File:H2+f ts distances pg.png

2018-05-18T13:37:25Z

Pg2116:

2018-05-17T17:42:36Z

Pg2116:

MRD:anythingpg2116

2018-05-17T17:42:18Z

Pg2116:

This potential surface is a function of two variables: r1 and r2. Functions of two variables can show three different points: minimum, maximum (but this one is not addressed in the question) and a saddle point. Local minimum is a point at which the function has the smallest value for the given r1 and r2 values. Therefore, the gradient of the tangent with respect to both variables at this point equals zero. The saddle point, which in this case is the location of the transition structure, is a point of the potential surface at which the function appears to reach a local minimum with respect to one of the variables and a local maximum with respect to the other variable. Therefore, the gradient of the tangent with respect to both variables is zero, too. Since all the gradients at the minimum and saddle point equal zero, these two points cannot be distinguished by the first derivative of the function. This needs to be done by looking at the curvature of the potential surface around these points, which is related to the second derivative of the function. At the minimum, the second derivative with respect to both variables is positive so the function appears to be concave up in both directions around this point. At the saddle point, the second derivative with respect to one variable is positive and negative with respect to the other one so the function appears to be concave up in one direction and concave down in the other.

Q2
The distance was estimated to be 0.9079 Å. At this point, there is very limited oscillation around the ridge which indicates that this distance is very close to the saddle point of the potential surface. Also, the limited oscillation is reflected in the internuclear distances vs. time plot which shows two straight lines (variation in A-B distance overlaps with B-C distance): this proves that the force acting on the system is very close to zero, as there is virtually no vibrating related to A-B or B-C, so the system must be very close to the saddle point of the potential energy surface, i.e. the transition state.

[[File:r_vs_t1_pg2116.png]]

Q3
The difference is that the "dynamics trajectory" is curvy so the B-C distance oscillates while the B-C molecule is getting more distant from atom A. This gives a more realistic view of the system as atoms in molecules are not static. The reaction path, on the other hand, does not take the vibrations into account as it corresponds to the minimum energy trajectory so it is a "non-curvy" line that goes along the valley floor. If it was curvy, it would not correspond to the minimum energy trajectory as the potential energy would oscillate and hence would not be minimal at every instant.

Q4

{| class="wikitable" border="1"
|+ Trajectory and total energy at different starting momenta (r1 = 0.74 Å, r2 = 2.0 Å)
! p1 !! p2 !! Total energy !! Trajectory
|-
| -1.25 || -2.5 || -99.018 || reactive
|-
| -1.5 || -2.0 || -100.456 || unreactive
|-
| -1.5 || -2.5 || -98.956 || reactive
|-
| -2.5 || -5.0 || -84.956 || unreactive
|-
| -2.5 || -5.2 || -83.416 || reactive
|}

Case1: H2 slightly vibrates (almost no vibrations) as it approaches H. Once they get close enough, an exchange of an atom takes place and the new H2 molecule leaves the "complex" with significant vibrations.

[[File:table_case1_pg.png]]

Case2: The vibrating H2 molecule is getting closer to the H atom but they do not have sufficiently high momentum to come close enough to reach and overcome the transition state so the reactants cannot interact and no substitution takes place.

[[File:table_case2_pg.png]]

Case3: The conditions are the same as in case 2 but now the H atom has a more negative momentum which gives it a velocity high enough to come close the to the H2 molecule and overcome the transition state so the substitution takes place.

[[File:table_case3_pg.png]]

Case4: The H2 molecule, barely vibrating, approaches the H atom and the transition state is overcome (so the substitution takes place) but the resulting conformation of the system is not stable enough so the system returns back to the reactants.

[[File:table_case4_pg.png]]

Case5: In this case, the reaction proceeds in the same way as in case 4 but now the system formed by the returning to the reactants is not stable enough either and collapses to the products where it reaches a more stable conformation and stays there so the substitution eventually does take place.

[[File:table_case5_pg.png]]

Q5
1.) There is a constant equilibrium between the reactants and the structure associated with the transition state.
2.) Boltzmann distribution can be used to model the energy distribution associated with the reacting particles.
3.) When the reactants form the transition state, the transition state conformation can only collapse towards the products.

As proved in the previous question in case4, the third assumption of the TST is not always true and the transition state can collapse back to the reactants. This means that a certain fraction of transition states does not result in the formation of products so the actual (experimental) rate can be lower than the one predicted by this theory.

