ChemWiki - User contributions [en]

MRD:AM9315 CP3MD

2018-05-18T17:00:06Z

Am9315:

==Exercise 1: H + H2==

===Initial coordinates===
{| class="wikitable" border=3
! r1 !! r2 !! p1 !! p2 !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

===Initial contour plot===
[[File:AM9315_initial_Cont_plt.JPG|thumbnail|left|Initial Contour Plot (dynamics)|150px]]

===Initial skew plot===
[[File:AM9315_initial_skew_plt.PNG|thumbnail|left|Initial Skew Plot (dynamics)|150px]]

===Initial surface plot===
[[File:AM9315_initial_surface_plt.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Distance vs time plot===
[[File:AM9315_int_dist_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Momenta vs Time plot===
[[File:AM9315_int_nuc_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Energy vs Time Plot===
[[File:AM9315_EN_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

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

====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.====

The value for the gradient at the potential energy surface is 0 both at the minimum and at the transition state as shown by the fact that the minima (see on graph below) is represented by ∂V/∂r1 = 0, and the transition state (see on graph below) is represented by ∂V/∂r2 = 0. However, the transition state can be distinguished from the minima via carrying out a 2nd derivative. A second derivative for minima would show that ∂2V/∂r12 > 0, thus demonstrating that a minima is a minimum energy point. A second derivative for the transition state would show that ∂2V/∂r12 < 0, thus demonstrating that the transition state is a maximum energy point, which would be expected. However, this is dependent on the direction, as the transition state can occassionaly be a minima. So long as the derivative in both directions shows one maxima, then the curve is a transition state.

===Trajectories from R1 = R2, locating the transition state===

====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.====

rts = 0.905653. This is demonstrated by the 'Internuclear Distance vs Time Plot' shown below where one can observe that the A-B and B-C curves are super imposed with one another, thus the averaging all the distances, and the transition state is the final value one gets.

[[File:AM9315_Mep_cont_plot_traj.PNG|thumbnail|left| Contour plot with transition state marked (dynamics)|150px]]

[[File:AM9315_Mep_dist_v._time_plot_traj.PNG ‎|thumbnail|left| Internuclear vs distance plot for r1 = r2 and p1 = p2 = 0 (dynamics)|150px]]

====Comment on how the mep and the trajectory you just calculated differ.====
The mep calculation depends solely on the trajectory corresponding to infinitely slow motion or in other words, the velocity is always reset to zero. The effect of this can be shown on the MEP contour plot shown below whereby the there are no fluctuations in the reaction pathway as the velocity is always considered to be reset to 0 prior to each step. Contrast this to the the step based dynamics contour plot, and it is clear to see how not resetting the velocity results in fluctuations for each step. What is also interesting note is that MEP requires nearly 32x the number steps that a regular dynamics process would require tor a similar trajectory. The dynamics oscillations are thus shown to be significantly faster. This is because it is only taking into account the vibrational motion in an MEP as compared to the vibrational and translational state motion.

[[File:AM9315_dyn_contour.PNG ‎|thumbnail|left| Contour plot (dynamics)|150px]]

[[File:AM9315_mep_contour.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]]

====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=3
! p1 !! p2 !! Total energy !! R or U !
|-
| -1.25 || -2.5 || -99.018 || R
|-
| -1.5 || -2.0 || -100.456 || U
|-
| -1.5 || -2.5 || -98.546 || R
|-
| -2.5 || -5.0 || -84.390 || U
|-
| -2.5 || -5.2 || -83.786 || U
|}

{| class="wikitable" border=3
! p1 !! p2 !! Surface Plots !! Detail !
|-
| -1.25 || -2.5 || [[File:AM9315_1st_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -1.5 || -2.0 || [[File:AM9315_2nd_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -1.5 || -2.5 || [[File:AM9315_3rd_surface_plt.PNG|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -2.5 || -5.0 || [[File:AM9315_4th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -2.5 || -5.2 || [[File:AM9315_5th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|}

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

Transition state theory states that a hypothetical transition state occurs between reactants and the products. This special type of transition state known as an activated transition state will have a special type of chemical equilibrium with the reactant state known as quasi equilibrium. The species found in the activated transition state is known as the activated complex. There are some basic theory points that one can make with regard to transition state theory:

1) Rates of reaction can be studied by examining activated complexes near the transition state of a potential energy surface.

2)The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3)The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

When applying transition state theory to all the examples applied above, given that only two of the reaction are reactive trajectories, it is likely that the activated complexes are all in quasi equilibrium with the reactants in at least these two trajectories. This is because Hammonds postulate states that the transition state resembles reactants for exothermic processes and resembles products for endothermic processes. Thus, it is unlikely that the unreactive trajectories be exothermic more likely that they are endothermic thus the transition states cannot be in quasi equilibrium with the reactants.

