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MRD:AMM416 Molecular Dynamics Y2

2018-05-11T13:43:13Z

Amm416: /* Reaction Dynamics and Polanyi's Empirical Rules */

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state (TS) and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot resting on the ridge (Figure 1), suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 2), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_H-H_PES_TS.png|thumb|Figure 1: Potential energy surface of H-H-H system at the TS: structure oscillating on the ridge.]]

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 2: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations as seen in Figure 3 and 4.

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 4: Trajectory from dynamics calculation: oscillation shown.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF reaction is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed by the system from the surroundings at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state will resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:

Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore by optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:

The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction, since H + HF is the reverse reaction of F + H2.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which are also retrieved from the program by separating the reactants by a large distance and keeping the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard bond lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1 (Figure 5):

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure 5: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from Figure 6; kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure 7), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure 6: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure 7: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy of the F atom and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.5 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure 8), apart from values close to the extreme values chosen: e.g. reactive trajectories at pH-H = -2.8 due to barrier recrossing.
However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure 9).

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy of the incoming H atom and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure 10).
Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions (Figure 11).

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|left|Figure 8: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|center|Figure 9: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|thumb|left|Figure 10: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

[[File:Amm416_H+HF_reactive_p1-0.1_polanyi.png|thumb|center|Figure 11: Reactive trajectory of H + HF, starting from TS: pH-H = -0.1, pH-F = 4.0]]

MRD:AMM416 Molecular Dynamics Y2

2018-05-11T13:41:57Z

Amm416: /* Reaction Dynamics and Polanyi's Empirical Rules */

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state (TS) and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot resting on the ridge (Figure 1), suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 2), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_H-H_PES_TS.png|thumb|Figure 1: Potential energy surface of H-H-H system at the TS: structure oscillating on the ridge.]]

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 2: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations as seen in Figure 3 and 4.

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 4: Trajectory from dynamics calculation: oscillation shown.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF reaction is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed by the system from the surroundings at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state will resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:

Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore by optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:

The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction, since H + HF is the reverse reaction of F + H2.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which are also retrieved from the program by separating the reactants by a large distance and keeping the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard bond lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1 (Figure 5):

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure 5: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from Figure 6; kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure 7), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure 6: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure 7: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy of the F atom and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.5 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure 8), apart from values close to the extreme values chosen: e.g. reactive trajectories at pH-H = -2.8 due to barrier recrossing.

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|left|Figure 8: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure 9).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|center|Figure 9: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy of the incoming H atom and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure 10).

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|thumb|left|Figure 10: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions (Figure 11).

[[File:Amm416_H+HF_reactive_p1-0.1_polanyi.png|thumb|center|Figure 11: Reactive trajectory of H + HF, starting from TS: pH-H = -0.1, pH-F = 4.0]]

MRD:AMM416 Molecular Dynamics Y2

2018-05-11T13:40:52Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state (TS) and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot resting on the ridge (Figure 1), suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 2), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_H-H_PES_TS.png|thumb|Figure 1: Potential energy surface of H-H-H system at the TS: structure oscillating on the ridge.]]

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 2: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations as seen in Figure 3 and 4.

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 4: Trajectory from dynamics calculation: oscillation shown.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF reaction is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed by the system from the surroundings at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state will resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:

Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore by optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:

The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction, since H + HF is the reverse reaction of F + H2.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which are also retrieved from the program by separating the reactants by a large distance and keeping the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard bond lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1 (Figure 5):

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure 5: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from Figure 6; kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure 7), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure 6: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure 7: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy of the F atom and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.5 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure 8), apart from values close to the extreme values chosen: e.g. reactive trajectories at pH-H = -2.8 due to barrier recrossing.

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|Figure 8: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure 9).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|Figure 9: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy of the incoming H atom and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure 10).

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|thumb|Figure 10: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions (Figure 11).

[[File:Amm416_H+HF_reactive_p1-0.1_polanyi.png|thumb|Figure 11: Reactive trajectory of H + HF, starting from TS: pH-H = -0.1, pH-F = 4.0]]

MRD:AMM416 Molecular Dynamics Y2

2018-05-11T13:39:39Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state (TS) and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot resting on the ridge (Figure 1), suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 2), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_H-H_PES_TS.png|thumb|Figure 1: Potential energy surface of H-H-H system at the TS: structure oscillating on the ridge.]]

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 2: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations as seen in Figure 3 and 4.

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 4: Trajectory from dynamics calculation: oscillation shown.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF reaction is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed by the system from the surroundings at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state will resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:

Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore by optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:

The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction, since H + HF is the reverse reaction of F + H2.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which are also retrieved from the program by separating the reactants by a large distance and keeping the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard bond lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1 (Figure 5):

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure 5: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from Figure 6; kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure 7), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure 6: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure 7: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy of the F atom and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.5 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure 8), apart from values close to the extreme values chosen: e.g. reactive trajectories at pH-H = -2.8 due to barrier recrossing.

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|Figure 8: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure 9).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|Figure 9: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy of the incoming H atom and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure 10).

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|Figure 10: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions (Figure 11).

