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Yw21218 MRD

2020-05-15T21:01:03Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition State Theory (TST) that once the molecular system has crossed the transition state, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected that the reaction should go to completion according to TST. However, the more realistic trajectory based on dynamics calculation shows it should bounce back from the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on the potential energy surface, hence may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, at small BC distance F + H2 is on the right, while HF + H is on the left at small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, and vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is set as MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction and it is similar to the product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm. After running the guessed position with MEP calculation, if the trajectory goes towards FH +H, we should increase the F-H distance, if it goes towards H2, we should decreases F-H distance. Adjust those two parameters until the trajectory is only a dot at the starting position. After several attempts, the transition state is located at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To find the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decreasing the FH distance and locate the last position.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted, and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground energy level is greater than that of the transition from the first energy level to the second. Besides, the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak absorption can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra frequently as the reaction is happening, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position of A + BC, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energies can play roles in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. However, as the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in a reactive path. Too much vibrational energy cannot help the reaction much to go towards the transition state, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:57:24Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition State Theory (TST) that once the molecular system has crossed the transition state, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected that the reaction should go to completion according to TST. However, the more realistic trajectory based on dynamics calculation shows it should bounce back from the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on the potential energy surface, hence may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, at small BC distance F + H2 is on the right, while HF + H is on the left at small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, and vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is set as MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction and it is similar to the product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm. After running the guessed position with MEP calculation, if the trajectory goes towards FH +H, we should increase the F-H distance, if it goes towards H2, we should decreases F-H distance. Adjust those two parameters until the trajectory is only a dot at the starting position. After several attempts, the transition state is located at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To find the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decreasing the FH distance and locate the last position.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted, and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground energy level is greater than that of the transition from the first energy level to the second. Besides, the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak absorption can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra frequently as the reaction is happening, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:53:07Z

Yw21218: /* Transition state and activation energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition State Theory (TST) that once the molecular system has crossed the transition state, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected that the reaction should go to completion according to TST. However, the more realistic trajectory based on dynamics calculation shows it should bounce back from the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on the potential energy surface, hence may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, at small BC distance F + H2 is on the right, while HF + H is on the left at small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, and vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is set as MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction and it is similar to the product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm. After running the guessed position with MEP calculation, if the trajectory goes towards FH +H, we should increase the F-H distance, if it goes towards H2, we should decreases F-H distance. Adjust those two parameters until the trajectory is only a dot at the starting position. After several attempts, the transition state is located at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To find the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decreasing the FH distance and locate the last position.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:48:14Z

Yw21218: /* classification of the reactions */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition State Theory (TST) that once the molecular system has crossed the transition state, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected that the reaction should go to completion according to TST. However, the more realistic trajectory based on dynamics calculation shows it should bounce back from the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on the potential energy surface, hence may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, at small BC distance F + H2 is on the right, while HF + H is on the left at small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, and vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:47:21Z

Yw21218: /* Transition state theory and experimental data */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition State Theory (TST) that once the molecular system has crossed the transition state, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected that the reaction should go to completion according to TST. However, the more realistic trajectory based on dynamics calculation shows it should bounce back from the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on the potential energy surface, hence may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:43:15Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.3 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:43:00Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The internuclear distance vs time diagram indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7.1 reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7.2 reactive F + H2 system, energy vs time]]
| [[File:Yw21218 Distance FHH.png | thumb | Figure 7.2 reactive F + H2 system, internuclear distance vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

File:Yw21218 Distance FHH.png

2020-05-15T20:41:39Z

Yw21218:

Yw21218 MRD

2020-05-15T20:36:51Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product. In the energy vs time diagram, it confirms that the kinetic energy has increased after reaction goes to the products and the total energy is conservative.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure 7. reactive F + H2 system, momenta vs time]]
| [[File:Yw21218 FHH energy.png| thumb | Figure 7. reactive F + H2 system, energy vs time]]
|}

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

File:Yw21218 FHH energy.png

2020-05-15T20:36:19Z

Yw21218:

Yw21218 MRD

2020-05-15T20:32:45Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure ]]

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:32:29Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 7. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 7, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure ]]

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:31:45Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

{|
| [[File:Yw21218 FHH moemta.png | thumb | Figure ]]

