the extreme of r<sub>AB</sub> bond length) from that of the transition state found, the activation energy of the F + H<sub>2</sub> reaction was found to be 696 J mol<sup>-1</sup> <br />
<br />
{| class="wikitable"<br />
! colspan="3" |System F + H<sub>2</sub>. Early ('reactant-like') transition state determined to be at r<sub>AB</sub> = 181.15 pm and r<sub>BC</sub> = 74.5 pm.<br />
|-<br />
|[[File:F H2 ts contour me1218.png|thumb|Figure 18: Contour diagram of the F + H<sub>2</sub> reaction - showing the early (reactant-like) transition state.]]<br />
|[[File:F H2 ts surface 1 me1218.png|thumb|Figure 19: Surface plot of the F + H<sub>2</sub> reaction showing the transition state at the 'saddle point'.]]<br />
|[[File:F H2 ts surface 2 me1218.png|thumb|Figure 20: Rotated surface plot of the F + H<sub>2</sub> reaction showing the transition state at the 'saddle point'.]]<br />
|} <br />
<br />
For the reaction of H with HF, the transition state is located at around the following: r<sub>AB</sub> = 74.5 pm, r<sub>BC</sub> = 181.15 pm. <br />
<br />
This transition state is closer to the products side of the potential energy surface and so the transition state can be said (by Hammond's postule) to be 'product like'. Therefore, the transition state is late and the reaction is endothermic.<br />
<br />
From the initial potential energies of the transition state and the reactants, the activation energy of the H + HF reaction was found to be 126.327 KJ mol<sup>-1 </sup>(126,327 J mol<sup>-1</sup>).<br />
<br />
'''''<u>Sensible estimates for the activation energies of both reactions. I'd say your F+H2 barrier is slightly too small, but given the size of the barrier, this is expected.</u>''''' <br />
{| class="wikitable"<br />
! colspan="3" |System H + HF. Late ('reactant-like') transition state determined to be at r<sub>AB</sub> = 74.5 pm and r<sub>BC</sub> = 181.15 pm.<br />
|-<br />
|[[File:H HF ts contour me1218.png|thumb|Figure 21: Contour diagram of the H + HF reaction - showing the late (product-like) transition state. ]]<br />
|[[File:H HF ts surface 1 me1218.png|thumb|Figure 22: Surface plot of the H + HF reaction showing the transition state at the 'saddle point'.]]<br />
|[[File:H HF ts surface 2 me1218.png|thumb|Figure 23: Rotated surface plot of the H + HF reaction showing the transition state at the 'saddle point'.]]<br />
|}<br />
<br />
The activation energy for the reaction of H + HF is much higher than that of the reaction of F and H<sub>2</sub>.<br />
<br />
Since the F + H<sub>2</sub> reaction is exothermic, there is more energy released to form the H-F bond than energy required to break the H-H bond. Also, as the H + HF reaction is endothermic, there is more energy required to break the HF bond than energy given back upon formation of the H-H bond. Therefore, it can be deduced that the H-F bond is stronger than the H-H bond. <br />
<br />
An reactive trajectory of the F + H<sub>2</sub> system was set up using the conditions in table 4.<br />
<br />
''Table 4: Conditions used for reactive F + H<sub>2</sub> trajectory.''<br />
{| class="wikitable"<br />
!<br />
!Distance (r) [pm]<br />
!Momentum (p) [g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|-<br />
|AB<br />
|225<br />
|<nowiki>-2.0</nowiki><br />
|-<br />
|BC<br />
|74.5<br />
|<nowiki>-0.5</nowiki><br />
|}<br />
A contour diagram of this reactive trajectory is shown in figure 24.<br />
[[File:Reactive contour F H2 me1218.png|none|thumb|Figure 24: Contour plot showing a reactive trajectory of F + H<sub>2</sub> system with conditions in table 4. ]]<br />
[[File:Reactive momenta F H2 me1218.png|none|thumb|Figure 25: Plot of momentum against time for a reactive trajectory of the F + H<sub>2</sub> system with conditions in table 4.. ]]From these plots using the reactive trajectory conditions, the mechanism of release of energy can be illustrated. <br />
<br />
From the contour plot in figure 24, it can be seen that the reactants start off with little vibrational energy. After surmounting the transition state, the products have much more vibrational energy (shown by the oscillations). Also, in the momentum against time plot for this process (figure 25), the oscillations (vibrational energy) in A-B and B-C are small before around 75 s (in the reactants) whereas in the products, the blue line (for A-B momentum) is high - the products have more vibrational energy. Therefore, in this example, the mechanism of release of energy is mostly via vibrational energy. A reactive trajectory for the reverse process (H + HF reaction) would have mostly translational energy released. <br />
<br />
The mechanism for release of a particular reaction can be measured experimentally. This can be done by using calorimetry and IR spectroscopy (to measure vibrational activity) in tandem. <br />
<br />
Taking the F+ H<sub>2</sub> reaction as an example, calorimetry will allow for the total energy released during the reaction to be measured. In the reactants, vibrational modes of H<sub>2</sub> molecule (being symmetrical and hence non-polar) are not IR active whereas those of the HF molecule formed in the products (being asymmetric and polar) are IR active. Therefore, the wavenumbers of the peaks in the IR spectrum of the products can be used to resolve the vibrational energy component to the total energy released in the reaction. <br />
<br />
'''''<u>How though? What features would you expect to see? More detail in your explanation is needed...</u>''''' <br />
<br />
''Table 5: Plots, conditions and descriptions to illustrate Polanyi's rules using the F + H<sub>2</sub> and the H + HF systems.''<br />
{| class="wikitable"<br />
!<br />
! colspan="2" |F + H<sub>2</sub> system<br />
! colspan="2" |H + HF system<br />
|-<br />
|Contour Plot <br />
showing <br />
<br />
trajectory.<br />
|[[File:Reactive contour F H2 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 1 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 2 me1218.png|thumb]]<br />
|[[File:Polanyi illustrate 3 me1218.png|thumb]]<br />
|-<br />
|Distances<br />
[pm]<br />
|r<sub>AB</sub>=225, r<sub>BC</sub>=74.5<br />
|r<sub>AB</sub>=225, r<sub>BC</sub>=74.5<br />
| colspan="2" |r<sub>AB</sub>=248, r<sub>BC</sub>=92<br />
|-<br />
|Momenta<br />
[g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>]<br />
|p<sub>AB</sub>= -2.0, p<sub>BC</sub>= -0.5<br />
|p<sub>AB</sub>= -0.5, p<sub>BC</sub>= -10<br />
|p<sub>AB</sub>= -20, p<sub>BC</sub>= -0.1<br />
|p<sub>AB</sub>= -10, p<sub>BC</sub>= -2.0<br />
|-<br />
|Description<br />
|Reactants (F + H<sub>2</sub>) have a lot of translational energy but little vibrational energy (little oscillation up/ down the 'valleys' of the potential energy surface). <br />
<br />
After surmounting the early transition state, the reaction proceeds to products. <br />
<br />
In the products (HF + H), there is a lot of vibrational energy (increased oscillations up/ down the potential energy surface).<br />
|For the same reaction with much more vibrational energy in the reactants but less translational energy, the early transition state is reached but 'barrier recrossing' occurs and system returns to the reactants.<br />
|Reactants (H + HF) have a lot of vibrational energy (large oscillation up/ down the sides of the potential energy surface). <br />
<br />
The system reaches the late transition state and proceeds to products. <br />
<br />
In the products, some of the vibrational energy is converted into translational energy. <br />
<br />
The vibrational energy in the products is lower than in the reactants.<br />
|For the same reaction with much less vibrational energy in the reactants, the late transition state does is not reached and no reaction proceeds.<br />
|}<br />
<br />
'''''<u>Most of your examples are good, but I'd say that one or two dont illustrate what you hope they do. For example, your demonstration of vibrational energy overcoming the barrier for the H+HF reaction also has a huge amount of translational energy.</u>''''' <br />
<br />
Using these plots in table 5, Polanyi's empirical rules can be illustrated. <br />
<br />
Total energy of the system is conserved in the process of moving from reactants to products. The distribution of this energy in different modes (vibrational and translational), however, can change. <br />
<br />
In terms of Polanyi's rules, the table 5 shows that, depending on the position of the transition state, the distribution of supplied energy in different modes affects whether the reaction will proceed or not. It shows that, for an exothermic reaction (with an early transition state) it is more efficient (in terms of reactants proceeding to products) to start with little vibrational energy and high translational energy in the reactants. For an endothermic reaction (with a late transition state), it is more efficient to start with high vibrational energy in the reactants so that the 'longer transition state bond length can be achieved'. It also shows that for an exothermic reaction, high translational energy in the reactants is converted to high vibrational energy in the products. Again, for an endothermic reaction, the converse is true. <br />
<br />
This rule can be demonstrated in real life observations - an early transition state gives greater vibrational energy out in the products - this increase is observed as a temperature increase upon formation of products. Hence exothermic reactions 'give out heat'.<ref>Polanyi, J. C. Some Concepts in Reaction Dyanmics. ''Science. ''[Online] 1987 https://science.sciencemag.org/content/236/4802/680/tab-pdf (accessed: May 15, 2020).</ref><br />
<br />
=== References ===<br />
<references /></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:hz7718&diff=812560MRD:hz77182020-06-01T15:54:36Z<p>Ng611: /* Translational energy VS Vibrational Energy */</p>
<hr />
<div>==Exercise 1: H + H<sub>2</sub> system==<br />
<br />
Grade: 3/5 - '''''<u>Some good features to this report. Some of your numerical estimates were a little off, and you failed to complete the final excercise. Other individual comments are provided throughout. </u>'''''<br />
<br />
===Transition state===<br />
<br />
<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. <br />
<br />
For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means Ī“V/Ī“r<sub>1</sub> = 0 and Ī“V/Ī“r<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) - Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) < 0.<br />
<br />
A minimum stationary point means in both directions, it is the minimum point with zero gradients (Ī“<sup>2</sup>V/Ī“x<sup>2</sup> > 0 and (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) -Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.<br />
<br />
'''''<u>Where do these directions lie on your PES though?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot<br />
|-<br />
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]] || [[file:hz7718_py3.png|500px|thumb|right]]<br />
|}<br />
<br />
===locating the transition state===<br />
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.<br />
<br />
[[file:hz7718_py4.png|300px]] <br />
<br />
<br />
'''<u>''A good estimate for the TS, well done!''</u>'''<br />
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+Ī“, r<sub>1</sub>=r<sub>ts</sub>===<br />
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.<br />
<br />
'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!! Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0<br />
|-<br />
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reactive and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. || [[file:hz7718_py7.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. || [[file:hz7718_py8.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating. || [[file:hz7718_py9.png|300px|thumb|right]]<br />
|- <br />
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. || [[file:hz7718_py10.png|300px|thumb|right]]<br />
|-<br />
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each. || [[file:hz7718_py11.png|300px|thumb|right]]<br />
|}<br />
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.<br />
<br />
<br />
===Transition State Theory===<br />
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup><br />
<br />
'''''<u>A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!</u>'''''<br />
<br />
==Exercise 2: F-H-H system==<br />
===PES inspection===<br />
{| class="wikitable" border="1"<br />
! potential energy surface graph !! Internuclear distance vs time plot <br />
|-<br />
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]] <br />
|}<br />
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. <br />
<br />
So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.<br />
<br />
When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.<br />
<br />
In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. <br />
<br />
Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).<br />
<br />
'''''<u>A sensible estimation. How did you confirm that this was a saddle point, and not a potnetial energy minimum though?</u>'''''<br />
<br />
===Activation EnergyFor finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F.===<br />
{| class="wikitable" border="1"<br />
! Contour Plot 1 !! Graph 1<br />
|-<br />
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]] <br />
|}<br />
<br />
When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.<br />
<br />
'''''<u>A sensible estimate of the activation energy for FH+H. your estimate for F+H2 is a bit too low, likely because you've not let the reaction go to completion. </u>'''''<br />
{| class="wikitable" border="1"<br />
! Contour Plot 2 !! Graph 2<br />
|-<br />
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reaction dynamics===<br />
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.<br />
<br />
The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy. <br />
<br />
'''''<u>No, this is incorrect. Simply looking at the peak of an IR spectrum is not enough? How would we detect vibrationally excited states in IR spectroscopy?</u>''''' <br />
{| class="wikitable" border="1"<br />
! Contour Plot !! Momentum vs Time<br />
|-<br />
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]] <br />
|}<br />
<br />
===Translational energy VS Vibrational Energy===<br />
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> <br />
<br />
Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup><br />
<br />
'''''<u>YOu need to provide some examples (i.e. trajectories for the F+h2 system) here to back up your assertion.</u>'''''<br />
{| class="wikitable" border="1"<br />
! Graph 1 !! Graph 2<br />
|-<br />
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]] <br />
|}<br />
<br />
==Reference==<br />
1.Veser, Gƶtz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.<br />
<br />
2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334ā338. doi:10.1021/ja01607a027.<br />
<br />
3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:hz7718&diff=812559MRD:hz77182020-06-01T15:50:58Z<p>Ng611: /* Activation Energy */</p>
<hr />
<div>==Exercise 1: H + H<sub>2</sub> system==<br />
===Transition state===<br />
<br />
<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. <br />
<br />
For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means Ī“V/Ī“r<sub>1</sub> = 0 and Ī“V/Ī“r<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) - Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) < 0.<br />
<br />
A minimum stationary point means in both directions, it is the minimum point with zero gradients (Ī“<sup>2</sup>V/Ī“x<sup>2</sup> > 0 and (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) -Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.<br />
<br />
'''''<u>Where do these directions lie on your PES though?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot<br />
|-<br />
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]] || [[file:hz7718_py3.png|500px|thumb|right]]<br />
|}<br />
<br />
===locating the transition state===<br />
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.<br />
<br />
[[file:hz7718_py4.png|300px]] <br />
<br />
<br />
'''<u>''A good estimate for the TS, well done!''</u>'''<br />
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+Ī“, r<sub>1</sub>=r<sub>ts</sub>===<br />
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.<br />
<br />
'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!! Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0<br />
|-<br />
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reactive and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. || [[file:hz7718_py7.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. || [[file:hz7718_py8.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating. || [[file:hz7718_py9.png|300px|thumb|right]]<br />
|- <br />
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. || [[file:hz7718_py10.png|300px|thumb|right]]<br />
|-<br />
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each. || [[file:hz7718_py11.png|300px|thumb|right]]<br />
|}<br />
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.<br />
<br />
<br />
===Transition State Theory===<br />
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup><br />
<br />
'''''<u>A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!</u>'''''<br />
<br />
==Exercise 2: F-H-H system==<br />
===PES inspection===<br />
{| class="wikitable" border="1"<br />
! potential energy surface graph !! Internuclear distance vs time plot <br />
|-<br />
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]] <br />
|}<br />
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. <br />
<br />
So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.<br />
<br />
When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.<br />
<br />
In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. <br />
<br />
Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).<br />
<br />
'''''<u>A sensible estimation. How did you confirm that this was a saddle point, and not a potnetial energy minimum though?</u>'''''<br />
<br />
===Activation EnergyFor finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F.===<br />
{| class="wikitable" border="1"<br />
! Contour Plot 1 !! Graph 1<br />
|-<br />
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]] <br />
|}<br />
<br />
When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.<br />
<br />
'''''<u>A sensible estimate of the activation energy for FH+H. your estimate for F+H2 is a bit too low, likely because you've not let the reaction go to completion. </u>'''''<br />
{| class="wikitable" border="1"<br />
! Contour Plot 2 !! Graph 2<br />
|-<br />
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reaction dynamics===<br />
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.<br />
<br />
The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy. <br />
<br />
'''''<u>No, this is incorrect. Simply looking at the peak of an IR spectrum is not enough? How would we detect vibrationally excited states in IR spectroscopy?</u>''''' <br />
{| class="wikitable" border="1"<br />
! Contour Plot !! Momentum vs Time<br />
|-<br />
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]] <br />
|}<br />
<br />
===Translational energy VS Vibrational Energy===<br />
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> <br />
<br />
Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup><br />
{| class="wikitable" border="1"<br />
! Graph 1 !! Graph 2<br />
|-<br />
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]] <br />
|}<br />
<br />
==Reference==<br />
1.Veser, Gƶtz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.<br />
<br />
2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334ā338. doi:10.1021/ja01607a027.<br />
<br />
3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:hz7718&diff=812558MRD:hz77182020-06-01T15:46:05Z<p>Ng611: /* PES inspection */</p>
<hr />
<div>==Exercise 1: H + H<sub>2</sub> system==<br />
===Transition state===<br />
<br />
<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. <br />
<br />
For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means Ī“V/Ī“r<sub>1</sub> = 0 and Ī“V/Ī“r<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) - Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) < 0.<br />
<br />
A minimum stationary point means in both directions, it is the minimum point with zero gradients (Ī“<sup>2</sup>V/Ī“x<sup>2</sup> > 0 and (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) -Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.<br />