==Exercise 2: H-F-F system==

====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? Locate the approximate position of the transition state. Report the activation energy for both reactions.====

{| class="wikitable" border=3
! Reaction !! Exo or Endo !! TS position (r1) (r2) !! Activation Energy !
|-
| F + H2 || Exothermic|| 1.810692, 0.745 || 0.175
|-
| H + HF || Endothermic || 1.810692, 0.745 || 30.103
|}

[[File:AM9315_TS_HF_F_moleclue.PNG ‎‎|thumbnail|left| Transition state for HH + HF (mep)|150px]]

====Initial conditions for reactive trajectory HH + F =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 1.824 || 0.745 || -2.5 || 0
|}

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

As the F and H2 molecules approach one another, there is an initial increase in kinetic energy as there is no potential energy barrier. As they collide dead center, the initial kinetic energy of both F and H2 is converted into potential energy as shown by the sudden spike in the potential energy as shown on the energy vs time graph.1 This is followed by a transition state between time 0.5 and 1.0, followed by continual vibrations as a result of the the HF bond vibrating thus influencing the potential curves. This can be confirmed by the Internuclear Momenta vs Time AB curve as the momentum is initially flat as the bond has not formed. The momenta drops initially as the direction is reversed and then increases significantly as the F and H2 molecules continue to colide. The momentum once again drops during the transition state before fluctuating indicating clear molecular vibration thus the formation of the HF molecule. Contrasts this to the BC curve where the momentum fluctuates throughout the transition state but after time 1.4, it proceeds to level out indicating that the energy that was released to produce the now fluctuating HF bond was used to break the HH bond thus the energy was conserved.

[[File:AM9315_int_mom_v._time_HF.PNG ‎‎|thumbnail|left| Internuclear distance vs time for release of reactoin energy|150px]]

[[File:AM9315_EN_v._time_HF.PNG ‎ ‎‎|thumbnail|left| Energy vs time graph for energy conservation|150px]]

====Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe?====

The internuclear momenta vs time graph below demonstrates that when the HH momenta is close to -3, the HH momenta initially fluctuates in a symmetric manner and then proceeds into a transition state whereby the fluctuations of the HH momenta decreases whilst the HF bond increases indicating the potential formation of a HF bond. After the transition state, HF fluctuations increase whereas the HH momenta eventually becomes constant thus indicating that the HF molecule forms. Thus, it can be concluded that when the HH momenta is close to -3 there is more than enough energy to overcome the transition state and form the relevant product.

A completely different observation is made for the internuclear vs time graph with a HH momenta close to 3, with there being an exceptionally small transition state is even smaller, indicating that there is not enough energy in the system, thus the reaction does not proceed to form a HF molecule.

However, in comparison, when the HH momenta equals 0, there is clearly enough enough energy for the reaction to form a HF molecule as demonstrated by the continual fluctuating HF momenta line following the transition state between periods.

The limit for momenta which seems to result in a continual transiotn state is when the HH momenta is equal to 0.48874 as shown in the graph below.

====For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?====
The molecule follows a similar fluctuation to what is shown for when pFH = -3 and pHH = 0, whereby there is a small transition state caused by the collision of the HH and F particles but there is not enough energy to overcome the transition state thus the HH continues to fluctuate whereas the HF becomes a constant momenta.

====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.====

====Initial conditions for reactive trajectory HF + H =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

The Polanyi Rules state that the vibrational energy used for a reaction is more efficient translational energy in promoting a late barrier/endothermic reaction i.e. where the transition state more closely resembles the products whereas the reverse is true for an earlier barrier/exothermic reaction i.e. where the transition state more closely resembles the reactants. This rule can be compared to our reaction because on the one hand, HH + F is exothermic and the reverse is endothermic. When viewing the exothermic process, it s clear to see that the forward reaction clearly depends more on translational energy than vibrational energy as whilst it is vibrationally active in that it crosses many vibrational energy barriers, it still appears to be mostly translational energy focused as it follows the reaction path one would expect to generate the HF molecule.

[[File:Skew_plot_for_fwd.PNG‎‎|thumbnail|left| Skew plot for forward reaction|150px]]

However, when viewing a contour plot for the reverse reaction, the reaction path seems to match the reaction path for the reverse reaction in that whilst it is vibrationally efficient, it is more efficient in a translational sense in that it follows the translational path. This actually disagrees with Polanyis rules. A potential reason could be down to the energies. The total energy for the reverse reaction was still found to be -76.008 kcal/mol which is significantly and theoretically, Polanyis rules hold unless the collisions occur at very low energies thus, this could be a potential reason as to why the revese reaction appears to be more translationally efficient than vibrationally efficient.