Q6
F + H2 is exothermic because there is a large drop in potential energy after the substitution proceeds. Therefore, H + HF, which is the reverse process, is endothermic. Whether a reaction is exothermic or endothermic depends on the bond strengths: exothermic reactions involve breaking of a weak bond and formation of a strong bond because the energy released upon the strong bond formation can offset the energy required to break the weak bond; the excess energy is lost as heat. In endothermic reactions, the opposite happens: the low amount of energy released when a weak bond is formed is not sufficient to form the strong bond so energy has to be supplied from the surroundings. In this case, the F-H bond is stronger than the H-H bond so the reaction F + H2 is exothermic.

Q7
r(H-H) 0.745 Å, r(H-F) 1.8107 Å.

Q8
To determine Ea for H2 + F, the H-H distance was increased by 0.01, 100000 steps were selected for the MEP calculation and energy vs. time was plotted. From the graph, the minimum corresponds to the total energy associated with the reactants that do not approach each other, and the maximum corresponds to the total energy in a system very close to the transition state. The activation energy is the difference between these two values.

[[File:Ea_f+h2_pg.png]]

Ea(F+H2)=-103.751-(-103.944)=0.193 kcal/mol

Ea for the reverse reaction was determined in a similar way: the H-H distance was decreased by 0.01 and also 100000 steps were selected. The activation energy was determined as the difference between the minimum (corresponding to HF and H that do not interact) and maximum (corresponding to a system near the transition state).

[[File:Ea_hf+h_pg.png]]

Ea(HF+H)=-103.673-(-133.885)=30.212 kcal/mol

Q9
This question was already mostly answered in Q6. When a strong bond is formed, certain amount of the energy released during this process can offset the energy required to break the weak bond in the reactants; the excess energy, due to the conservation of energy principle, has to be lost into the surroundings in a certain form, in most cases as heat. This can be confirmed by calorimetry which studies the flow of heat throughout a reaction by measuring the change in temperature.

Q10
Polanyi's rules state that reactions with late transitions states are more efficiently promoted by vibrational energy while early transition state reactions are more efficiently promoted by transitional energy (https://pubs.acs.org/doi/pdf/10.1021/jz301649w). The F + H2 reaction is exothermic so has an early transition state. Therefore, based on the Polanyi's rules, it can be expected that this reaction will be more likely to take place with higher translational energy. This was tested with two sets of parameters:

1.) The distances selected were r(F-H)=2.4 Å and r(H-H)=0.74 Å so that the reaction starts at the side of reactants at the bottom of the well. The momentum p(HH) was set to -2.3 which results in a relatively high vibartional energy while the momentum p(FH) was kept low at -0.5 which corresponds to a very low translational motion. The resulting trajectory was unreactive:

[[File:polanyi_exo_case1_pg.png]]

2.) In this set, the following parameters were selected: r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=0.1 (low vibrational energy) and p(HF)=-0.8 (slight increase in translational energy compared to the first set). These parameters resulted in a reactive trajectory:

[[File:polanyi_exo_case2_pg.png]]

These results are in accordance with the Polanyi's rules: in the first set, the translational motion was kept minimal so that the reaction was governed by vibrational energy and it was found that the products were not formed. In the second set, the translational energy was slightly increased while the vibrational energy was decreased dramatically and the resulting trajectory was reactive. This means that the vibrational energy does not have a significant impact on the reaction whereas translational energy affects the reaction significantly as only a small increase could lead to reactants to products. This matched the prediction based on Polanyi's rules.

A similar experiment was done for the reverse reaction which is endothermic, therefore having a late transition state. According to the rules, the exact opposite should be found than in the first experiment. The following sets of parameters were selected:

1.) r(F-H)=2.4 Å, r(H-H)=0.74 Å (these distances correspond to the valley floor at the side of the reactants), p(HH)=-5 (high translational energy) and p(HF)=-0.4 (low vibrational energy). The trajectory was unreactive:

[[File:polanyi_endo_case1_pg.png]]

2.) r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=-1 (low translational energy) and p(HF)=-0.8 (small increase in vibrational energy). The trajectory was reactive:

[[File:polanyi_endo_case2_pg.png]]

The result shows that the vibrational energy is more important for endothermic reactions: When very high translational energy and low vibrational energy were set, the reaction did not take place but when the vibrational energy was slightly increased while keeping the translational energy relatively low, the products were formed. Therefore, the prediction based on Polanyi's rules was confirmed.