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

File:Yw21218 FHH moemta.png

2020-05-15T20:30:43Z

Yw21218:

Yw21218 MRD

2020-05-15T20:29:49Z

Yw21218: /* Identifying reactive trajectories */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibrations before reaching the transition state and the vibration amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy in the system is slightly higher than that in the first example, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four times as much as the activation energy. It overcomes the energy barrier and the vibration is significant, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the products. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:26:36Z

Yw21218: /* Identifying reactive trajectories */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame. With high kinetic energy, energy barrier is no longer the factor to determine the reaction reactivity.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:24:42Z

Yw21218: /* MEP and trajectory */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier. The extra kinetic energy is from the energy difference between the transition state and the reactant.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:23:41Z

Yw21218: /* MEP and trajectory */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2. The dynamics calculation gives a wavy trajectory, indication the bond vibration after overcome the energy barrier.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:22:23Z

Yw21218: /* MEP and trajectory */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero at every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:19:54Z

Yw21218: /* Location of transition state */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate the effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. A decent guess can start from the saddle point region discussed above. The ideal transition state position is found when the trajectory is just a dot in the contour plot, even with step number is as large as 5000 (if the initial condition is at transition state, no reaction should occur over the time). Another way to confirm the transition state position is to check if the calculated forces are close to zero, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven by the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:12:29Z

Yw21218: /* Definition of transition state */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It has the maximum energy of the minimum energy path, hence it is also called saddle point. This is its special characteristics to separate it from other local minima. In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross sections by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:09:36Z

Yw21218: /* Reference */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

== Reference ==

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:09:03Z

Yw21218: /* Transition state theory and experimental data */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier. <ref name="first" />

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction. <ref name="first" />

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

=== Reference ===

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:07:51Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

=== Reference ===

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:07:25Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

<references>
<ref name="first">Laidler, K., 1987. Chemical Kinetics. New Delhi: Pearson.</ref>
</references>

Yw21218 MRD

2020-05-15T20:03:16Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

<references>
<ref name="first"></ref>
</references>

Yw21218 MRD

2020-05-15T20:02:23Z

Yw21218: /* Dynamics from the transition state region */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics<ref name = "first" />, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T20:01:46Z

Yw21218: /* Definition of transition state */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ==== <ref name = "first" />
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T19:59:18Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest. The absorption of energy to transmit the from ground level is greater than that of the transition from first level to second level, besides the population at ground level is much more than that in first energy level. Therefore, transition (0→1) gives a much more intense peak at higher energy in IR spectrum, where the intensity of the peak can be used to calculate the amount of vibrational energy. Experimentally, stop flow can be used to collect IR spectra at a frequency as the reaction undergoes, which can monitor the changes of vibrational energy of the reaction.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T19:43:48Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy in product.

To measure kinetic energy, calorimeter can be used, because higher kinetic energy will lead to higher temperature. To measure the vibrational energy, IR spectrometer can be used to measure the emission or the absorption of transition. Because the population of molecules at a specific vibrational energy level is proportional to the transition energy absorbed or emitted and the vibrational frequency can indicate the type of bond, we can therefore find out the vibrational energy of the bond of interest.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T18:14:47Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -0 g.mol-1.pm.fs-1]]

As seen in figure 6, there is little vibrational energy in when the reaction starts. After overcome the energy barrier, trajectory goes down the hill and form products. Since the potential energy of the products is a lot less than the reactant, so the extra energy is converted to vibrational energy. This is why we can see an amplified waves in product region. The momenta vs time diagram shows the small vibrations of HH bond at the beginning of the reaction, and nearly no vibrations between F and H. As reaction goes along the reaction coordination, the vibrational energy is transferred to the FH bond with a greater amount. The animation indicates the same result of the vibration migration and the increase of vibrational energy.