[[File:Skew_plot_for_reverse.PNG ‎‎|thumbnail|left| Skew plot for reversereaction|150px]]

File:Skew plot for reverse.PNG

2018-05-18T16:59:28Z

Am9315:

MRD:AM9315 CP3MD

2018-05-18T16:59:05Z

Am9315:

==Exercise 1: H + H2==

===Initial coordinates===
{| class="wikitable" border=3
! r1 !! r2 !! p1 !! p2 !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

===Initial contour plot===
[[File:AM9315_initial_Cont_plt.JPG|thumbnail|left|Initial Contour Plot (dynamics)|150px]]

===Initial skew plot===
[[File:AM9315_initial_skew_plt.PNG|thumbnail|left|Initial Skew Plot (dynamics)|150px]]

===Initial surface plot===
[[File:AM9315_initial_surface_plt.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Distance vs time plot===
[[File:AM9315_int_dist_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Momenta vs Time plot===
[[File:AM9315_int_nuc_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Energy vs Time Plot===
[[File:AM9315_EN_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

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

====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.====

The value for the gradient at the potential energy surface is 0 both at the minimum and at the transition state as shown by the fact that the minima (see on graph below) is represented by ∂V/∂r1 = 0, and the transition state (see on graph below) is represented by ∂V/∂r2 = 0. However, the transition state can be distinguished from the minima via carrying out a 2nd derivative. A second derivative for minima would show that ∂2V/∂r12 > 0, thus demonstrating that a minima is a minimum energy point. A second derivative for the transition state would show that ∂2V/∂r12 < 0, thus demonstrating that the transition state is a maximum energy point, which would be expected. However, this is dependent on the direction, as the transition state can occassionaly be a minima. So long as the derivative in both directions shows one maxima, then the curve is a transition state.

===Trajectories from R1 = R2, locating the transition state===

====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.====

rts = 0.905653. This is demonstrated by the 'Internuclear Distance vs Time Plot' shown below where one can observe that the A-B and B-C curves are super imposed with one another, thus the averaging all the distances, and the transition state is the final value one gets.

[[File:AM9315_Mep_cont_plot_traj.PNG|thumbnail|left| Contour plot with transition state marked (dynamics)|150px]]

[[File:AM9315_Mep_dist_v._time_plot_traj.PNG ‎|thumbnail|left| Internuclear vs distance plot for r1 = r2 and p1 = p2 = 0 (dynamics)|150px]]

====Comment on how the mep and the trajectory you just calculated differ.====
The mep calculation depends solely on the trajectory corresponding to infinitely slow motion or in other words, the velocity is always reset to zero. The effect of this can be shown on the MEP contour plot shown below whereby the there are no fluctuations in the reaction pathway as the velocity is always considered to be reset to 0 prior to each step. Contrast this to the the step based dynamics contour plot, and it is clear to see how not resetting the velocity results in fluctuations for each step. What is also interesting note is that MEP requires nearly 32x the number steps that a regular dynamics process would require tor a similar trajectory. The dynamics oscillations are thus shown to be significantly faster. This is because it is only taking into account the vibrational motion in an MEP as compared to the vibrational and translational state motion.

[[File:AM9315_dyn_contour.PNG ‎|thumbnail|left| Contour plot (dynamics)|150px]]

[[File:AM9315_mep_contour.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]]

====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=3
! p1 !! p2 !! Total energy !! R or U !
|-
| -1.25 || -2.5 || -99.018 || R
|-
| -1.5 || -2.0 || -100.456 || U
|-
| -1.5 || -2.5 || -98.546 || R
|-
| -2.5 || -5.0 || -84.390 || U
|-
| -2.5 || -5.2 || -83.786 || U
|}

{| class="wikitable" border=3
! p1 !! p2 !! Surface Plots !! Detail !
|-
| -1.25 || -2.5 || [[File:AM9315_1st_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -1.5 || -2.0 || [[File:AM9315_2nd_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -1.5 || -2.5 || [[File:AM9315_3rd_surface_plt.PNG|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -2.5 || -5.0 || [[File:AM9315_4th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -2.5 || -5.2 || [[File:AM9315_5th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|}

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

Transition state theory states that a hypothetical transition state occurs between reactants and the products. This special type of transition state known as an activated transition state will have a special type of chemical equilibrium with the reactant state known as quasi equilibrium. The species found in the activated transition state is known as the activated complex. There are some basic theory points that one can make with regard to transition state theory:

1) Rates of reaction can be studied by examining activated complexes near the transition state of a potential energy surface.