<br />
'''''<u>Where do these directions lie on your PES though?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot<br />
|-<br />
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]] || [[file:hz7718_py3.png|500px|thumb|right]]<br />
|}<br />
<br />
===locating the transition state===<br />
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.<br />
<br />
[[file:hz7718_py4.png|300px]] <br />
<br />
<br />
'''<u>''A good estimate for the TS, well done!''</u>'''<br />
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+Ī“, r<sub>1</sub>=r<sub>ts</sub>===<br />
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.<br />
<br />
'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!! Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0<br />
|-<br />
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reactive and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. || [[file:hz7718_py7.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. || [[file:hz7718_py8.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating. || [[file:hz7718_py9.png|300px|thumb|right]]<br />
|- <br />
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. || [[file:hz7718_py10.png|300px|thumb|right]]<br />
|-<br />
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each. || [[file:hz7718_py11.png|300px|thumb|right]]<br />
|}<br />
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.<br />
<br />
<br />
===Transition State Theory===<br />
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup><br />
<br />
'''''<u>A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!</u>'''''<br />
<br />
==Exercise 2: F-H-H system==<br />
===PES inspection===<br />
{| class="wikitable" border="1"<br />
! potential energy surface graph !! Internuclear distance vs time plot <br />
|-<br />
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]] <br />
|}<br />
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. <br />
<br />
So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.<br />
<br />
When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.<br />
<br />
In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. <br />
<br />
Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).<br />
<br />
'''''<u>A sensible estimation. How did you confirm that this was a saddle point, and not a potnetial energy minimum though?</u>'''''<br />
<br />
===Activation Energy===<br />
For finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F. <br />
{| class="wikitable" border="1"<br />
! Contour Plot 1 !! Graph 1<br />
|-<br />
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]] <br />
|}<br />
<br />
When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.<br />
<br />
Y<br />
{| class="wikitable" border="1"<br />
! Contour Plot 2 !! Graph 2<br />
|-<br />
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reaction dynamics===<br />
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.<br />
<br />
The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy. <br />
{| class="wikitable" border="1"<br />
! Contour Plot !! Momentum vs Time<br />
|-<br />
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]] <br />
|}<br />
<br />
===Translational energy VS Vibrational Energy===<br />
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> <br />
<br />
Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup><br />
{| class="wikitable" border="1"<br />
! Graph 1 !! Graph 2<br />
|-<br />
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]] <br />
|}<br />
<br />
==Reference==<br />
1.Veser, Gƶtz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.<br />
<br />
2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334ā338. doi:10.1021/ja01607a027.<br />
<br />
3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:hz7718&diff=812557MRD:hz77182020-06-01T15:41:22Z<p>Ng611: /* Transition State Theory */</p>
<hr />
<div>==Exercise 1: H + H<sub>2</sub> system==<br />
===Transition state===<br />
<br />
<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. <br />
<br />
For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means Ī“V/Ī“r<sub>1</sub> = 0 and Ī“V/Ī“r<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) - Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) < 0.<br />
<br />
A minimum stationary point means in both directions, it is the minimum point with zero gradients (Ī“<sup>2</sup>V/Ī“x<sup>2</sup> > 0 and (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) -Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.<br />
<br />
'''''<u>Where do these directions lie on your PES though?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot<br />
|-<br />
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]] || [[file:hz7718_py3.png|500px|thumb|right]]<br />
|}<br />
<br />
===locating the transition state===<br />
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.<br />
<br />
[[file:hz7718_py4.png|300px]] <br />
<br />
<br />
'''<u>''A good estimate for the TS, well done!''</u>'''<br />
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+Ī“, r<sub>1</sub>=r<sub>ts</sub>===<br />
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.<br />
<br />
'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!! Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0<br />
|-<br />
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reactive and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. || [[file:hz7718_py7.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. || [[file:hz7718_py8.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating. || [[file:hz7718_py9.png|300px|thumb|right]]<br />
|- <br />
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. || [[file:hz7718_py10.png|300px|thumb|right]]<br />
|-<br />
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each. || [[file:hz7718_py11.png|300px|thumb|right]]<br />
|}<br />
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.<br />
<br />
<br />
===Transition State Theory===<br />
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup><br />
<br />
'''''<u>A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!</u>'''''<br />
<br />
==Exercise 2: F-H-H system==<br />
===PES inspection===<br />
{| class="wikitable" border="1"<br />
! potential energy surface graph !! Internuclear distance vs time plot <br />
|-<br />
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]] <br />
|}<br />
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. <br />
<br />
So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.<br />
<br />
When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.<br />
<br />
In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. <br />
<br />
Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).<br />
<br />
===Activation Energy===<br />
For finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F. <br />
{| class="wikitable" border="1"<br />
! Contour Plot 1 !! Graph 1<br />
|-<br />
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]] <br />
|}<br />
<br />
When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.<br />
{| class="wikitable" border="1"<br />
! Contour Plot 2 !! Graph 2<br />
|-<br />
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reaction dynamics===<br />
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.<br />
<br />
The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy. <br />
{| class="wikitable" border="1"<br />
! Contour Plot !! Momentum vs Time<br />
|-<br />
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]] <br />
|}<br />
<br />
===Translational energy VS Vibrational Energy===<br />
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> <br />
<br />
Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup><br />
{| class="wikitable" border="1"<br />
! Graph 1 !! Graph 2<br />
|-<br />
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]] <br />
|}<br />
<br />
==Reference==<br />
1.Veser, Gƶtz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.<br />
<br />
2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334ā338. doi:10.1021/ja01607a027.<br />
<br />
3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:hz7718&diff=812554MRD:hz77182020-06-01T15:38:10Z<p>Ng611: /* Transition state */</p>
<hr />
<div>==Exercise 1: H + H<sub>2</sub> system==<br />
===Transition state===<br />
<br />
<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. <br />
<br />
For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means Ī“V/Ī“r<sub>1</sub> = 0 and Ī“V/Ī“r<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) - Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) < 0.<br />
<br />
A minimum stationary point means in both directions, it is the minimum point with zero gradients (Ī“<sup>2</sup>V/Ī“x<sup>2</sup> > 0 and (Ī“<sup>2</sup>V/Ī“r<sub>1</sub><sup>2</sup>)*(Ī“<sup>2</sup>V/Ī“r<sub>2</sub><sup>2</sup>) -Ī“<sup>2</sup>V/(Ī“r<sub>1</sub>*Ī“r<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.<br />
<br />
'''''<u>Where do these directions lie on your PES though?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot<br />
|-<br />
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]] || [[file:hz7718_py3.png|500px|thumb|right]]<br />
|}<br />
<br />
===locating the transition state===<br />
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.<br />
<br />
[[file:hz7718_py4.png|300px]] <br />
<br />
<br />
'''<u>''A good estimate for the TS, well done!''</u>'''<br />
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+Ī“, r<sub>1</sub>=r<sub>ts</sub>===<br />
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.<br />
<br />
'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''<br />
<br />
{| class="wikitable" border="1"<br />
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!! Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0<br />
|-<br />
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reactive and unreactive trajectories===<br />
{| class="wikitable" border="1"<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. || [[file:hz7718_py7.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. || [[file:hz7718_py8.png|300px|thumb|right]]<br />
|-<br />
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating. || [[file:hz7718_py9.png|300px|thumb|right]]<br />
|- <br />
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. || [[file:hz7718_py10.png|300px|thumb|right]]<br />
|-<br />
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each. || [[file:hz7718_py11.png|300px|thumb|right]]<br />
|}<br />
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.<br />
<br />
<br />
===Transition State Theory===<br />
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup><br />
<br />
==Exercise 2: F-H-H system==<br />
===PES inspection===<br />
{| class="wikitable" border="1"<br />
! potential energy surface graph !! Internuclear distance vs time plot <br />
|-<br />
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]] <br />
|}<br />
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. <br />
<br />
So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.<br />
<br />
When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.<br />
<br />
In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. <br />
<br />
Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).<br />
<br />
===Activation Energy===<br />
For finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F. <br />
{| class="wikitable" border="1"<br />
! Contour Plot 1 !! Graph 1<br />
|-<br />
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]] <br />
|}<br />
<br />
When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.<br />
{| class="wikitable" border="1"<br />
! Contour Plot 2 !! Graph 2<br />
|-<br />
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]] <br />
|}<br />
<br />
===Reaction dynamics===<br />
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.<br />
<br />
The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy. <br />
{| class="wikitable" border="1"<br />
! Contour Plot !! Momentum vs Time<br />
|-<br />
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]] <br />
|}<br />
<br />
===Translational energy VS Vibrational Energy===<br />
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> <br />
<br />
Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup><br />
{| class="wikitable" border="1"<br />
! Graph 1 !! Graph 2<br />
|-<br />
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]] <br />
|}<br />
<br />
==Reference==<br />
1.Veser, Gƶtz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.<br />
<br />
2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334ā338. doi:10.1021/ja01607a027.<br />
<br />
3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810858MRD:yh130182020-05-22T18:03:15Z<p>Ng611: /* Explanations and possible answers to the questions */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 19:03, 22 May 2020 (BST) Overall: 4/5 - Generally correct, but with one or two bits missing. I'd maybe also consider making your answers more succinct.<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
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Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
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===EXERCISE 2: F - H - H system===<br />
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====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
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The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
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====<u>question 7: Locate the approximate position of the transition state</u>====<br />
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[[File:DFH.png]]<br />
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The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:56, 22 May 2020 (BST) A good estimate and a sensible piece of evidence provided, well done! How did you confirm this was a TS and not an energetic minimum?<br />
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====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
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The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:57, 22 May 2020 (BST) Excellent responses and correct estimates, well done!<br />
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====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
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The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:59, 22 May 2020 (BST) Calorimetry is insensitive to whether energy is released as translational or vibrational energy though. Can you think of a method that is sensitive to this difference?<br />
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====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
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Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 19:01, 22 May 2020 (BST) Again, correct answer but somewhat unclear in your explanation. You also needed to provide some examples to back up your assertions here as well.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810852MRD:yh130182020-05-22T18:01:31Z<p>Ng611: /* question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. */</p>
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<div>== Explanations and possible answers to the questions ==<br />
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=== EXERCISE 1: H + H2 system ===<br />
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====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
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The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
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Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
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All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
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====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
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The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
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====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
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The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
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====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
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The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
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{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
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The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:56, 22 May 2020 (BST) A good estimate and a sensible piece of evidence provided, well done! How did you confirm this was a TS and not an energetic minimum?<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
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<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:57, 22 May 2020 (BST) Excellent responses and correct estimates, well done!<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
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The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:59, 22 May 2020 (BST) Calorimetry is insensitive to whether energy is released as translational or vibrational energy though. Can you think of a method that is sensitive to this difference?<br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 19:01, 22 May 2020 (BST) Again, correct answer but somewhat unclear in your explanation. You also needed to provide some examples to back up your assertions here as well.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810845MRD:yh130182020-05-22T17:59:06Z<p>Ng611: /* question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */</p>
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<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
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====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
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The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
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Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
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All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
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====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
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The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
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====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
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The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
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====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
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The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
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{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
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Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
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====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
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However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
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===EXERCISE 2: F - H - H system===<br />
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====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
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The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
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====<u>question 7: Locate the approximate position of the transition state</u>====<br />
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[[File:DFH.png]]<br />
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The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:56, 22 May 2020 (BST) A good estimate and a sensible piece of evidence provided, well done! How did you confirm this was a TS and not an energetic minimum?<br />
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====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