MRD:anythingpg2116

2018-05-17T17:30:59Z

Pg2116:

This potential surface is a function of two variables: r1 and r2. Functions of two variables can show three different points: minimum, maximum (but this one is not addressed in the question) and a saddle point. Local minimum is a point at which the function has the smallest value for the given r1 and r2 values. Therefore, the gradient of the tangent with respect to both variables at this point equals zero. The saddle point, which in this case is the location of the transition structure, is a point of the potential surface at which the function appears to reach a local minimum with respect to one of the variables and a local maximum with respect to the other variable. Therefore, the gradient of the tangent with respect to both variables is zero, too. Since all the gradients at the minimum and saddle point equal zero, these two points cannot be distinguished by the first derivative of the function. This needs to be done by looking at the curvature of the potential surface around these points, which is related to the second derivative of the function. At the minimum, the second derivative with respect to both variables is positive so the function appears to be concave up in both directions around this point. At the saddle point, the second derivative with respect to one variable is positive and negative with respect to the other one so the function appears to be concave up in one direction and concave down in the other.

Q2
The distance was estimated to be 0.9079 Å. At this point, there is very limited oscillation around the ridge which indicates that this distance is very close to the saddle point of the potential surface. Also, the limited oscillation is reflected in the internuclear distances vs. time plot which shows two straight lines (variation in A-B distance overlaps with B-C distance): this proves that the force acting on the system is very close to zero, as there is virtually no vibrating related to A-B or B-C, so the system must be very close to the saddle point of the potential energy surface, i.e. the transition state.

[[File:r_vs_t1_pg2116.png]]

Q3
The difference is that the "dynamics trajectory" is curvy so the B-C distance oscillates while the B-C molecule is getting more distant from atom A. This gives a more realistic view of the system as atoms in molecules are not static. The reaction path, on the other hand, does not take the vibrations into account as it corresponds to the minimum energy trajectory so it is a "non-curvy" line that goes along the valley floor. If it was curvy, it would not correspond to the minimum energy trajectory as the potential energy would oscillate and hence would not be minimal at every instant.

Q4

{| class="wikitable" border="1"
|+ Trajectory and total energy at different starting momenta (r1 = 0.74 Å, r2 = 2.0 Å)
! p1 !! p2 !! Total energy !! Trajectory
|-
| -1.25 || -2.5 || -99.018 || reactive
|-
| -1.5 || -2.0 || -100.456 || unreactive
|-
| -1.5 || -2.5 || -98.956 || reactive
|-
| -2.5 || -5.0 || -84.956 || unreactive
|-
| -2.5 || -5.2 || -83.416 || reactive
|}

Case1: H2 slightly vibrates (almost no vibrations) as it approaches H. Once they get close enough, an exchange of an atom takes place and the new H2 molecule leaves the "complex" with significant vibrations.

[[File:table_case1_pg.png]]

Case2: The vibrating H2 molecule is getting closer to the H atom but they do not have sufficiently high momentum to come close enough to reach and overcome the transition state so the reactants cannot interact and no substitution takes place.

[[File:table_case2_pg.png]]

Case3: The conditions are the same as in case 2 but now the H atom has a more negative momentum which gives it a velocity high enough to come close the to the H2 molecule and overcome the transition state so the substitution takes place.

[[File:table_case3_pg.png]]

Case4: The H2 molecule, barely vibrating, approaches the H atom and the transition state is overcome (so the substitution takes place) but the resulting conformation of the system is not stable enough so the system returns back to the reactants.

[[File:table_case4_pg.png]]

Case5: In this case, the reaction proceeds in the same way as in case 4 but now the system formed by the returning to the reactants is not stable enough either and collapses to the products where it reaches a more stable conformation and stays there so the substitution eventually does take place.

[[File:table_case5_pg.png]]

Q5
1.) There is a constant equilibrium between the reactants and the structure associated with the transition state.
2.) Boltzmann distribution can be used to model the energy distribution associated with the reacting particles.
3.) When the reactants form the transition state, the transition state conformation can only collapse towards the products.

As proved in the previous question in case4, the third assumption of the TST is not always true and the transition state can collapse back to the reactants. This means that a certain fraction of transition states does not result in the formation of products so the actual (experimental) rate can be lower than the one predicted by this theory.