[[File:Amm416_H+HF_reactive_p1-0.1_polanyi.png|thumb|Figure 11: Reactive trajectory of H + HF, starting from TS: pH-H = -0.1, pH-F = 4.0]]

File:Amm416 H+HF reactive p1-0.1 polanyi.png

2018-05-11T13:32:40Z

Amm416:

File:Amm416 H-H PES TS.png

2018-05-11T13:16:57Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-11T13:08:08Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations as seen in Figure 2 and 3.

[[File:Amm416_MEP_trajectory.png|thumb|Figure 2: Trajectory from MEP calculation: vibrational motion not observed.]]

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 3: Trajectory from dynamics calculation: oscillation shown.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.8 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure A).

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|Figure A: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure B).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|Figure B: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure C).

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|Figure C: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions.

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T17:05:30Z

Amm416: /* F-H-H System */

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.8 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure A).

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|Figure A: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure B).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|Figure B: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

Comparatively, in the endothermic reverse reaction H + HF, the pH-H represents the translational energy and pH-F the vibrational energy in the H-F bond. If the translational energy of the H colliding is very large compared to the vibrational energy of HF, the resulting trajectory is unreactive as the H atom bounces of HF (Figure C).

[[File:Amm416_H+HF_unreactive_p1-3_polanyi.png|Figure C: Unreactive trajectory of H + HF: pH-H = -3.0, pH-F = -0.1]]

Since this is an endothermic reaction, if the translational energy is decreased and the vibrational energy of the H-F bond is increased, the reaction is successful as vibrational energy promotes endothermic reactions.

File:Amm416 H+HF unreactive p1-3 polanyi.png

2018-05-10T16:42:34Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T16:23:55Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'' '''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''' ''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'' '''3. Comment on how the mep and the trajectory you just calculated differ.''' ''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

=== Reactive and unreactive trajectories ===

'' '''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''' ''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'' '''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'' '''1. 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?''' ''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'' '''2. Locate the approximate position of the transition state.''' ''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

'' '''3. Report the activation energy for both reactions. ''' ''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

'' '''4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

'' '''5. 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. ''' ''

Polanyi's empirical rules state that higher amounts of vibrational energy, compared to translational energy, promote reactions with a late transition state, i.e. endothermic reactions. On the other hand, translational energy promotes reactions with an early transition state, i.e. exothermic reactions.

For instance, in the exothermic reaction F + H2, the pF-H represents the translational energy and pH-H the vibrational energy in the H-H bond. If pF-H is kept constant at a low value of -0.8 and pH-H is varied using relatively high values between such as -3 and 3, very few reactive trajectories were found (Figure A).

[[File:Amm416_F+H2_unreactive_p2-2_9_polanyi.png|thumb|Figure A: Unreactive trajectory of F + H2: pF-H = -0.5, pH-H = -2.9]]

However, if the translational energy is slightly increased to -0.8 and the vibrational energy decreased to -0.1, the resulting trajectory is reactive since more translational energy promotes exothermic reactions (Figure B).

[[File:Amm416_F+H2_reactive_polanyi.png|thumb|Figure B: Reactive trajectory of F + H2: pF-H = -0.8, pH-H = -0.1]]

File:Amm416 F+H2 unreactive p2-2 9 polanyi.png

2018-05-10T16:12:28Z

Amm416:

File:Amm416 F+H2 reactive polanyi.png

2018-05-10T16:12:12Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T15:20:16Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'' '''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'''3. Comment on how the mep and the trajectory you just calculated differ.'''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

=== Reactive and unreactive trajectories ===

'''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'''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?'''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'''Locate the approximate position of the transition state.'''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

''' Report the activation energy for both reactions. '''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

''' In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T15:19:20Z

Amm416:

== H + H2 System ==

=== Dynamics from the transition state region ===

'''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'''3. Comment on how the mep and the trajectory you just calculated differ.'''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

=== Reactive and unreactive trajectories ===

'''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'''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?'''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'''Locate the approximate position of the transition state.'''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

''' Report the activation energy for both reactions. '''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics and Polanyi's Empirical Rules ===

''' In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.
To confirm it experimentally, IR spectroscopy can be employed to measure the vibrational energy of the reactants compared to the products.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T14:31:37Z

Amm416:

== H + H2 System ==

=== Questions ===

'''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'''3. Comment on how the mep and the trajectory you just calculated differ.'''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

'''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'''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?'''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'''Locate the approximate position of the transition state.'''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

''' Report the activation energy for both reactions. '''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved, which is retrieved from the program by separating the reactants by a large distance and keep the bond lengths of H-H (in F + H2) and H-F (in H + HF) as the standard lengths. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104.020) = 0.277 kcalmol-1

Ea(H + HF) = -103.743 - (-133.954) = 30.211 kcalmol-1

[[File:Amm416_F+H2_potential_e.png|thumb|Figure X: Potential energy vs. time plot showing the energy of the transition state.]]

=== Reaction Dynamics ===

For the F + H2 reaction, a reactive trajectory was found by setting the initial conditions as following:

rF-H = 2.3

rH-H = 0.74

pF-H = -1

pH-H = -5

Energy is conserved at the end of the reaction as seen from the Energy vs. Time graph (Figure Y); kinetic and potential energies are exchanged throughout the reaction.
As seen from the Internuclear momenta vs. Time graph (Figure Z), the H-H bond in the reactants oscillates with less maximum displacement than the H-F bond in the products. Therefore, it suggests that the energy released in the exothermic reaction is used to make H-F vibrate with a greater amplitude.