2)The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3)The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

When applying transition state theory to all the examples applied above, given that only two of the reaction are reactive trajectories, it is likely that the activated complexes are all in quasi equilibrium with the reactants in at least these two trajectories. This is because Hammonds postulate states that the transition state resembles reactants for exothermic processes and resembles products for endothermic processes. Thus, it is unlikely that the unreactive trajectories be exothermic more likely that they are endothermic thus the transition states cannot be in quasi equilibrium with the reactants.

==Exercise 2: H-F-F system==

====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? Locate the approximate position of the transition state. Report the activation energy for both reactions.====

{| class="wikitable" border=3
! Reaction !! Exo or Endo !! TS position (r1) (r2) !! Activation Energy !
|-
| F + H2 || Exothermic|| 1.810692, 0.745 || 0.175
|-
| H + HF || Endothermic || 1.810692, 0.745 || 30.103
|}

[[File:AM9315_TS_HF_F_moleclue.PNG ‎‎|thumbnail|left| Transition state for HH + HF (mep)|150px]]

====Initial conditions for reactive trajectory HH + F =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 1.824 || 0.745 || -2.5 || 0
|}

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

As the F and H2 molecules approach one another, there is an initial increase in kinetic energy as there is no potential energy barrier. As they collide dead center, the initial kinetic energy of both F and H2 is converted into potential energy as shown by the sudden spike in the potential energy as shown on the energy vs time graph.1 This is followed by a transition state between time 0.5 and 1.0, followed by continual vibrations as a result of the the HF bond vibrating thus influencing the potential curves. This can be confirmed by the Internuclear Momenta vs Time AB curve as the momentum is initially flat as the bond has not formed. The momenta drops initially as the direction is reversed and then increases significantly as the F and H2 molecules continue to colide. The momentum once again drops during the transition state before fluctuating indicating clear molecular vibration thus the formation of the HF molecule. Contrasts this to the BC curve where the momentum fluctuates throughout the transition state but after time 1.4, it proceeds to level out indicating that the energy that was released to produce the now fluctuating HF bond was used to break the HH bond thus the energy was conserved.

[[File:AM9315_AM9315_int_mom_v._time_HF.PNG ‎‎|thumbnail|left| Internuclear distance vs time for release of reactoin energy|150px]]

[[File:AM9315_EN_v._time_HF.PNG ‎ ‎‎|thumbnail|left| Energy vs time graph for energy conservation|150px]]

====Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe?====

The internuclear momenta vs time graph below demonstrates that when the HH momenta is close to -3, the HH momenta initially fluctuates in a symmetric manner and then proceeds into a transition state whereby the fluctuations of the HH momenta decreases whilst the HF bond increases indicating the potential formation of a HF bond. After the transition state, HF fluctuations increase whereas the HH momenta eventually becomes constant thus indicating that the HF molecule forms. Thus, it can be concluded that when the HH momenta is close to -3 there is more than enough energy to overcome the transition state and form the relevant product.

A completely different observation is made for the internuclear vs time graph with a HH momenta close to 3, with there being an exceptionally small transition state is even smaller, indicating that there is not enough energy in the system, thus the reaction does not proceed to form a HF molecule.

However, in comparison, when the HH momenta equals 0, there is clearly enough enough energy for the reaction to form a HF molecule as demonstrated by the continual fluctuating HF momenta line following the transition state between periods.

The limit for momenta which seems to result in a continual transiotn state is when the HH momenta is equal to 0.48874 as shown in the graph below.

====For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?====
The molecule follows a similar fluctuation to what is shown for when pFH = -3 and pHH = 0, whereby there is a small transition state caused by the collision of the HH and F particles but there is not enough energy to overcome the transition state thus the HH continues to fluctuate whereas the HF becomes a constant momenta.

====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.====

====Initial conditions for reactive trajectory HF + H =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

The Polanyi Rules state that the vibrational energy used for a reaction is more efficient translational energy in promoting a late barrier/endothermic reaction i.e. where the transition state more closely resembles the products whereas the reverse is true for an earlier barrier/exothermic reaction i.e. where the transition state more closely resembles the reactants. This rule can be compared to our reaction because on the one hand, HH + F is exothermic and the reverse is endothermic. When viewing the exothermic process, it s clear to see that the forward reaction clearly depends more on translational energy than vibrational energy as whilst it is vibrationally active in that it crosses many vibrational energy barriers, it still appears to be mostly translational energy focused as it follows the reaction path one would expect to generate the HF molecule.

[[File:Skew_plot_for_fwd.PNG‎‎|thumbnail|left| Skew plot for forward reaction|150px]]

However, when viewing a contour plot for the reverse reaction, the reaction path seems to match the reaction path for the reverse reaction in that whilst it is vibrationally efficient, it is more efficient in a translational sense in that it follows the translational path. This actually disagrees with Polanyis rules. A potential reason could be down to the energies. The total energy for the reverse reaction was still found to be -76.008 kcal/mol which is significantly and theoretically, Polanyis rules hold unless the collisions occur at very low energies thus, this could be a potential reason as to why the revese reaction appears to be more translationally efficient than vibrationally efficient.

