= '''Molecular Dynamics Lab''' =

== '''Exercise 1ː H - H - H System''' ==

=== '''Dynamics from the transition state region''' ===
On a potential energy surface diagram, the transition state is mathematically defined as the maximum on the minimum energy path, so ∂V(ri)/∂ri=0 (the gradient of the potential, or the first derivative, is zero). However, it is not a simple local maximum. The transition state is a saddle point which means that while it is a local maximum in potential energy with respect to reaction coordinates, it is also a local minima with respect to the orthogonal coordinates. Therefore, in order to distinguish the transition state saddle point from a local minimum of maximum, the second partial derivative should be computed: H = fxx(x0,y0)fyy(x0,y0)-fxy(x0,y0)2. If H < 0, then (x0,y0) is a saddle point but if H > 0, then (x0,y0) is either a local maximum or minimum.

==== Trajectories from r1 = r2 : Locating the Transition State ====
The three atoms are the same (H) and therefore the transition state will be symmetric and the internuclear distances equal (r1 = r2). At the transition state point, as described above, the gradient of the potential is zero, and consequently a trajectory started exactly at the transition state, with no initial momentum, will remain there forever - this can be seen in the Contour Plot, where the trajectory is a single point. This means that the Internuclear Distances will not change (seen in the Internuclear Distance vs Time plot, showing distance as a constant) and the forces along r1 and r2 will be exactly 0 kJ mol-1 pm-1. Therefore to find the transition state position, different initial conditions with r1 = r2 and p1 = p2 = 0 g mol-1 pm fs-1 were tested to meet these provisions. The transition state was found when r1 = r2 = rts = 90.777 pm.
<nowiki> </nowiki><gallery>
File:Ms12218_Ex_1_TS_Contour.png|Contour Plot of the Transition State
File:Ms12218_Ex_1_TS_IND.png|Internuclear Distance vs Time for the Transition State
File:Ms12218_Ex_1_TS_A.png|The Transition State
</gallery>

==== Trajectories from r1 = rts + 1, r2 = rtsː The Reaction Path ====

The reaction or minimum energy path (mep) is a trajectory that corresponds to infinitely slow motion from the transition state to H1 + H2-H3. However, it does not provide a realistic account of the motion of atoms during a reaction.

With the MEP calculation, the trajectory of r1 is approximately constant, but with the dynamics calculation, it oscillates. This is because MEP does not take account of the vibrational energy of the forming molecule H2-H3 - real atoms do not move in infinitely slow motion, they vibrate.

Also with the MEP calculation, the trajectory of r2 (H1 atom moving away from the H2-H3 molecule) increases to ~ 200 pm, but increases infinitely with the dynamics calculation. This is MEP does not account for the fact that atoms have mass, and as a result, in the gas phase their motion will be inertial - constant motion.

<gallery>
File:Ms12218_Ex_1_Dynamic.png|Contour Plot of the Reaction Path (Dynamics)
File:Ms12218_Ex_1_mep.png|Contour Plot of the Reaction Path (MEP)
</gallery>

If the initial conditions were changed to make r1 = rts, r2 = rts + 1, then the same process would occur, but H1-H2 would form the molecule, with H3 as the atom moving away.

By setting up a dynamics calculation where the initial positions correspond to the final positions of the previous trajectory with the same final momenta values but their signs reversed:
* r1 = 352.598374019175 p1 = - 5.06794069218826
* r2 = 74.0408880639704. p2 = - 3.2038548324798097
The trajectory then shows the opposite of the reaction path - the formation of the transition state from H1 and H2-H3. The contour plot shows this trajectory clearly, and the Internuclear Distances vs Time plot shows the formation of the transition state as r1 and r2 come to be equal.

<gallery>
File:Ms12218_Ex_1_Reverse_C.png|Contour Plot of the Reaction Path (Dynamics)
File:Ms12218_Ex_1_Reverse_IND.png|Internuclear Distances vs Time for the Reverse of the Reaction Path
</gallery>

=== Reactive and Unreactive Trajectories ===
{| class="wikitable"

| '''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'''

|-

| -2.56
| -5.1
| -414.280
| Yes
|The trajectory shows the reaction path from H1-H2 + H3, through the transition state, to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T1.png|200px]]

|-

| -3.1
| -4.1
| -420.077
| No
| The trajectory shows H3 approaching H1-H2, however they do not react and H3 moves away again. The transition state is never formed.
| [[File:Ms12218_Ex_1_T2.png|200px]]

|-

| -3.1
| -5.1
| -413.977
| Yes
|The H3 atom has more momenta than the previous example.

