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= Molecular Reaction Dynamics: Applications to Triatomic systems =

{{fontcolor1|green| Overall, your report was perfect from start to end. You have clearly explained how each step was conducted and provided full discussions. It is obvious you have understood the concepts throughout. Your presentation should also be praised. I would advise you begin adding in references to your report and any piece of course work it would definitely help you! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:00, 24 June 2020 (BST)}}

==H + H2 system==

=== Introduction ===
In this wiki, the potential energy surface is applied to study the a triatomic system.[[File:1amolabcazp2518.png|centre|thumb|Figure.1. Example triatomic system]]
In this triatomic system, an H atom A collide with the H<sub>2 </sub>molecule BC. Their separations and momentums are represented by r<sub>1, </sub> r<sub>2</sub>, p<sub>1</sub> and p<sub>2</sub>. The the example shown in the figure 1 have the following initial condition.
{| class="wikitable"
!r<sub>1 </sub>/ pm
!r<sub>2 </sub>/ pm
!p<sub>1 </sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>
!p<sub>2 </sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>
|-
|74
|230
|0
|<nowiki>-5.2</nowiki>
|}
The trajectory of this example system was calculated on the potential energy surface diagram with following setting.(Fig.2)
{| class="wikitable"
!Calculation type
!Step number
!Step size / fs
|-
|Dynamic 
|500
|0.1
|}
[[File:2asurfaceplotazp2518.png|centre|thumb|Figure.2. the trajectory of the system with the example initial condition in a potential energy surface]]

{{fontcolor1|green|Great introduction [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:45, 24 June 2020 (BST)}}

===locating the transition state===
[[File:3aTSazp2518.png|centre|thumb|Figure.3. The position of a transition state in a potential energy surface.]]

In a potential energy surface diagram, the transition state is defined as the maximum along the minimum energy path that links the reactant and the product.(Fig.3) Therefore, it is also a saddle point on the potential energy surface diagram.

The transition state and a local minimum both have zero gradient of potential on an energy surface diagram (∂V<sub>(ri)</sub>/∂r<sub>i</sub>=0). However, saddle points could be distinguished from a local minimum by calculating the hessian matrix and second partial derivative of potential. A saddle point has indefinite hessian matrix with the eigenvalues being one positive and one negative, while a local minimum have Hessian that is positive definite with all eigenvalues positive being positive an a second partial derivative that is positive.

{{fontcolor1|green|Excellent! But some more details are needed. the second derivative will not automatically churn our one positive and one negative value, that is wrt. the direction in matter. If one is positive in a direction the orthogonal direction will be negative. Whereas for the local minimum it is positive in all directions.[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:48, 24 June 2020 (BST)}}

Because the the transition state is a local maximum in the reaction path, the gradient of potential energy surface and therefore the rate of change in momentum are both equal zero (∂p<sub>i</sub>/∂t=∂V<sub>(ri)</sub>/∂r<sub>i</sub>=0). Due to the rate of change in momentum is zero, the force along the vector direction of r<sub>1 </sub>and r<sub>2 </sub>should also be zero, in other word, the r<sub>1 </sub>and r<sub>2 </sub>should be fixed in time.

The transition state of the system H+H<sub>2 </sub>have the geometry that r<sub>1</sub>=r<sub>2 </sub>because system is symetric. The geometry of the transition state that r<sub>1</sub>=r<sub>2</sub>=r is found by trying different initial values of r with zero initial momentum to find the value of r that results stationary trajectory. 