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T18:00:39Z

Yw21218: /* Reaction dynamics */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. a F + H2 system, rFH = 170pm rHH = 74pm, pFH = 0 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T17:56:13Z

Yw21218: /* Mechanism of release of the reaction energy */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

[[File:Yw21218 Contour FHH 200.png | thumb | Figure 6. ]]

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

File:Yw21218 Contour FHH 200.png

2020-05-15T17:55:03Z

Yw21218:

Yw21218 MRD

2020-05-15T17:43:50Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. With low translational energy, sufficient vibrational energy can lead to a reactive path. As the kinetic energy increases, higher translational energy doesn't necessarily ensure a successful reaction if the vibrational energy is not sufficient. On the other hand, H + H2 prefers translational energy in reactive path. Too much vibrational energy cannot help the reaction going towards the transition state much, and sometimes may even make the trajectory bounce back to reactants after overcome the energy barrier.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:50:38Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction. As shown in the table below, with similar amount of kinetic energy, FH + H system prefers vibrational energy to complete the reaction. Large H + H2 prefers translational energy. However, it doesn't

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:35:17Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{| class="wikitable" border="1"
!
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.

Yw21218 MRD

2020-05-15T16:34:53Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{| class="wikitable" border="1"
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E'''
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.

Yw21218 MRD

2020-05-15T16:33:44Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E''' |
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.

Yw21218 MRD

2020-05-15T16:33:04Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
|-
| '''Translational E > Vibrational E''' |
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]] Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:32:21Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)!! H2 + H (endothermic)
| '''Translational E > Vibrational E''' |
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]] Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:32:10Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
!! H2 + H (endothermic)
| '''Translational E > Vibrational E''' |
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]] Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:31:30Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
| '''Translational E > Vibrational E''' |
| [[File:Yw21218 Contour FH H trans.png | thumb| rFH = 91pm rHH = 200pm, pFH = -2 g.mol-1.pm.fs-1 rHH = -15 g.mol-1.pm.fs-1]] Even if the kinetic energy is large enough to overcome the barrier, it still may not go to completion.
| [[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
|-
| '''Translational E <Vibrational E'''
| [[File:Yw21218 Contour FH+H vib.png | thumb | rFH = 91pm rHH = 200pm, pFH = 14 g.mol-1.pm.fs-1 rHH = -2 g.mol-1.pm.fs-1]]
| [[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

File:Yw21218 Contour FH+H vib.png

2020-05-15T16:29:51Z

Yw21218:

File:Yw21218 Contour FH H trans.png

2020-05-15T16:23:43Z

Yw21218:

Yw21218 MRD

2020-05-15T16:19:18Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
| '''Translational E > Vibrational E''' || ||[[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
| '''Translational E <Vibrational E''' || ||[[File:Yw21218 Contour HH F vib.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = 0 g.mol-1.pm.fs-1]]
|}

Yw21218 MRD

2020-05-15T16:18:30Z

Yw21218: /* Distribution of kinetic energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
| '''Translational E > Vibrational E''' || ||[[File:Yw21218 Contour HH F trans.png | thumb| rFH = 200pm rHH = 74pm, pFH = -1 g.mol-1.pm.fs-1 rHH = -6 g.mol-1.pm.fs-1]]
| '''Translational E <Vibrational E''' || ||[[File:Yw21218 Contour HH F vib.png]]
|}

File:Yw21218 Contour HH F vib.png

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Yw21218:

File:Yw21218 Contour HH F trans.png

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Yw21218:

Yw21218 MRD

2020-05-15T16:07:51Z

Yw21218: /* Distribution of energy between translation and vibration */

== H + H2 System ==

=== Dynamics from the transition state region ===
==== Definition of transition state ====
According to Chemical Kinetics and Dynamics, the transition state is a special minimum. It is at the maximum energy of the minimum energy path, hence it is also called saddle point. This is used to separate it from any local minimum In a symmetrical potential energy surface (where reactants have the same energy as the products), the saddle point is at the intercept of S and ξ as shown in Figure 1. S = r2 + r1 and ξ = r2 - r1. The cross section by cutting the surface energy plot along S and ξ are shown in figure1.a and 1.b respectively, where E0 is the energy of the transition state. Mathematically, it is the solution of the following equations: <math>\frac{\partial^2 U}{\partial s} = 0 </math>, <math>\frac{dU}{ds} < 0</math> , <math>\frac{\partial^2 U}{\partial \xi} = 0 </math> , <math>\frac{dU}{d\xi} > 0</math> [[File:Yw21218 saddle.png|250px| thumb | Figure 1. Saddle point region of symmetrical potential energy surface]] [[File:Yw21218 sections saddle.png | 250px | thumb | Figure 1.a, 1.b. Cross Sections of potential energy surface]]

==== Location of transition state ====
[[File:Yw21218 HHH.png | centre | 200px | thumb | Figure 2. H-H-H system ]]