[[File:AM9315_AM9315_int_mom_v._time_HF.PNG ‎‎|thumbnail|left| Skew plot for reversereaction|150px]]

File:Skew plot for fwd.PNG

2018-05-18T16:58:45Z

Am9315:

File:AM9315 int mom v. time HF.PNG

2018-05-18T16:54:33Z

Am9315:

File:AM9315 EN v. time HF.PNG

2018-05-18T16:54:01Z

Am9315:

File:AM9315 TS HF F moleclue.PNG

2018-05-18T16:48:04Z

Am9315:

File:AM9315 5th surface plt.PNG

2018-05-18T16:44:30Z

Am9315:

File:AM9315 4th surface plt.PNG

2018-05-18T16:44:14Z

Am9315:

File:AM9315 3rd surface plt.PNG

2018-05-18T16:44:03Z

Am9315:

File:AM9315 2nd surface plt.PNG

2018-05-18T16:43:53Z

Am9315:

File:AM9315 1st surface plt.PNG

2018-05-18T16:38:01Z

Am9315:

File:AM9315 mep contour.PNG

2018-05-18T16:28:47Z

Am9315:

File:AM9315 dyn contour.PNG

2018-05-18T16:27:16Z

Am9315:

File:AM9315 Mep dist v. time plot traj.PNG

2018-05-18T16:18:21Z

Am9315:

File:AM9315 Mep cont plot traj.PNG

2018-05-18T16:14:05Z

Am9315:

File:AM9315 EN v. time.PNG

2018-05-18T16:05:43Z

Am9315:

File:AM9315 int dist v. time.PNG

2018-05-18T16:02:07Z

Am9315:

File:AM9315 initial surface plt.PNG

2018-05-18T16:00:13Z

Am9315:

File:AM9315 initial skew plt.PNG

2018-05-18T15:55:46Z

Am9315:

File:AM9315 initial Cont plt.JPG

2018-05-18T15:52:21Z

Am9315:

MRD:AM9315 CP3MD

2018-05-18T15:25:51Z

Am9315:

==Exercise 1: H + H2==

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

====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.====

The value for the gradient at the potential energy surface is 0 both at the minimum and at the transition state as shown by the fact that the minima (see on graph below) is represented by ∂V/∂r1 = 0, and the transition state (see on graph below) is represented by ∂V/∂r2 = 0. However, the transition state can be distinguished from the minima via carrying out a 2nd derivative. A second derivative for minima would show that ∂2V/∂r12 > 0, thus demonstrating that a minima is a minimum energy point. A second derivative for the transition state would show that ∂2V/∂r12 < 0, thus demonstrating that the transition state is a maximum energy point, which would be expected.

====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.====

rts = 0.905653. This is demonstrated by the 'Internuclear Distance vs Time Plot' shown below where one can observe that the A-B and B-C curves are super imposed with one another, thus the averaging all the distances, and the transition state is the final value one gets.

====Comment on how the mep and the trajectory you just calculated differ.====
The mep calculation depends solely on the trajectory corresponding to infinitely slow motion or in other words, the velocity is always reset to zero. The effect of this can be shown on the MEP contour plot shown below whereby the there are no fluctuations in the reaction pathway as the velocity is always considered to be reset to 0 prior to each step. Contrast this to the the step based dynamics contour plot, and it is clear to see how not resetting the velocity results in fluctuations for each step. What is also interesting note is that MEP requires nearly 32x the number steps that a regular dynamics process would require tor a similar trajectory. The dynamics oscillations are thus shown to be significantly faster. This is because it is only taking into account the vibrational motion in an MEP as compared to the vibrational and translational state motion.

====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=3
! p1 !! p2 !! Total energy !! R or U !
|-
| -1.25 || -2.5 || -99.018 || R
|-
| -1.5 || -2.0 || -100.456 || U
|-
| -1.5 || -2.5 || -98.546 || R
|-
| -2.5 || -5.0 || -84.390 || U
|-
| -2.5 || -5.2 || -83.786 || U
|}

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

Transition state theory states that a hypothetical transition state occurs between reactants and the products. This special type of transition state known as an activated transition state will have a special type of chemical equilibrium with the reactant state known as quasi equilibrium. The species found in the activated transition state is known as the activated complex. There are some basic theory points that one can make with regard to transition state theory:

1) Rates of reaction can be studied by examining activated complexes near the transition state of a potential energy surface.

2)The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3)The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

To be continued
''Will add relevant figures on friday''

==Exercise 2: H-F-F system==

====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? Locate the approximate position of the transition state. Report the activation energy for both reactions.====

{| class="wikitable" border=3
! Reaction !! Exo or Endo !! TS position (r1) (r2) !! Activation Energy !
|-
| F + H2 || Exothermic|| 1.810692, 0.745 || 0.175
|-
| H + HF || Endothermic || 1.810692, 0.745 || 30.103
|}

====Initial conditions for reactive trajectory HH + F =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 1.824 || 0.745 || -2.5 || 0
|}

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

As the F and H2 molecules approach one another, there is an initial increase in kinetic energy as there is no potential energy barrier. As they collide dead center, the initial kinetic energy of both F and H2 is converted into potential energy as shown by the sudden spike in the potential energy as shown on the energy vs time graph. This is followed by a transition state between time 0.5 and 1.0, followed by continual vibrations as a result of the the HF bond vibrating thus influencing the potential curves. This can be confirmed by the Internuclear Momenta vs Time AB curve as the momentum is initially flat as the bond has not formed. The momenta drops initially as the direction is reversed and then increases significantly as the F and H2 molecules continue to colide. The momentum once again drops during the transition state before fluctuating indicating clear molecular vibration thus the formation of the HF molecule. Contrasts this to the BC curve where the momentum fluctuates throughout the transition state but after time 1.4, it proceeds to level out indicating that the energy that was released to produce the now fluctuating HF bond was used to break the HH bond thus the energy was conserved.

====Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe?====

The internuclear momenta vs time graph below demonstrates that when the HH momenta is close to -3, the HH momenta initially fluctuates in a symmetric manner and then proceeds into a transition state whereby the fluctuations of the HH momenta decreases whilst the HF bond increases indicating the potential formation of a HF bond. After the transition state, HF fluctuations increase whereas the HH momenta eventually becomes constant thus indicating that the HF molecule forms. Thus, it can be concluded that when the HH momenta is close to -3 there is more than enough energy to overcome the transition state and form the relevant product.

A completely different observation is made for the internuclear vs time graph with a HH momenta close to 3, with there being an exceptionally small transition state is even smaller, indicating that there is not enough energy in the system, thus the reaction does not proceed to form a HF molecule.

However, in comparison, when the HH momenta equals 0, there is clearly enough enough energy for the reaction to form a HF molecule as demonstrated by the continual fluctuating HF momenta line following the transition state between periods.

The limit for momenta which seems to result in a continual transiotn state is when the HH momenta is equal to 0.48874 as shown in the graph below.

====For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?====
The molecule follows a similar fluctuation to what is shown for when pFH = -3 and pHH = 0, whereby there is a small transition state caused by the collision of the HH and F particles but there is not enough energy to overcome the transition state thus the HH continues to fluctuate whereas the HF becomes a constant momenta.

====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.====

====Initial conditions for reactive trajectory HF + H =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

The Polanyi Rules state that the vibrational energy used for a reaction is more efficient translational energy in promoting a late barrier/endothermic reaction i.e. where the transition state more closely resembles the products whereas the reverse is true for an earlier barrier/exothermic reaction i.e. where the transition state more closely resembles the reactants. This rule can be compared to our reaction because on the one hand, HH + F is exothermic and the reverse is endothermic. When viewing the exothermic process, it s clear to see that the forward reaction clearly depends more on translational energy than vibrational energy as whilst it is vibrationally active in that it crosses many vibrational energy barriers, it still appears to be mostly translational energy focused as it follows the reaction path one would expect to generate the HF molecule. However, when viewing a contour plot for the reverse reaction, one can see how the reaction pathway passed through a lot of vibrational states meaning the vibrational energy used for the reaction appears to be more efficient in regards to contributing to the process.

MRD:AM9315 CP3MD

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Am9315: Created page with "==Exercise 1: H + H2== ===Dynamics from the transition state region=== ====What value do the different components of the gradient of the potential energy surface..."

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===EX3 Section===

====BH3====

[[File:BH3_image.JPG|thumbnail|left|A BH3 molecule|150px]]

''Note that all BH3 molecules had bond angles of exactly 120.000o regardless of the level of optimisation and frequency.''

'''BH3 energy and key information (3-21G)'''

[[File:BH3_opti_321G.JPG|thumbnail|left|Key information for BH3 molecule after 3-21G optimisation|250px]]

Bond length for optimised 3-21G BH3 = 1.19453 A

Bond angle = 120.000o

[[File:Log_of_information_3-21G.JPG|thumbnail|left|Log of information for BH3 optimised using the method 3-21G|400px]]

Above, the log of information contains information such as the gradient of the slope for the energy vs distance graphs through the force items.