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The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:57, 22 May 2020 (BST) Excellent responses and correct estimates, well done!<br />
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====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
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The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:59, 22 May 2020 (BST) Calorimetry is insensitive to whether energy is released as translational or vibrational energy though. Can you think of a method that is sensitive to this difference?<br />
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====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
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Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810838MRD:yh130182020-05-22T17:57:46Z<p>Ng611: /* question 8: Report the activation energy for both reactions. */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
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The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
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Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
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All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
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<br />
<br />
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The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
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====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
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The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
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====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
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The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
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{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
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Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:56, 22 May 2020 (BST) A good estimate and a sensible piece of evidence provided, well done! How did you confirm this was a TS and not an energetic minimum?<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:57, 22 May 2020 (BST) Excellent responses and correct estimates, well done!<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810835MRD:yh130182020-05-22T17:56:58Z<p>Ng611: /* question 7: Locate the approximate position of the transition state */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:56, 22 May 2020 (BST) A good estimate and a sensible piece of evidence provided, well done! How did you confirm this was a TS and not an energetic minimum?<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810832MRD:yh130182020-05-22T17:55:56Z<p>Ng611: /* Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Excellent!<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810829MRD:yh130182020-05-22T17:55:20Z<p>Ng611: /* Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:55, 22 May 2020 (BST) Both tunelling and TS recrossing both represent deficiencies of TS theory, however, TS recrossin is far more significant than tunelling.<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF. <br />
<br />
<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810827MRD:yh130182020-05-22T17:54:06Z<p>Ng611: /* Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the tab...</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:54, 22 May 2020 (BST) These all look sensible, but a contour or surface plot would have been better as it would enable you to observe the position of the atoms relative to the TS you found?<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF. <br />
<br />
<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810823MRD:yh130182020-05-22T17:52:25Z<p>Ng611: /* Question 3: Comment on how the mep and the trajectory you just calculated differ. */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:52, 22 May 2020 (BST) Good!<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF. <br />
<br />
<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810820MRD:yh130182020-05-22T17:51:46Z<p>Ng611: /* Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory. */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:51, 22 May 2020 (BST) Excellent reasoning, and a sensible answer for the TS position. Well done!<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
<br />
<br />
<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF. <br />
<br />
<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:yh13018&diff=810815MRD:yh130182020-05-22T17:50:52Z<p>Ng611: /* Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? */</p>
<hr />
<div>== Explanations and possible answers to the questions ==<br />
<br />
<br />
=== EXERCISE 1: H + H2 system ===<br />
<br />
====<u>Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?</u>====<br />
[[File:Surface_PlotQ1.png]] [[File:Q1 2.png]]<br />
<br />
<br />
The potential energy surface (PES) above shows that when the reactant molecules are closer, the energy of the system increases, and the structure of maximum energy is the transition state. The energy then decreases until the product is formed. The transition state is the first-order saddle point on PES. It shows that the energy is maximum in one direction in this configuration space but in other directions, the energy is minimum. In this case, the gradient of the function is set to zero and the stationary point of the function might be regarded as the transition state (āV(ri)/āri=0).<br />
<br />
Firstly, saddle points generally appear in a multivariable function.In one direction, if the slope of the function is zero, (āV(ri)/ārAB)rBC=0 and (āV(ri)/ārBC)rAB=0. At the minima, both (ā^2V(ri)/ārAB^2)rBC=>0 and (ā^2V(ri)/ārBC^2)rAB=>0 due to the positive curvature. However, at the saddle point, (ā^2V(ri)/ārAB^2)rBC=>0 but (āV^2(ri)/ārBC^2)rAB=<0 because one of the curvature is negative. Therefore, the function is differentiated first to obtain the minima and then it undergoes partial differentiation to check the curvature in different directions. If one is positive and one is negative, the saddle point is found and it is the transition state. Or if f(r1r1)-f^2(r2r2)<0, there will be a saddle point; if it is bigger than zero and f(r1r1)<0, there will be a local minimum, but if f(r1r1)>0, there will be local maximum.<br />
<br />
All in all, minima has only positive curvature and it shows a positive partial differentiation while saddle point has both positive and negative curvature. The energy goes down along the minimum energy pathway, and if a trajectory stays on transition state with zero initial momentum, it will never fall off.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:50, 22 May 2020 (BST) Your answer, whilst absolutely correct, is a bit muddled. Think about how you could distill your answer further into something a bit more concise. Otherwise, this is a good response, well done!<br />
<br />
====<u>Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.</u>====<br />
[[File:T&D+10.png]]<br />
[[File:T&D.png]]<br />
[[File:T&D-10.png]]<br />
<br />
<br />
<br />
<br />
The estimate of the transition state position is about 90.8pm. The triatomic molecule is the transition state because it shows the highest energy, and in this reaction, it should be H-H-H. The trajectory which stays on the ridge is always keeping oscillating, but if it moves slightly, it will fall off. As can be seen from the three plots, when r=90.8pm, the three atoms do not move while when r=100.8pm or r=80.8pm, the largest distance that the three atoms move and the time they use is almost the same. If the distance increases slightly, the atoms will be closer first, while the atoms move further first when the distance decreases. It shows that the triatomic molecule is formed at about 90.8pm and the trajectory formed at transition state will not fall off because the gradient there is zero and it is a stationary point.<br />
<br />
<br />
====<u>Question 3: Comment on how the mep and the trajectory you just calculated differ.</u>====<br />
[[File:thermodynamic.png]]<br />
[[File:mep22.png]]<br />
<br />
The two graphs show a dynamic trajectory and a minimum energy pathway (MEP) with the same condition which are the positions r1 = 91.8pm, r2 = 90.8pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. Any point on MEP is at the energy minimum in all directions perpendicular to the path. An infinitely slow motion can be achieved. From the graph above, MEP one shows that atom B and atom C become slightly closer, while atom A and atom B separate from each other and reach a relatively constant separation and zero velocity. Dynamic one shows that atom A and atom B stay reasonably stable first and then continue moving further away from each other. In this case, as we reach the zero velocity in MEP, the kinetic energy and momentum always keep zero according to KE=1/2mv^2 and p=mv. Moreover, as the distance between atom B and atom C increases compared with the transition state, the reaction pathway is prone to the product, which is BC. The dynamic graph also shows that the atoms continue vibrating because the distance between B and C is changing. However, if we use r2=91.8, the pathway will be prone to the reactant, which is AB. Other properties do not change too much. If the momenta have a different sign, the velocity will be in the opposite direction. The atoms move in different directions can determine which product is formed.<br />
<br />
<br />
<br />
<br />
<br />
====<u>Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?</u>====<br />
<br />
The momentum graphs show the relative momentum between atoms and the gradient is the force that the atoms experience.<br />
<br />
{| class="wikitable" border="1" <br />
<br />
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.28 || Reactive ||[[File:M1.png]] From the graph, we can see that atom A experiences large force because it is close to B, which allows attraction present. However, the force continues decreasing because atom moves further away from both B and C, and finally the force becomes zero and the momentum is constant. The change in sign means that atom A moves to the opposite direction. The atom C experiences low force at first as it is further away from AB, but the interaction goes up later. When BC is formed, the momentum is always changing because the two atoms are vibrating. They undergo attraction if they move slightly further away and they experience repulsion if they are too close. The change in sign also shows that atom C moves to opposite directions. The velocity can also be analysed. When BC is formed at about 30fs, atom A moves faster than both atom B and atom C which both have similar velocity. The positive sign of the relative velocity means that atom A and atom B move in different directions, while atom B and atom C are oscillating so the direction of movement keeps changing.<br />
|| [[File:D11.png]]<br />
Initially, A and B are closer together while B and C are further. A and B move to C, which allows BC to form and A to continue moving away. All atoms are vibrating.||<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Uneactive || [[File:M2.png]] The reaction do not happen thus no BC is formed. The atom C moves closer to AB first but the KE and the attraction are not high enough to allow the interaction between B and C to form. In this case, atom C moves further away from AB after about 20fs. The force between atom C and AB continues going down until disappearing, thus the momentum is constant after about 40fs. The force between atom A and atom B is always changing from repulsion to attraction and changing back so there is an oscillation. The atom C has the largest velocity when they move further away, and the positive relative velocity shows that the distance between atom C and AB increases, vice versa.||[[File:D2.png]] It is unreactive so no product is formed. The separation between A and B is reasonably constant but C continue moving further away from both A and B.||<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || [[File:M33.png]] The situation is similar to the first one. The difference is the initial momentum used, which causes the atom A and atom B to oscillate slightly before they approach atom C. Because the higher KE contributes to the higher energy of the atoms. In this case, the velocity varies as well. || [[File:d3.png]] A and B are closer initially and they move together closer to C. The product BC is formed, which shows that A keeps moving away. It shows a smooth motion. ||<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Reactive || [[File:M4.png]] The much higher momentum means that more KE is provided for atoms to oscillate. In this case, when three atoms are close together, the oscillation happens between all of them. The force between atom A and atom B goes down initially thus the force between atom B and atom C is higher. However, the KE does not allow the stable force to exist between atom B and atom C. The atom C moves away from AB. The force between atom A and atom B varies significantly and the distance between them changes a lot during oscillation. Because the KE is relatively large, which causes distablisation of the atoms. Finally, the reactant is reverted back. || [[File:D4.png]] The final product is AB but before about 40fs, the distance between B and C is smaller and BC is formed. However, B and C are vibrating relatively vigorously and finally A and B approach to each other||<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reactive || [[File:M5.png]] It has the similar situation as the one above. The system crosses the transition state region, thus once the product BC is formed, the reactant AB is prone to form again because the force between atom A and atom B increases, and the energy is not stable. The product BC is formed in this case. The oscillation is relatively large and the atom which is far away experience zero force. || [[File:D5.png]] It has a similar situation with the one above but finally BC is formed. B vibrates between A and C so it approaches C first and after one oscillation, B and C move away from A. <br />
|}<br />
<br />
Conclusion:<br />
1. When two atoms are close together, they are oscillating. If they are too close, there will be a repulsion; if they are slightly further away, there will be an attraction. The higher the KE, the stronger the oscillation.<br />
2. A reaction can not happen without enough energy. However, if the KE is out of a specific range, the products and reactants will be unstable, which may contribute to the barrier recrossing.<br />
3. The atom which is far away from the other two atoms experience zero forces, which leads to constant momentum.<br />
4. The lower value of kinetic energy is still able to overcome the potential energy.<br />
<br />
====<u>Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?</u>====<br />
Compared with experimental result, transition state theory might overestimate rate values. Transition state theory assumes that the transition state structure does not go back to the reactants once the transition state is formed and an equilibrium is present between reactant and transition state. In this case, it ignores the effect from barrier recrossing.<br />
<br />
However, the theory is treated classically, and thus the tunneling effect, which is the quantum mechanical phenomenon, is ignored. In this case, particle with lower kinetic energy is able to overcome the higher potential energy in experimental values. Transition state theory may underestimate the rate because it requires more energy and higher barrier.<br />
<br />
<br />
<br />
===EXERCISE 2: F - H - H system===<br />
<br />
<br />
<br />
====<u>Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?</u>====<br />
[[File:F+H2 reaction.png]]<br />
<br />
<br />
<br />
The initial conditions of the PES are rFH=150 pm and rHH=80pm at minimum energy pathway. It clearly shows that when the distance between H and F decreases and the distance between H and H increases, the energy is lower, which is more negative. In order to form HF in the F+H2 reaction, the energy is released, which is exothermic. If the distance between F and H decreases and the distance between H atoms goes up, relatively higher energy is needed to absorb, thus HF + H reaction is endothermic. An exothermic means that the energy of bond-forming is higher than that of bond breaking, in this case, the energy needed to form HF is higher than the energy to break H2. For another reaction which is endothermic, the energy of the formation of H2 is lower than the energy of breaking HF. In conclusion, the bond strength of H2 is higher than that of HF. <br />
<br />
<br />
<br />
====<u>question 7: Locate the approximate position of the transition state</u>====<br />
<br />
[[File:DFH.png]]<br />
<br />
The transition state estimated is at about rFH=74.5pm and rHH=181.1pm. Hammond postulate implies that the transition state is closer to the reactants than to the products in energy In an exothermic reaction, and vice versa. Therefore the transition state is prone to the H2 and F, thus the distance between H atoms is relatively small. From the graph, we can see that the three atomic molecule is formed and no product or reactant is present. When potential energy goes up, the product (HF) is formed and vice versa.<br />
<br />
<br />
====<u>question 8: Report the activation energy for both reactions.</u>====<br />
[[File:EH2.png]] [[File:ETS reaction.png]] [[File:EHF reaction.png]]<br />
<br />
<br />
The activation of F+H2 reaction is about 1.12 KJ mol-1 and the activation energy of HF+H reaction is about 126.424KJ mol-1. The first plot of energy shows the potential energy as well as the total energy of the reactant(F+H2), and the second plot shows the one of the product(HF+H). The activation energy is the energy that the reactant needed to overcome so that the reaction can happen. It is the difference between the potential energy of transition state and their initial state.<br />
<br />
<br />
====<u>question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.</u>====<br />
[[File:D1.1.png]] [[File:D20.png]] <br />
<br />
The initial condition set is rHF=190pm and rHH=74 for F+H2 reaction. It is an exothermic reaction thus the energy released for foming HF. From the first graph, the momentum is set at pHF=-0.1g.mol-1.pm.fs-1 and pHH=-1.1g.mol-1.pm.fs-1. It is obvious that the HF is not formed and the kinetic energy is only about +1.105KJ mol-1, which shows that not so much potential energy is transfered to kinetic energy. However, in the second graph, the activation energy is overcomed and HF is formed. The momentum now is pHF=-0.1g.mol-1.pm.fs-1 and pHH=-20g.mol-1.pm.fs-1. More potential energy is transfered to kinetic energy, which is about +398.005KJ mol-1. In conclusion, when the HF bond is formed, the energy released shows that potential energe is transfered to kinetic energy. Moreover, temperature is the average kinetic energy, in this case, the increase in kinetic energy can be proved by the increase in temperature. We can use calorimetry to investigate the change in temperature during reaction. <br />
<br />
====<u>question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.</u>====<br />
<br />
<br />
Translational energy is due to the change in the locations of atoms or molecules and vibrational energy is from the change in the shape of molecule, like stretching and bending. The rate of reaction can sometimes be detemined by the properties of transition state. Transition state is always located close to the top of a potential barrier, which requires high energy to overcome it. Translational energy and vibrational energy have the influence on the rate of the reaction, which often depends on the position of transition state. The translational energy is more effective than vibrational energy when there is a reactant-like transition state, while the vibrational energy will be more effective if a transtion state is closer to the product. Generally, an exothermic reaction gives the reactant-like transition state and thus higher kinetic energy, which can comes from the energy chaning in position or location. Moreover, symmetric vibration sometimes has higher rate and reactivity than antisymmetry vibration.</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cd1618&diff=810792MRD:cd16182020-05-22T17:46:19Z<p>Ng611: </p>