The estimated position of the transition state is at r<sub>1</sub>=r<sub>2</sub>=90.77 pm. This is the position of the TS  that result stationary trajectory as we predicted above. Which is proven by the Internuclear Distances vs Time plot (Fig.4). This plot clear shows that the  r<sub>AB</sub>(r<sub>2</sub>) and r<sub>BC</sub>(r<sub>1</sub>) is fixed in time. 
[[File:4aadatapazp2518.png|centre|thumb|Figure.4. The distance vs time plot for the detriment transition state gerometry.png]]

{{fontcolor1|green|Nicely shown![[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:48, 24 June 2020 (BST)}}

=== Calculating the reaction path ===
The reaction path (Fig.5) was calculated with following initial conditions and calculation setting.
{| class="wikitable"
!Calculation type
!Step number
!step size / fs
!r<sub>1</sub> / pm
!r<sub>2</sub> / pm
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|MEP
|1500
|0.15
|91.77
|90.77
|0.0
|0.0
|}
[[File:5amepazp2518.png|centre|thumb|Fig.5. Reaction path calculated with MEP]]

=== Trajectories from r<sub>1</sub> = r<sub>ts</sub>+δ, r<sub>2</sub> = r<sub>ts</sub> ===
The trajectories (Fig.6) were calculated with following initial conditions and calculation setting.
{| class="wikitable"
!Calculation type
!Step number
!step size / fs
!r<sub>1</sub> / pm
!r<sub>2</sub> / pm
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|Dynamics
|500
|0.1
|91.77
|90.77
|0.0
|0.0
|}
[[File:6adynamicazp2518.png|centre|thumb|Fig.6. Trajectory calculated with dynamics<p>(r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>
]]
By comparing the reaction path calculated from the MEP (Fig.5) and the trajectories calculated from the dynamics (Fig.6). We found that the reaction path do not take the vibration of bond in a molecule in to account because it reset the moment to the initial condition in each time step. It represent the path of the reaction from the reactant to the product, but not the motion of the atoms. The elimination of the momentum also make it takes longer to the obtain a complete trajectory (225 fs for the calculation in MEP and 50 fs for the calculation in dynamics). While the trajectories calculated with the dynamics is shown as a sinuous curve on the potential energy surface. Which indicate that the dynamic calculation takes the motion of atoms or the vibration of the molecule bond into account.  

{{fontcolor1|green|Perfect and that makes dynamic more realistic![[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:49, 24 June 2020 (BST)}}

The plots of “Internuclear Distances vs Time” and “Momenta vs Time” are shown in Fig.7 and Fig.8. 
[[File:7advtazp2518.png|centre|thumb|Figure.7. Internuclear Distances vs Time plot<p>(r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>
]]
[[File:8amvtazp2518.png|centre|thumb|Figure.8. Momenta vs Time<p>(r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>
]]The final values of the positions and momentums are shown below
{| class="wikitable"
!r<sub>1</sub> / pm
!r<sub>2</sub> / pm
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|352.2080
|73.8902
|5.05855
|3.21350
|}

=== Trajectories from r<sub>1</sub> = r<sub>ts</sub>, r<sub>2</sub> = r<sub>ts</sub>+δ ===
The trajectories (Fig.9) were calculated with following initial conditions and calculation setting.
{| class="wikitable"
!Calculation type
!Step number
!step size / fs
!r<sub>1</sub> / pm
!r<sub>2</sub> / pm
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|Dynamics
|500
|0.1
|90.77
|91.77
|0.0
|0.0
|}
[[File:7adynamic2azp2518.png|centre|thumb|Figure.9. Trajectories calculated with dynamics<p>(r<sub>1</sub> = r<sub>ts</sub>, r<sub>2</sub> = r<sub>ts</sub>+1pm)</p>
]]
The plots of “Internuclear Distances vs Time” and “Momenta vs Time” are shown in Fig.10 and Fig.11.
[[File:10advtazp2518.png|centre|thumb|Figure.10. Internuclear Distances vs Time plot
(r<sub>1</sub> = r<sub>ts</sub>, r<sub>2</sub> = r<sub>ts</sub>+1pm)
]]
[[File:11amvtazp2518.png|centre|thumb|Figure.11. Momenta vs Time Plot
(r<sub>1</sub> = r<sub>ts</sub>, r<sub>2</sub> = r<sub>ts</sub>+1pm)
]]
By comparing the "Internuclear Distances vs Time" plots with two different initial condition(fig.8 and fig.10), we found that with the initial geometry that r<sub>1</sub> = r<sub>ts</sub>+1pm and r<sub>2</sub> = r<sub>ts</sub>. The value of r<sub>1 </sub>(or r<sub>bc</sub>) increase to infinity as times pass and the value of r<sub>2 </sub>(or r<sub>ab</sub>) decrease first and then fluctuate around a constant. Which indicate that the H<sub>A</sub>-H<sub>B</sub>-H<sub>C </sub>system decompose to an atom H<sub>C </sub>that leave with constant velocity and a molecule H<sub>2(AB)</sub> staying and vibrating with a momentum. 

While the trend reversed with value of r<sub>2</sub> increase to infinity as times pass and the value of r<sub>1 </sub>decrease first and be a constant if we use the initial geometry that r<sub>1</sub> = r<sub>ts</sub>+1pm and r<sub>2</sub> = r<sub>ts</sub>. Which indicate that  H<sub>A</sub>-H<sub>B</sub>-H<sub>C </sub>system decompose to an atom H<sub>A </sub>and a molecule H<sub>2(BC)</sub>.

The "Momenta vs Time" plots with two different initial condition(fig.9 and fig.11) support our statement above. Fig.9  indicates that the H<sub>C</sub> leaves with constant velocity and the H<sub>2(AB)</sub> vibrating with a constant momentum. Fig.11 indicates that the H<sub>C</sub> leaves with constant velocity and the H<sub>2(AB)</sub> vibrating with a constant momentum.

=== The reverse of the Trajectories from r<sub>1</sub> = r<sub>ts</sub>+δ, r<sub>2</sub> = r<sub>ts</sub> ===
The trajectories (Fig.12) were calculated with following initial conditions and calculation setting.
{| class="wikitable"
!Calculation type
!Step number
!step size / fs
!r<sub>1</sub> / pm
!r<sub>2</sub> / pm
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|Dynamics
|1000
|0.1
|352.2080
|73.8902
|<nowiki>-5.05855</nowiki>
|<nowiki>-3.21350</nowiki>
|}
[[File:14areverseazp2518.png|centre|thumb|Fig.12. Trajectory calculated with dynamics<p>The reveres of (r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>]]

The plots of “Internuclear Distances vs Time” and “Momenta vs Time” for this calculation are shown in Fig.13 and Fig.14.
[[File:12advtazp2518.png|centre|thumb|Figure.13. Internuclear Distances vs Time plot<p>The reserve of (r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>]]
[[File:13amvtazp2518.png|centre|thumb|Figure.14. Momenta vs Time<p>the reverse of (r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts)</sub></p>]]

As we seem from these plots of the trajectories which starting from the finial condition at time 50 fs of the trajectories with the initial condition of r<sub>1</sub> = r<sub>ts</sub>+1pm, r<sub>2</sub> = r<sub>ts </sub>and p<sub>1</sub> = p<sub>2</sub> = 0. The internuclear distance between H<sub>B </sub>and H<sub>C</sub> constantly decrease to the distance of 91.77 pm first, then this value increase after a plateau period and the The internuclear distance between H<sub>A </sub>and H<sub>B</sub> slightly increased when H<sub>C</sub> approach but drop back to its original state later. The momentum shows that the the two momentum values is unchanged in value but the sign was reversed. These plots indicate that H<sub>C</sub> atom approaches the H<sub>2(AB)</sub> molecule first to form a system near to the transition state that H<sub>A</sub>-H<sub>B</sub>-H<sub>C</sub>, but the H<sub>C</sub> atom do no have enough initial kinetic energy to form the molecule H<sub>2(BC), </sub>so, the H<sub>C </sub>atom then leave with a opposite momentum that have same numeric value as its initial momentum. The molecule H<sub>2(AB) </sub>remain in their original position after this collision and vibrate with a momentum of an opposite sign.

=== Reactive and unreactive trajectories ===
For the initial position r<sub>1</sub> = 74 pm and r<sub>2</sub> = 200 pm, five trajectories with the following momenta combinations were calculated
{| class="wikitable"
!p<sub>1</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!Etot / kJ.mol<sup>-1</sup>
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The reactant have high kinetic energy to overcome the activation barrier.

The reaction path go over the TS and product formed.

|[[File:15a1azp2518.png|centre|thumb|Figure.15]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactant have do not have enough kinetic energy to overcome the activation barrier.

The reaction path turns back to the reactant side and the reactant reformed.

|[[File:16a2azp2518.png|thumb|Figure.16]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactant have high kinetic energy to overcome the activation barrier.

The reaction path go over the TS and product formed.

|[[File:17a3azp2518.png|centre|thumb|Figure.17]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The reactant have higher kinetic energy.

But the kinetic energy was not distributed correctly in the translational and vibrational modes. It shows that on the PES that the molecule vibrate strongly.  As a result, although the kinetic energy is very high, the translational energy was not high.

The reaction path recross the activation barrier. The bond in the product is actually formed but it revers back to the reactant.

|[[File:18a4azp2518.png|centre|thumb|Figure.18]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|The reactant have high kinetic energy, and the translation energy was larger than that in the previous scenario.  

The reaction path path the activation barrier and multiple times and the product formed. 
|[[File:19a5azp2518.png|centre|thumb|Figure.19]]
|}
The tables shows us that for trajactories to be reactive, a high kinetic energy that can overcome the activation barrier is not enough. The kinetic energy should be distributed in the right translational and vibrational mode, otherwise the reaction path for may recross the energy barrier and reactant will be reformed, even if the system have a high initial energy.


{{fontcolor1|green|Excellent! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:50, 24 June 2020 (BST)}}

=== Transition State Theory ===
There will be a large derivation from the experimental value, it assumes that all trajectories with kinetic energy larger than the activation energy will be reactive but it did not take account the possibility that the trajectories with high initial kinetic energy may also go back to the reactant if it recross the barrier (see example in fig.18), so the rate will be underestimated using this this theory. This theory also fails to take account the effect of quantum tunnelling that enable the reactant to go directly to the product without having energy excessing the activation energy. This theory also assume that only reactant are in equilibrium with the transition state while it is not realistic in many case (equilibrium reaction in thermodynamic condition). Furthermore, not all the kinetic energy will follow the Boltzmann distribution.

{{fontcolor1|green|Great job, remember TST vs. experimental so I think that's what you meant. TST overestimates the rate of reaction. You could go aead and reference all those assumptions. Even name a few more in your answer. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:52, 24 June 2020 (BST)}}

==F-H-H system==

=== PES inspection ===
By inspecting the valley floor in a potential energy floor of a F-H-H (Fig.20) system, we find potential of the part that correspond with the a HF molecule and a separate H atom is lower than the potential of the part that correspond with a H<sub>2</sub> molecule and a separate F atom. Therefore, we classify that the H<sub>2</sub>+F reaction is exothermic while H+HF reaction is endothermic.

This could be explained by the fact that the H-F bond is much stronger than a H-H bond due to the large electronegativity difference between H and F. H-F is a strong ionic bond while H-H is only a weak covalent bond. As a result, the reaction from F + H<sub>2</sub> to H + HF involved with low energy taken to break a weak H-H bond and high energy released in forming a H-F bond, hence exothermic. The reaction from H + HF to F + H<sub>2</sub> involved with high energy taken to break a strong F-H bond and low energy released in forming a H-H bond, hence endothermic
[[File:20aFHHazp2518.png|centre|thumb|Figure.20. Potential energy surface for a F-H-H system
A=F, B=H,C=H
]]

{{fontcolor1|green|Fantastic discussion! You could even justify your answer one step further by searching literature for the bond energies of H2 and HF to experimentally prove indeed HF has a greater value than H2. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:54, 24 June 2020 (BST)}}

=== Transition state. ===
The transition state geometry is determined by trying various value to find the the value of r<sub>HF</sub> and r<sub>HH</sub> that results a stationary trajectory when the initial momentums are set to zero. The Hammond's postulate is used to  find the position of the transition. As the reaction from F + H<sub>2</sub> to H + HF is exothermic reaction and a early TS one, the transition state structure will resemble the structure of the reactant, so r<sub>HH</sub> will be smaller than r<sub>HF</sub>.

The predict position of the the transition state (fig.21) is r<sub>HF</sub> = 181.1pm and r<sub>HH</sub> =74.5 pm, which is proven by the internuclear distance vs time plot in this geometry with zero initial momentum (fig.22). Potential energy in the transition state was found to be -433.980 kJ/mol.
[[File:20aFHHazp2518.png|centre|thumb|Figure.21. position of the TS]]
[[File:20aFHHadvtazp2518.png|centre|thumb|Figure.22. Internuclear distance vs time plot
(r<sub>HF</sub>= 181.1pm and r<sub>HH</sub>=74.5 pm)
]]

{{fontcolor1|green|Values look as expected. Good unit inclusion. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:55, 24 June 2020 (BST)}}

=== Activating energy ===
The activation energy is shown in the table below.
{| class="wikitable"
!Reaction
!The TS energy  / kJ.mol<sup>-1</sup>
!The reactants energy  / kJ.mol<sup>-1</sup>
!The activation energy energy  / kJ.mol<sup>-1</sup>
|-
|F+H<sub>2</sub>
|<nowiki>-433.980</nowiki>
|<nowiki>-435.851</nowiki>
|<nowiki>+1.870</nowiki>
|-
|H+HF
|<nowiki>-433.980</nowiki>
|<nowiki>-560.172</nowiki>
|<nowiki>+126.192</nowiki>
|}
The activation energy is the energy difference between the reactant and the TS state. The energy of the reactant was estimated by plotting the reaction path on a potential energy surface. 

=== Reactive trajectory for the F + H2, ===
The trajectories (Fig.23) were calculated with following initial conditions and calculation setting.
{| class="wikitable"
!Calculation type
!Step number
!step size / fs
!r<sub>FH</sub> / pm
!r<sub>H2</sub> / pm
!p<sub>FH</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>H2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
|-
|Dynamics
|3000
|0.1
|170
|74
|<nowiki>-1.6</nowiki>
|0.2
|}
[[File:23aFHHasurfacetazp2518.png|centre|thumb|Figure.23. Reactive trajectory for F + H2 ]]

The energy vs time plot for this reaction is shown below.(Fig.24)
[[File:24aFHHaenergyvtazp2518.png|centre|thumb|Figure.24. Energy vs time plot for the system F + H2]]

<nowiki> </nowiki>In this reaction we could clearly see from the energy vs time plot that the potential energy was first converted to the  kinetic energy of the vibration of the bonds in the transition state. This vibrational energy was periodically covert back to the potential energy in which is indicated by the the fluctuating line in the (Fig.24).After the time pass 37 fs. Some part of this vibrational energy was converted to the translational energy of the leaving atom. Dnd the potential energy permanently drops. Which indicate that the potential energy release in this reaction is converted to vibration energy and then the vibration energy is converted the transitional energy.  

To confirm this conclusion, we could use IR spectroscopy to determine the intensity of the overtone peaks and the the main peak. The potential energy that converted to the vibrational energy will cause the higher level vibrational modes to be excited, and therefore in overtone peak intensity in the IR spectrum will increase and the main peak will decrease. Then, due to the vibrational energy is converted to the thermal energy, the overtone peak will decrease in intensity and main peak will then increase in intensity.

{{fontcolor1|green| How would you measure thermal changes experimentally? You can even check out  'chemiluminescence' which may be useful in monitoring the emission of infrared radiation from the sample. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:56, 24 June 2020 (BST)}}

=== illustration of Polanyi's empirical rules ===
The Polanyi's rule suggest that the reactant vibrational energy is more effective than the transitional energy to overcome the barrier of a later transition state reaction and the transitional energy is more effective than the vibrational energy to overcome the barrier of a early transition state reaction. 

For the initial position r<sub>HH</sub> = 74 pm, r<sub>FH</sub> = 230 pm and p<sub>FH</sub> = -1.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, several trajectories with the following momenta were calculated.
{| class="wikitable"
!p<sub>H2</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!Etot / kJ.mol<sup>-1</sup>
!Reacative?
!Illustration of the trajectory
|-
|<nowiki>-6</nowiki>
|<nowiki>-404.014</nowiki>
|Reactive
|[[File:25a-6azp2518.png|centre|thumb]]
|-
|<nowiki>-5</nowiki>
|<nowiki>-414.014</nowiki>
|Unreactive
|[[File:26a-5azp2518.png|centre|thumb]]
|-
|<nowiki>-4</nowiki>
|<nowiki>-422.014</nowiki>
|Unreactive
|[[File:27a-4azp2518.png|thumb]]
|-
|3
|<nowiki>-422.014</nowiki>
|Unreactive
|[[File:31a3azp2518.png|thumb]]
|-
|4
|<nowiki>-414.014</nowiki>
|Unreactive
|[[File:30a4azp2518.png|thumb]]
|-
|5
|<nowiki>-404.014</nowiki>
|Reactive
|[[File:29a5azp2518.png|thumb]]
|-
|6
|<nowiki>-392.014</nowiki>
|Unreactive
|[[File:28a6azp2518.png|thumb]]
|}

We could clearly see that these trajectories have high initial kinetic energy, and most of the kinetic energy is distributed in the vibration mode. A high kinetic energy distributed in this way is need for the trajectory to be reactive. But, if we distribute more initial kinetic energy in the translational mode instead of the vibrational mode, a lower total kinetic energy is needed for the reaction to take place(see below ).

{| class="wikitable"
!r<sub>FH</sub> / pm
!r<sub>HH</sub> / pm
!p<sub>FH</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>HH</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!Etot / kJ.mol<sup>-1</sup>
!reactive
!illustration
|-
|230
|74
|<nowiki>-5.05855</nowiki>
|<nowiki>-3.21350</nowiki>
|<nowiki>-432.833</nowiki>
|Reactive
|[[File:31a3azp2518.png|thumb]]
|}
These trajectories represent the reaction from F + H<sub>2</sub> to HF + H, which is a exothermic reaction, and the position of the transition  state is in the early part of the reaction profile. These cases agree with the Polanyi's empirical rule that the transitional energy is more effective than the vibrational energy to overcome the barrier of a early transition state reaction.

Other trajectories that represent the reaction from HF + H to F + H<sub>2</sub> were calculated below with r<sub>HH</sub> = 230 pm, r<sub>FH</sub> = 74 pm，step numebr = 3000 and step size = 0.1 fs.
{| class="wikitable"
!p<sub>FH</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!p<sub>HH</sub> / gmol<sup>-1</sup>pmfs<sup>-1</sup>
!Etot / kJ.mol<sup>-1</sup>
!reactive
!illustration
|-
|<nowiki>-0.1</nowiki>
|<nowiki>-5</nowiki>
|<nowiki>-392.600</nowiki>
|Unreactive
|[[File:34a5azp2518.png|thumb]]
|-
|<nowiki>-0.1</nowiki>
|<nowiki>-8</nowiki>
|<nowiki>-353.900</nowiki>
|Unreactive
|[[File:35a8azp2518.png|thumb]]
|-
|<nowiki>-0.1</nowiki>
|<nowiki>-10</nowiki>
|<nowiki>-318.100</nowiki>
|Reactive
|[[File:36a10azp2518.png|thumb]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-1</nowiki>
|<nowiki>-414.369</nowiki>
|Reactive
|[[File:33areactiveazp2518.png|thumb]]
|}


These trajectories represent the reaction from HF + H to F + H<sub>2</sub>, which is a endothermic reaction, and the position of the transition state is in the late part of the reaction profile. As we seem from the first three trajectories with high initial kinetic energy with most of the kinetic energy distributed in the in the translational mode, a high total initial kinetic energy is needed for the trajectories to be reactive. But if we distribute more of initial kinetic energy in the vibration mode (higher P<sub>FH</sub>), a lower total kinetic energy is needed for the trajectories to be reactive. These cases agree with the Polanyi's empirical rule that the vibrational energy is more effective than the translational energy to overcome the barrier of a later transition state reaction.

{{fontcolor1|green| This is a perfect discussion with obvious understanding of Polanyi's rule! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 22:58, 24 June 2020 (BST)}}