Because the potential energy surface is symmetrical, initial condition can be set as r1 = r2. To eliminate effect of external forces and the bond oscillation, p1 = 0 g.mol-1.pm.fs-1, p2 = 0 g.mol-1.pm.fs-1, and MEP calculation should be used. Some decent guess can be started around the saddle point region discussed above. The ideal transition state position is found when the trajectory calculated is just a dot in the contour plot, even with step number as large as 5000. Another way to check the transition state position is to make the calculated forces as close to zero as possible, because force is the negative derivative of the potential energy. Finally, rts = r1 = r2 = 91 pm, with potential energy of -415.369 KJ mol-1. The reactant potential energy is -434.705 KJ mol-1, hence the activation energy is the energy difference 19.336 KJ mol-1.

At this point, rAB and rBC are equal and constant, which has been proven from the 'Internuclear distance vs Time' plot (Figure 3). [[File:Yw21218 HHH distance ts.png | 150px | thumb| figure 3. 'Internuclear distance vs Time' plot.(The atom notations can be found in Figure 2)]] Transition state is at the point where slope equals to zero, without any external forces it doesn't have preference to form products or reactants, hence the atoms are still relative to each other.

==== MEP and trajectory ====
Minimum energy path is in theory the trajectory that reaction should follow, corresponding to the lowest energy valley of the potential energy surface. MEP calculation sets the momenta to zero in every step in the trajectory, therefore, it doesn't take into account the vibrational energy of molecules. However, 'Dynamics' calculation can generate a more realistic trajectory by keeping the momenta as what they are in the previous step, so it shows the bond oscillation. To compare those two calculations, their corresponding trajectories were produced under the same initial conditions: r1 = rts+ 1, r2 = rts, p1 = p2 = 0 g.mol-1.pm.fs-1, and are shown in Figure 4.1 and Figure 4.2.
{|
|-
| [[File:Yw21218 Mep ts.png | 300px | centre |thumb | Figure 4.1. MEP calculation trajectory]]
| [[File:Yw21218 Dynamics ts.png | 300px | thumb | centre |Figure 4.2. Dynamics calculation trajectory]]
|}

=== Reactive and unreactive trajactory ===

==== Identifying reactive trajectories ====

The amplitude of the trajectory waves indicates the amount of vibrational energy. The greater amplitude the waves have, the more vibrational energy the system has at that stage.

When the kinetic energy (KE) is lower than the activation energy (Ea), the reaction will not complete. When KE reaches the Ea, the extra energy will become vibrational energy to the molecules. However, if the vibrational energy is too high, it can cause the reaction reform the reactants even if the collision occurs and energy barrier is overcame.

{| class="wikitable" border="1"
|+ Table 1. Reactive and unreactive trajectories
! 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 || style="text-align: center;" | yes || The molecule H2 has small vibration before reaching the transition state and the oscillation amplifies slightly after it overcomes the energy barrier and forms product || [[File:Ag7518Test1contour.png | 200px]]
|-
| -3.1 || -4.1 || -420.077 || style="text-align: center;" | no || The kinetic energy 13.710 KJ mol-1 input in the reaction is less than the activation energy, so it cannot overcome the energy barrier to form products. The reactants have a certain amount of vibrational energy when reaction starts and it is the same amount at the end. || [[File:Ag7518Test2contour.png | 200px]]
|-
| -3.1 || -5.1 || -413.977 || style="text-align: center;" | yes || The kinetic energy put acted on the system is slightly higher than that of the first case, so the amplitude of the waves of the trajectory is greater, indicating more energy used to vibrate. Hence the product has a greater oscillation. || [[File:Ag7518Test3contour.png | 200px]]
|-
| -5.1 || -10.1 || -357.277 || style="text-align: center;" | no || The kinetic energy is nearly four time as much as the activation energy. It overcomes the energy barrier and the collision happens, but due to vibrational motion, it reforms the reactants at the end. Because when kinetic energy is too high, energy barrier is not playing in a role in achieving reactive path, instead, it depends more on the trajectory with the given initial conditions. || [[File:Ag7518Test4contour.png | 200px]]
|-
| -5.1 || -10.6 || -349.477 || style="text-align: center;" | Yes || Although in this case the kinetic energy is quite large, it does manage to reach the product after some chaos around saddle point region. In this case, the trajectory has managed to reach the product. || [[File:Ag7518Test5contour.png | 200px]]
|}