[[File:Energy_against_optimisation_and_gradient.JPG|thumbnail|left|The first graph gives the energy of BH3 at each step of the optimisation. The second graph gives the gradient of the energy of the optimisation at each step of the optimisation |450px]]

As is shown in the graph above, the optimisation allows the BH3 molecule to reach its equilibrium state, thus being able to stabilise the molecule. When nuclei and and electrons are not in equilibrium their mutual interactions, i.e. nuclear-nuclear repulsion, nuclear-electron attraction, electron-electron interactions are not stable, these interactions cause forces to be exerted on the nuclei and electrons making them shift into better positions. That is, the molecule will experience forces F(R) as long as a change in position (delta R) causes a change in energy, delta E(R).

'''BH3 energy and key information (6-31G)'''

[[File:BH3_opti_6-31G.JPG|thumbnail|left|Key information for BH3 molecule after 6-31G optimisation|350px]]

Bond length = 1.19273 A

''Note how the bond length changes simply by using a higher level basis set.''

Bond angle = 120.000o

Also, note how the energy differs by using the a different basis method.

E(3-21G) = -26.61532342 a.u.

E(6-31G) = -26.61532342 a.u.

ΔE = 0.1530593 a.u. = 402 kJ/mol

[[File:Frequency_analysis.JPG|thumbnail|left|Key information for BH3 molecule after frequency analysis|350px]]

''Note that the symmetry label after frequency analysis is C3H which cannot be explained as the bond angles are still exactly 120.000o.

[[File:Frequency_key_info.JPG|thumbnail|left|Log of information for BH3 optimised using the method 3-21G and then frequency analysed|350px]]

'''Association energies: Ammonia and Borane'''

''NH3''

[[File:NH3.JPG|thumbnail|left|A NH3 molecule|150px]]

[[File:NH3_info.JPG|thumbnail|left|Key info for NH3 molecule|150px]]

''NH3BH3''
[[File:NH3BH3.JPG|thumbnail|left|A NH3BH3 molecule|150px]]

[[File:NH3BH3_info.JPG|thumbnail|left|Key info for NH3BH3 molecule|150px]]

E(NH3) = -56.55664134 a.u.

E(BH3) = -26.61532342 a.u.

E(NH3BH3) = -83.22469026 a.u.

ΔE = E(NH3BH3) - [E(NH3)+E(BH3)] = -0.0527255 a.u. = -138.4 kJ/mol.

The dative BN bond is weaker than other Boron based bonds such as BCl which has a bond formation of -456 kJ/mol or a nitrogen based bond such as NCl which has a bond formation of -313kJ/mol.

'''MO and LCAO diagram for BH3'''
[[File:MO_diagram_for_BH3_AM9315.jpg|thumbnail|left|An MO diagram for BH3|450px]]

As is seen in the Figure above, there are no significant differences between the LCAO MOs (i.e. black and white images) and the and the real MOs. This demonstrates that qualitative MO theory is very accurate and useful in regards to building and describing Molecular Orbitals in a Chemistry sense.

===BBr3===
[[File:BBr3.JPG|thumbnail|left|A BBr3 molecule|150px]]

[[File:BBr3_info.JPG|thumbnail|left|Key info for BBr3 molecule|150px]]

{{DOI|10042/202399}}

===Ionic Liquids: Designer Solvents===

====[N(CH3)4]+====
[[File:AM9315_N(CH3)4-+_image_2.JPG|thumbnail|left|An [N(CH3)4]+ molecule |300px]]

Dative N-C Bond length = 1.50936 A

Standard N-C Bond length = 1.50934(2-5) A

Tetrahedral Bond Angle = 109.47o

[[File:-N(CH3)4-+_opt_info.JPG|thumbnail|left|Key info for [N(CH3)4]+ after optimisation|300px]]

[[File:-N(CH3)4-+_freq.JPG|thumbnail|left|Key info for [N(CH3)4]+ after optimisation and frequency analysis|300px]]

[[File:-N(CH3)4-+_freq_key_info.JPG|thumbnail|left|Log of information for [N(CH3)4]+ optimised using the method 6-31G and then frequency analysed|300px]]

[[File:-N(CH3)4-+_energy_opt.JPG|thumbnail|left|The first graph gives the energy of [N(CH3)4]+ at each step of the optimisation. The second graph gives the gradient of the energy of the optimisation at each step of the optimisation|300px]]

The first graph which shows the effects of optimisation against energy shows how the optimisation of the cation succeeds in achieving an energy minima where the cation is its most stable configuration. The frequency analysis further stabilizes the cation as can be shown when comparing the energy before and after frequency analysis.