Q6
F + H2 is exothermic because there is a large drop in potential energy after the substitution proceeds. Therefore, H + HF, which is the reverse process, is endothermic. Whether a reaction is exothermic or endothermic depends on the bond strengths: exothermic reactions involve breaking of a weak bond and formation of a strong bond because the energy released upon the strong bond formation can offset the energy required to break the weak bond; the excess energy is lost as heat. In endothermic reactions, the opposite happens: the low amount of energy released when a weak bond is formed is not sufficient to form the strong bond so energy has to be supplied from the surroundings. In this case, the F-H bond is stronger than the H-H bond so the reaction F + H2 is exothermic.

Q7
r(H-H) 0.745 Å, r(H-F) 1.8107 Å.

Q8
To determine Ea for H2 + F, the H-H distance was increased by 0.01, 100000 steps were selected for the MEP calculation and energy vs. time was plotted. From the graph, the minimum corresponds to the total energy associated with the reactants that do not approach each other, and the maximum corresponds to the total energy in a system very close to the transition state. The activation energy is the difference between these two values.

[[File:Ea_f+h2_pg.png]]

Ea(F+H2)=-103.751-(-103.944)=0.193 kcal/mol

Ea for the reverse reaction was determined in a similar way: the H-H distance was decreased by 0.01 and also 100000 steps were selected. The activation energy was determined as the difference between the minimum (corresponding to HF and H that do not interact) and maximum (corresponding to a system near the transition state).

[[File:Ea_hf+h_pg.png]]

Ea(HF+H)=-103.673-(-133.885)=30.212 kcal/mol

Q9
This question was already mostly answered in Q6. When a strong bond is formed, certain amount of the energy released during this process can offset the energy required to break the weak bond in the reactants; the excess energy, due to the conservation of energy principle, has to be lost into the surroundings in a certain form, in most cases as heat. This can be confirmed by calorimetry which studies the flow of heat throughout a reaction by measuring the change in temperature.

Q10
Polanyi's rules state that reactions with late transitions states are more efficiently promoted by vibrational energy while early transition state reactions are more efficiently promoted by transitional energy (https://pubs.acs.org/doi/pdf/10.1021/jz301649w). The F + H2 reaction is exothermic so has an early transition state. Therefore, based on the Polanyi's rules, it can be expected that this reaction will be more likely to take place with higher translational energy. This was tested with two sets of parameters:

1.) The distances selected were r(F-H)=2.4 Å and r(H-H)=0.74 Å so that the reaction starts at the side of reactants at the bottom of the well. The momentum p(HH) was set to -2.3 which results in a relatively high vibartional energy while the momentum p(FH) was kept low at -0.5 which corresponds to a very low translational motion. The resulting trajectory was unreactive:

[[File:polanyi_exo_case1_pg.png]]

2.) In this set, the following parameters were selected: r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=0.1 (low vibrational energy) and p(HF)=-0.8 (slight increase in translational energy compared to the first set). These parameters resulted in a reactive trajectory:

[[File:polanyi_exo_case2_pg.png]]

These results are in accordance with the Polanyi's rules: in the first set, the translational motion was kept minimal so that the reaction was governed by vibrational energy and it was found that the products were not formed. In the second set, the translational energy was slightly increased while the vibrational energy was decreased dramatically and the resulting trajectory was reactive. This means that the vibrational energy does not have a significant impact on the reaction whereas translational energy affects the reaction significantly as only a small increase could lead to reactants to products. This matched the prediction based on Polanyi's rules.

A similar experiment was done for the reverse reaction which is endothermic, therefore having a late transition state. According to the rules, the exact opposite should be found than in the first experiment. The following sets of parameters were selected:

1.) r(F-H)=2.4 Å, r(H-H)=0.74 Å (these distances correspond to the valley floor at the side of the reactants), p(HH)=-5 (high translational energy) and p(HF)=-0.4 (low vibrational energy). The trajectory was unreactive:

[[File:polanyi_endo_case1_pg.png]]

2.) r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=-1 (low translational energy) and p(HF)=-0.8 (small increase in vibrational energy). The trajectory was reactive:

[[File:polanyi_endo_case2_pg.png]]

The result shows that the vibrational energy is more important for endothermic reactions: When very high translational energy and low vibrational energy were set, the reaction did not take place but when the vibrational energy was slightly increased while keeping the translational energy relatively low, the products were formed. Therefore, the prediction based on Polanyi's rules was confirmed.