<hr />
<div>== Questions and Answers ==<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:46, 22 May 2020 (BST) Overall: 3/5 - Your answers, whilst mostly correct, were not nearly detailed enough. See the individual comments below for examples. <br />
<br />
1.On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?<br />
<br />
A saddle point is a point on the potential surface that the slope in orthogonal directions are zero and is not a local minimum or maximum. <br />
It can be identified by finding the first-order saddle point on potential energy surface.<br />
To differentiate saddle point from a local minimum, second partial derivative test can be done. Taking two orthogonal curves on potential surface both orthogonal to z axis, if the product of the second derivatives of the is negative, then it is a saddle point. However for a local minimum, the product of second derivatives and the curvature are positive, which indicates the energy goes up in all directions.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:38, 22 May 2020 (BST) What are the two othogonal directions on the PES you take the second derivatives with respect to? A diagram or some equations would be helpful. <br />
<br />
2.Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.<br />
<br />
r<sub>ts</sub>= 91 pm All atoms in this system are hydrogen atoms. In this case, a symmetric transition state is expected, the Internuclear Distances vs Time graph confirmed that the distances between AB and BC are the same.<br />
<br />
[[File:TScd1618.png|250px]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) I still see minor oscillations in your distances vs time graph. What does that tell you about the TS position?<br />
<br />
3.Comment on how the mep and the trajectory you just calculated differ.<br />
<br />
Dynamic calculation makes trajectory vibrate more at the valley, because it considers the vibration of the bond.<br />
Minimum energy path calculation resets momenta to zero in each step, so there is no bond vibration, the trajectory is more like a smooth curve.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Some trajectories would have been useful here. <br />
<br />
4.Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?<br />
<br />
{| class="wikitable" border="1"<br />
|+ Reactive and unreactive trajectories<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Reactive || after TS, more bond vibration is observed || [[File:-2.56-5.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Unreactive || fail to reach TS || [[File:-3.1-4.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || a reactive trajectory with more vibrational energy || [[File:-3.1-5.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Unreactive || energy is too high, barrier recrossed || [[File:-5.1-10.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reavtive || energy high enough to pass TS after recrossing || [[File:-5.1-10.6cd1618.png|150px]]<br />
|}<br />
<br />
Even with high enough energy, a trajectory is not always reactive. Right amount of energy is required for the reaction to happen. The system can go beyond the transition state and recross the energy barrier back to reactant.<br />
<br />
5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br />
<br />
TST will overestimate reaction rate, because it assumes any trajectory with kinetic energy higher than activation energy will be reactive, but from the simulation, it can be found that even with higher energy can recross the transition state and goes back to reactant. So experimentally, the rate will be lower than the estimated value.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Good answer. What are the other assumptions of TS theory though?<br />
<br />
6.By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?<br />
<br />
Exothermic for F + H<sub>2</sub> and endo thermic for H + HF, this indicates that the bond strength for H-F is higher than H-H. So breaking a weak H-H bond to form a strong H-F bond will be exothermic and vice versa <br />
<br />
[[File:PEScd1618.png|250px]]<br />
<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) You are absolutely correct, but what features of the TS would support your answer? If, for example, you only had the TS to go on, how could you confirm the exo/endothermicity of the reaction?<br />
<br />
7.Locate the approximate position of the transition state.<br />
<br />
F-H distance 181 pm H-H distance 75 pm<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) How did you confirm this was the TS?<br />
<br />
8.Report the activation energy for both reactions.<br />
<br />
-433.940 kJmol<sup>-1</sup> for transition state<br />
0.066 kJmol<sup>-1</sup> for F + H<sub>2</sub> It is very close to 0 and hard to determine.<br />
125.367 kJmol<sup>-1</sup> for H + HF<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) Again, how was this calculated?<br />
<br />
9.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.<br />
<br />
The energy of bond forming is transferred to the vibration of F-H bond, shown in the Momenta vs Time diagram. The Momentum of F-H experienced a huge increase when the system goes over the transition state. In the experiment, this can be confirmed by infrared chemiluminiscence.<br />
Also, the internal vibration will be converted to translational energy and transferred to the surrendering molecules.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) IR chemiluminesence is absolutely correct, well done. I'm unsure for your second part though. How would this help?<br />
<br />
10.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.<br />
<br />
In the exothermic process examined here, F + H<sub>2</sub>, translational energy is more effective in overcoming the activation energy barrier. This can be confirmed by calculation, the total kinetic energy is much lower when more energy in the system is the translational mode. Because the early transition state can be more easily overcomed by translational energy.<ref name="RefTS1" /> This is realated to the position of transition state.<br />
<br />
[[File:-10.6cd1618.png|250px]]<br />
p<sub>1</sub>= -1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +1.486 kJmol<sup>-1</sup><br />
<br />
[[File:-1.050cd1618.png|250px]]<br />
p<sub>1</sub>= -1.05 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +0.580 kJmol<sup>-1</sup><br />
<br />
In the endothermic process, FH + H, vibrational energy is more effective to promote reaction. The initial condition with higher vibrational energy requies less total kinetic energy to reach TS.<br />
<br />
[[File:-0.2-19gcd1618.png|250px]]<br />
p<sub>1</sub>= -0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -19 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +357.221 kJmol<sup>-1</sup><br />
<br />
[[File:13.2-2.4gcd1618.png|250px]]<br />
p<sub>1</sub>= 13.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -2.4 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +129.145 kJmol<sup>-1</sup><br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:45, 22 May 2020 (BST) A good set of examples. Well done!<br />
<br />
== References ==<br />
<references><br />
<ref name="RefTS1">Polanyi, J. C. "Some Concepts in Reaction Dynamics." Science 236.4802 (1987): 680-90.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cd1618&diff=810785MRD:cd16182020-05-22T17:45:14Z<p>Ng611: /* Questions and Answers */</p>
<hr />
<div>== Questions and Answers ==<br />
<br />
1.On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?<br />
<br />
A saddle point is a point on the potential surface that the slope in orthogonal directions are zero and is not a local minimum or maximum. <br />
It can be identified by finding the first-order saddle point on potential energy surface.<br />
To differentiate saddle point from a local minimum, second partial derivative test can be done. Taking two orthogonal curves on potential surface both orthogonal to z axis, if the product of the second derivatives of the is negative, then it is a saddle point. However for a local minimum, the product of second derivatives and the curvature are positive, which indicates the energy goes up in all directions.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:38, 22 May 2020 (BST) What are the two othogonal directions on the PES you take the second derivatives with respect to? A diagram or some equations would be helpful. <br />
<br />
2.Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.<br />
<br />
r<sub>ts</sub>= 91 pm All atoms in this system are hydrogen atoms. In this case, a symmetric transition state is expected, the Internuclear Distances vs Time graph confirmed that the distances between AB and BC are the same.<br />
<br />
[[File:TScd1618.png|250px]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) I still see minor oscillations in your distances vs time graph. What does that tell you about the TS position?<br />
<br />
3.Comment on how the mep and the trajectory you just calculated differ.<br />
<br />
Dynamic calculation makes trajectory vibrate more at the valley, because it considers the vibration of the bond.<br />
Minimum energy path calculation resets momenta to zero in each step, so there is no bond vibration, the trajectory is more like a smooth curve.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Some trajectories would have been useful here. <br />
<br />
4.Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?<br />
<br />
{| class="wikitable" border="1"<br />
|+ Reactive and unreactive trajectories<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Reactive || after TS, more bond vibration is observed || [[File:-2.56-5.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Unreactive || fail to reach TS || [[File:-3.1-4.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || a reactive trajectory with more vibrational energy || [[File:-3.1-5.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Unreactive || energy is too high, barrier recrossed || [[File:-5.1-10.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reavtive || energy high enough to pass TS after recrossing || [[File:-5.1-10.6cd1618.png|150px]]<br />
|}<br />
<br />
Even with high enough energy, a trajectory is not always reactive. Right amount of energy is required for the reaction to happen. The system can go beyond the transition state and recross the energy barrier back to reactant.<br />
<br />
5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br />
<br />
TST will overestimate reaction rate, because it assumes any trajectory with kinetic energy higher than activation energy will be reactive, but from the simulation, it can be found that even with higher energy can recross the transition state and goes back to reactant. So experimentally, the rate will be lower than the estimated value.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Good answer. What are the other assumptions of TS theory though?<br />
<br />
6.By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?<br />
<br />
Exothermic for F + H<sub>2</sub> and endo thermic for H + HF, this indicates that the bond strength for H-F is higher than H-H. So breaking a weak H-H bond to form a strong H-F bond will be exothermic and vice versa <br />
<br />
[[File:PEScd1618.png|250px]]<br />
<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) You are absolutely correct, but what features of the TS would support your answer? If, for example, you only had the TS to go on, how could you confirm the exo/endothermicity of the reaction?<br />
<br />
7.Locate the approximate position of the transition state.<br />
<br />
F-H distance 181 pm H-H distance 75 pm<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) How did you confirm this was the TS?<br />
<br />
8.Report the activation energy for both reactions.<br />
<br />
-433.940 kJmol<sup>-1</sup> for transition state<br />
0.066 kJmol<sup>-1</sup> for F + H<sub>2</sub> It is very close to 0 and hard to determine.<br />
125.367 kJmol<sup>-1</sup> for H + HF<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) Again, how was this calculated?<br />
<br />
9.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.<br />
<br />
The energy of bond forming is transferred to the vibration of F-H bond, shown in the Momenta vs Time diagram. The Momentum of F-H experienced a huge increase when the system goes over the transition state. In the experiment, this can be confirmed by infrared chemiluminiscence.<br />
Also, the internal vibration will be converted to translational energy and transferred to the surrendering molecules.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) IR chemiluminesence is absolutely correct, well done. I'm unsure for your second part though. How would this help?<br />
<br />
10.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.<br />
<br />
In the exothermic process examined here, F + H<sub>2</sub>, translational energy is more effective in overcoming the activation energy barrier. This can be confirmed by calculation, the total kinetic energy is much lower when more energy in the system is the translational mode. Because the early transition state can be more easily overcomed by translational energy.<ref name="RefTS1" /> This is realated to the position of transition state.<br />
<br />
[[File:-10.6cd1618.png|250px]]<br />
p<sub>1</sub>= -1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +1.486 kJmol<sup>-1</sup><br />
<br />
[[File:-1.050cd1618.png|250px]]<br />
p<sub>1</sub>= -1.05 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +0.580 kJmol<sup>-1</sup><br />
<br />
In the endothermic process, FH + H, vibrational energy is more effective to promote reaction. The initial condition with higher vibrational energy requies less total kinetic energy to reach TS.<br />
<br />
[[File:-0.2-19gcd1618.png|250px]]<br />
p<sub>1</sub>= -0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -19 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +357.221 kJmol<sup>-1</sup><br />
<br />
[[File:13.2-2.4gcd1618.png|250px]]<br />
p<sub>1</sub>= 13.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -2.4 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +129.145 kJmol<sup>-1</sup><br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:45, 22 May 2020 (BST) A good set of examples. Well done!<br />
<br />
== References ==<br />
<references><br />
<ref name="RefTS1">Polanyi, J. C. "Some Concepts in Reaction Dynamics." Science 236.4802 (1987): 680-90.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cd1618&diff=810778MRD:cd16182020-05-22T17:44:03Z<p>Ng611: /* Questions and Answers */</p>
<hr />
<div>== Questions and Answers ==<br />
<br />
1.On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?<br />
<br />
A saddle point is a point on the potential surface that the slope in orthogonal directions are zero and is not a local minimum or maximum. <br />
It can be identified by finding the first-order saddle point on potential energy surface.<br />
To differentiate saddle point from a local minimum, second partial derivative test can be done. Taking two orthogonal curves on potential surface both orthogonal to z axis, if the product of the second derivatives of the is negative, then it is a saddle point. However for a local minimum, the product of second derivatives and the curvature are positive, which indicates the energy goes up in all directions.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:38, 22 May 2020 (BST) What are the two othogonal directions on the PES you take the second derivatives with respect to? A diagram or some equations would be helpful. <br />
<br />
2.Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.<br />
<br />
r<sub>ts</sub>= 91 pm All atoms in this system are hydrogen atoms. In this case, a symmetric transition state is expected, the Internuclear Distances vs Time graph confirmed that the distances between AB and BC are the same.<br />
<br />
[[File:TScd1618.png|250px]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) I still see minor oscillations in your distances vs time graph. What does that tell you about the TS position?<br />
<br />
3.Comment on how the mep and the trajectory you just calculated differ.<br />
<br />
Dynamic calculation makes trajectory vibrate more at the valley, because it considers the vibration of the bond.<br />
Minimum energy path calculation resets momenta to zero in each step, so there is no bond vibration, the trajectory is more like a smooth curve.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Some trajectories would have been useful here. <br />
<br />
4.Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?<br />
<br />
{| class="wikitable" border="1"<br />
|+ Reactive and unreactive trajectories<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Reactive || after TS, more bond vibration is observed || [[File:-2.56-5.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Unreactive || fail to reach TS || [[File:-3.1-4.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || a reactive trajectory with more vibrational energy || [[File:-3.1-5.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Unreactive || energy is too high, barrier recrossed || [[File:-5.1-10.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reavtive || energy high enough to pass TS after recrossing || [[File:-5.1-10.6cd1618.png|150px]]<br />
|}<br />
<br />
Even with high enough energy, a trajectory is not always reactive. Right amount of energy is required for the reaction to happen. The system can go beyond the transition state and recross the energy barrier back to reactant.<br />
<br />
5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br />
<br />
TST will overestimate reaction rate, because it assumes any trajectory with kinetic energy higher than activation energy will be reactive, but from the simulation, it can be found that even with higher energy can recross the transition state and goes back to reactant. So experimentally, the rate will be lower than the estimated value.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Good answer. What are the other assumptions of TS theory though?<br />
<br />
6.By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?<br />
<br />
Exothermic for F + H<sub>2</sub> and endo thermic for H + HF, this indicates that the bond strength for H-F is higher than H-H. So breaking a weak H-H bond to form a strong H-F bond will be exothermic and vice versa <br />
<br />
[[File:PEScd1618.png|250px]]<br />
<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) You are absolutely correct, but what features of the TS would support your answer? If, for example, you only had the TS to go on, how could you confirm the exo/endothermicity of the reaction?<br />
<br />
7.Locate the approximate position of the transition state.<br />
<br />
F-H distance 181 pm H-H distance 75 pm<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) How did you confirm this was the TS?<br />
<br />
8.Report the activation energy for both reactions.<br />
<br />
-433.940 kJmol<sup>-1</sup> for transition state<br />
0.066 kJmol<sup>-1</sup> for F + H<sub>2</sub> It is very close to 0 and hard to determine.<br />
125.367 kJmol<sup>-1</sup> for H + HF<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) Again, how was this calculated?<br />
<br />
9.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.<br />
<br />
The energy of bond forming is transferred to the vibration of F-H bond, shown in the Momenta vs Time diagram. The Momentum of F-H experienced a huge increase when the system goes over the transition state. In the experiment, this can be confirmed by infrared chemiluminiscence.<br />
Also, the internal vibration will be converted to translational energy and transferred to the surrendering molecules.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:44, 22 May 2020 (BST) IR chemiluminesence is absolutely correct, well done. I'm unsure for your second part though. How would this help?<br />
<br />
10.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.<br />
<br />