The trajectory shows the reaction path from H1-H2 + H3, through the transition state, to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T3.png|200px]]

|-

| -5.1
| -10.1
| -357.277
| No
|The trajectory shows H3 approaching H1-H2 and even forming the transition state. However, though H2 and H3 interact, H1-H2 reforms once more, and H3 moves away again.
| [[File:Ms12218_Ex_1_T4.png|200px]]

|-

| -5.1
| -10.6
| -349.477
| Yes
|The H3 atom has more momenta than the previous example. The trajectory shows the reaction path from H1-H2 + H3, through the transition state (with some interaction between H1 and H2), to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T5.png|200px]]

|}
From the table, it is clear that it is not sufficient for just the momenta (kinetic energy) of the molecule (p1) to increase (i.e. from -3.1 to -5.1), the momenta of the atom (p2) needs to also be sufficiently high for a reaction to occur (i.e. -5.1 vs -5.6 and -10.1 vs -10.6). The atom must also have a larger momentum (more kinetic energy) than the molecule for a reaction to take place.

=== Transition State Theory ===
''Given the results you have obtained, how will TST predictions for reaction rate values compare with experimental values?''

Transition State Theory is a powerful theory to rationalise and calculate the rate of chemical reactions based on the properties of the reactants and the transition state structure.

A crucial assumption of TST is that all trajectories with a kinetic energy along the reaction coordinate greater than the activation energy will be reactive.

The penultimate set of conditions in the above table (p1 = - 5.1 and p2 = - 10.1) show a case of barrier recrossing - the system crosses the transition state region and a bond in the product forms, but then the system reverts back to the reactants.

== '''Exercise 2: F - H - H System''' ==

=== PES Inspection ===
By inspecting the potential energy surfaces, the reactions of a F - H - H system can be classified as follows:
* F + H2 → HF + H is exothermic: the reactants are higher energy than the products.

* HF + H → H2 + F is endothermic: the products are higher energy than the reactants.
From this, it can be concluded that HF is a stronger bond than H2. In the exothermic reaction, more energy is released forming H-F than used to break H-H. In the endothermic reaction, more energy is used to break H-F than is released forming H-H.

<gallery>
File:Ms12218_Ex_2_ExoEndo.png| Surface Plot of the F-H-H System where A=F, B=C=H.
</gallery>

<nowiki>--------------------------------------------------------------------------------------</nowiki>

As the F - H - H system is not symmetric, the transition state is not simply found where r1 = r2. However, it is still the case that at the transition state, the forces along r1 and r2 equal zero. The Internuclear Distances that satisfy these conditions are r1 = 180.1 and r2 = 74.488 where r1 is the distance between F and H r2 is the distance between H and H. At this point, with no initial momentum, the trajectory is a single point.

Hammond Postulate:
* Exothermic = early TS, reactant-like.
* Endothermic = late TS, product-like.

<gallery>
File:Ms12218_Ex_2_TS_C.png| Surface Plot of the Transition State Point
File:Ms12218_Ex_2_TS_IND.png| Internuclear Distances vs Time for the Transition State
</gallery>

As seen in the surface plot, the transition state point is much closer to the reactants H2 + F. This can be explained with Hammond's Postulate: an exothermic reaction will have an early transition state that is reactant-like, and an endothermic reaction will have a late transition state that is product-like. H2 + F are the reactants of F + H2 → HF + H which is exothermic, and the products of HF + H → H2 + F which is endothermic.

<nowiki>--------------------------------------------------------------------------------------</nowiki>

The activation energy for each reaction can be found by performing an MEP calculation from a structure neighbouring the transition state to find the trajectory that corresponds to infinitely slow motion from the transition state to the products.