[[File:Amm416_F+H2_conservation_energy.png|thumb|Figure Y: Energy vs. time graph.]]

[[File:Amm416_F+H2_internuclear_momenta.png|thumb|Figure Z: Internuclear momenta vs. time plot for F + H2 reaction: B-C = H-H , A-B = H-F.]]

File:Amm416 F+H2 conservation energy.png

2018-05-10T14:30:12Z

Amm416:

File:Amm416 F+H2 internuclear momenta.png

2018-05-10T14:19:21Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T14:00:17Z

Amm416:

File:Amm416 F+H2 potential e.png

2018-05-10T13:58:39Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T13:44:15Z

Amm416:

== H + H2 System ==

=== Questions ===

'''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'''3. Comment on how the mep and the trajectory you just calculated differ.'''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

'''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy/kcalmol-1 !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'''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?'''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'''Locate the approximate position of the transition state.'''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.740 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å, as in the F + H2 reaction.

''' Report the activation energy for both reactions. '''

The activation energy (Ea) is equivalent to the energy difference between the maximum of the transition state and the minimum of the reactants. The energy of the transition state is calculated by the program, while the energy of the reactants is simply the sum of the bond enthalpies of the species involved. For both reactions, the energy of the transition state was found to be -103.743 kcalmol-1.

Ea(F + H2) = -103.743 - (-104) = 0.257 kcalmol-1

Ea(H + HF) = -103.743 - (-136) = 32.257 kcalmol-1

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T13:26:26Z

Amm416:

== H + H2 System ==

=== Questions ===

'''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

At both transition state and minimum structures, which are saddle points or stationary points, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

'''2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

From the initial conditions used in the exercise before, the TS was found approximately at r=0.75. Hence, values greater than that were tried while displaying the surface potential for each. At r=0.91, the black trajectory in the surface potential was just a dot, suggesting that at this point the structure was stable at the TS. As seen in the inter-nuclear distance vs. time graph below (Figure 1), there is no formation of the A-B graph, which is the product of the collision, hence suggesting that the reactants are oscillating at the transition state.

[[File:Amm416_internuclear_dist_vs_time_at_TS.png|thumb|Figure 1: Internuclear vs. Time plot at the transition state, rts = 0.91.]]

'''3. Comment on how the mep and the trajectory you just calculated differ.'''

In MEP, the trajectory does not show any vibrational motion, while the trajectory calculated with Dynamics shows oscillations.

[[File:Amm416_dynamics_trajectory.png|thumb|Figure 2: Trajectory from dynamics calculation: oscillation shown.]]

[[File:Amm416_MEP_trajectory.png|thumb|Figure 3: Trajectory from MEP calculation: vibrational motion not observed.]]

'''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border=1
! Entry !! p1 !! p2 !! Total Energy !! Contour Plot !! Reactivity !!
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:Amm416_p1-1.25_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:Amm416_p1-1.5_p2-2.0.png|100px|]] || Unreactive: trajectory reaches transition state but turns around to exit reactants channel. ||
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:Amm416_p1-1.5_p2-2.5.png|100px|]] || Reactive: trajectory goes through transition state and exits through product channel. ||
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:Amm416_p1-2.5_p2-5.0.png|100px|]] || Unreactive: trajectory reaches transition state and recrosses the barrier. However, it turns around to exit reactants channel. ||
|-
| 5 || -2.5 || -5.2 || - 83.416 || [[File:Amm416_p1-2.5_p2-5.2.png|100px|]] || Reactive: trajectory reaches transition state and recrosses the barrier. It manages to go through the transition state and to exit the products channel. ||
|}

'''5. State what are the main assumptions of Transition State Theory (TST). Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

TST is based on three main assumptions:

1. The activated complex is in equilibrium with the reactants.

2. The reactant nuclei behave like classical masses.

3. The reaction pathway proceeds by lowest energy transition state on the PES.

Nevertheless, from the results obtained in Table 1, the assumptions made in TST fail to represent the actual reality. First of all, if the nuclei behaved classically, all the collisions with enough total energy will be reactive. Therefore, Reaction 5 should be reactive but barrier recrossing occurs, which makes the reaction actually unreactive. Barrier recrossing is not predicted by TST, and neither are the other reaction pathways such as quantum tunneling. Hence, the rates of reaction calculated using TST will be larger than the experimental ones due to the negligence of the other pathways.

== F-H-H System ==

=== PES Inspection ===

'''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?'''

The F + H2 reaction is exothermic. The product is HF and the H-F bond is relatively strong due to the large difference in electronegativity of H and F, therefore the energy of the products is more negative than the energy of the reactants and energy is released at the end of the reaction. On the other hand, the H + HF is endothermic due to the weaker H-H bond formed as the product. Therefore, the energy of the products is more positive than the energy of the reactants and energy is absorbed at the end of the reaction.

'''Locate the approximate position of the transition state.'''

From Hammond's postulate, if the reaction is exothermic, the structure and energy of the transition state resemble the reactants. While for endothermic reactions, the transition state will resemble the products.

F + H2:
Since the reaction is exothermic, the transition state will resemble the reactants. So the H-H bond won't be too different from the standard H-H bond in H2. The H-H bond length is 0.74 Å, therefore from optimising the initial conditions, it was found that the structure oscillates at the transition state when the H-F distance is 1.81 Å.

H + HF:
The reaction is endothermic so the transition state will resemble the products. Thus, the distance between the H atoms in the transition state will be similar to the H-H bond length in H2 and the transition state is found when the H-F bond length is 1.81 Å.

'''

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T11:35:16Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-10T11:13:58Z

Amm416: /* H + H2 System */

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:37:00Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:32:59Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:32:09Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:31:14Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:30:48Z

Amm416:

File:Amm416 p1-2.5 p2-5.2.png

2018-05-08T16:26:07Z

Amm416:

File:Amm416 p1-2.5 p2-5.0.png

2018-05-08T16:25:40Z

Amm416:

File:Amm416 p1-1.25 p2-2.5.png

2018-05-08T16:24:53Z

Amm416:

File:Amm416 p1-1.5 p2-2.5.png

2018-05-08T16:24:39Z

Amm416:

File:Amm416 p1-1.5 p2-2.0.png

2018-05-08T16:24:27Z

Amm416:

File:Amm416 MEP trajectory.png

2018-05-08T16:21:23Z

Amm416:

File:Amm416 dynamics trajectory.png

2018-05-08T16:19:10Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T16:13:27Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T15:27:15Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:42:11Z

Amm416:

File:Amm416 internuclear dist vs time at TS.png

2018-05-08T14:41:04Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:39:07Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:38:49Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:33:57Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:15:21Z

Amm416:

MRD:AMM416 Molecular Dynamics Y2

2018-05-08T14:14:52Z

Amm416: Created page with "== H + H2 System == === Questions === 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transit..."

== H + H2 System ==

=== Questions ===

1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

At both transition state and minimum structures, the gradient of the surface potential is equal to zero. To distinguish them, the second derivative of the gradient has to be calculated. If its value is greater than 0, the structure is at a minimum energy. However, if the value is less than zero, the structure is at a maximum energy, hence at the transition state.

2.

Rep:Mod:Computational Y2 amm416

2018-05-04T14:37:39Z

Amm416: /* MO Analysis of [N(CH3)4]+ */

== EX3 ==

=== BH3 ===

==== 3-21G Pre-optimisation Calculation ====

[[File:Amm416 summary table 3-21G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000217 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000919 0.001800 YES
RMS Displacement 0.000441 0.001200 YES
Predicted change in Energy=-1.635268D-07
Optimization completed.
-- Stationary point found.

==== 6-31G(d,p) (D3h symmetry constraint applied) Optimisation Calculation ====

[[File:Amm416 summary table 6-31G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000192 0.000450 YES
RMS Force 0.000126 0.000300 YES
Maximum Displacement 0.000763 0.001800 YES
RMS Displacement 0.000500 0.001200 YES
Predicted change in Energy=-2.201780D-07
Optimization completed.
-- Stationary point found.

==== Frequency Calculation ====

[[Media:AMM416 BH3 FREQ.LOG| BH3 Frequency Calculation .log File]]

Low frequencies --- -0.2263 -0.1037 -0.0054 47.9770 49.0378 49.0383
Low frequencies --- 1163.7209 1213.6704 1213.6731

<jmol><jmolApplet>
<title>BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== IR of BH3 ====

[[File:Amm416_BH3_computed_IR_spectrum.PNG|thumb|Figure 1: Computed IR Spectrum of BH3.]]

[[File:Amm416_BH3_IR_table2.PNG|thumb|Table 1: Table of Computed IR Frequencies and Intensities of BH3, also showing vibrational modes]]

In the IR of BH3, only 3 peaks are present even though there are 6 vibrations. As seen in the Table of IR frequencies, there are 2 sets of 2 generate vibrations: odes 2 and 3 are degenerate (scissoring and rocking) as well as modes 5 and 6 (both asymmetric stretches). Therefore, the degenerate signals will overlap forming a single peak, generating two peaks in the spectrum. The third peak is due to the wagging motion (mode 1). The remaining peak would be the one due to the symmetric stretching, but in this vibrational mode there is no change in net dipole moment; therefore, it is IR inactive.

==== MO Diagram ====

[[File:MO_diagram_BH3.PNG|thumb|Figure 2: MO Diagram of BH3.]]

The computed MOs are very similar to the qualitative LCAO analysis. The only MO that is not immediately recongnisable is one of the 2e' antibonding orbitals; the computed MO on the left shows more lobe repulsions than predicted in the qualitative LCAO. Nevertheless, predicting MO shapes and relative sizes qualitatively by linearly combining the AOs gives a very good approximation of the real MOs.

=== Association Energies: Ammonia-Borane ===

==== NH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000016 0.001800 YES
RMS Displacement 0.000011 0.001200 YES
Predicted change in Energy=-1.228228D-10
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3_FREQ.LOG| NH3 Frequency Calculation .log File]]

Low frequencies --- -0.0138 -0.0032 -0.0015 7.0783 8.0932 8.0937
Low frequencies --- 1089.3840 1693.9368 1693.9368

<jmol><jmolApplet>
<title>NH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== NH3-BH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3-BH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000121 0.000450 YES
RMS Force 0.000057 0.000300 YES
Maximum Displacement 0.000505 0.001800 YES
RMS Displacement 0.000294 0.001200 YES
Predicted change in Energy=-1.610954D-07
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3-BH3_FREQ.LOG| NH3BH3 Frequency Calculation .log File]]

Low frequencies --- -0.0252 -0.0033 -0.0012 17.0405 17.0427 36.9265
Low frequencies --- 265.7534 632.2124 639.3376

<jmol><jmolApplet>
<title>NH3-BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3-BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== Association Energy Calculations ====

E(NH3)= -26.61532 a.u.

E(BH3)= -56.55777 a.u.

E(NH3BH3)= -83.22469 a.u.

ΔE=E(NH3BH3)-[E(NH3)+E(BH3)], where ΔE is the dissociation energy of NH3BH3

∴ ΔE = -0.05160 a.u. = - 135 kJmol-1

Hence, the N-B bond energy is 135 kJmol-1. This value is 3 times smaller than the energy of the C-C bond in the corresponding molecule of ethane, 402 kJmol-1.

=== BBr3 Optimisation and Frequency Calculation using SCAN Server ===

==== Optimisation ====

[[File:Amm416_summary_table_BBr3_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000008 0.000450 YES
RMS Force 0.000005 0.000300 YES
Maximum Displacement 0.000036 0.001800 YES
RMS Displacement 0.000024 0.001200 YES
Predicted change in Energy=-4.190601D-10
Optimization completed.
-- Stationary point found.

==== Frequency ====

[[Media:AMM416_BBr3_SYM_FREQ_3.log| BBr3 Frequency Calculation .log File]]

Low frequencies --- -0.0136 -0.0064 -0.0046 2.4367 2.4367 4.8447
Low frequencies --- 155.9631 155.9651 267.7048

<jmol><jmolApplet>
<title>BBr3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BBr3_SYM_FREQ_3.log</uploadedFileContents>
</jmolApplet></jmol>

D-Space Link: {{DOI|10042/202296}}

== Ionic Liquids: Designer Solvents ==

=== Optimisation and Frequency Calculations for [N(CH3)4]+ and [P(CH3)4]+ ===

==== [N(CH3)4]+ ====

===== Optimisation =====

[[File:amm416_summary_table_NMe4+_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000074 0.000450 YES
RMS Force 0.000027 0.000300 YES
Maximum Displacement 0.000362 0.001800 YES
RMS Displacement 0.000111 0.001200 YES
Predicted change in Energy=-9.316300D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NME4+_FREQ.LOG|[N(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -7.5520 -0.0011 -0.0009 0.0003 6.8978 7.9666
Low frequencies --- 184.2924 289.3429 289.8709

<jmol><jmolApplet>
<title>[N(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NME4+_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== [P(CH3)4]+ ====

===== Optimisation =====

[[File:Amm416_summary_table_PMe4+_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000030 0.000450 YES
RMS Force 0.000012 0.000300 YES
Maximum Displacement 0.000107 0.001800 YES
RMS Displacement 0.000044 0.001200 YES
Predicted change in Energy=-1.742375D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_PME4+_SYM_FREQ.LOG|[P(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -0.0032 0.0017 0.0018 25.3058 25.3058 25.3058
Low frequencies --- 161.2512 195.7467 195.7467

<jmol><jmolApplet>
<title>[P(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_PME4+_SYM_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge Distribution Analysis ===

[[File:Amm416_charge_table_NMe4+.PNG|thumb|Table 2: Table showing atoms' charges contributions in [N(CH3)4]+]]
[[File:Amm416_charges_NMe4+.PNG|thumb|Figure 3: Diagram showing atoms' charges contributions in [N(CH3)4]+: negative to positive = red to green]]

[[File:Amm416_charge_table_PMe4+.PNG|thumb|Table 3: Table showing atoms' charges contributions in [P(CH3)4]+]]
[[File:Amm416_charges_PMe4+.PNG|thumb|Figure 4: Diagram showing atoms' charges contributions in [P(CH3)4]+: negative to positive = red to green]]

From Table 2, it can be seen that nitrogen is partially negatively charged as well as the carbon atoms. Therefore, the classical model in which nitrogen in a tetrahedral arrangement has a positive charge is actually wrong. The assumed positive charge arises from the fact that nitrogen can form a dative covalent bond and hence become more electrophilic. However, from the charge distribution analysis, the positive charge is entirely located on the hydrogen atoms.

In [P(CH3)4]+, all the H atoms and the central P atom are partially positive, with the C atoms bearing all the negative charge, as seen from Table 3. Compared to N in [N(CH3)4]+, P is positively charged because its electronegativity is lower than N's, hence attracts less electron density.

=== MO Analysis of [N(CH3)4]+ ===

[[File:Amm416_Ligand_FOs.PNG]]
[[File:Amm416_LCAO_MOs.PNG]]

Rep:Mod:Computational Y2 amm416

2018-05-04T14:37:18Z

Amm416: /* MO Analysis of [N(CH3)4]+ */

File:Amm416 LCAO MOs.PNG

2018-05-04T14:36:56Z

Amm416:

File:Amm416 Ligand FOs.PNG

2018-05-04T14:36:00Z

Amm416:

Rep:Mod:Computational Y2 amm416

2018-05-04T14:34:40Z

Amm416: /* MO Analysis of [N(CH3)4]+ */

== EX3 ==

=== BH3 ===

==== 3-21G Pre-optimisation Calculation ====

[[File:Amm416 summary table 3-21G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000217 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000919 0.001800 YES
RMS Displacement 0.000441 0.001200 YES
Predicted change in Energy=-1.635268D-07
Optimization completed.
-- Stationary point found.

==== 6-31G(d,p) (D3h symmetry constraint applied) Optimisation Calculation ====

[[File:Amm416 summary table 6-31G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000192 0.000450 YES
RMS Force 0.000126 0.000300 YES
Maximum Displacement 0.000763 0.001800 YES
RMS Displacement 0.000500 0.001200 YES
Predicted change in Energy=-2.201780D-07
Optimization completed.
-- Stationary point found.

==== Frequency Calculation ====

[[Media:AMM416 BH3 FREQ.LOG| BH3 Frequency Calculation .log File]]

Low frequencies --- -0.2263 -0.1037 -0.0054 47.9770 49.0378 49.0383
Low frequencies --- 1163.7209 1213.6704 1213.6731

<jmol><jmolApplet>
<title>BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== IR of BH3 ====

[[File:Amm416_BH3_computed_IR_spectrum.PNG|thumb|Figure 1: Computed IR Spectrum of BH3.]]

[[File:Amm416_BH3_IR_table2.PNG|thumb|Table 1: Table of Computed IR Frequencies and Intensities of BH3, also showing vibrational modes]]

In the IR of BH3, only 3 peaks are present even though there are 6 vibrations. As seen in the Table of IR frequencies, there are 2 sets of 2 generate vibrations: odes 2 and 3 are degenerate (scissoring and rocking) as well as modes 5 and 6 (both asymmetric stretches). Therefore, the degenerate signals will overlap forming a single peak, generating two peaks in the spectrum. The third peak is due to the wagging motion (mode 1). The remaining peak would be the one due to the symmetric stretching, but in this vibrational mode there is no change in net dipole moment; therefore, it is IR inactive.

==== MO Diagram ====

[[File:MO_diagram_BH3.PNG|thumb|Figure 2: MO Diagram of BH3.]]

The computed MOs are very similar to the qualitative LCAO analysis. The only MO that is not immediately recongnisable is one of the 2e' antibonding orbitals; the computed MO on the left shows more lobe repulsions than predicted in the qualitative LCAO. Nevertheless, predicting MO shapes and relative sizes qualitatively by linearly combining the AOs gives a very good approximation of the real MOs.

=== Association Energies: Ammonia-Borane ===

==== NH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000016 0.001800 YES
RMS Displacement 0.000011 0.001200 YES
Predicted change in Energy=-1.228228D-10
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3_FREQ.LOG| NH3 Frequency Calculation .log File]]

Low frequencies --- -0.0138 -0.0032 -0.0015 7.0783 8.0932 8.0937
Low frequencies --- 1089.3840 1693.9368 1693.9368

<jmol><jmolApplet>
<title>NH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== NH3-BH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3-BH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000121 0.000450 YES
RMS Force 0.000057 0.000300 YES
Maximum Displacement 0.000505 0.001800 YES
RMS Displacement 0.000294 0.001200 YES
Predicted change in Energy=-1.610954D-07
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3-BH3_FREQ.LOG| NH3BH3 Frequency Calculation .log File]]

Low frequencies --- -0.0252 -0.0033 -0.0012 17.0405 17.0427 36.9265
Low frequencies --- 265.7534 632.2124 639.3376

<jmol><jmolApplet>
<title>NH3-BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3-BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== Association Energy Calculations ====

E(NH3)= -26.61532 a.u.

E(BH3)= -56.55777 a.u.

E(NH3BH3)= -83.22469 a.u.

ΔE=E(NH3BH3)-[E(NH3)+E(BH3)], where ΔE is the dissociation energy of NH3BH3

∴ ΔE = -0.05160 a.u. = - 135 kJmol-1

Hence, the N-B bond energy is 135 kJmol-1. This value is 3 times smaller than the energy of the C-C bond in the corresponding molecule of ethane, 402 kJmol-1.

=== BBr3 Optimisation and Frequency Calculation using SCAN Server ===

==== Optimisation ====

[[File:Amm416_summary_table_BBr3_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000008 0.000450 YES
RMS Force 0.000005 0.000300 YES
Maximum Displacement 0.000036 0.001800 YES
RMS Displacement 0.000024 0.001200 YES
Predicted change in Energy=-4.190601D-10
Optimization completed.
-- Stationary point found.

==== Frequency ====

[[Media:AMM416_BBr3_SYM_FREQ_3.log| BBr3 Frequency Calculation .log File]]

Low frequencies --- -0.0136 -0.0064 -0.0046 2.4367 2.4367 4.8447
Low frequencies --- 155.9631 155.9651 267.7048

<jmol><jmolApplet>
<title>BBr3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BBr3_SYM_FREQ_3.log</uploadedFileContents>
</jmolApplet></jmol>

D-Space Link: {{DOI|10042/202296}}

== Ionic Liquids: Designer Solvents ==

=== Optimisation and Frequency Calculations for [N(CH3)4]+ and [P(CH3)4]+ ===

==== [N(CH3)4]+ ====

===== Optimisation =====

[[File:amm416_summary_table_NMe4+_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000074 0.000450 YES
RMS Force 0.000027 0.000300 YES
Maximum Displacement 0.000362 0.001800 YES
RMS Displacement 0.000111 0.001200 YES
Predicted change in Energy=-9.316300D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NME4+_FREQ.LOG|[N(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -7.5520 -0.0011 -0.0009 0.0003 6.8978 7.9666
Low frequencies --- 184.2924 289.3429 289.8709

<jmol><jmolApplet>
<title>[N(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NME4+_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== [P(CH3)4]+ ====

===== Optimisation =====

[[File:Amm416_summary_table_PMe4+_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000030 0.000450 YES
RMS Force 0.000012 0.000300 YES
Maximum Displacement 0.000107 0.001800 YES
RMS Displacement 0.000044 0.001200 YES
Predicted change in Energy=-1.742375D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_PME4+_SYM_FREQ.LOG|[P(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -0.0032 0.0017 0.0018 25.3058 25.3058 25.3058
Low frequencies --- 161.2512 195.7467 195.7467

<jmol><jmolApplet>
<title>[P(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_PME4+_SYM_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge Distribution Analysis ===

[[File:Amm416_charge_table_NMe4+.PNG|thumb|Table 2: Table showing atoms' charges contributions in [N(CH3)4]+]]
[[File:Amm416_charges_NMe4+.PNG|thumb|Figure 3: Diagram showing atoms' charges contributions in [N(CH3)4]+: negative to positive = red to green]]

[[File:Amm416_charge_table_PMe4+.PNG|thumb|Table 3: Table showing atoms' charges contributions in [P(CH3)4]+]]
[[File:Amm416_charges_PMe4+.PNG|thumb|Figure 4: Diagram showing atoms' charges contributions in [P(CH3)4]+: negative to positive = red to green]]

From Table 2, it can be seen that nitrogen is partially negatively charged as well as the carbon atoms. Therefore, the classical model in which nitrogen in a tetrahedral arrangement has a positive charge is actually wrong. The assumed positive charge arises from the fact that nitrogen can form a dative covalent bond and hence become more electrophilic. However, from the charge distribution analysis, the positive charge is entirely located on the hydrogen atoms.

In [P(CH3)4]+, all the H atoms and the central P atom are partially positive, with the C atoms bearing all the negative charge, as seen from Table 3. Compared to N in [N(CH3)4]+, P is positively charged because its electronegativity is lower than N's, hence attracts less electron density.

=== MO Analysis of [N(CH3)4]+ ===

[[File:amm416_NMe4+_MO6.PNG]],
[[File:amm416_NMe4+_MO12.PNG]],
[[File:amm416_NMe4+_MO19.PNG]]
.

Rep:Mod:Computational Y2 amm416

2018-05-04T13:16:33Z

Amm416: /* IR of BH3 */

== EX3 ==

=== BH3 ===

==== 3-21G Pre-optimisation Calculation ====

[[File:Amm416 summary table 3-21G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000217 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000919 0.001800 YES
RMS Displacement 0.000441 0.001200 YES
Predicted change in Energy=-1.635268D-07
Optimization completed.
-- Stationary point found.

==== 6-31G(d,p) (D3h symmetry constraint applied) Optimisation Calculation ====

[[File:Amm416 summary table 6-31G opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000192 0.000450 YES
RMS Force 0.000126 0.000300 YES
Maximum Displacement 0.000763 0.001800 YES
RMS Displacement 0.000500 0.001200 YES
Predicted change in Energy=-2.201780D-07
Optimization completed.
-- Stationary point found.

==== Frequency Calculation ====

[[Media:AMM416 BH3 FREQ.LOG| BH3 Frequency Calculation .log File]]

Low frequencies --- -0.2263 -0.1037 -0.0054 47.9770 49.0378 49.0383
Low frequencies --- 1163.7209 1213.6704 1213.6731

<jmol><jmolApplet>
<title>BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== IR of BH3 ====

[[File:Amm416_BH3_computed_IR_spectrum.PNG|thumb|Figure 1: Computed IR Spectrum of BH3.]]

[[File:Amm416_BH3_IR_table2.PNG|thumb|Table 1: Table of Computed IR Frequencies and Intensities of BH3, also showing vibrational modes]]

In the IR of BH3, only 3 peaks are present even though there are 6 vibrations. As seen in the Table of IR frequencies, there are 2 sets of 2 generate vibrations: odes 2 and 3 are degenerate (scissoring and rocking) as well as modes 5 and 6 (both asymmetric stretches). Therefore, the degenerate signals will overlap forming a single peak, generating two peaks in the spectrum. The third peak is due to the wagging motion (mode 1). The remaining peak would be the one due to the symmetric stretching, but in this vibrational mode there is no change in net dipole moment; therefore, it is IR inactive.

==== MO Diagram ====

[[File:MO_diagram_BH3.PNG|thumb|Figure 2: MO Diagram of BH3.]]

The computed MOs are very similar to the qualitative LCAO analysis. The only MO that is not immediately recongnisable is one of the 2e' antibonding orbitals; the computed MO on the left shows more lobe repulsions than predicted in the qualitative LCAO. Nevertheless, predicting MO shapes and relative sizes qualitatively by linearly combining the AOs gives a very good approximation of the real MOs.

