2020-05-08T13:49:45Z

At6318: /* Distribution of energy between different modes */

== EXERCISE 1: H + H2 system ==

=== Transition State ===

On a potential energy diagram the transition state can mathematically be defined as a '''saddle point''' and as the ''maximum'' on the '''minimum energy path''' between the reactants and products. The transition state can be identified by looking for where the gradient of the potential is zero <math>\frac{\mathrm \delta}{\mathrm \delta r_i} \big( V(r_i) \big) =0 </math>. It can be distinguished from a local minimum by taking the second derivates and ensuring they are opposite signs as can be seen in the mathematical definition in '''''figure 1''''' whereas for a local minimum both second derivatives would be positive. The opposite signs of the second derivatives taken from orthogonal vectors show that the saddle point (transition state) is a minimum for one vector and a maximum for the other.

'''<math>\frac{\mathrm \delta}{\mathrm \delta r_1} \big( V(r_1) \big) \frac{\mathrm \delta}{\mathrm \delta r_2} \big( V(r_2) \big) = 0</math>
and
<math>\frac{\mathrm \delta^2}{\mathrm \delta q_1^2} \big( V(q_1) \big) >0</math>
and
<math>\frac{\mathrm \delta^2}{\mathrm \delta q_2^2} \big( V(q_2) \big) <0</math>'''

'''''figure 1'''''-''The mathematical definition of a transition state on a potential energy surface ''

Two more vectors can be defined from <math>r_1</math> and <math>r_2</math> these are diagonal vectors called <math>q_1</math> and <math>q_2</math> and the second derivatives of these will indicate a minimum and a maximum (one more than zero, the other less than zero respectively).

This can be represented on a surface plot, '''''figure 2''''':

[[File:transition state surface plot EDITEDat6318.png|400px|Figure 2]]

=== Estimation of the transition state position ===

By using the internuclear distance plot as shown in '''''figure 2''''' with the initial distances being the same and the initial momentum is zero. The initial distance is changed until there is no more oscillations and the gradient on the internuclear distance v time plot is zero and the forces are 0. This occurs at <math>r_1 =r_2 = 90.775 pm</math>.

Using the hessian to get an estimate for the ts bond lengths.

[[File:rts 90.775at6318.png|400px]]

'''''figure 2'''''

=== Comparison of the mep and dynamic trajectories===

When the initial conditions are set to <math>r_1=r_{ts}+1 pm</math> , <math>r_2=r_{ts}</math> and <math>p_1=p_2=0</math> .

For mep (minimum energy pathway) the trajectory decreases smoothly in potential energy from the transition state and to the floor of the surface plot. For the trajectory with the same initial conditions in dynamics it travels the same path as the mep but oscillates, showing vibrations. The main differences observed in these plots are that there are no oscillations in the mep and the path is shorter this highlights how kinetic energy is zero in the mep as it's infinitely slow and doesn't allow the atom to build up momentum over each time step therefore doesn't oscillate, whereas in the dynamics which is a physical representation of the atoms can build up kinetic energy so therefore starts to oscillate. This is also highlighted in the "Momenta vs Time" plot where in mep it is zero for the whole of time and in dynamic it rises and oscillates around <math>2.5 gmol^{-1}pm.fs^{-1}</math> for the <math>H_2</math> and rises to <math>5 gmol^{-1}pm.fs^{-1}</math> for the solo <math>H</math> atom. As seen below with the first image being due to dynamics caluclations and the second due to mep calculations:

[[File:Momentavtimedynamicsat6318.png|400px]]

[[File:Momentavtimemepat6318.png|400px]]

The difference is highlighted in the contour plots below with the first using dynamics calculation type and the second with mep calculation type:

[[File:dynamicscontourat6318.png|400px]]

[[File:mepconourat6318.png|400px]]

=== Reactive and unreactive trajectories ===

* For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.
{| class="wikitable" border=1
! 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 ||-414.28 ||Yes ||Reacts as the molecules have sufficient energy to overcome the activation energy barrier and pass through the transition state to form the product. ||[[File:table1at6318.png|400px]]
|-
| -3.1 || -4.1 ||-420.08 ||No ||Doesn't react as the molecules don't have sufficient energy to overcome the activation energy barrier and pass through the transition state and instead rolls back down the potential after the collision as reactants. ||[[File:table2at6318.png|400px]]
|-
| -3.1 || -5.1 ||-413.98 ||Yes ||Reacts as the molecules have sufficient energy to overcome the activation energy barrier and pass through the transition state to form the product. The velocity of the product after the collision and the velocity of the reactant is higher than in row 1 as the total energy in this system is higher. ||[[File:table3at6318.png|400px]]
|-
| -5.1 || -10.1 ||-357.28 ||No ||Doesn't react as although the reactants have sufficient energy to overcome the activation energy barrier and pass through the transition state and form the product after this the product still has excess energy which is sufficient to pass back over the barrier and through the transition state to form reactants again. ||[[File:table4at6318.png|400px]]
|-
| -5.1 || -10.6 ||-349.48 ||Yes ||Reacts as the reactants have sufficient energy to overcome the potential barrier and pass through the transition state to form the product and then pass back over the potential barrier to form the reactants again and then to finally pass back over the barrier to form the products. ||[[File:table5at6318.png|400px]]
|}

This table concludes that there must be sufficient energy to overcome the potential barrier and pass through the transition state otherwise the product won't form as in row 2, however if the reactants have an excess of energy the products are able to pass back over the barrier and form reactants again and not produce the product as in row 4. This shows how having sufficient kinetic energy isn't the only factor in whether a collision leads to a successful reaction.

=== Transition State Theory ===

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

Transition state theory predictions for reaction rate values don't include other factors that would prevent a successful reaction from occurring whereas transition state theory states that if the reactants have more kinetic energy than a certain value (energy of the transition state) then a reaction will successfully occur which is not the case in real experimental values. This therefore shows that the predictions for reaction rate from transition state theory will be much larger than experimental values.

== EXERCISE 2: F - H - H system ==

=== PES inspection ===

By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

For F + H2 the reaction is exothermic as on the potential energy surface it can be seen that the reactant energy is higher than the product energy therefore indicating a loss of energy in the reaction through heat. Therefore for H + HF reaction it's reactants are the products of the F +H2 reaction and it's products are F + H2 therefore is an endothermic reaction, with the opposite direction of the figure below, which shows the F + H2 exothermic reaction:

[[File:exothermicat6318.png]]

This therefore shows that the total energy of H2 and F are higher than HF and H so the bond energy (strength) of the HF bond is much larger than that of H2 as energy is required to be inputted for the reaction to occur (endothermic) as more energy is required to break the HF bond than used to form the bond of H2.

=== Approximate position of the transition state ===

The position of the transition state was found using the method above for H + H2; by adjusting the initial positions until the force on the atoms are zero and the gradient for internuclear distances v time graph is zero. It is shown that the transition state is when AB distance = 74.5 pm and BC distance= 181 pm with A=H, B=H and C=F. This is confirmed by the hessian which confirms a saddle point. The internuclear distance-time graph is shown below:

[[File:internucleardistanceHHFat6318.png]]

=== Activation energy for both reactions ===

This can be found by finding the energy at the transition state whose position was found above, as -433.98 kJ.mol-1. The activation energy can then be deduced from the difference between the reactants energy and the transition state energy. The reactants energy was found by putting atoms in a molecule their bond length apart and the atom not in the molecule 1000 pm away so that there would be no interaction between them. The bond lengths were found by trial and error until the forces between the atoms were 0. The bond length for H2 was found to be 74 pm and the bond length for HF was found to be 92 pm. Therefore the activation energy for H2 + F is 1.12 kJmol-1 as the reactants energy is -435.10 kJ.mol-1. The activation energy for HF + H is 126.72 kJ.mol-1 as the reactants energy is -560.70 kJ.mol-1.

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

By setting the initial conditions so that the H2 + F reaction occurs with AB= 74.5 pm BC= 200 pm and the initial momenta as -0.5 g.mol-1.pm.fs-1 and -2 g.mol-1.pm.fs-1 are respectively, where A and B are H and C is F. The reaction is shown to occur as below in the contour plot:

[[File:countourreactionenergyat6318.png]]

This reaction is then considered by using the energy v time graph as shown below:

[[File:energytimeHHFat6318.png]]

This shows that as total energy is a straight horizontal line (with gradient 0) that total energy is conserved and also shows the kinetic energy and potential energy oscillating out of phase so that kinetic energy + potential energy = total energy. After the reaction has taken place the potential energy's overall oscillation has shifted downward (decreased in energy) and the kinetic energy's overall oscillation has shifted upward (increased in energy). This shows the exothermic nature of the reaction as some of the potential energy is transferred to kinetic energy which in a real life experiment would be released as heat. Experimentally this can be measured through calorimetry which measures the heat and therefore the energy released from the reaction. These shifts are shown in the figures below:

[[File:potentialenergydecreaseat6318.png]]
[[File:kineticenergyincreaseat6318.png]]

In a momenta v time graph as shown below in figure it can be seen that the energy released from the reaction (difference of energies between reactants and products) is translated into vibrational energy in HF oscillations. This is seen as after the reaction occurs the BC (HF) oscillations increases and the AB plateaus to a constant momentum of ~ 5 g.mol-1.pm.fs-1, this is because energy is released from the formation of the HF bond which increases it's vibrational energy (which is measured as heat in calorimetry); the HH plateaus as the HF molecule and H atom move away from each other.

[[File:momentatimehhfat6318.png]]

=== Distribution of energy between different modes ===

The table below shows the reactivity and energy of the reaction H2 + F with differing initial momenta. The initial positions are set to AB = 74 pm and BC= 200 pm where A and B are H and C is F.

{| class="wikitable" border=1
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot/ '''kJ.mol-1''' !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.0 || -6.1 ||-402.5 ||||||[[File:2table1at6318.png|400px]]
|-
| -1.0 || -4.5 ||-417.9 || || ||[[File:2table2at6318.png|400px]]
|-
| -1.0 || -1.2 ||-433.4 || ||||[[File:2table3at6318.png|400px]]
|-
| -1.0 || 0 ||-433.6 || || ||[[File:2table4at6318.png|400px]]
|-
| -1.0 || 1.2 ||-431.9 || || ||[[File:2table5at6318.png|400px]]
|-
| -1.0 || 4.5 ||-408.9 || || ||[[File:2table6at6318.png|400px]]
|-
| -1.0 || 6.1 ||-390.3 || || ||[[File:2table7at6318.png|400px]]
|-
| -1.6 || 0.2 ||-432.4 || || ||[[File:2table8at6318.png|400px]]
|}

Let us now focus on the reverse reaction, H + HF.
Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).
Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

References?

MRD:at6318

2020-05-08T13:49:17Z

At6318: /* Distribution of energy between different modes */

== EXERCISE 1: H + H2 system ==

=== Transition State ===

On a potential energy diagram the transition state can mathematically be defined as a '''saddle point''' and as the ''maximum'' on the '''minimum energy path''' between the reactants and products. The transition state can be identified by looking for where the gradient of the potential is zero <math>\frac{\mathrm \delta}{\mathrm \delta r_i} \big( V(r_i) \big) =0 </math>. It can be distinguished from a local minimum by taking the second derivates and ensuring they are opposite signs as can be seen in the mathematical definition in '''''figure 1''''' whereas for a local minimum both second derivatives would be positive. The opposite signs of the second derivatives taken from orthogonal vectors show that the saddle point (transition state) is a minimum for one vector and a maximum for the other.

'''<math>\frac{\mathrm \delta}{\mathrm \delta r_1} \big( V(r_1) \big) \frac{\mathrm \delta}{\mathrm \delta r_2} \big( V(r_2) \big) = 0</math>
and
<math>\frac{\mathrm \delta^2}{\mathrm \delta q_1^2} \big( V(q_1) \big) >0</math>
and
<math>\frac{\mathrm \delta^2}{\mathrm \delta q_2^2} \big( V(q_2) \big) <0</math>'''

'''''figure 1'''''-''The mathematical definition of a transition state on a potential energy surface ''

Two more vectors can be defined from <math>r_1</math> and <math>r_2</math> these are diagonal vectors called <math>q_1</math> and <math>q_2</math> and the second derivatives of these will indicate a minimum and a maximum (one more than zero, the other less than zero respectively).

This can be represented on a surface plot, '''''figure 2''''':

[[File:transition state surface plot EDITEDat6318.png|400px|Figure 2]]

=== Estimation of the transition state position ===

By using the internuclear distance plot as shown in '''''figure 2''''' with the initial distances being the same and the initial momentum is zero. The initial distance is changed until there is no more oscillations and the gradient on the internuclear distance v time plot is zero and the forces are 0. This occurs at <math>r_1 =r_2 = 90.775 pm</math>.

Using the hessian to get an estimate for the ts bond lengths.

[[File:rts 90.775at6318.png|400px]]

'''''figure 2'''''

=== Comparison of the mep and dynamic trajectories===

When the initial conditions are set to <math>r_1=r_{ts}+1 pm</math> , <math>r_2=r_{ts}</math> and <math>p_1=p_2=0</math> .

For mep (minimum energy pathway) the trajectory decreases smoothly in potential energy from the transition state and to the floor of the surface plot. For the trajectory with the same initial conditions in dynamics it travels the same path as the mep but oscillates, showing vibrations. The main differences observed in these plots are that there are no oscillations in the mep and the path is shorter this highlights how kinetic energy is zero in the mep as it's infinitely slow and doesn't allow the atom to build up momentum over each time step therefore doesn't oscillate, whereas in the dynamics which is a physical representation of the atoms can build up kinetic energy so therefore starts to oscillate. This is also highlighted in the "Momenta vs Time" plot where in mep it is zero for the whole of time and in dynamic it rises and oscillates around <math>2.5 gmol^{-1}pm.fs^{-1}</math> for the <math>H_2</math> and rises to <math>5 gmol^{-1}pm.fs^{-1}</math> for the solo <math>H</math> atom. As seen below with the first image being due to dynamics caluclations and the second due to mep calculations:

[[File:Momentavtimedynamicsat6318.png|400px]]

[[File:Momentavtimemepat6318.png|400px]]

The difference is highlighted in the contour plots below with the first using dynamics calculation type and the second with mep calculation type:

[[File:dynamicscontourat6318.png|400px]]

[[File:mepconourat6318.png|400px]]

=== Reactive and unreactive trajectories ===

* For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.
{| class="wikitable" border=1
! 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 ||-414.28 ||Yes ||Reacts as the molecules have sufficient energy to overcome the activation energy barrier and pass through the transition state to form the product. ||[[File:table1at6318.png|400px]]
|-
| -3.1 || -4.1 ||-420.08 ||No ||Doesn't react as the molecules don't have sufficient energy to overcome the activation energy barrier and pass through the transition state and instead rolls back down the potential after the collision as reactants. ||[[File:table2at6318.png|400px]]
|-
| -3.1 || -5.1 ||-413.98 ||Yes ||Reacts as the molecules have sufficient energy to overcome the activation energy barrier and pass through the transition state to form the product. The velocity of the product after the collision and the velocity of the reactant is higher than in row 1 as the total energy in this system is higher. ||[[File:table3at6318.png|400px]]
|-
| -5.1 || -10.1 ||-357.28 ||No ||Doesn't react as although the reactants have sufficient energy to overcome the activation energy barrier and pass through the transition state and form the product after this the product still has excess energy which is sufficient to pass back over the barrier and through the transition state to form reactants again. ||[[File:table4at6318.png|400px]]
|-
| -5.1 || -10.6 ||-349.48 ||Yes ||Reacts as the reactants have sufficient energy to overcome the potential barrier and pass through the transition state to form the product and then pass back over the potential barrier to form the reactants again and then to finally pass back over the barrier to form the products. ||[[File:table5at6318.png|400px]]
|}

This table concludes that there must be sufficient energy to overcome the potential barrier and pass through the transition state otherwise the product won't form as in row 2, however if the reactants have an excess of energy the products are able to pass back over the barrier and form reactants again and not produce the product as in row 4. This shows how having sufficient kinetic energy isn't the only factor in whether a collision leads to a successful reaction.

=== Transition State Theory ===

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

Transition state theory predictions for reaction rate values don't include other factors that would prevent a successful reaction from occurring whereas transition state theory states that if the reactants have more kinetic energy than a certain value (energy of the transition state) then a reaction will successfully occur which is not the case in real experimental values. This therefore shows that the predictions for reaction rate from transition state theory will be much larger than experimental values.

== EXERCISE 2: F - H - H system ==

=== PES inspection ===

By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

For F + H2 the reaction is exothermic as on the potential energy surface it can be seen that the reactant energy is higher than the product energy therefore indicating a loss of energy in the reaction through heat. Therefore for H + HF reaction it's reactants are the products of the F +H2 reaction and it's products are F + H2 therefore is an endothermic reaction, with the opposite direction of the figure below, which shows the F + H2 exothermic reaction:

[[File:exothermicat6318.png]]

This therefore shows that the total energy of H2 and F are higher than HF and H so the bond energy (strength) of the HF bond is much larger than that of H2 as energy is required to be inputted for the reaction to occur (endothermic) as more energy is required to break the HF bond than used to form the bond of H2.

=== Approximate position of the transition state ===

The position of the transition state was found using the method above for H + H2; by adjusting the initial positions until the force on the atoms are zero and the gradient for internuclear distances v time graph is zero. It is shown that the transition state is when AB distance = 74.5 pm and BC distance= 181 pm with A=H, B=H and C=F. This is confirmed by the hessian which confirms a saddle point. The internuclear distance-time graph is shown below:

[[File:internucleardistanceHHFat6318.png]]

=== Activation energy for both reactions ===

This can be found by finding the energy at the transition state whose position was found above, as -433.98 kJ.mol-1. The activation energy can then be deduced from the difference between the reactants energy and the transition state energy. The reactants energy was found by putting atoms in a molecule their bond length apart and the atom not in the molecule 1000 pm away so that there would be no interaction between them. The bond lengths were found by trial and error until the forces between the atoms were 0. The bond length for H2 was found to be 74 pm and the bond length for HF was found to be 92 pm. Therefore the activation energy for H2 + F is 1.12 kJmol-1 as the reactants energy is -435.10 kJ.mol-1. The activation energy for HF + H is 126.72 kJ.mol-1 as the reactants energy is -560.70 kJ.mol-1.

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

By setting the initial conditions so that the H2 + F reaction occurs with AB= 74.5 pm BC= 200 pm and the initial momenta as -0.5 g.mol-1.pm.fs-1 and -2 g.mol-1.pm.fs-1 are respectively, where A and B are H and C is F. The reaction is shown to occur as below in the contour plot:

[[File:countourreactionenergyat6318.png]]

This reaction is then considered by using the energy v time graph as shown below:

[[File:energytimeHHFat6318.png]]

This shows that as total energy is a straight horizontal line (with gradient 0) that total energy is conserved and also shows the kinetic energy and potential energy oscillating out of phase so that kinetic energy + potential energy = total energy. After the reaction has taken place the potential energy's overall oscillation has shifted downward (decreased in energy) and the kinetic energy's overall oscillation has shifted upward (increased in energy). This shows the exothermic nature of the reaction as some of the potential energy is transferred to kinetic energy which in a real life experiment would be released as heat. Experimentally this can be measured through calorimetry which measures the heat and therefore the energy released from the reaction. These shifts are shown in the figures below:

[[File:potentialenergydecreaseat6318.png]]
[[File:kineticenergyincreaseat6318.png]]

In a momenta v time graph as shown below in figure it can be seen that the energy released from the reaction (difference of energies between reactants and products) is translated into vibrational energy in HF oscillations. This is seen as after the reaction occurs the BC (HF) oscillations increases and the AB plateaus to a constant momentum of ~ 5 g.mol-1.pm.fs-1, this is because energy is released from the formation of the HF bond which increases it's vibrational energy (which is measured as heat in calorimetry); the HH plateaus as the HF molecule and H atom move away from each other.

[[File:momentatimehhfat6318.png]]

=== Distribution of energy between different modes ===

The table below shows the reactivity and energy of the reaction H2 + F with differing initial momenta. The initial positions are set to AB = 74 pm and BC= 200 pm where A and B are H and C is F.

{| class="wikitable" border=1
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot/ '''kJ.mol-1''' !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.0 || -6.1 ||-402.5 ||||||[[File:2table1at6318.png|200px]]
|-
| -1.0 || -4.5 ||-417.9 || || ||[[File:2table2at6318.png|200px]]
|-
| -1.0 || -1.2 ||-433.4 || ||||[[File:2table3at6318.png|200px]]
|-
| -1.0 || 0 ||-433.6 || || ||[[File:2table4at6318.png|200px]]
|-
| -1.0 || 1.2 ||-431.9 || || ||[[File:2table5at6318.png|200px]]
|-
| -1.0 || 4.5 ||-408.9 || || ||[[File:2table6at6318.png|200px]]
|-
| -1.0 || 6.1 ||-390.3 || || ||[[File:2table7at6318.png|200px]]
|-
| -1.6 || 0.2 ||-432.4 || || ||[[File:2table8at6318.png|200px]]
|}

Let us now focus on the reverse reaction, H + HF.
Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).
Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

References?

2020-05-08T10:32:38Z

At6318: /* Activation energy for both reactions */