====Transition state theory and experimental data====

It is assumed in Transition state theory (TST) that once the molecular system has crossed the transition state towards the direction of the product, it can not go back to the reactants. However, it is not always the case in experiments. In the second last case mentioned in the table above, the kinetic energy should be high enough to overcome the energy barrier, it is expected the reaction to go to completion according to TST. However, the more realistic trajectory bounces on the energy surface and reform the reactants at the end. With high kinetic energy, energy barrier is no longer the key to have a reactive path. It depends more on how vibrational and translational motion interact with the energy surface to form a trajectory.

Besides, TST simplifies the motion of molecules to one dimensional motion in reaction coordinate and separates molecular motion into electronic and nuclear part. In reality, molecules also have other normal reaction motions like vibration. Hence, the reaction trajectory may be above the minimum energy configuration on potential energy surface and may require higher activation energy than the energy barrier.

Another limitation of the TST is that it can predict reasonably accurate saddle point via harmonic potential at low temperature, but when the temperature is high the potential harmonic requires some anharmonic correction.

== F-H-H System ==

=== PES inspection ===

==== classification of the reactions ====
[[File:Yw21218 FHH annotation.png | 200px | thumb | centre | Figure 5. F-H-H system]]

[[File:Yw21218 Surface Plot FHH.png | 300Px | thumb | Figure 6. Energy surface plot of FHH system]]

As shown in Figure 6, with small BC distance F + H2 is on the right, while HF + H is on the left with small AB distance. The energy graph indicates F + H2 is going to have exothermic reaction, but FH + H system will undergo an endothermic reaction. It is consistent with the bond strength of HF and H2. 𝛥HHF = 181.5 JK mol-1, 𝛥HH2 = 74.5 JK mol-1. Bond breaking is endothermic and bond making is exothermic. By looking at the reaction HF + H → F + H2, the enthalpy is 107 JK mol-1, indicating it is endothermic reaction, vice versa.

==== Transition state and activation energy====

As mentioned in H-H-H system, calculation is MEP, p1 = p2 = 0 g.mol-1.pm.fs-1, step number = 5000. However, the energy surface is not symmetrical, so r1 is not equal to r2. According to Hammond Postulate, the transition state is more similar to the reactant in exothermic reaction or similar to product in endothermic reaction. Therefore, the transition state in FHH system should be closer to H2 + F side.

Starting with the condition rHH = 94 pm, rFH = 200 pm, adjust those two parameters until the trajectory is only a dot at the starting position. If the trajectory goes towards FH +H, increase the F-H distance, if it goes towards H2, decreases F-H distance. After several attempts, the transition state is at rHH = 94 pm, rFH = 181.35 pm, with potential energy at -433.942 KJ mol-1.

To the activation energy of the reaction, the difference of the potential energy of reactants and transition state is what we are looking for (still keep momenta as zero, so no kinetic energy). Based on the settings of the transition state, one can increase the FH distance to find the most stable position for H + H2, hence its potential energy. The energy of FH + H can be found by decrease the FH distance.

H + H2 potential energy is -434.536 KJ mol-1 at rFH = 230 pm and rHH = 74 pm, so Ea = 0.594 KJ mol-1

HF + H potential energy is -560.417 KJ mol-1 at rFH = 92 pm and rHH = 262 pm, so Ea = 126.475 KJ mol-1

=== Reaction dynamics ===

==== Mechanism of release of the reaction energy ====

For a reactive trajectory, the initial conditions can be

==== Distribution of kinetic energy between translation and vibration ====

According to Polanyi's empirical rules, vibrational energy is more efficient in promoting later barrier reaction (endothermic reaction) than translational energy, while for early barrier reaction (exothermic reaction) translational energy is more efficient. To illustrate the rules, several trajectories with different initial settings were plotted. Kinetic energy is the sum of the translational and vibrational energy, so if one is small, the other must be dominant. Starting from the reactant position, the momentum on the bond of the reactant molecule BC reflects the vibrational energy, while the momentum between the single atom A and the middle atom B can reflect the translational energy. The relative amount of those two energy can play a role in the completion of the reaction.

{|
! HF +H (exothermic)
! H2 + H (endothermic)
|
|
|}