In the exothermic process examined here, F + H<sub>2</sub>, translational energy is more effective in overcoming the activation energy barrier. This can be confirmed by calculation, the total kinetic energy is much lower when more energy in the system is the translational mode. Because the early transition state can be more easily overcomed by translational energy.<ref name="RefTS1" /> This is realated to the position of transition state.<br />
<br />
[[File:-10.6cd1618.png|250px]]<br />
p<sub>1</sub>= -1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +1.486 kJmol<sup>-1</sup><br />
<br />
[[File:-1.050cd1618.png|250px]]<br />
p<sub>1</sub>= -1.05 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +0.580 kJmol<sup>-1</sup><br />
<br />
In the endothermic process, FH + H, vibrational energy is more effective to promote reaction. The initial condition with higher vibrational energy requies less total kinetic energy to reach TS.<br />
<br />
[[File:-0.2-19gcd1618.png|250px]]<br />
p<sub>1</sub>= -0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -19 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +357.221 kJmol<sup>-1</sup><br />
<br />
[[File:13.2-2.4gcd1618.png|250px]]<br />
p<sub>1</sub>= 13.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -2.4 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +129.145 kJmol<sup>-1</sup><br />
<br />
== References ==<br />
<references><br />
<ref name="RefTS1">Polanyi, J. C. "Some Concepts in Reaction Dynamics." Science 236.4802 (1987): 680-90.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cd1618&diff=810767MRD:cd16182020-05-22T17:40:20Z<p>Ng611: /* Questions and Answers */</p>
<hr />
<div>== Questions and Answers ==<br />
<br />
1.On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?<br />
<br />
A saddle point is a point on the potential surface that the slope in orthogonal directions are zero and is not a local minimum or maximum. <br />
It can be identified by finding the first-order saddle point on potential energy surface.<br />
To differentiate saddle point from a local minimum, second partial derivative test can be done. Taking two orthogonal curves on potential surface both orthogonal to z axis, if the product of the second derivatives of the is negative, then it is a saddle point. However for a local minimum, the product of second derivatives and the curvature are positive, which indicates the energy goes up in all directions.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:38, 22 May 2020 (BST) What are the two othogonal directions on the PES you take the second derivatives with respect to? A diagram or some equations would be helpful. <br />
<br />
2.Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.<br />
<br />
r<sub>ts</sub>= 91 pm All atoms in this system are hydrogen atoms. In this case, a symmetric transition state is expected, the Internuclear Distances vs Time graph confirmed that the distances between AB and BC are the same.<br />
<br />
[[File:TScd1618.png|250px]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) I still see minor oscillations in your distances vs time graph. What does that tell you about the TS position?<br />
<br />
3.Comment on how the mep and the trajectory you just calculated differ.<br />
<br />
Dynamic calculation makes trajectory vibrate more at the valley, because it considers the vibration of the bond.<br />
Minimum energy path calculation resets momenta to zero in each step, so there is no bond vibration, the trajectory is more like a smooth curve.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Some trajectories would have been useful here. <br />
<br />
4.Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?<br />
<br />
{| class="wikitable" border="1"<br />
|+ Reactive and unreactive trajectories<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Reactive || after TS, more bond vibration is observed || [[File:-2.56-5.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Unreactive || fail to reach TS || [[File:-3.1-4.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || a reactive trajectory with more vibrational energy || [[File:-3.1-5.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Unreactive || energy is too high, barrier recrossed || [[File:-5.1-10.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reavtive || energy high enough to pass TS after recrossing || [[File:-5.1-10.6cd1618.png|150px]]<br />
|}<br />
<br />
Even with high enough energy, a trajectory is not always reactive. Right amount of energy is required for the reaction to happen. The system can go beyond the transition state and recross the energy barrier back to reactant.<br />
<br />
5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br />
<br />
TST will overestimate reaction rate, because it assumes any trajectory with kinetic energy higher than activation energy will be reactive, but from the simulation, it can be found that even with higher energy can recross the transition state and goes back to reactant. So experimentally, the rate will be lower than the estimated value.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:40, 22 May 2020 (BST) Good answer. What are the other assumptions of TS theory though?<br />
<br />
6.By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?<br />
<br />
Exothermic for F + H<sub>2</sub> and endo thermic for H + HF, this indicates that the bond strength for H-F is higher than H-H. So breaking a weak H-H bond to form a strong H-F bond will be exothermic and vice versa <br />
<br />
[[File:PEScd1618.png|250px]]<br />
<br />
7.Locate the approximate position of the transition state.<br />
<br />
F-H distance 181 pm H-H distance 75 pm<br />
<br />
8.Report the activation energy for both reactions.<br />
<br />
-433.940 kJmol<sup>-1</sup> for transition state<br />
0.066 kJmol<sup>-1</sup> for F + H<sub>2</sub> It is very close to 0 and hard to determine.<br />
125.367 kJmol<sup>-1</sup> for H + HF<br />
<br />
9.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.<br />
<br />
The energy of bond forming is transferred to the vibration of F-H bond, shown in the Momenta vs Time diagram. The Momentum of F-H experienced a huge increase when the system goes over the transition state. In the experiment, this can be confirmed by infrared chemiluminiscence.<br />
Also, the internal vibration will be converted to translational energy and transferred to the surrendering molecules.<br />
<br />
10.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.<br />
<br />
In the exothermic process examined here, F + H<sub>2</sub>, translational energy is more effective in overcoming the activation energy barrier. This can be confirmed by calculation, the total kinetic energy is much lower when more energy in the system is the translational mode. Because the early transition state can be more easily overcomed by translational energy.<ref name="RefTS1" /> This is realated to the position of transition state.<br />
<br />
[[File:-10.6cd1618.png|250px]]<br />
p<sub>1</sub>= -1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +1.486 kJmol<sup>-1</sup><br />
<br />
[[File:-1.050cd1618.png|250px]]<br />
p<sub>1</sub>= -1.05 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +0.580 kJmol<sup>-1</sup><br />
<br />
In the endothermic process, FH + H, vibrational energy is more effective to promote reaction. The initial condition with higher vibrational energy requies less total kinetic energy to reach TS.<br />
<br />
[[File:-0.2-19gcd1618.png|250px]]<br />
p<sub>1</sub>= -0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -19 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +357.221 kJmol<sup>-1</sup><br />
<br />
[[File:13.2-2.4gcd1618.png|250px]]<br />
p<sub>1</sub>= 13.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -2.4 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +129.145 kJmol<sup>-1</sup><br />
<br />
== References ==<br />
<references><br />
<ref name="RefTS1">Polanyi, J. C. "Some Concepts in Reaction Dynamics." Science 236.4802 (1987): 680-90.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:cd1618&diff=810760MRD:cd16182020-05-22T17:38:06Z<p>Ng611: /* Questions and Answers */</p>
<hr />
<div>== Questions and Answers ==<br />
<br />
1.On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?<br />
<br />
A saddle point is a point on the potential surface that the slope in orthogonal directions are zero and is not a local minimum or maximum. <br />
It can be identified by finding the first-order saddle point on potential energy surface.<br />
To differentiate saddle point from a local minimum, second partial derivative test can be done. Taking two orthogonal curves on potential surface both orthogonal to z axis, if the product of the second derivatives of the is negative, then it is a saddle point. However for a local minimum, the product of second derivatives and the curvature are positive, which indicates the energy goes up in all directions.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:38, 22 May 2020 (BST) What are the two othogonal directions on the PES you take the second derivatives with respect to? A diagram or some equations would be helpful. <br />
<br />
2.Report your best estimate of the transition state position (r<sub>ts</sub>) and explain your reasoning illustrating it with a āInternuclear Distances vs Timeā plot for a relevant trajectory.<br />
<br />
r<sub>ts</sub>= 91 pm All atoms in this system are hydrogen atoms. In this case, a symmetric transition state is expected, the Internuclear Distances vs Time graph confirmed that the distances between AB and BC are the same.<br />
<br />
[[File:TScd1618.png|250px]]<br />
<br />
3.Comment on how the mep and the trajectory you just calculated differ.<br />
<br />
Dynamic calculation makes trajectory vibrate more at the valley, because it considers the vibration of the bond.<br />
Minimum energy path calculation resets momenta to zero in each step, so there is no bond vibration, the trajectory is more like a smooth curve.<br />
<br />
4.Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?<br />
<br />
{| class="wikitable" border="1"<br />
|+ Reactive and unreactive trajectories<br />
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| -2.56 || -5.1 || -414.280 || Reactive || after TS, more bond vibration is observed || [[File:-2.56-5.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -4.1 || -420.077 || Unreactive || fail to reach TS || [[File:-3.1-4.1cd1618.png|150px]]<br />
|-<br />
| -3.1 || -5.1 || -413.977 || Reactive || a reactive trajectory with more vibrational energy || [[File:-3.1-5.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.1 || -357.277 || Unreactive || energy is too high, barrier recrossed || [[File:-5.1-10.1cd1618.png|150px]]<br />
|-<br />
| -5.1 || -10.6 || -349.477 || Reavtive || energy high enough to pass TS after recrossing || [[File:-5.1-10.6cd1618.png|150px]]<br />
|}<br />
<br />
Even with high enough energy, a trajectory is not always reactive. Right amount of energy is required for the reaction to happen. The system can go beyond the transition state and recross the energy barrier back to reactant.<br />
<br />
5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?<br />
<br />
TST will overestimate reaction rate, because it assumes any trajectory with kinetic energy higher than activation energy will be reactive, but from the simulation, it can be found that even with higher energy can recross the transition state and goes back to reactant. So experimentally, the rate will be lower than the estimated value.<br />
<br />
6.By inspecting the potential energy surfaces, classify the F + H<sub>2</sub> and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?<br />
<br />
Exothermic for F + H<sub>2</sub> and endo thermic for H + HF, this indicates that the bond strength for H-F is higher than H-H. So breaking a weak H-H bond to form a strong H-F bond will be exothermic and vice versa <br />
<br />
[[File:PEScd1618.png|250px]]<br />
<br />
7.Locate the approximate position of the transition state.<br />
<br />
F-H distance 181 pm H-H distance 75 pm<br />
<br />
8.Report the activation energy for both reactions.<br />
<br />
-433.940 kJmol<sup>-1</sup> for transition state<br />
0.066 kJmol<sup>-1</sup> for F + H<sub>2</sub> It is very close to 0 and hard to determine.<br />
125.367 kJmol<sup>-1</sup> for H + HF<br />
<br />
9.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.<br />
<br />
The energy of bond forming is transferred to the vibration of F-H bond, shown in the Momenta vs Time diagram. The Momentum of F-H experienced a huge increase when the system goes over the transition state. In the experiment, this can be confirmed by infrared chemiluminiscence.<br />
Also, the internal vibration will be converted to translational energy and transferred to the surrendering molecules.<br />
<br />
10.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.<br />
<br />
In the exothermic process examined here, F + H<sub>2</sub>, translational energy is more effective in overcoming the activation energy barrier. This can be confirmed by calculation, the total kinetic energy is much lower when more energy in the system is the translational mode. Because the early transition state can be more easily overcomed by translational energy.<ref name="RefTS1" /> This is realated to the position of transition state.<br />
<br />
[[File:-10.6cd1618.png|250px]]<br />
p<sub>1</sub>= -1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +1.486 kJmol<sup>-1</sup><br />
<br />
[[File:-1.050cd1618.png|250px]]<br />
p<sub>1</sub>= -1.05 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +0.580 kJmol<sup>-1</sup><br />
<br />
In the endothermic process, FH + H, vibrational energy is more effective to promote reaction. The initial condition with higher vibrational energy requies less total kinetic energy to reach TS.<br />
<br />
[[File:-0.2-19gcd1618.png|250px]]<br />
p<sub>1</sub>= -0.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -19 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +357.221 kJmol<sup>-1</sup><br />
<br />
[[File:13.2-2.4gcd1618.png|250px]]<br />
p<sub>1</sub>= 13.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> p<sub>2</sub>= -2.4 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> E<sub>k</sub>= +129.145 kJmol<sup>-1</sup><br />
<br />
== References ==<br />
<references><br />
<ref name="RefTS1">Polanyi, J. C. "Some Concepts in Reaction Dynamics." Science 236.4802 (1987): 680-90.</ref><br />
</references></div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810749MRD:014973842020-05-22T17:34:30Z<p>Ng611: </p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:34, 22 May 2020 (BST) Overall 5/5 - An excellent (really, near perfect) piece of work. Well done. One or two minor comments are outlined throughout. <br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:29, 22 May 2020 (BST) These are good estimates of the activation barriers. Your F+H2 reaction barrier is about 0.2 kJ/mol too small (as you can see you still have a negative gradient in your energy vs time plot) but overall this is a minor thing so don't worry too much about it. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:31, 22 May 2020 (BST) Good response. IR chemiluminescence was indeed how this mechanism of energy release was confirmed. Well done!<br />
<br />
====Polanyi's empirical rule====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:33, 22 May 2020 (BST) A very neat and well thought-out discussion of Polanyi's rules. Well done!<br />
<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810745MRD:014973842020-05-22T17:33:15Z<p>Ng611: /* Polanyi's empirical rule */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:29, 22 May 2020 (BST) These are good estimates of the activation barriers. Your F+H2 reaction barrier is about 0.2 kJ/mol too small (as you can see you still have a negative gradient in your energy vs time plot) but overall this is a minor thing so don't worry too much about it. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:31, 22 May 2020 (BST) Good response. IR chemiluminescence was indeed how this mechanism of energy release was confirmed. Well done!<br />
<br />
====Polanyi's empirical rule====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:33, 22 May 2020 (BST) A very neat and well thought-out discussion of Polanyi's rules. Well done!<br />
<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810740MRD:014973842020-05-22T17:31:49Z<p>Ng611: /* Mechanism of energy release */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:29, 22 May 2020 (BST) These are good estimates of the activation barriers. Your F+H2 reaction barrier is about 0.2 kJ/mol too small (as you can see you still have a negative gradient in your energy vs time plot) but overall this is a minor thing so don't worry too much about it. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:31, 22 May 2020 (BST) Good response. IR chemiluminescence was indeed how this mechanism of energy release was confirmed. Well done!<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810731MRD:014973842020-05-22T17:30:35Z<p>Ng611: /* Activation energy */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:29, 22 May 2020 (BST) These are good estimates of the activation barriers. Your F+H2 reaction barrier is about 0.2 kJ/mol too small (as you can see you still have a negative gradient in your energy vs time plot) but overall this is a minor thing so don't worry too much about it. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810726MRD:014973842020-05-22T17:29:44Z<p>Ng611: /* Activation energy */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:29, 22 May 2020 (BST) These are good estimates of the activation barriers. Your F+H2 reaction barrier is about 0.2 kJ/mol too small but overall this is a minor thing so don't worry too much about it. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810721MRD:014973842020-05-22T17:28:04Z<p>Ng611: /* Approximate TS position */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:28, 22 May 2020 (BST) Excellent response. I'd also maybe also do a small experiment where you displace the system a little way from the TS and allow it to roll down either side of the barrier. The reason i'd recommend this is that for such a small energy barrier, the forces may disappear due to rounding, but they'll still have an effect on the trajectory.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810712MRD:014973842020-05-22T17:25:08Z<p>Ng611: /* Transition State Theory */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:25, 22 May 2020 (BST) An exemplary answer to the question and an exemplary section 1 overall. Well done!<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
<br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810704MRD:014973842020-05-22T17:23:08Z<p>Ng611: /* Comparison between Dynamics and MEP trajectories */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:23, 22 May 2020 (BST) Excellent!!<br />
<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
<br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810699MRD:014973842020-05-22T17:21:59Z<p>Ng611: /* Locating the transition state */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:21, 22 May 2020 (BST) Good estimate and a good rationale for the TS location, well done!<br />
<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
<br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
<br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01497384&diff=810676MRD:014973842020-05-22T17:18:22Z<p>Ng611: /* Dynamics from the transition state region */</p>
<hr />
<div><br />
= Molecular Reaction Dynamics: Applications to Triatomic systems =<br />
<br />
== H + H<sub>2</sub>Ā system ==<br />
<br />
=== Dynamics from the transition state region ===<br />
The transition state on a potential energy surface (PES) is neither the local maximum nor the local minimum, it is the configuration corresponding to the maximum (which is also termed a 'first-order saddle point') in the direction of the minimum energy path, and a minimum in all other directions perpendicular to the path. The minimum energy path is highlighted by the oscillating black line in Figure 1<br />
and 2.<br />
<gallery mode=packed class=center widths="285px" heights="285px"><br />
File:xjgTSmin.png|'''Figure 1. A Potential Energy Surface plot showing the minimum point in the direction orthogonal to the minimum energy path.'''<br />
File:xjgTSmax.png|'''Figure 2. A Potential Energy Surface plot showing the maximum point along the minimum energy path.'''<br />
</gallery><br />
<br />
<br />
<br />
====A mathematical view====<br />