=== Association Energies: Ammonia-Borane ===

==== NH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000016 0.001800 YES
RMS Displacement 0.000011 0.001200 YES
Predicted change in Energy=-1.228228D-10
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3_FREQ.LOG| NH3 Frequency Calculation .log File]]

Low frequencies --- -0.0138 -0.0032 -0.0015 7.0783 8.0932 8.0937
Low frequencies --- 1089.3840 1693.9368 1693.9368

<jmol><jmolApplet>
<title>NH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== NH3-BH3 Optimisation and Frequency Calculation ====

===== Optimisation =====

[[File:Amm416_summary_table_NH3-BH3_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000121 0.000450 YES
RMS Force 0.000057 0.000300 YES
Maximum Displacement 0.000505 0.001800 YES
RMS Displacement 0.000294 0.001200 YES
Predicted change in Energy=-1.610954D-07
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NH3-BH3_FREQ.LOG| NH3BH3 Frequency Calculation .log File]]

Low frequencies --- -0.0252 -0.0033 -0.0012 17.0405 17.0427 36.9265
Low frequencies --- 265.7534 632.2124 639.3376

<jmol><jmolApplet>
<title>NH3-BH3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NH3-BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== Association Energy Calculations ====

E(NH3)= -26.61532 a.u.

E(BH3)= -56.55777 a.u.

E(NH3BH3)= -83.22469 a.u.

ΔE=E(NH3BH3)-[E(NH3)+E(BH3)], where ΔE is the dissociation energy of NH3BH3

∴ ΔE = -0.05160 a.u. = - 135 kJmol-1

Hence, the N-B bond energy is 135 kJmol-1. This value is 3 times smaller than the energy of the C-C bond in the corresponding molecule of ethane, 402 kJmol-1.

=== BBr3 Optimisation and Frequency Calculation using SCAN Server ===

==== Optimisation ====

[[File:Amm416_summary_table_BBr3_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000008 0.000450 YES
RMS Force 0.000005 0.000300 YES
Maximum Displacement 0.000036 0.001800 YES
RMS Displacement 0.000024 0.001200 YES
Predicted change in Energy=-4.190601D-10
Optimization completed.
-- Stationary point found.

==== Frequency ====

[[Media:AMM416_BBr3_SYM_FREQ_3.log| BBr3 Frequency Calculation .log File]]

Low frequencies --- -0.0136 -0.0064 -0.0046 2.4367 2.4367 4.8447
Low frequencies --- 155.9631 155.9651 267.7048

<jmol><jmolApplet>
<title>BBr3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_BBr3_SYM_FREQ_3.log</uploadedFileContents>
</jmolApplet></jmol>

D-Space Link: {{DOI|10042/202296}}

== Ionic Liquids: Designer Solvents ==

=== Optimisation and Frequency Calculations for [N(CH3)4]+ and [P(CH3)4]+ ===

==== [N(CH3)4]+ ====

===== Optimisation =====

[[File:amm416_summary_table_NMe4+_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000074 0.000450 YES
RMS Force 0.000027 0.000300 YES
Maximum Displacement 0.000362 0.001800 YES
RMS Displacement 0.000111 0.001200 YES
Predicted change in Energy=-9.316300D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_NME4+_FREQ.LOG|[N(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -7.5520 -0.0011 -0.0009 0.0003 6.8978 7.9666
Low frequencies --- 184.2924 289.3429 289.8709

<jmol><jmolApplet>
<title>[N(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_NME4+_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

==== [P(CH3)4]+ ====

===== Optimisation =====

[[File:Amm416_summary_table_PMe4+_sym_opt.PNG]]

Item Value Threshold Converged?
Maximum Force 0.000030 0.000450 YES
RMS Force 0.000012 0.000300 YES
Maximum Displacement 0.000107 0.001800 YES
RMS Displacement 0.000044 0.001200 YES
Predicted change in Energy=-1.742375D-08
Optimization completed.
-- Stationary point found.

===== Frequency =====

[[Media:AMM416_PME4+_SYM_FREQ.LOG|[P(CH3)4]+ Frequency Calculation .log File]]

Low frequencies --- -0.0032 0.0017 0.0018 25.3058 25.3058 25.3058
Low frequencies --- 161.2512 195.7467 195.7467

<jmol><jmolApplet>
<title>[P(CH3)4]+</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>AMM416_PME4+_SYM_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge Distribution Analysis ===

[[File:Amm416_charge_table_NMe4+.PNG|thumb|Table 2: Table showing atoms' charges contributions in [N(CH3)4]+]]
[[File:Amm416_charges_NMe4+.PNG|thumb|Figure 3: Diagram showing atoms' charges contributions in [N(CH3)4]+: negative to positive = red to green]]

[[File:Amm416_charge_table_PMe4+.PNG|thumb|Table 3: Table showing atoms' charges contributions in [P(CH3)4]+]]
[[File:Amm416_charges_PMe4+.PNG|thumb|Figure 4: Diagram showing atoms' charges contributions in [P(CH3)4]+: negative to positive = red to green]]

From Table 2, it can be seen that nitrogen is partially negatively charged as well as the carbon atoms. Therefore, the classical model in which nitrogen in a tetrahedral arrangement has a positive charge is actually wrong. The assumed positive charge arises from the fact that nitrogen can form a dative covalent bond and hence become more electrophilic. However, from the charge distribution analysis, the positive charge is entirely located on the hydrogen atoms.

In [P(CH3)4]+, all the H atoms and the central P atom are partially positive, with the C atoms bearing all the negative charge, as seen from Table 3. Compared to N in [N(CH3)4]+, P is positively charged because its electronegativity is lower than N's, hence attracts less electron density.

=== MO Analysis of [N(CH3)4]+ ===

[[File:amm416_NMe4+_MO6.PNG]]
[[File:amm416_NMe4+_MO12.PNG]]
[[File:amm416_NMe4+_MO19.PNG]]