View source for MRD:av4217

==Exercise 1: H + H<sub>2<sub>==
===Transition State===
[[File:Surface_Plot_H+H2-av4217.png|thumb|right|Figure 1 - A potential energy surface diagram showing the maxima at the minimum energy pathway, which is the transition state.]]

====Potential energy surface diagram====
The transition state in a potential energy surface diagram is when the  partial differentiation of the potential energy with respect to r<sub>i</sub> is 0. The transition state is located at the saddle point of the potential energy surface diagram.At the transition state ∂V(r<sub>1</sub>)/∂r<sub>1</sub> >0 and ∂V(r<sub>2</sub>)/∂r<sub>2</sub> <0, or vice versa. When you travel along the minimum valley there will be a point where it is the maximum of the valley, this is the transition state. If a trajectory is started at a transition state with no initial momentum it will remain there, with a small change in geometry it will roll toward the reactants or products depending on the direction. The transition state can be identified in this way. At a local minimum it will not roll toward the products or reactants with a small change as it is a stable point. The maximum at the transition state is what allows the small change in geometry to cause the roll towards the products or reactants. The local minimum wont be a saddle point and will be a lower energy than the transition state.
{{fontcolor1|pink|Good explanation, but a good way to distinguish between the local maximum and minimum is by the second partial derivative which will be less than or more than zero, respectively. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:13, 15 May 2019 (BST)}}

====Locating the Transition State====
[[File:TS_dvst_graph-av4217.png|thumb|left|Figure 2 - Internuclear distance vs Time graph at transition state (r=0.903)]]

The best estimate for r<sub>ts</sub> is 0.903. At this distance the atoms oscillate a little but stay at equal distances to each other over a period of time as shown by figure two. The plot doesn't show the r<sub>AB</sub> getting smaller indicating they formed a bond and r<sub>BC</sub> getting larger showing C is a separate atom, or vice versa. The products or reactants are not formed so this is the transition state position, where no bonds are formed.
{{fontcolor1|pink|Good, but how did you get to this value for the position of the transition state?. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:16, 15 May 2019 (BST)}}

===Trajectory===
====MEP vs Dynamic====
When the calculation method is dynamic instead of MEP, the oscillation of the atom and molecules can be seen. While, for MEP it can be seen as a straight line following the valley. It doesnt take into account any instantaneous momentum.
[[File:Mep_trajectory_valley-av4217.png|thumb|right|Figure 3 - Minimum energy reaction path]]
[[File:Dynamic_trajectory-av4217.png|thumb|left|Figure 4 - Trajectory near transition state]]
{{fontcolor1|pink|Good comparison. Also, you can see that the momentum is zero for the MEP calculation using the momenta vs time plot. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:21, 15 May 2019 (BST)}}

====Changing initial conditions====
When the initial conditions are changed so r<sub>1</sub> = r<sub>ts</sub> and r<sub>2</sub> = r<sub>ts</sub> + 0.01, then reaction is the opposite. Atom C reacts with molecule AB to form BC and atom C.

[[File:Final_pos_mom-av4217.png|thumb|right|Figure 5 - Inter nuclear Distance vs Time Plot when the initial data is replaced with the final data]]


When the initial positions are swapped with the final positions and the momenta are changed with the opposite signs no reaction takes place. However the distance between the atom and the molecule is getting closer so in time they may collide and a reaction takes place

====Reactivity of Trajectories====
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! p<sub>1</sub> !! p<sub>2</sub> !! E<sub>tot</sub> !! Reactive? !! Description of dynamics
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:T1_traj-av4217.png|centre|300px]] The molecules don't oscillate much at the beginning. It passes the transition state and then collides to form the products. Once the products are formed there is more oscillating.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:T2_traj-av4217.png|center|300px]] The molecules dont reach the transition state, they don't go over the maximum and instead reflect back to their original path. They come close together but dont have enough energy react so move away from each other.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:T3_traj-av4217.png|center|300px]] The molecules oscillate more as they have greater momentum, they reach the transition state, follow the trajectory and react to form the products.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:T4_traj-av4217.png|center|300px]] The molecules reach the transition state and the product is formed. However the molecules pass back over the transition state and form the reactants again and remain that way.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:T51_traj-av4217.png|center|300px]] The molecules reach the transition state and collide to almost form products but then goes back to the transition state. It partially forms the reactants and then passes the transition state again and forms the products and remains that way.
|}
{{fontcolor1|pink|Good, but what can you summarise from this table in terms of reaction completion. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:38, 15 May 2019 (BST)}}

====Transition State Theory====
The transition state theory assumes the energies of the molecules follows the Boltzmann distribution and the reactants are in equilibrium with the transition state structure. Also, once the transition state is achieved the molecules do not collapse back into the reactant structures.<ref name="TST"/> As seen in penultimate example this is not true and the reactants can be formed again once the transition state is achieved. This suggests the transition state theory will predict a higher rate of reaction than the experimental values as it does not take into account the reactants reforming as seen in the last example. The reactants reforming will mean more time taken to form the products so the rate of reaction will be longer for the experimental values. If it is assumed the reactants are not reformed and the reaction only goes forward, the predicted rate would be higher than the experimental values.

== Exercise 2: F-H-H ==
===PES Inspection===

[[File:Exo_HHF-av4217.png|thumb|right|Figure 6 - surface plot of the reaction of H<sub>2</sub> with F.]]
The reaction of H<sub>2</sub> with F is exothermic. As seen in figure 6 the reactants are of a higher energy than the products so the reaction is exothermic.

[[File:Endo_FHH-av4217.png|thumb|left|Figure 7 - Surface plot of the reaction of HF with F.]]


























The reaction of HF with F is endothermic as seen by the surface plot as the products are of a higher energy than the reactants. So energy is absorbed.

This shows the HF bond is stronger than the H<sub>2</sub> bond as a large amount of energy is required to break it meaning the reaction is endothermic. Also a when the bond is made a large amount of energy is released causing the reaction to be exothermic. The large release of energy is greater then the energy required to break the H<sub>2</sub> since the reaction is exothermic. So it's stronger as it has greater bond energy.
{{fontcolor1|pink|Good explanation, next time you should include the bond energies for HF and H<sub>2</sub>. Also, when you use an acronym you should define it first, ie potential energy surface (PES). [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:45, 15 May 2019 (BST)}}


The transition state of both reactions are at r<sub>BC</sub>=0.745 and r<sub>AB</sub>=1.84 where atom A is F and atom B and C are H. Also, momentum is at 0

[[File:Ea_endo-av4217.png|thumb|left|Figure 8 - Energy vs Time diagram (MEP).]]


The activation energy for reaction of F + H<sub>2</sub> is around 0 as it's a spontaneous reaction. The the energy levels of the reactants and transition state are almost the same. The activation energy for the reaction of HF + H is around 29.6 kcalmol<sup>-1</sup>, it has a high activation energy since it's endothermic.
{{fontcolor1|pink|Good, but how did you get to those values for the position of the transition state? You should refer to Hammond postulate. Also, you should of attempted to calculate the activation energy for the exothermic reaction as you did for the endothermic reaction. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:51, 15 May 2019 (BST)}}

===Reaction Dynamics===
[[File:Momvst_con-av4217.png|thumb|right|Figure 9 -Momentum vs Time Graph]]


The energy released from the reaction is absorbed by the molecule and causes it to oscillate, changing its momentum. The molecule vibrates at a specific frequency therefore wavelength, which can be determined using emission spectroscopy.

When changing the momentum from -3<p<sub>HH</sub><3 while keeping the other values constant at a specific value it can be seen all the possibilities approach the transition state. This could be because as said energy above the activation is put into the system. Therefore, the system can overcome the  energy barrier and reach the transition state and produce the products. However in many cases the molecules collapse back into the reactants, after reaching the transition state again.

When increasing p<sub>FH</sub> to -0.8 and reducing p<sub>HH</sub> to 0.1 the reaction takes place. The molecules reach the transition state form the products and then collapse back to the reactant. Then the molecules form the products and remain that way following the trajectory to the valley.

For the HF + H reaction initial conditions of r<sub>HF</sub>=1.2, r<sub>HH</sub>=1.5, p<sub>HF</sub>=-0.9, and p<sub>HH</sub>=-0.1 will allow a reaction to take place and form the products.


Polyani discusses when the reaction is endothermic the transition state is close to the the products, so there is a late energy barrier. Therefore, greater vibrational energy in the reactants would more likely lead to a reaction and formation of products. With just translational energy, the atom gets too close to the molecule and repulsive forces causes the atom repel back. For an exothermic reaction translational energy is enough to overcome the early energy barrier (transition state closer to the reactants). The reactants have enough energy to cause a reaction and form products. This shows the mode of energy in the reactants is important for a successful reaction and this effected by the position of the transition state.<ref name="Polyani"/>
{{fontcolor1|pink|Good, but there are some things you need to explain further; how does emission spectroscopy determine the change in frequency for this reaction?, you didn't address the conservation of energy clearly and you should of shown your work for the reaction processes with different initial values as an example to Polanyi's rules. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:05, 15 May 2019 (BST)}}

==References==
<references>

<ref name="TST">P. Atkins, J. de Paula, Elements of Physical Chemistry, 5th ed.; Oxford University Press, 2009. </ref>

<ref name="Polyani">Jeffrey I.Seinfeld, Joseph S.Francisco, William L.Hase, Chemical Kinetics and Dynamics, Prentice Hall Inc, 1989.</ref>

</references>