The transition state is defined mathematically as having a partial derivative of 0 with respect to each of its axes on the PES, given by āV(rAB)/ārAB= āV(rBC)/ārBC= 0, which is characterised by a zero gradient. A simple criterion for distinguishing between a saddle point and a local minima is to compute the Hessian Matrix at the point of the PES function. A local minima would have a negative Hessian matrix determinant while a saddle point would have a positive Hessian determinant.<br />
<br />
<div style="text-align: center;"><math>\mathbf H = \begin{bmatrix}<br />
\dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x\partial y} \\[2.2ex]<br />
\dfrac{\partial^2 f}{\partial y\partial x} & \dfrac{\partial^2 f}{\partial y^2}<br />
<br />
\end{bmatrix}</math> (1)</div><br />
<br />
If :<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{\partial{V^{2}(rAB)}}{\partial{rAB}^{2}}>0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local minima.<br />
<br />
<br />
<math display=>\frac{ \partial{Vr}}{\partial{r}}=0, \frac{ \partial{V^{2}(rAB)}}{\partial{rAB}^{2}}<0 </math>, AND <math display=>det(H) > 0 </math>, the point is a local maxima. <br />
<br />
<br />
<math display=> \frac{ \partial{Vr}}{\partial{r}}=0, det(H) <0 </math>, the point is a saddle point.<br />
<br />
The Hessian is defined along the AB and BC direction. The eigenvalues of the Hessian matrix correspond to the vibrational frequencies and determines the curvature along its eigenvectors. A local minima only has '''positive eigenvalues''' as the curvature at the point in all directions are positive. The saddle point of the transition state has '''one''' (and only one) '''negative eigenvalue''' in its Hessian, as the point is a maximum in one direction along the reaction path and a minimum in all other orthogonal directions. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 18:18, 22 May 2020 (BST) An excellent answer, well done!<br />
<br />
====Locating the transition state====<br />
If a trajectory is set at the exact point of the transition state with no initial momentum, there will be no force acting on the atoms (gradient on the PES is zero), thus it will remain there indefinitely. The transition state bond length can be located by starting trajectories near the transition state and adjust accordingly upon observation of the signs of forces along on the atoms. According to the Hammond's Postulate, the transition state is symmetrical and the PES is symmetric. Thus it is expected that rAB=rBC at the transition state.<br />
<br />
An initial estimation is made for the transition state bond length (r<sub>TS</sub>) by initiating a trajectory with rAB and rBC at 90.0 pm. The force along AB and BC are +0.132 kJ/mol/pm and slight oscillations of rBC could be seen on the Internuclear Distance vs Time plot shown in Figure 3. This suggests that 90 pm is not the exact transition state bond length. By expanding in to Figure 3, an average position of r<sub>TS</sub> was given to be 90.8 pm. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_90_.png|'''Figure 3. Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
File:xjg18_90_expanded.png|'''Figure 4. Expanded Internuclear Distance vs Time plot when r<sub>TS</sub>=90 pm'''<br />
</gallery><br />
<br />
Further estimation of a 90.8 pm r<sub>TS</sub> showed the forces to be -0.004 kJ/mol/pm, showing that the forces acting upon the atoms are in the opposite direction from the first estimnation.<br />
<br />
This allowed a satisfactory estimate of the r<sub>TS</sub> of 90.775 pm, where the forces are -0.000 kJ/mol/pm and straight horizontal lines with no oscillations are shown in the Internuclear Distance vs Time plot in Figure 5, indicating a zero potential gradient.<br />
<br />
<gallery mode=packed class=center widths="400px" heights="400px"><br />
File:xjg18_90.8.png|'''Figure 5. Internuclear Distance vs Time plot when r<sub>TS</sub>=90.775 pm<br />
</gallery>'''<br />
<br />
Thus, the best estimate of the transition state position (rts) is '''90.775 pm'''. At this TS position, the Hessian has one positive and one negative eigenvalue, corresponding to a negative curvature in one direction and a positive curvature in the orthogonal direction.<br />
<br />
<br />
<br />
====Comparison between Dynamics and MEP trajectories====<br />
The minimum energy path (MEP) is defined as the lowest energy reaction path with infinitely slow motion, such that at each time step, the velocities of the atoms are reset to 0, thus the atoms have no oscillations.<br />
A trajectory is initiated at a position of rAB= 90.775 pm and rBC= 91.775 pm with each atom having a zero initial momenta, resulting in a downhill trajectory forming the product of AB molecule. Figure 6.0 and 6.1 correspond to the MEP and Dynamics surface respectively, The difference observed can be seen in the oscillatory motion in the Dynamics Calculation which is not seen in the MEP calculation. The oscillatory motion observed in the Dynamics plot is due to the gain in momenta of the atoms, allowing them to be at positions with higher potential energies which result in their vibrational motions. The absence of oscillatory motion in MEP shows that the molecule is not vibrating, it simply follows the valley floor of the PES. This is because the inertial effect of the atoms are removed in a MEP calculation, thus does not gain any vibrational energy.This is not the case in reality.<br />
<br />
<gallery mode=packed class=center widths="270px" heights="270px"><br />
File:xjg_dynamics.png|'''Figure 6.0. Dynamics Calculation of a contour plot.'''<br />
File:xjg_MEP.png|'''Figure 6.1 MEP calculation of a contour plot.'''<br />
</gallery><br />
<br />
In Figure 7.0, the dynamics calculation shows oscillating momenta over time allowing the molecule to vibrate and have oscillating internuclear distance as shown in Figure 8.0. Whereas in Figure 7.1. of the MEP calculation, the momenta is constant over time showing no vibrational motions, thus giving a constant internuclear distance over time as shown in Figure 8.1.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsM.png|'''Figure 7.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
File:xjg18_MEPM.png|'''Figure 7.1. MEP calculation of H + H<sub>2</sub> system with trajectory from AB=r<sub>TS</sub> and BC=r<sub>TS</sub>+ 1 pm..'''<br />
</gallery><br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamicsID.png|'''Figure 8.0. Dynamics calculation of Internuclear Distance vs Time.'''<br />
File:xjg18_MEPID.png|'''Figure 8.1 MEP calculation of Internuclear Distance vs TIme'''.<br />
</gallery><br />
<br />
<br />
<br />
====Changing the initial conditions of the trajectory====<br />
Changing the initial conditions by swapping the initial values of rAB and rBC would reverse the reaction, in which the trajectory would travel in the opposite direction forming the molecule BC instead of AB, illustrated in Figure 9.0 and 9.1.<br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_dynamics_swapinitials.png|'''Figure 9.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:xjg18_MEP_swapinitials.png|'''Figure 9.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
For a Dynamics calculation, initiating a trajectory with the final coordinates and the same values of momentum with inverted signs obtained from the calculations done above forms a pathway back to the same initial coordinates and momentum values. For an MEP calculation, the reaction pathway continues down the valley along the lowest energy on the PES as all of the atoms have zero momentum.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_dynamics_reverse.png|'''Figure 10.0. Dynamics calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm.'''<br />
File:01497384_MEPreverse.png|'''Figure 10.1. MEP calculation of H + H<sub>2</sub> system with trajectory from BC=r<sub>TS</sub> and AB=r<sub>TS</sub>+ 1 pm. .'''<br />
</gallery><br />
<br />
<br />
<br />
====Reactive and unreactive trajectories====<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+<br />
! Reaction !! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactivity !! Contour Plot !! Description of pathway <br />
|-<br />
| A || -2.56 || -5.1 || -414.3 ||Reactive || [[File:xjg18_1.png|400 px]] || H<sub>A</sub> atom approaches a non-vibrating H<sub>BC</sub> molecule. The atoms/molecule have enough momentum to overcome the energy barrier at the TS, resulting in a new vibrating molecule H<sub>AB</sub> and H<sub>C</sub> departs.<br />
|-<br />
| B || -3.1 || -4.1 || -420.1 || Unreactive || [[File:xjg18_2.png|400 px]] || There is insufficient momentum/kinetic energy to overcome the energy barrier, hence molecule H<sub>AB</sub> does not form and H<sub>A</sub> is rebounded.<br />
|-<br />
| C || -3.1 || -5.1 || -414.0|| Reactive|| [[File:xjg18_3.png|400 px]] || Atom H<sub>A</sub> approaches a slightly oscillating H<sub>BC</sub> molecule with sufficient kinetic energy to react and form the product of H<sub>AB</sub> molecule while H<sub>C</sub> departs.<br />
|-<br />
| D || -5.1 || -10.1 ||-357.3 || Unreactive|| [[File:xjg18_4.png|400 px]] || The H<sub>BC</sub> molecule was initially formed in the reaction. However, the excess kinetic energy resulted in recrossing of the barrier and the reactants are reformed.<br />
|- <br />
| E || -5.1 || -10.6 || -349.5 || Reactive || [[File:xjg18_5.png|400 px]] || The H<sub>A</sub> atom approaches a non-oscillating H<sub>BC</sub> molecule with high kinetic energy, subsequently forming a product with high vibrational energy.<br />
|}<br />
<br />
In conclusion, it is shown that a system with sufficient momentum i.e kinetic energy alone is not enough for a reaction to be reactive. This is because not every oscillation along the reaction coordinate takes the complex through the transition state and a molecule might be rotating about the wrong axis. The energy must be in the right vibrational modes and the reactants have to be in the correct orientation for a successful outcome of the reaction.<br />
<br />
<br />
<br />
====Transition State Theory====<br />
The transition state theory (TST) provide a means of calculating the rate constant of a reaction. It considers a critical dividing surface separating the reactants and the products and relies on a few assumptions:<br />
<br />
1. A system that has crossed the TS (the dividing surface) in the direction of the product cannot recross the barrier and reform the reactants.<br />
<br />
2. The energy among the reactants are distributed according to the Maxwell-Boltzmann law. <br />
<br />
3. At the TS, any motion along the reaction coordinates can be treated classically as translation, any quantum tunnelling effects are neglected.<br />
<br />
4. The Born-Oppenheimer approximation is applied.<br />
<br />
<br />
Assumption 3 of the TST might lead to an underestimation of the rate constant as the theory neglects quantum tunnelling effect which will lead to the formation of products. This leads to a lower predicted rate constant as some particles with insufficient energy are able to overcome the barrier due to quantum tunnelling. However, this effect is negligible compared to assumption 1 which is more significant in the prediction of rate constants. This theory does not predict the recrossing of barrier in which the products reform the reactants, as seen in reaction D. Thus, the assumptions from the transition state theory would provide an overestimation of the rate constants in comparison with experimental values.<br />
<br />
== F - H - H system==<br />
<br />
===PES inspection===<br />
<br />
====Energetics of the reactions====<br />
<br />
Figures 11.0 and 11.1 show the PES of a F + H<sub>2</sub> and a HF + H system respectively. The F + H<sub>2</sub> is an '''exothermic''' reaction while the HF + H is an '''endothermic''' reaction. Both reactions are backward reactions of the other, thus they share an identical PES in opposite directions. <br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_exo.png|'''Figure 11.0. Potential Energy Surface of F + H<sub>2</sub>, where A=F and BC=H<sub>2</sub>'''<br />
File:xjg18_endo.png|'''Figure 11.1. Potential Energy Surface of H + HF, where A=H and BC=HF .'''<br />
</gallery><br />
<br />
In Figure 11.0, the reactants (H<sub>2</sub> + F) have a higher potential energy than the products (HF + H), where r<sub>AB</sub> denotes the distance between H-F and r<sub>BC</sub> is the distance between the reactant atoms H<sub>2</sub>. This can be related to the stronger bond strength of the product H-F compared to the weaker H-H bond. The energy released from the formation of H-F bond is higher than the energy needed to break the H-H bond as a result of their bond strengths. Thus, the enthalpy change of reaction would be negative, suggesting exothermic reaction with a release in energy. Similarly for the reaction in the reverse direction (HF + H), the reactants (HF and H) have a higher potential energy than the products (H<sub>2</sub> + F). For the same reason, a positive enthalpy change of reaction suggests formation of the weaker H-H bond and the dissociation of the stronger H-F bond, leading to an endothermic reaction, where energy is being taken into the system.<br />
<br />
<br />
<br />
====Approximate TS position====<br />
Since both reactions are reverse reactions of one another, they have the same transition state. According to the Hammond's postulate, the structure of a transition state would resemble that which is closer in energy to the TS. The exothermic F + H<sub>2</sub> would have an early transition state, thus the H-H bond length in the TS would be expected to be similar the the reactant bond length, which is around 74 pm. Thus a BC distance of 74 pm would be a good starting point for the approximation of the TS position. <br />
<br />
The TS can be found by finding a position at the PES where there are no net forces acting on the particles with zero initial momentum. The position of the F-H-H transition state was approximated, where the '''F-H distance''' is '''181.1 pm''' and the '''H-H distance''' is '''74.5 pm'''. At this position, there is zero net force on the particles, indicating a saddle point with zero gradient.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_TS.png|'''Figure 12.0. Contour plot showing the saddle point (TS) of F + H<sub>2</sub> reaction'''<br />
File:xjg18_H2_TS.png|'''Figure 12.1. Contour plot showing the saddle point (TS) of H + HF reaction '''<br />
</gallery><br />
<br />
<br />
<br />
====Activation energy====<br />
The activation energies (E<sub>a</sub>) of the reactant can be estimated by slightly displacing the particles from the transition state in the direction of the reactants and the products in the F + H<sub>2</sub> system and plotting an MEP calculation of the total energy vs time. The total energy in an MEP calculation corresponds to the potential energy in the system, allowing us to calculate the E<sub>a</sub>. The E<sub>a</sub> is then taken as the difference in potential energy between the transition state and the respective energies of the reactants. <br />
<br />
Figure 13.0 shows the Energy vs Time plot in the direction of the HF formation. The energy of the transition state is -434.0 kJ/mol while the energy of the products H + HF is -560.5 kJ/mol. Figure 13.1 shows the Energy vs Time plot in the direction of the H<sub>2</sub> formation. The energy of the products F + H<sub>2</sub> is approximately -434.9 kJ/mol.<br />
<br />
<gallery mode=packed class=center widths="280 px" heights="280 px"><br />
File:xjg18_HF_Ea.png|'''Figure 13.0. Energy vs Time plot of the formation of HF+ H from the TS'''<br />
File:xjg18_H2_Ea.png|'''Figure 13.1. Energy vs Time plot of the formation of H<sub>2</sub>+ F from the TS'''<br />
</gallery><br />
<br />
From Figures 13.0 and 13.1, the estimated activation energies are:<br />
<br />
'''0.9 kJ/mol''' for the F + H<sub>2</sub> -> HF + F reaction and<br />
<br />
'''126.5 kJ/mol''' for the H + HF -> F + H<sub>2</sub> reaction.<br />
<br />
<br />
<br />
===Reaction dynamics===<br />
====Mechanism of energy release====<br />
From the Momenta vs Time plot (Figure 14) of the reactive trajectories for the F+ H<sub>2</sub> system, it is observed that the system moves faster and has greater oscillations. This shows that the loss in potential energy associated with the reaction is converted to translational and vibrational kinetic energy as shown in the oscillating momentum. The kinetic energy gained in the system would then be converted to heat and released to its surroundings. Experimentally, this could be determined by measuring a change in temperature. The bomb calorimetric method- although useful in direct measurement of the increase in temperature as a result of gain in kinetic energy- is unable to distinguish between the 2 forms of kinetic energy.<br />
<br />
A better alternative would be to perform an infrared chemiluminescence experiment. The intensities of the IR emission lines in the emission spectrum from the vibrationally excited molecules can then be used to measure the relative populations of the vibrational states of the product molecules. IR absorption spectroscopy would also be useful in analysing the vibrational states of the products. From the figures shown below, the products have higher oscillations than the reactants, showing that they are more highly vibrating. Thus, overtones could be observed in the absorption spectroscopy as a result of an increased population of the vibrational excited states of the products. <br />
<br />
<gallery mode=packed class=center widths="300 px" heights="300 px"><br />
File:xjg18_momentaRD.png|'''Figure 14 Momenta vs Time plot of F + H<sub>2</sub> with initial conditions r<sub>FH</sub>= 194 pm, r<sub>HH</sub>= 74 pm, p<sub>FH</sub>= -2&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, p<sub>HH</sub>= 0&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>'''<br />
</gallery><br />
<br />
<br />
<br />
====Polanyi's empirical rule====<br />
The Polanyi's empirical rule states that for a reaction with early transition state (i.e an exothermic reaction), translational kinetic energy is more effective than the vibrational energy in overcoming the activation barrier, and vice versa for a reaction with late transition state (i.e. an endothermic reaction), provided that the system has enough total energy to overcome the barrier. This provides a better understanding of the dependance of reaction rate constants on the distribution of energy over the different modes motions of the reactants. <br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ F + H<sub>2</sub> --> HF + H <br />
|-<br />
! Case !! R<sub>HF</sub> !! R<sub>HH</sub> !! Ļ<sub>HF</sub> !! Ļ<sub>HH</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 74 pm || -1.0 || -3 || [[File:xjg18_case1.png|400px]] <br />
|-<br />
| 2 || 190 pm || 74 pm || -1.0 || 5.6 || [[File:xjg18_case2.png|400px]] <br />
|-<br />
| 3 || 190 pm || 74 pm || -1.6 || 0.2 || [[File:xjg18_case3.png|400px]] <br />
|}<br />
<br />
For the F + H<sub>2</sub> exothermic reaction, several trajectories have been generated as shown in the table above, where Ļ<sub>HF</sub> is the translational momentum of atom F approaching the H<sub>2</sub> molecule, and Ļ<sub>HH</sub> is the vibrational momentum of the H<sub>2</sub> molecule. In case 1 and 2, the singular F atom has a relatively low translational energy while the H<sub>2</sub> molecule posses high vibrational energies. In both cases, the trajectories are unreactive. In case 3, the kinetic energy of the approaching F atom is increased while the vibrational motion of the H<sub>2</sub> molecule is significantly lowered. The initial conditions in case 3 led to a reactive trajectory successfully forming products. Thus, the results shown are in accordance with Polanyi's rule. This reaction is an exothermic reaction with an early transition state, therefore an increase in translational energy of a system is more effective in allowing the crossing of the barrier to form products.<br />
<br />
{| class="wikitable" style="margin: 1em auto 1em auto"<br />
|+ H + HF --> H<sub>2</sub> + F<br />
|-<br />
! Case !! R<sub>HH</sub> !! R<sub>HF</sub> !! Ļ<sub>HH</sub> !! Ļ<sub>HF</sub> !! Contour Plot <br />
|-<br />
| 1 || 190 pm || 90 pm || -10 || -0.1 || [[File:xjg18_case4.png|400px]] <br />
|-<br />
| 2 || 190 pm || 90 pm || -1.0 || 21 || [[File:xjg18_case5.png|400px]] <br />