Energy after optimisation = -214.18110186

Energy after optimisation and frequency analysis = -214.18128359

[[File:-N(CH3)4-+_charge_distribution.JPG|thumbnail|left|Charge distribution for [N(CH3)4]+ molecule|300px]]

Charge of nitrogen = -0.295.

Charge of carbon atoms of methane = -0.483

Charge of hydrogen atoms of methane = 0.259

It is interesting to observe that the nitrogen (central species) has a higher charge, thus lower electron density surrounding it compared to the carbon (ligand). One would not expect that as nitrogen has an electronegativity of 3.04 and carbon has an electronegativity of 2.55, that the charge centered around the nitrogen. An explanation as to why this is not the case is that the fact that the molecule is [N(CH3)4]+, thus the one of the nitrogen bonds is dative meaning that a lone electron pair is being donated into the methyl group. This is shown by one bond being significantly shorter than the others. However, the Charge distribution shows that all carbon atoms have the same charge which would not be the case if one was receiving electron density. It could be that due to the effectiveness of nitrogen as a Lewis base i.e. a strong donor and the electronegativity difference in favour of nitrogen result in the electron density of the NC bond would be concentrated at the center of the bond. The carbon atoms would then also receive additional electron density from the methane groups thus making it the group with the highest charge density in this cation.

This theory slightly support the theory that in an ionic structure, the ligand would have the charge concentrated on the central atom (as shown below). The charge seems to represent the relative charge of the central species compared to its ligand and in this instance, appears to be valid as the charge distribution of the carbon ligands is more negative than that of the nitrogen central species. However, the image shown below is not entirely accurate as the electron density surrounding the nitrogen species. A more accurate image would rather show the positive charge for each individual hydrogen atom since this is really where the electron density is lowest, thus the charge value is the highest.

[[File:R_image.jpg|thumbnail|left|The theoretical charge distribution of an NR4+|150px]]

'''MO diagrams (LCAO and real)'''

====[P(CH3)4]+====
[[File:AM9315_N(CH3)4-+_image_2.JPG|thumbnail|left|A [P(CH3)4]+ molecule |300px]]

P-C Bond length = 1.816(69-72) A

Tetrahedral Bond Angle = 109.468o

[[File:-P(CH3)4-+_opt_info.JPG|thumbnail|left|Key info for [P(CH3)4]+ after optimisation|300px]]

[[File:-P(CH3)4-+_freq.JPG |thumbnail|left|Key info for [P(CH3)4]+ after optimisation and frequency analysis|300px]]

[[File:-P(CH3)4-+_freq_key_info.JPG|thumbnail|left|Log of information for [P(CH3)4]+ optimised using the method 6-31G and then frequency analysed|300px]]

[[File:-P(CH3)4-+_energy_opt.JPG|thumbnail|left|The first graph gives the energy of [P(CH3)4]+ at each step of the optimisation. The second graph gives the gradient of the energy of the optimisation at each step of the optimisation|300px]]

The first graph which shows the effects of optimisation against energy shows how the optimisation of the cation succeeds in achieving an energy minima where the cation is its most stable configuration. The frequency analysis further stabilizes the cation as can be shown when comparing the energy before and after frequency analysis.

Energy after optimisation = -500.82292956

Energy after optimisation and frequency analysis = -500.82699271

[[File:-P(CH3)4-+_charge_distribution.JPG|thumbnail|left|Charge distribution for [P(CH3)4]+ molecule|300px]]

Charge of phosphorus= 1.667.

Charge of carbon atoms of methane = -1.050

Charge of hydrogen atoms of methane = 0.298

Unlike the nitrogen example, one would expect that phosphorous would have a higher charge as it is less electronegative than carbon thus one would expect that the electron density to be closer to the carbons of methane hence why they have a lower charge. The effect of this is that the positive charge is more likely to be found closer to the phosphorus, meaning this will more accurately represent what the theoretical image shown above. The fact that the phosphonium appears more ionic means that it can more effectivley couple to anionic liquids thus making it have a higher melting point compared to an equivalent ammonium salt e.g. tetramethlyl ammonium iodide has a mp of 280oC[1] and tetramethylphosphonium iodide has a melting point of around 325oC[2]. This is probably due to the fact that phosphorus is softer than nitrogen, thus forms stronger dative bonds, and as the charge is more dispersed, it means that it is a more effective cationic solvent.

===References===
[1] E. Wait and H. M. Powell (1958). "The crystal and molecular structure of tetraethylammonium iodide." J. Chem. Soc. 1872-1875.
[2] JPETAB Journal of Pharmacology and Experimental Therapeutics. (Williams & Wilkins Co., 428 E. Preston St., Baltimore, MD 21202) V.1- 1909/10- Volume(issue)/page/year: 25,315,1925

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