MRD:anythingpg2116

2018-05-17T17:24:53Z

Pg2116:

This potential surface is a function of two variables: r1 and r2. Functions of two variables can show three different points: minimum, maximum (but this one is not addressed in the question) and a saddle point. Local minimum is a point at which the function has the smallest value for the given r1 and r2 values. Therefore, the gradient of the tangent with respect to both variables at this point equals zero. The saddle point, which in this case is the location of the transition structure, is a point of the potential surface at which the function appears to reach a local minimum with respect to one of the variables and a local maximum with respect to the other variable. Therefore, the gradient of the tangent with respect to both variables is zero, too. Since all the gradients at the minimum and saddle point equal zero, these two points cannot be distinguished by the first derivative of the function. This needs to be done by looking at the curvature of the potential surface around these points, which is related to the second derivative of the function. At the minimum, the second derivative with respect to both variables is positive so the function appears to be concave up in both directions around this point. At the saddle point, the second derivative with respect to one variable is positive and negative with respect to the other one so the function appears to be concave up in one direction and concave down in the other.

Q2
The distance was estimated to be 0.9079 Å. At this point, there is very limited oscillation around the ridge which indicates that this distance is very close to the saddle point of the potential surface. Also, the limited oscillation is reflected in the internuclear distances vs. time plot which shows two straight lines (variation in A-B distance overlaps with B-C distance): this proves that the force acting on the system is very close to zero, as there is virtually no vibrating related to A-B or B-C, so the system must be very close to the saddle point of the potential energy surface, i.e. the transition state.

[[File:r_vs_t1_pg2116.png]]

Q3
The difference is that the "dynamics trajectory" is curvy so the B-C distance oscillates while the B-C molecule is getting more distant from atom A. This gives a more realistic view of the system as atoms in molecules are not static. The reaction path, on the other hand, does not take the vibrations into account as it corresponds to the minimum energy trajectory so it is a "non-curvy" line that goes along the valley floor. If it was curvy, it would not correspond to the minimum energy trajectory as the potential energy would oscillate and hence would not be minimal at every instant.

Q4

{| class="wikitable" border="1"
|+ Trajectory and total energy at different starting momenta (r1 = 0.74 Å, r2 = 2.0 Å)
! p1 !! p2 !! Total energy !! Trajectory
|-
| -1.25 || -2.5 || -99.018 || reactive
|-
| -1.5 || -2.0 || -100.456 || unreactive
|-
| -1.5 || -2.5 || -98.956 || reactive
|-
| -2.5 || -5.0 || -84.956 || unreactive
|-
| -2.5 || -5.2 || -83.416 || reactive
|}

Case1: H2 slightly vibrates (almost no vibrations) as it approaches H. Once they get close enough, an exchange of an atom takes place and the new H2 molecule leaves the "complex" with significant vibrations.

[[File:table_case1_pg.png]]

Case2: The vibrating H2 molecule is getting closer to the H atom but they do not have sufficiently high momentum to come close enough to reach and overcome the transition state so the reactants cannot interact and no substitution takes place.

[[File:table_case2_pg.png]]

Case3: The conditions are the same as in case 2 but now the H atom has a more negative momentum which gives it a velocity high enough to come close the to the H2 molecule and overcome the transition state so the substitution takes place.

[[File:table_case3_pg.png]]

Case4: The H2 molecule, barely vibrating, approaches the H atom and the transition state is overcome (so the substitution takes place) but the resulting conformation of the system is not stable enough so the system returns back to the reactants.

[[File:table_case4_pg.png]]

Case5: In this case, the reaction proceeds in the same way as in case 4 but now the system formed by the returning to the reactants is not stable enough either and collapses to the products where it reaches a more stable conformation and stays there so the substitution eventually does take place.

[[File:table_case5_pg.png]]

Q5
1.) There is a constant equilibrium between the reactants and the structure associated with the transition state.
2.) Boltzmann distribution can be used to model the energy distribution associated with the reacting particles.
3.) When the reactants form the transition state, the transition state conformation can only collapse towards the products.

As proved in the previous question in case4, the third assumption of the TST is not always true and the transition state can collapse back to the reactants. This means that a certain fraction of transition states does not result in the formation of products so the actual (experimental) rate can be lower than the one predicted by this theory.