|-<br />
| 3 || 190 pm || 90 pm || -1.0 || -26 || [[File:xjg18_case6.png|400px]] <br />
|}<br />
<br />
For the H + HF endothermic reaction, the Ļ<sub>HH</sub> is the translational momentum of atom H approaching the HF molecule, and Ļ<sub>HF</sub> is the vibrational momentum of the HF molecule. In case 1, the singular H atom has a relatively high translational energy while the H<sub>2</sub> molecule posses low vibrational energies. This results in an unreactive trajectory. In case 2 and 3, the kinetic energy of the approaching H atom is decreased while the vibrational motion of the H<sub>2</sub> molecule is significantly increased. The initial conditions in both cases led to reactive trajectories, successfully forming products. These results, again, follow Polanyi's rule. This reaction is an endothermic reaction with a late transition state, therefore vibrational kinetic energy plays a more important role in overcoming of the activation barrier to form products.<br />
<br />
<br />
==Bibliography==<br />
1. H. Bernhard Schlegel, Optimization of equilibrium geometries and transition structures, J. Comput. Chem., 1982, 3(2), pp. 214-218.<br />
<br />
2. I. N. Levine, ''Physical Chemistry'', McGraw-Hill, 6th edition, 2009, ch. 22.<br />
<br />
3. J. C. Polanyi and W. H. Wong, ''J. Chem. Phys.'', 1969, '''51'''(4), pp 1439-1450.<br />
<br />
4. K. J. Laidler, Chemical Kinetics, 1951, 55 (5), pp 759-760</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01340400&diff=806214MRD:013404002020-05-18T20:44:25Z<p>Ng611: /* Modelling a H + H2 system */</p>
<hr />
<div><br />
== Modelling a H + H<sub>2</sub> system ==<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:44, 18 May 2020 (BST) 3/5 - in general, a good piece of work. It's a shame that you didn't manage to do the final section of the report though. More specific feedback is given throughout.<br />
<br />
=== Potential Energy Surfaces ===<br />
Chemical reactions can be modelled by using a potential energy surface diagram, which relates the potential energy to the positions of atoms in space. For a system involving three atoms, like the H + H<sub>2</sub> system, the distances between the atoms can be described by the bond lengths between atoms A-B and B-C. This is shown for a successful collision of such a system by figure 1. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) It is only possible to represent this system as a surface if we make one key assumption. What is it? Think about the angles of collision...<br />
<br />
[[file:Successful_Reaction_HHH01340400.png|thumb|center|Figure 1: A potential energy surface diagram describing a successful reaction between a H atom and a H<sub>2</sub> molecule. The path taken by the reaction is denoted by the black line and goes over the Transition State where bond lengths A-B and B-C are equal]]<br />
====Transition States on Potential Energy Surface Diagrams====<br />
The potential energy of a transition state can be defined mathematically by the highest point along the lowest energy route between reactants and products on a potential energy surface. That is, it is a saddle point on a 3-dimensional surface where the relative distances between atoms make up the XY plane and the Z axis defines the potential energy of the system. The saddle point of a surface has the property āV(r<sub>i</sub>)/ār<sub>i</sub>=0 or that it has a gradient of 0. In order to distinguish the transition state from local minima by differentiating the potential energy with respect to the x axis and then with respect to the y axis and multiplying them together. If it is a saddle point the results should be above and below zero respectively. When simulating a chemical reaction, the transition state can be easily estimated for symmetric systems, that is systems where the transition state lies upon a mirror plane of the potential surface between the reactant region and the product region. Whilst using this method, it was estimated that the TS bond length of each bond in H-H-H was 91 pm.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) What directions on the PES do X and Y correspond to here?<br />
<br />
[[file:HHH_TS_Estimate.png|thumb|center|Figure 2: A distance vs. time plot of the H + H<sub>2</sub> system near the transition state with no initial forces acting on the system. This estimate describes an initial bond length of 91 pm for both A-B and B-C where A, B and C are the respective Hydrogen atoms of the system. This estimate is not exact as at the exact position of the transition state there should be no vibrational energy in the system, and therefore the distance should be unchanging with time.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:29, 18 May 2020 (BST) Very very true. You're off by <1 pm though, so it's not a significant issue. How could you use the MEP calculation to more accurately estimate the TS?<br />
<br />
====Minimum Energy Path (MEP)====<br />
The minimum energy path (MEP) is a path along the valley floor of the potential energy surface. The MEP disregards any vibrational energy the molecule may have and so it outlines the minimum point at each step along the reaction path. The Dynamic pathway is distinct from the MEP as it includes the vibrations of a molecule. Therefore the pathways shown by each simulation differ, as the dynamic pathway will oscillate up and down the valley walls whilst the MEP will stay strictly at the lowest local point at all times (figure 3). Whilst simulating the MEP, a comparison between the internuclear distance and momenta against time shows that although the internuclear distance is increasing over time, the momenta of the atoms is constant at zero. This is because the MEP resets the momentum of the atoms to zero at each step of the simulation, which is how the program will negate any vibrational energy in the system. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:31, 18 May 2020 (BST) Good!<br />
<br />
{| class="wikitable" border="1"<br />
|+ Figure 3<br />
! colspan="2" style="text-align: center;"|Comparison of the MEP and Dynamic pathways with slight displacement of the transition state<br />
|-<br />
! MEP !! Dynamic Pathway<br />
|-<br />
| [[file:MEP_HHH01340400.png|thumb|center|The Minimum Energy Pathway, MEP, shows the reaction pathway with no vibrational energy. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] || [[file:Dynamic_Pathway_HHH01340400.png|thumb|center|The Dynamic pathway, here shown on a contour plot, shows the molecule has vibrational energy and thus will oscillate along the valley walls as the system. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] <br />
|}<br />
<br />
====Reactive and Unreactive Trajectories====<br />
Not all collisions between molecular Hydrogen and atomic Hydrogen will result in a successful reaction. Whether a collision is successful or not is dependant on the relative magnitudes of momentum (kinetic energy) of each species. Below is a series of different initial trajectories from the same coordinates (figure 4). This table demonstrates how the total energy of the system is not a factor in the successful collision between the reacting species. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:36, 18 May 2020 (BST) Slightly misleading here. The total energy *is* a factor here, but it isn't the *only* factor. In general, not having enough energy means you absolutely will not cross the TS (neglecting quantum effects), but having enough energy doesn't guarantee that you will cross the TS. <br />
<br />
{| class="wikitable" border=1<br />
|+ Figure 4<br />
! p<sub>1</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> / kJmol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| style="text-align: center;"|-2.56 || style="text-align: center;"|-5.1 || style="text-align: center;"|-414.280 || style="text-align: center;"|Yes || style="text-align: center;"|There is little to no visible vibrational energy in the system until after the system passes through the transition state. || [[file:Contour_Plot1_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-4.1 || style="text-align: center;"|-420.077 || style="text-align: center;"|No || style="text-align: center;"|The system begins with vibrational energy as shown by the oscillation of the trajectory along the walls of the potential well, however the Hydrogen atom (Atom A) didnt have enough kinetic energy for a successful reaction to occur. Even though the total energy is higher than the previous simulation, the relative directions and magnitudes of the momenta of each of the reacting species means a reaction is not possible. The Hydrogen molecule, H<sub>2 BC</sub> has momentum in the same direction as the Hydrogen atom, H<sub>A</sub>, and therefore the relative total colliding momentum is diminished. || [[file:Contour_Plot2_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-5.1 || style="text-align: center;"|-413.977 || style="text-align: center;"| Yes|| style="text-align: center;"| There is an initial vibrational energy to the H<sub>2 AB</sub> molecule and after the the transition state has been crossed, the resultant H<sub>2 BC</sub> molecule has more vibrational energy, as indicated by the larger oscillations. || [[file:Contour_Plot3_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.1 || style="text-align: center;"| -357..277|| style="text-align: center;"| No|| style="text-align: center;"| Initially the molecular Hydrogen species, H<sub>2 AB</sub>, has little to no visible vibrational energy as the dynamic simulation shows no oscillation along the potential energy surface. The system then crosses over the transition state and forms a new bond between H<sub>B</sub> and H<sub>C</sub>. However the system reverts back to moleculat H<sub>2 AB</sub> and H<sub>C</sub>. || [[file:Contour_Plot4_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.6 || style="text-align: center;"|-349.477 || style="text-align: center;"| Yes|| style="text-align: center;"| Initially the molecular Hydrogen species H<sub>2 AB</sub> has little to no visible vibrational energy. After the system crosses the transition state, the system proceeds to vibrate rigorously and break the BC bond to temporarily reform the original AB bond. After that the system again crosses the transition state to form molecular H<sub>2 BC</sub>. || [[file:Contour_Plot5_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|}<br />
<br />
====Transition State Theory====<br />
Transition State Theory (TST) is a powerful theory which helps predict the reaction rate for a given process. Unlike the Arrhenius Equation, TST takes into account temperature dependancy in the form of the pre-exponential values R<sub>Cl</sub> and the partition functions. TST is an imperfect approximation of the reaction dynamics of a system, as it makes the following assumptions: the kinetic energy along the reaction coordinate follows the Boltzman Distribution; all reaction trajectories with an energy greater than the activation energy will be reactive; the motion of the system over the transition state is considered classically, and therefore doesn't take into account quantum effects such as tunnelling or quantisation of energy states; once a reaction has crossed over the transition state it cannot cross back over again; and finally that the reactants are in equilibrium with the transition state structure. <br />
<br />
These assumptions mean that TST will overestimate the rate of a reaction when compared to experimental results, since a reaction system can cross over the transition state multiple times. Also, collisions which do initially cross over the transition state are counted as successful in TST but they may be counted towards the rate as a successful collision.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:38, 18 May 2020 (BST) Excellent, and absolutely correct! If we were to include tunelling effects, how would it alter the rate and how significant would it be compared to TS recrossing?<br />
<br />
==Modelling F-H-H Systems==<br />
The reaction of a Fluorine atom with a H<sub>2</sub> molecule to form HF + H is exothermic, whilst the reaction of Hydrogen Fluoride with a Hydrogen atom to form F + H<sub>2</sub> is endothermic. This is seen in figure 5, which shows a potential energy surface for the F-H-H system. The lower potential energy seen at small AB distances (F-H bond formed) indicates that the F-H bond is stronger and more stable than that of the H-H bond. The formation of a more stable species will result in energy being dispersed to the surroundings, whilst the formation of a less stable species will require energy to promote that species to the higher energy level. <br />
<br />
The energy barrier of the transition state in this diagram is relatively small, and is therefore difficult to discern by eye. However, using the information provided by the Hessian it can be estimated that the transition state occurs at F-H 181 pm and H-H 75 pm. Using Hammond's Postulate, we can infer that the structure of the transition state is similar to the structure of H<sub>2</sub>, since they are energetically so similar. The energy barrier for the reaction of H<sub>2</sub> with F is 0.071 kJ.mol<sup>-1</sup> and the energy barrier for the reaction of HF with H is 124.618 kJ.mol<sup>-1</sup>. <br />
<br />
[[file:FHH_Surface_Plot_01340400.png|thumb|center|Figure 5: A potential energy surface diagram for the F-H-H system.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:41, 18 May 2020 (BST) Your F+H2->H+HF barrier is somewhat too small, which suggests either your estimate for the TS is off, or that you've not separated the molecules enough. <br />
<br />
<br />
====Reaction Dynamics====<br />
A successful reaction between F and H<sub>2</sub> forms Hydrogen Fluoride and a Hydrogen atom can result in the HF molecule having a large vibrational energy. The molecule disperses this energy by emission of radiation to its surroundings via radiative decay. Figure 6 shows how the HF molecule (AB) retains a significant amount of momentum after the reaction is complete. <br />
[[file:F_H2_Reaction_01340400.png|thumb|center|Figure 6: Momentum vs Time graph for a Fluorine atom successfully reacting with a H<sub>2</sub> molecule. Initial conditions are: AB (F-H) Distance 230 pm, BC (H-H) Distance 74 pm, AB momentum -2 g.mol<sup>-1</sup>pm.fs<sup>-1</sup>.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:43, 18 May 2020 (BST) Good start. How would you confirm this experimentally?<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:43, 18 May 2020 (BST) Missing the final section on Polanyi's rules unfortunately...</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01340400&diff=806213MRD:013404002020-05-18T20:43:07Z<p>Ng611: /* Reaction Dynamics */</p>
<hr />
<div><br />
== Modelling a H + H<sub>2</sub> system ==<br />
=== Potential Energy Surfaces ===<br />
Chemical reactions can be modelled by using a potential energy surface diagram, which relates the potential energy to the positions of atoms in space. For a system involving three atoms, like the H + H<sub>2</sub> system, the distances between the atoms can be described by the bond lengths between atoms A-B and B-C. This is shown for a successful collision of such a system by figure 1. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) It is only possible to represent this system as a surface if we make one key assumption. What is it? Think about the angles of collision...<br />
<br />
[[file:Successful_Reaction_HHH01340400.png|thumb|center|Figure 1: A potential energy surface diagram describing a successful reaction between a H atom and a H<sub>2</sub> molecule. The path taken by the reaction is denoted by the black line and goes over the Transition State where bond lengths A-B and B-C are equal]]<br />
====Transition States on Potential Energy Surface Diagrams====<br />
The potential energy of a transition state can be defined mathematically by the highest point along the lowest energy route between reactants and products on a potential energy surface. That is, it is a saddle point on a 3-dimensional surface where the relative distances between atoms make up the XY plane and the Z axis defines the potential energy of the system. The saddle point of a surface has the property āV(r<sub>i</sub>)/ār<sub>i</sub>=0 or that it has a gradient of 0. In order to distinguish the transition state from local minima by differentiating the potential energy with respect to the x axis and then with respect to the y axis and multiplying them together. If it is a saddle point the results should be above and below zero respectively. When simulating a chemical reaction, the transition state can be easily estimated for symmetric systems, that is systems where the transition state lies upon a mirror plane of the potential surface between the reactant region and the product region. Whilst using this method, it was estimated that the TS bond length of each bond in H-H-H was 91 pm.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) What directions on the PES do X and Y correspond to here?<br />
<br />
[[file:HHH_TS_Estimate.png|thumb|center|Figure 2: A distance vs. time plot of the H + H<sub>2</sub> system near the transition state with no initial forces acting on the system. This estimate describes an initial bond length of 91 pm for both A-B and B-C where A, B and C are the respective Hydrogen atoms of the system. This estimate is not exact as at the exact position of the transition state there should be no vibrational energy in the system, and therefore the distance should be unchanging with time.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:29, 18 May 2020 (BST) Very very true. You're off by <1 pm though, so it's not a significant issue. How could you use the MEP calculation to more accurately estimate the TS?<br />
<br />
====Minimum Energy Path (MEP)====<br />
The minimum energy path (MEP) is a path along the valley floor of the potential energy surface. The MEP disregards any vibrational energy the molecule may have and so it outlines the minimum point at each step along the reaction path. The Dynamic pathway is distinct from the MEP as it includes the vibrations of a molecule. Therefore the pathways shown by each simulation differ, as the dynamic pathway will oscillate up and down the valley walls whilst the MEP will stay strictly at the lowest local point at all times (figure 3). Whilst simulating the MEP, a comparison between the internuclear distance and momenta against time shows that although the internuclear distance is increasing over time, the momenta of the atoms is constant at zero. This is because the MEP resets the momentum of the atoms to zero at each step of the simulation, which is how the program will negate any vibrational energy in the system. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:31, 18 May 2020 (BST) Good!<br />
<br />
{| class="wikitable" border="1"<br />
|+ Figure 3<br />
! colspan="2" style="text-align: center;"|Comparison of the MEP and Dynamic pathways with slight displacement of the transition state<br />
|-<br />
! MEP !! Dynamic Pathway<br />
|-<br />
| [[file:MEP_HHH01340400.png|thumb|center|The Minimum Energy Pathway, MEP, shows the reaction pathway with no vibrational energy. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] || [[file:Dynamic_Pathway_HHH01340400.png|thumb|center|The Dynamic pathway, here shown on a contour plot, shows the molecule has vibrational energy and thus will oscillate along the valley walls as the system. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] <br />
|}<br />
<br />
====Reactive and Unreactive Trajectories====<br />
Not all collisions between molecular Hydrogen and atomic Hydrogen will result in a successful reaction. Whether a collision is successful or not is dependant on the relative magnitudes of momentum (kinetic energy) of each species. Below is a series of different initial trajectories from the same coordinates (figure 4). This table demonstrates how the total energy of the system is not a factor in the successful collision between the reacting species. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:36, 18 May 2020 (BST) Slightly misleading here. The total energy *is* a factor here, but it isn't the *only* factor. In general, not having enough energy means you absolutely will not cross the TS (neglecting quantum effects), but having enough energy doesn't guarantee that you will cross the TS. <br />
<br />
{| class="wikitable" border=1<br />
|+ Figure 4<br />
! p<sub>1</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> / kJmol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| style="text-align: center;"|-2.56 || style="text-align: center;"|-5.1 || style="text-align: center;"|-414.280 || style="text-align: center;"|Yes || style="text-align: center;"|There is little to no visible vibrational energy in the system until after the system passes through the transition state. || [[file:Contour_Plot1_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-4.1 || style="text-align: center;"|-420.077 || style="text-align: center;"|No || style="text-align: center;"|The system begins with vibrational energy as shown by the oscillation of the trajectory along the walls of the potential well, however the Hydrogen atom (Atom A) didnt have enough kinetic energy for a successful reaction to occur. Even though the total energy is higher than the previous simulation, the relative directions and magnitudes of the momenta of each of the reacting species means a reaction is not possible. The Hydrogen molecule, H<sub>2 BC</sub> has momentum in the same direction as the Hydrogen atom, H<sub>A</sub>, and therefore the relative total colliding momentum is diminished. || [[file:Contour_Plot2_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-5.1 || style="text-align: center;"|-413.977 || style="text-align: center;"| Yes|| style="text-align: center;"| There is an initial vibrational energy to the H<sub>2 AB</sub> molecule and after the the transition state has been crossed, the resultant H<sub>2 BC</sub> molecule has more vibrational energy, as indicated by the larger oscillations. || [[file:Contour_Plot3_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.1 || style="text-align: center;"| -357..277|| style="text-align: center;"| No|| style="text-align: center;"| Initially the molecular Hydrogen species, H<sub>2 AB</sub>, has little to no visible vibrational energy as the dynamic simulation shows no oscillation along the potential energy surface. The system then crosses over the transition state and forms a new bond between H<sub>B</sub> and H<sub>C</sub>. However the system reverts back to moleculat H<sub>2 AB</sub> and H<sub>C</sub>. || [[file:Contour_Plot4_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.6 || style="text-align: center;"|-349.477 || style="text-align: center;"| Yes|| style="text-align: center;"| Initially the molecular Hydrogen species H<sub>2 AB</sub> has little to no visible vibrational energy. After the system crosses the transition state, the system proceeds to vibrate rigorously and break the BC bond to temporarily reform the original AB bond. After that the system again crosses the transition state to form molecular H<sub>2 BC</sub>. || [[file:Contour_Plot5_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|}<br />
<br />
====Transition State Theory====<br />
Transition State Theory (TST) is a powerful theory which helps predict the reaction rate for a given process. Unlike the Arrhenius Equation, TST takes into account temperature dependancy in the form of the pre-exponential values R<sub>Cl</sub> and the partition functions. TST is an imperfect approximation of the reaction dynamics of a system, as it makes the following assumptions: the kinetic energy along the reaction coordinate follows the Boltzman Distribution; all reaction trajectories with an energy greater than the activation energy will be reactive; the motion of the system over the transition state is considered classically, and therefore doesn't take into account quantum effects such as tunnelling or quantisation of energy states; once a reaction has crossed over the transition state it cannot cross back over again; and finally that the reactants are in equilibrium with the transition state structure. <br />
<br />
These assumptions mean that TST will overestimate the rate of a reaction when compared to experimental results, since a reaction system can cross over the transition state multiple times. Also, collisions which do initially cross over the transition state are counted as successful in TST but they may be counted towards the rate as a successful collision.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:38, 18 May 2020 (BST) Excellent, and absolutely correct! If we were to include tunelling effects, how would it alter the rate and how significant would it be compared to TS recrossing?<br />
<br />
==Modelling F-H-H Systems==<br />
The reaction of a Fluorine atom with a H<sub>2</sub> molecule to form HF + H is exothermic, whilst the reaction of Hydrogen Fluoride with a Hydrogen atom to form F + H<sub>2</sub> is endothermic. This is seen in figure 5, which shows a potential energy surface for the F-H-H system. The lower potential energy seen at small AB distances (F-H bond formed) indicates that the F-H bond is stronger and more stable than that of the H-H bond. The formation of a more stable species will result in energy being dispersed to the surroundings, whilst the formation of a less stable species will require energy to promote that species to the higher energy level. <br />
<br />
The energy barrier of the transition state in this diagram is relatively small, and is therefore difficult to discern by eye. However, using the information provided by the Hessian it can be estimated that the transition state occurs at F-H 181 pm and H-H 75 pm. Using Hammond's Postulate, we can infer that the structure of the transition state is similar to the structure of H<sub>2</sub>, since they are energetically so similar. The energy barrier for the reaction of H<sub>2</sub> with F is 0.071 kJ.mol<sup>-1</sup> and the energy barrier for the reaction of HF with H is 124.618 kJ.mol<sup>-1</sup>. <br />
<br />
[[file:FHH_Surface_Plot_01340400.png|thumb|center|Figure 5: A potential energy surface diagram for the F-H-H system.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:41, 18 May 2020 (BST) Your F+H2->H+HF barrier is somewhat too small, which suggests either your estimate for the TS is off, or that you've not separated the molecules enough. <br />
<br />
<br />
====Reaction Dynamics====<br />
A successful reaction between F and H<sub>2</sub> forms Hydrogen Fluoride and a Hydrogen atom can result in the HF molecule having a large vibrational energy. The molecule disperses this energy by emission of radiation to its surroundings via radiative decay. Figure 6 shows how the HF molecule (AB) retains a significant amount of momentum after the reaction is complete. <br />
[[file:F_H2_Reaction_01340400.png|thumb|center|Figure 6: Momentum vs Time graph for a Fluorine atom successfully reacting with a H<sub>2</sub> molecule. Initial conditions are: AB (F-H) Distance 230 pm, BC (H-H) Distance 74 pm, AB momentum -2 g.mol<sup>-1</sup>pm.fs<sup>-1</sup>.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:43, 18 May 2020 (BST) Good start. How would you confirm this experimentally?<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:43, 18 May 2020 (BST) Missing the final section on Polanyi's rules unfortunately...</div>Ng611https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01340400&diff=806212MRD:013404002020-05-18T20:41:00Z<p>Ng611: /* Modelling F-H-H Systems */</p>
<hr />
<div><br />
== Modelling a H + H<sub>2</sub> system ==<br />
=== Potential Energy Surfaces ===<br />
Chemical reactions can be modelled by using a potential energy surface diagram, which relates the potential energy to the positions of atoms in space. For a system involving three atoms, like the H + H<sub>2</sub> system, the distances between the atoms can be described by the bond lengths between atoms A-B and B-C. This is shown for a successful collision of such a system by figure 1. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) It is only possible to represent this system as a surface if we make one key assumption. What is it? Think about the angles of collision...<br />
<br />
[[file:Successful_Reaction_HHH01340400.png|thumb|center|Figure 1: A potential energy surface diagram describing a successful reaction between a H atom and a H<sub>2</sub> molecule. The path taken by the reaction is denoted by the black line and goes over the Transition State where bond lengths A-B and B-C are equal]]<br />
====Transition States on Potential Energy Surface Diagrams====<br />
The potential energy of a transition state can be defined mathematically by the highest point along the lowest energy route between reactants and products on a potential energy surface. That is, it is a saddle point on a 3-dimensional surface where the relative distances between atoms make up the XY plane and the Z axis defines the potential energy of the system. The saddle point of a surface has the property āV(r<sub>i</sub>)/ār<sub>i</sub>=0 or that it has a gradient of 0. In order to distinguish the transition state from local minima by differentiating the potential energy with respect to the x axis and then with respect to the y axis and multiplying them together. If it is a saddle point the results should be above and below zero respectively. When simulating a chemical reaction, the transition state can be easily estimated for symmetric systems, that is systems where the transition state lies upon a mirror plane of the potential surface between the reactant region and the product region. Whilst using this method, it was estimated that the TS bond length of each bond in H-H-H was 91 pm.<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:27, 18 May 2020 (BST) What directions on the PES do X and Y correspond to here?<br />
<br />
[[file:HHH_TS_Estimate.png|thumb|center|Figure 2: A distance vs. time plot of the H + H<sub>2</sub> system near the transition state with no initial forces acting on the system. This estimate describes an initial bond length of 91 pm for both A-B and B-C where A, B and C are the respective Hydrogen atoms of the system. This estimate is not exact as at the exact position of the transition state there should be no vibrational energy in the system, and therefore the distance should be unchanging with time.]]<br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:29, 18 May 2020 (BST) Very very true. You're off by <1 pm though, so it's not a significant issue. How could you use the MEP calculation to more accurately estimate the TS?<br />
<br />
====Minimum Energy Path (MEP)====<br />
The minimum energy path (MEP) is a path along the valley floor of the potential energy surface. The MEP disregards any vibrational energy the molecule may have and so it outlines the minimum point at each step along the reaction path. The Dynamic pathway is distinct from the MEP as it includes the vibrations of a molecule. Therefore the pathways shown by each simulation differ, as the dynamic pathway will oscillate up and down the valley walls whilst the MEP will stay strictly at the lowest local point at all times (figure 3). Whilst simulating the MEP, a comparison between the internuclear distance and momenta against time shows that although the internuclear distance is increasing over time, the momenta of the atoms is constant at zero. This is because the MEP resets the momentum of the atoms to zero at each step of the simulation, which is how the program will negate any vibrational energy in the system. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:31, 18 May 2020 (BST) Good!<br />
<br />
{| class="wikitable" border="1"<br />
|+ Figure 3<br />
! colspan="2" style="text-align: center;"|Comparison of the MEP and Dynamic pathways with slight displacement of the transition state<br />
|-<br />
! MEP !! Dynamic Pathway<br />
|-<br />
| [[file:MEP_HHH01340400.png|thumb|center|The Minimum Energy Pathway, MEP, shows the reaction pathway with no vibrational energy. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] || [[file:Dynamic_Pathway_HHH01340400.png|thumb|center|The Dynamic pathway, here shown on a contour plot, shows the molecule has vibrational energy and thus will oscillate along the valley walls as the system. Simulation begins at 1 pm displacement from the estimated transition state in the AB direction.]] <br />
|}<br />
<br />
====Reactive and Unreactive Trajectories====<br />
Not all collisions between molecular Hydrogen and atomic Hydrogen will result in a successful reaction. Whether a collision is successful or not is dependant on the relative magnitudes of momentum (kinetic energy) of each species. Below is a series of different initial trajectories from the same coordinates (figure 4). This table demonstrates how the total energy of the system is not a factor in the successful collision between the reacting species. <br />
<br />
[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:36, 18 May 2020 (BST) Slightly misleading here. The total energy *is* a factor here, but it isn't the *only* factor. In general, not having enough energy means you absolutely will not cross the TS (neglecting quantum effects), but having enough energy doesn't guarantee that you will cross the TS. <br />
<br />
{| class="wikitable" border=1<br />
|+ Figure 4<br />
! p<sub>1</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub> /&nbsp;g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub> / kJmol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory<br />
|-<br />
| style="text-align: center;"|-2.56 || style="text-align: center;"|-5.1 || style="text-align: center;"|-414.280 || style="text-align: center;"|Yes || style="text-align: center;"|There is little to no visible vibrational energy in the system until after the system passes through the transition state. || [[file:Contour_Plot1_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-4.1 || style="text-align: center;"|-420.077 || style="text-align: center;"|No || style="text-align: center;"|The system begins with vibrational energy as shown by the oscillation of the trajectory along the walls of the potential well, however the Hydrogen atom (Atom A) didnt have enough kinetic energy for a successful reaction to occur. Even though the total energy is higher than the previous simulation, the relative directions and magnitudes of the momenta of each of the reacting species means a reaction is not possible. The Hydrogen molecule, H<sub>2 BC</sub> has momentum in the same direction as the Hydrogen atom, H<sub>A</sub>, and therefore the relative total colliding momentum is diminished. || [[file:Contour_Plot2_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-3.1 || style="text-align: center;"|-5.1 || style="text-align: center;"|-413.977 || style="text-align: center;"| Yes|| style="text-align: center;"| There is an initial vibrational energy to the H<sub>2 AB</sub> molecule and after the the transition state has been crossed, the resultant H<sub>2 BC</sub> molecule has more vibrational energy, as indicated by the larger oscillations. || [[file:Contour_Plot3_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.1 || style="text-align: center;"| -357..277|| style="text-align: center;"| No|| style="text-align: center;"| Initially the molecular Hydrogen species, H<sub>2 AB</sub>, has little to no visible vibrational energy as the dynamic simulation shows no oscillation along the potential energy surface. The system then crosses over the transition state and forms a new bond between H<sub>B</sub> and H<sub>C</sub>. However the system reverts back to moleculat H<sub>2 AB</sub> and H<sub>C</sub>. || [[file:Contour_Plot4_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
|-<br />
| style="text-align: center;"|-5.1 || style="text-align: center;"|-10.6 || style="text-align: center;"|-349.477 || style="text-align: center;"| Yes|| style="text-align: center;"| Initially the molecular Hydrogen species H<sub>2 AB</sub> has little to no visible vibrational energy. After the system crosses the transition state, the system proceeds to vibrate rigorously and break the BC bond to temporarily reform the original AB bond. After that the system again crosses the transition state to form molecular H<sub>2 BC</sub>. || [[file:Contour_Plot5_01340400.png|thumb|center|200px|Contour plot of a dynamic simulation of H + H<sub>2</sub>.]]<br />
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====Transition State Theory====<br />
Transition State Theory (TST) is a powerful theory which helps predict the reaction rate for a given process. Unlike the Arrhenius Equation, TST takes into account temperature dependancy in the form of the pre-exponential values R<sub>Cl</sub> and the partition functions. TST is an imperfect approximation of the reaction dynamics of a system, as it makes the following assumptions: the kinetic energy along the reaction coordinate follows the Boltzman Distribution; all reaction trajectories with an energy greater than the activation energy will be reactive; the motion of the system over the transition state is considered classically, and therefore doesn't take into account quantum effects such as tunnelling or quantisation of energy states; once a reaction has crossed over the transition state it cannot cross back over again; and finally that the reactants are in equilibrium with the transition state structure. <br />
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These assumptions mean that TST will overestimate the rate of a reaction when compared to experimental results, since a reaction system can cross over the transition state multiple times. Also, collisions which do initially cross over the transition state are counted as successful in TST but they may be counted towards the rate as a successful collision.<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:38, 18 May 2020 (BST) Excellent, and absolutely correct! If we were to include tunelling effects, how would it alter the rate and how significant would it be compared to TS recrossing?<br />
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==Modelling F-H-H Systems==<br />
The reaction of a Fluorine atom with a H<sub>2</sub> molecule to form HF + H is exothermic, whilst the reaction of Hydrogen Fluoride with a Hydrogen atom to form F + H<sub>2</sub> is endothermic. This is seen in figure 5, which shows a potential energy surface for the F-H-H system. The lower potential energy seen at small AB distances (F-H bond formed) indicates that the F-H bond is stronger and more stable than that of the H-H bond. The formation of a more stable species will result in energy being dispersed to the surroundings, whilst the formation of a less stable species will require energy to promote that species to the higher energy level. <br />
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The energy barrier of the transition state in this diagram is relatively small, and is therefore difficult to discern by eye. However, using the information provided by the Hessian it can be estimated that the transition state occurs at F-H 181 pm and H-H 75 pm. Using Hammond's Postulate, we can infer that the structure of the transition state is similar to the structure of H<sub>2</sub>, since they are energetically so similar. The energy barrier for the reaction of H<sub>2</sub> with F is 0.071 kJ.mol<sup>-1</sup> and the energy barrier for the reaction of HF with H is 124.618 kJ.mol<sup>-1</sup>. <br />
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[[file:FHH_Surface_Plot_01340400.png|thumb|center|Figure 5: A potential energy surface diagram for the F-H-H system.]]<br />
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[[User:Ng611|Ng611]] ([[User talk:Ng611|talk]]) 21:41, 18 May 2020 (BST) Your F+H2->H+HF barrier is somewhat too small, which suggests either your estimate for the TS is off, or that you've not separated the molecules enough. <br />
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====Reaction Dynamics====<br />
A successful reaction between F and H<sub>2</sub> forms Hydrogen Fluoride and a Hydrogen atom can result in the HF molecule having a large vibrational energy. The molecule disperses this energy by emission of radiation to its surroundings via radiative decay. Figure 6 shows how the HF molecule (AB) retains a significant amount of momentum after the reaction is complete. <br />
[[file:F_H2_Reaction_01340400.png|thumb|center|Figure 6: Momentum vs Time graph for a Fluorine atom successfully reacting with a H<sub>2</sub> molecule. Initial conditions are: AB (F-H) Distance 230 pm, BC (H-H) Distance 74 pm, AB momentum -2 g.mol<sup>-1</sup>pm.fs<sup>-1</sup>.]]</div>Ng611