To find the activation energy of the endothermic HF + H → H2 + F:
* r1 = rts - 1, r2 = rts so r1 = 180.1 and r2 = 74.488.
* The trajectory shows the reaction path proceeding from the transition state to HF + H.
* By plotting energy against time for this trajectory, the potential energy difference can be found between HF + H (the reactants) and the transition state.
* Transition state potential = 433.981 kJ mol-1.
* Reactants (HF + H) potential = - 560.036 kJ mol-1.
* Therefore the activation energy for HF + H → H2 + F is 126.055 kJ mol-1.
<gallery>
File:Ms12218_Ex_2_HFH_C.png| Surface Plot
File:Ms12218_Ex_2_HFH_Ea.png| Energy vs Time
</gallery>

To find the activation energy of the exothermic H2 + F → HF + H:
* r1 = rts + 1, r2 = rts so r1 = 182.1 and r2 = 74.488.
* The trajectory shows the reaction path proceeding from the transition state to H2 + F.
* By plotting energy against time for this trajectory, the potential energy difference can be found between H2 + F (the reactants) and the transition state.
* Transition state potential = 433.981 kJ mol-1.
* Reactants (H2 + F) potential = - 435.013 kJ mol-1.
* Therefore the activation energy for H2 + F → HF + H is 1.032 kJ mol-1.
* This is much smaller than the above reaction () reflecting the fact that this is an exothermic vs an endothermic reaction.
<gallery>
File:Ms12218_Ex_2_H2H_C.png| Surface Plot
File:Ms12218_Ex_2_H2F_Ea.png| Energy vs Time
File:Ms12218_Ex_2_H2F_EaZ.png| Energy vs Time (Zoomed In)
</gallery>

=== Reaction Dynamics ===
==== Mechanism for the release of reaction energy for: F + H2 → HF + H ====
The initial conditions r1 = 175 pm, r2 = 74 pm, p1 = -1.6 g mol-1 pm fs-1, and p2 = 0.2 g mol-1 pm fs-1 resulted in a reactive trajectory for this reaction as seen in the contour plot below.

<gallery>
File:Ms12218_Ex_2_RD_1.png| Contour Plot of F + H2 → HF + H
</gallery>

Energy in the reaction is conserved, and the reaction energy is released into the vibrational mode of newly formed HF molecule - potential energy is converted into kinetic (vibrational energy) energy. This is clearly seen in the contour plot above by the extent of oscillation in HF which corresponds to vibration of the molecule.

This mechanism of release of reaction energy into vibrational energy can be confirmed experimentally with IR absorption spectroscopy. The formation of the H-F bond and release of reaction energy means that the HF molecule is vibrationally excited: a proportion of the molecules are in the 1st excited level. Therefore, there are two vibrational transitions available: 0 → 1, and 1 → 2, and we expect to see two peaks in the IR spectrum - a fundamental intensity (reflecting the 0 → 1 transition), and a hot band (reflecting the 1 → 2 transition between excited levels) at a lower frequency. The overtone has a lower frequency due to the anharmonicity of the interaction. As the molecule reverts back to the relaxed state, the overtone intensity will become weaker and the fundamental peak will increase in intensity. If the reaction energy was not released as vibrational kinetic energy, the overtone would not be seen and there would be only one peak in the IR spectra.

Another method of confirming the method experimentally is with IR emission spectroscopy (e.g. gas-phase FTIR). As the vibrationally excited molecules revert to the relaxed state, photons are emitted as electrons fall to the ground state - the energy of the photon is equal to the difference in energy levels, and therefore can be used to probe the electronic structure of the molecule (and confirm vibrational excitation).

-----------------------------------
{| class="wikitable"

| '''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'''

|-

| -1.0
| -6.1
| -402.845
| No
|
| [[File:Ms12218_Ex_2_T1.png|200px]]

|-

| -1.0
| -4.1
| -421.245
| Yes
|
| [[File:Ms12218_Ex_2_T2.png|200px]]

|-

| -1.0
| -2.1
| -431.645
| No
|
| [[File:Ms12218_Ex_2_T3.png|200px]]

|-

| -1.0
| 0
| -433.955
| No
|
| [[File:Ms12218_Ex_2_T4.png|200px]]

|-

| -1.0
| 2.1
| -427.445
| No
|
| [[File:Ms12218_Ex_2_T5.png|200px]]

|-

| -1.0
| 4.1
| -390.645
| Yes
|
| [[File:Ms12218_Ex_2_T6.png|200px]]

|-

| -1.0
| 6.1
| -349.477
| No
|
| [[File:Ms12218_Ex_2_T7.png|200px]]

|}

{| class="wikitable"

| '''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'''

|-

| -1.0
| -6.1
| -434.482
| Yes
|
| [[File:Ms12218_Ex_2_T8.png|200px]]

|}

File:Ms12218 Ex 2 T8.png

2020-05-22T19:40:37Z

Ms12218:

MRD:Ms12218

2020-05-22T19:36:10Z

MRD:Ms12218

2020-05-22T08:13:39Z

Ms12218: /* Molecular Dynamics Lab */