Q6
F + H2 is exothermic because there is a large drop in potential energy after the substitution proceeds. Therefore, H + HF, which is the reverse process, is endothermic. Whether a reaction is exothermic or endothermic depends on the bond strengths: exothermic reactions involve breaking of a weak bond and formation of a strong bond because the energy released upon the strong bond formation can offset the energy required to break the weak bond; the excess energy is lost as heat. In endothermic reactions, the opposite happens: the low amount of energy released when a weak bond is formed is not sufficient to form the strong bond so energy has to be supplied from the surroundings. In this case, the F-H bond is stronger than the H-H bond so the reaction F + H2 is exothermic.

Q7
r(H-H) 0.745 Å, r(H-F) 1.8107 Å.

Q8
To determine Ea for H2 + F, the H-H distance was increased by 0.01, 100000 steps were selected for the MEP calculation and energy vs. time was plotted. From the graph, the minimum corresponds to the total energy associated with the reactants that do not approach each other, and the maximum corresponds to the total energy in a system very close to the transition state. The activation energy is the difference between these two values.

[[File:Ea_f+h2_pg.png]]

Ea(F+H2)=-103.751-(-103.944)=0.193 kcal/mol

Ea for the reverse reaction was determined in a similar way: the H-H distance was decreased by 0.01 and also 100000 steps were selected. The activation energy was determined as the difference between the minimum (corresponding to HF and H that do not interact) and maximum (corresponding to a system near the transition state).

[[File:Ea_hf+h_pg.png]]

Ea(HF+H)=-103.673-(-133.885)=30.212 kcal/mol

Q9
This question was already mostly answered in Q6. When a strong bond is formed, certain amount of the energy released during this process can offset the energy required to break the weak bond in the reactants; the excess energy, due to the conservation of energy principle, has to be lost into the surroundings in a certain form, in most cases as heat. This can be confirmed by calorimetry which studies the flow of heat throughout a reaction by measuring the change in temperature.

Q10
Polanyi's rules state that reactions with late transitions states are more efficiently promoted by vibrational energy while early transition state reactions are more efficiently promoted by transitional energy (https://pubs.acs.org/doi/pdf/10.1021/jz301649w). The F + H2 reaction is exothermic so has an early transition state. Therefore, based on the Polanyi's rules, it can be expected that this reaction will be more likely to take place with higher translational energy. This was tested with two sets of parameters:

1.) The distances selected were r(F-H)=2.4 Å and r(H-H)=0.74 Å so that the reaction starts at the side of reactants at the bottom of the well. The momentum p(HH) was set to -2.3 which results in a relatively high vibartional energy while the momentum p(FH) was kept low at -0.5 which corresponds to a very low translational motion. The resulting trajectory was unreactive:

[[File:polanyi_exo_case1_pg.png]]

2.) In this set, the following parameters were selected: r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=0.1 (low vibrational energy) and p(HF)=-0.8 (slight increase in translational energy compared to the first set). These parameters resulted in a reactive trajectory:

[[File:polanyi_exo_case2_pg.png]]

These results are in accordance with the Polanyi's rules: in the first set, the translational motion was kept minimal so that the reaction was governed by vibrational energy and it was found that the products were not formed. In the second set, the translational energy was slightly increased while the vibrational energy was decreased dramatically and the resulting trajectory was reactive. This means that the vibrational energy does not have a significant impact on the reaction whereas translational energy affects the reaction significantly as only a small increase could lead to reactants to products. This matched the prediction based on Polanyi's rules.

A similar experiment was done for the reverse reaction which is endothermic, therefore having a late transition state. According to the rules, the exact opposite should be found than in the first experiment. The following sets of parameters were selected:

1.) r(F-H)=2.4 Å, r(H-H)=0.74 Å (these distances correspond to the valley floor at the side of the reactants), p(HH)=-5 (high translational energy) and p(HF)=-0.4 (low vibrational energy). The trajectory was unreactive:

[[File:polanyi_endo_case1_pg.png]]

2.) r(F-H)=2.4 Å, r(H-H)=0.74 Å, p(HH)=-1 (low translational energy) and p(HF)=-0.8 (small increase in vibrational energy). The trajectory was reactive:

[[File:polanyi_endo_case2_pg.png]]

File:Polanyi exo case2 pg.png

2018-05-17T17:22:44Z

Pg2116:

File:Polanyi exo case1 pg.png

2018-05-17T17:22:30Z

Pg2116:

File:Polanyi endo case2 pg.png

2018-05-17T17:22:14Z

Pg2116:

2018-05-17T13:27:13Z

Pg2116:

MRD:anythingpg2116

2018-05-17T13:26:35Z

Pg2116: