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MRD:MLN

2018-05-25T16:57:33Z

Ml5816: /* H + HF → F + H2 */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
Consider the following system of three hydrogen atoms, atom A, and molecule BC: [[File:Mln_atoms.PNG|300px]]

===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|right|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:56:23Z

Ml5816: /* H-H-H System */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
Consider the following system of three hydrogen atoms, atom A, and molecule BC: [[File:Mln_atoms.PNG|300px]]

===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:55:58Z

Ml5816: /* H-H-H System */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
Consider the following system: [[File:Mln_atoms.PNG|300px]]

===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:55:50Z

Ml5816: /* H-H-H System */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
Consider the following system: [[File:Mln_atoms.PNG|100px]]

===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:55:37Z

Ml5816: /* H-H-H System */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
Consider the following system: [[File:Mln_atoms.PNG]]

===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

File:Mln atoms.PNG

2018-05-25T16:54:55Z

Ml5816:

MRD:MLN

2018-05-25T16:52:51Z

Ml5816: /* Transfer of Reaction Energy */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time plot for the F + H2 → HF + H reaction; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:51:40Z

Ml5816: /* F-H-H System */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right|Successful trajectory surface plot for the F + H2 → HF + H reaction]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left|Energy vs Time plot for the F + H2 → HF + H reaction, showing the difference in potential energy between the transition state and products.]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left|Energy vs Time plot for the HF + H → F + H2 reaction, showing the difference in potential energy between the transition state and products.]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:49:05Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|steelblue|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:48:32Z

Ml5816: /* Transfer of Reaction Energy */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|right|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:48:18Z

Ml5816: /* Transfer of Reaction Energy */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:Mln_momenta.PNG|thumb|500px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

File:Mln momenta.PNG

2018-05-25T16:48:00Z

Ml5816:

MRD:MLN

2018-05-25T16:47:19Z

Ml5816: /* Transfer of Reaction Energy */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|steelblue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 

====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:44:11Z

Ml5816: /* Transition State Theory */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|steelblue|'''''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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:43:21Z

Ml5816: /* Molecular Reaction Dynamics: Applications to Triatomic Systems */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:43:02Z

Ml5816: /* Minimum Energy Path (mep) */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|steelblue|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:41:55Z

Ml5816: /* Minimum Energy Path (mep) */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the mep and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]
For the ''mep'' when all momenta are zero and ''rBC'' is slightly displaced by 0.01 Å, the molecule AB will form. When ''rBC'' is slightly displaced by -0.01 Å, the molecule BC will form. This is because for ''mep'' the system just follows the lowest energy pathway towards the lower potential energy as there is no momentum and therefore no kinetic energy.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 0.0014, ''pAB(t)'' = 0.000065.

The dynamic calculation will have oscillations due to the transfer of potential energy to kinetic energy as the molecule is formed and the system moves away from the transition state position.
For the ''mep'' calculation the final values were: ''rBC(t)'' > 9.00 Å, ''rAB(t)'' = 0.74 Å; ''pBC(t)'' = 2.50, ''pAB(t)'' = 1.32.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:24:47Z

Ml5816: /* Transition State Theory */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms, TSS is most reliable and will produce the predicted reaction rates that are closest to the experimental reaction rates.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:22:45Z

Ml5816: /* Transition State Theory */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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 is used to explain the reaction rates of chemical reactions by using several assumptions. The main assumptions of TSS assume that:
*Classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*All trajectories that cross the transitions state will react <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*An equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*The system will follow the lowest energy pathway (over the lowest energy saddle point) on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

As the theory assumes that all crossings of the transition state are a successful reaction, it neglects any recrossing of the system and would account for the recrossing as multiple successful reactions: An overall unreactive recrossing trajectory will cross the transition state an even number of times, and an overall reactive recrossing trajectory will cross the transition state an odd number of times. Therefore, TSS will overestimate the reaction rate of a system as it assumes that more successful reactions are occurring in a given time frame than will actually occur.

Furthermore, due to the assumption of lowest energy pathway, any higher energy pathway trajectories that are successful will not be taken into account. Therefore at high temperatures, when higher energy levels will be more occupied and higher energy trajectories are possible, the theory will underestimate the number of successful reactions occurring, thus predicting a lower reaction rate that experimentally seen.

In conclusion, at temperatures where thermal energy is less than the transition barrier, recrossing is less likely to occur so the assumption will be more valid. As the temperature is lowe, higher energy states will not be occupied and so the lowest energy pathway assumption is also valid. Thus, TSS is most reliable at temperatures where thermal energy is lower than the activation energy of the reaction and the excitation energy of the reactant atoms.

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T16:04:28Z

Ml5816: /* Transition State Theory */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

*Assumes that all trajectories that cross the transitions state will react, thus neglecting any recrossing. <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 

==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:58:49Z

Ml5816: /* High Energy Trajectories */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
F + H2 reaction.
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 

====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:57:10Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

*For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

*For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

Thus, in the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:56:47Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>:

For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

In the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:56:06Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>
:

For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.

In the F-H-H system reactions above, the F + H2 reaction is exothermic, therefore it is more likely for a given trajectory to be reactive if it has a large ''pFH'' (more translational energy) and smaller ''pHH''. The HF + H reaction is endothermic, so a given trajectory is also more likely to be reactive if it has a larger ''pHF'' (more vibrational energy) and smaller ''pHH''.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:49:33Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions. The trajectories can be made reactive by giving the reactants kinetic energy, however reactions that are energetically different will need to be kinetically excited in different ways. The empirical rules lead to the following conclusions<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>
:

For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using a greater internuclear momentum of the reactant molecule to increase its initial vibrational energy.

For an early-barrier reaction, such as an exothermic reaction, the trajectory will be more likely to be reactive if the reactants are translationally excited. This can be simulated by using a greater initial internuclear momentum between the unbound atom and the atom that it collides with in the molecule.
 

==References==
<references/>

MRD:MLN

2018-05-25T15:40:35Z

Ml5816: /* Reactant Energy Distribution */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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 describe the most efficient way of activating a reaction, by differentiating the trajectories needed for endothermic and exothermic reactions.
For a late-barrier reaction, i.e. an endothermic reaction, a reaction is most effectively activated by using vibrationally excited reactants. This condition can be simulated by using greater internuclear momenta of the reactant molecule in the trajectories above, which are more likely to be reactive.

<ref>Levine, R. (2005). Molecular Reaction Dynamics. Chap. 5</ref>
 

==References==
<references/>

MRD:MLN

2018-05-25T15:30:15Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
====Reactive and Unreactive Trajectories====
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

====Transfer of Reaction Energy====
'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
====High Energy Trajectories====
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surface plot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
====Reactant Energy Distribution====
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

File:Mln traj 12.PNG

2018-05-25T15:20:29Z

Ml5816:

MRD:MLN

2018-05-25T15:20:22Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.8:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| 0.1 || Reacts || [[File:Mln_traj_12.PNG|thumb|200px]]
|}

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

MRD:MLN

2018-05-25T15:18:22Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| -3.0 || No reaction || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || Reacts || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || No reaction || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || No reaction || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || No reaction || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || No reaction || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || Reacts || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || Reacts || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || No reaction || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || No reaction || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || No reaction || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

MRD:MLN

2018-05-25T15:17:06Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot !! Trajectory surface plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| -3.0 || || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

File:Mln traj 11.PNG

2018-05-25T15:16:24Z

Ml5816:

File:Mln traj 10.PNG

2018-05-25T15:16:14Z

Ml5816:

File:Mln traj 9.PNG

2018-05-25T15:16:04Z

Ml5816:

File:Mln traj 8.PNG

2018-05-25T15:15:53Z

Ml5816:

File:Mln traj 7.PNG

2018-05-25T15:15:41Z

Ml5816:

File:Mln traj 6.PNG

2018-05-25T15:15:33Z

Ml5816:

File:Mln traj 5.PNG

2018-05-25T15:15:25Z

Ml5816:

File:Mln traj 4.PNG

2018-05-25T15:15:16Z

Ml5816:

File:Mln traj 3.PNG

2018-05-25T15:14:52Z

Ml5816:

File:Mln traj 2.PNG

2018-05-25T15:14:44Z

Ml5816:

File:Mln traj 1.PNG

2018-05-25T15:14:30Z

Ml5816:

MRD:MLN

2018-05-25T15:14:12Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| -3.0 || || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || || [[File:Mln_traj_3.PNG|thumb|200px]]
|-
| -1.5 || || [[File:Mln_traj_4.PNG|thumb|200px]]
|-
| 0 || || [[File:Mln_traj_5.PNG|thumb|200px]]
|-
| 0.5 || || [[File:Mln_traj_6.PNG|thumb|200px]]
|-
| 1.5 || || [[File:Mln_traj_7.PNG|thumb|200px]]
|-
| 2.5 || || [[File:Mln_traj_8.PNG|thumb|200px]]
|-
| 3 || || [[File:Mln_traj_9.PNG|thumb|200px]]
|-
| 3.2 || || [[File:Mln_traj_10.PNG|thumb|200px]]
|-
| -3.2 || || [[File:Mln_traj_11.PNG|thumb|200px]]
|}

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

MRD:MLN

2018-05-25T15:12:02Z

Ml5816: /* Reaction dynamics */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

That the reaction is exothermic can be confirmed experimentally by measuring the change in temperature, as the release of energy causes temperature to increase. The greater vibrational energy of the products can be confirmed by IR spectroscopy. As the product is more vibrationally excited, more higher level vibrational energy states are occupied so overtones will begin to appear in the IR spectrum at a lower frequency than the main band (energy spacing decreases for higher energy levels). If the products were further vibrationally excited the intensity of the overtones would increase.

 
Initial conditions: ''rHH'' = 0.74, ''rHF'' = 2.3, ''pHF'' = -0.5:
{| class="wikitable" border=1
! ''pHH'' !! Observations !! Trajectory surfaceplot
|-
| -3.0 || || [[File:Mln_traj_1.PNG|thumb|200px]]
|-
| -2.5 || || [[File:Mln_traj_2.PNG|thumb|200px]]
|-
| -2.0 || || [[File:Mln_rxndyn_3.PNG|thumb|200px]]
|-
| -1.5 || || [[File:Mln_rxndyn_4.PNG|thumb|200px]]
|-
| 0 || || [[File:|thumb|200px]]
|-
| 0.5 || || [[File:|thumb|200px]]
|-
| 1.5 || || [[File:|thumb|200px]]
|-
| 2.5 || || [[File:|thumb|200px]]
|-
| 3 || || [[File:|thumb|200px]]
|-
| 3.2 || || [[File:|thumb|200px]]
|-
| -3.2 || || [[File:|thumb|200px]]
|}

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

MRD:MLN

2018-05-25T14:38:15Z

Ml5816: /* H + HF → F + H2 */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.25 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

This could be confirmed experimentally by measuring the change in temperature, because the reaction is exothermic so there will be a release of energy causing the temperature to increase. This can also be seen in the greater vibrational energy of the products compared to reactants, so heat is generated by higher energy vibrations.

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

File:Mln HF H acten 4.PNG

2018-05-25T14:37:36Z

Ml5816:

MRD:MLN

2018-05-25T14:37:21Z

Ml5816: /* H + HF → F + H2 */

=Molecular Reaction Dynamics: Applications to Triatomic Systems=
*new graphs for table w/ increased cut off and black line
*draw and label H-H-H system with ABC labels
*upload new 2nd activation energy picture with obvious final value
*transition state theory question
*caption pictures

==H-H-H System==
===Dynamics from the transition state region===
{{fontcolor|steelblue|'''''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.'''''}}
[[File:Mln_3H_surface_1.PNG|thumb|400px|right|Surface plot showing the energy minima and saddle point of the triatomic reaction.]]

At the minima of the potential plot, the first partial derivative (with respect to either distance), which is the gradient, will be equal to zero:
<math> \frac{\partial V}{\partial r} = 0 </math>

The second partial derivative, now with respect to both distances, will be positive (greater than zero) as a minimum turning point has a positive curvature:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} > 0</math>

For a saddle point, the first derivative is again zero, however the second derivative is now also equal to zero, because a saddle point will have no curvature and therefore be completely flat at that point:
<math>\frac{\partial^2 V}{\partial r_1 \partial r_2} = 0</math>

The saddle point can be seen in the surface plot, where the energy of the minima increase to a maximum. It can be seen on the contour plot (the 2D lines at the bottom of the surface plot), where the two contours of the entrance and exit channels come to turning points. Each contour defines a specific energy value, thus these adjacent turning points (saddle point) signify the maximum of this energy path.

 
{{fontcolor|steelblue|'' '''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.''' ''}}
[[File:Mln_3H_ts_position.PNG|thumb|350px|right|Plot of Internuclear Distances vs Time at the transition state position, ''rts'' = 0.9075 Å.]]

The transition state position is estimated by finding the ''rBC = rAB'' at which the Internuclear Distance vs Time plot has a gradient of zero, because if there is no gradient, the trajectory will oscillate at this point and will not move to a different position, or 'fall off' the maximum. The gradient signifies a change in energy, so if there is no gradient, then there is no change in energy at the point which the system is oscillating. Thus it does not oscillate over a lower energy position like it would if the gradient was non-zero and does not fall off. The plot shown is calculated at ''r'' = 0.9075 Å. As the gradient is zero, ''rts'' = 0.9075 Å.

 
===Minimum Energy Path (''mep'')===
{{fontcolor1|red|'' '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ''}}
[[File:Mln_3H_mep_comp.PNG|thumb|700px|right|Contour plots showing the lowest energy trajectory. Left: The MEP. Right: The dynamic calculation corresponding to the mep trajectory.]]

for mep: dist v time: ''rBC(t)'' > 9.00, ''rAB(t)'' = 0.74; mom v time: the average momenta ''pBC(t)'' = 2.50, 1.57 > ''pAB(t)'' > 0.92 at large ''t''.

 
{{fontcolor1|steelblue|'''''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''''
'''''What do you observe?'''''}}

Distance graphs are a reflection of each other in the y axis and the momentum graph is a reflection in the line y=-x, as the momentum is negative, bringing the atoms together to form the transitions structure at the transition state separation, so the separation decreases and negative momentum increases to zero.

 

===Reactive and Unreactive Trajectories===
{{fontcolor|steelblue|'''''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
! ''pBC'' !! ''pAB'' !! Energy /kcal !! Reactive? !! Trajectory contour plot !! Trajectory surface plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Mln_3H_traj11.PNG|thumb|300px]] || [[File:Mln_3H_traj12.PNG|thumb|300px]] || The hydrogen atom, A, approaches the hydrogen molecule, BC, colliding and going through the transition state, then forming a new hydrogen molecule, AB, leaving hydrogen C unbound.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mln_3H_traj21.PNG|thumb|300px]] || [[File:Mln_3H_traj22.PNG|thumb|300px]] || Hydrogen A approaches molecule BC and collides. Atom A rebounds off of the molecule, reversing all atoms' momenta, but no reaction occurs and the BC distance remains the same before and after collision.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mln_3H_traj31.PNG|thumb|300px]] || [[File:Mln_3H_traj32.PNG|thumb|300px]] || This trajectory is very similar to the first, except due to the larger ''pBC'', the oscillations have greater magnitude, so the BC distance is very slightly larger after the collision.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mln_3H_traj41.PNG|thumb|300px]] || [[File:Mln_3H_traj42.PNG|thumb|300px]] || Molecule BC approaches atom A, colliding and passing through the transition state. As the momentum is large, the system has enough energy to recrosses the transition state, back to the reactants, exchanging energy between translation and vibration.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mln_3H_traj51.PNG|thumb|300px]] || [[File:Mln_3H_traj52.PNG|thumb|300px]] || BC collides with A, passing the transition state, then recrosses back to entrance channel as above. As the momentum, and therefore energy, is greater than above, the system recrosses again into the exit channel, producing molecule AB and unbound atom C.
|}

 
===Transition State Theory===
{{fontcolor|red|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?}}
*Assumes classical mechanics can adequately describe the system <ref> B. H. Mahan J. Chem. Educ. 51(11):709 (1974), http://dx.doi.org/10.1021/ed051p709 </ref>

*Assumes that an equilibrium exists between the reactants and the transition state<ref>Fowler, R., and Guggenheim, E. A., “Statistical Thermodynamics,” Cambridge University Press, London, 1939. Chap. 12.</ref>

*Assumes the system will follow the lowest energy pathway, over the lowest energy saddle point, on the energy surface <ref>Pineda, J. R., Schwartz, S. D., "Protein dynamics and catalysis: The problems of transition state theory and the subtlety of dynamic control," Phil. Trans. R. Soc. B. 361(1472):1433–1438 (2006), http://dx.doi.org/10.1098/rstb.2006.1877.</ref>

''compare values''

''high temp - more states occupied so less accurate - not always lowest energy pathway...''

 
==F-H-H System==

===PES inspection===
[[File:Mln_F_H2.PNG|thumb|300px|right]]
*'''''{{fontcolor|steelblue|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?}}'''''
*'''''{{fontcolor|steelblue|Locate the approximate position of the transition state.}}'''''
*'''''{{fontcolor|steelblue|Report the activation energy for both reactions.}}'''''
====''F + H2 → HF + H''====
[[File:Mln_F_H2_acten.PNG|thumb|300px|left]]
The reaction is exothermic because the products are lower in energy than the reactants, due to the HF bond being stronger (565 kJmol-1) than the HH bond (432 kJmol-1).<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref> Because of this, more energy is released when the HF bond forms than is taken in to break the HH bond, thus it is classified as exothermic due to an overall release of energy.

Hammond's postulate can be used to estimate the structure of the TS, as it states that the transition state look most like the structure that it is closest to in energy: i.e. an exothermic reaction TS structure will be more similar to the reactants, an endothermic reaction TS will be more similar to the products structure.
Therefore, the TS for this reaction will be more similar to an F atom and an H2 molecule. So the HH bond distance, 0.74 Å<ref>Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958</ref>, can be used as a starting point to find the transition state position.

Using the same methods as for the H-H-H system above, the transition state position is found to be where the gradient of the Distance vs Time plot is zero, as explained above, which is at ''r1,ts'' = 1.811 Å, ''r2,ts'' = 0.745 Å.

By slightly displacing one distance from the transition state position (1.811 Å → 1.801 Å) and following the ''mep'', the change in potential energy of the transition state to the products can be seen on the Energy vs Time plot. This difference is equal to the activation energy, which is found to be 30 kcalmol-1.

====''H + HF → F + H2''====
[[File:Mln_HF_H_acten_4.PNG|thumb|300px|left]]
This reaction is endothermic as the products are higher in energy than the reactants, due to the HH bond being weaker than the HF bond. The bond energies can be seen above.

Following the same thought process and using the same methods as the previous reaction, the transition state position and activation energy can be found. The TS position is found to be at ''r1,ts'' = 0.745 Å, ''r2,ts'' = 1.811 Å, the same as the previous reaction. This is expected as they are the inverse reactions of each other, so will have the same transition state structure.

The activation energy was found (by displacing 1.811 Å → 1.801 Å) to be 0.2 kcalmol-1.

====''H' + HF → H'F + H''====
This is the reaction that would occur from the same reactants as the second reaction, but forming different products (which are equivalent to the reactants).

This is a proton transfer reaction and as the bond strengths are the same in both reactants and products, the activation energy will be 0 kcalmol-1, so the system is in equilibrium and the reaction will be constantly occurring in both directions.

 

=== Reaction dynamics ===
For the initial conditions ''rAB'' = 2.3, ''rBC'' = 0.74:
{| class="wikitable" border=1
! ''pAB'' !! ''pBC'' !! Reactive? !! Trajectory contour plot
|-
| -2.5 || 0 || Yes || [[File:Mln_rxndyn_1.PNG|thumb|200px]] || [[File:Mln_rxndyn_11.PNG|thumb|200px]]
|-
| -2.0 || 0 || No || [[File:Mln_rxndyn_2.PNG|thumb|200px]] || [[File:Mln_rxndyn_22.PNG|thumb|200px]]
|-
| -2.0 || -1.5 || Yes || [[File:Mln_rxndyn_3.PNG|thumb|200px]] || [[File:Mln_rxndyn_33.PNG|thumb|200px]]
|-
| 0 || -1.0 || No || [[File:Mln_rxndyn_4.PNG|thumb|200px]] || [[File:Mln_rxndyn_44.PNG|thumb|200px]]
|-
| -0.6 || -1.0 || No || [[File:Mln_rxndyn_5.PNG|thumb|200px]] || [[File:Mln_rxndyn_55.PNG|thumb|200px]]
|-
| -0.65 || -1.0 || Yes || [[File:Mln_rxndyn_6.PNG|thumb|200px]] || [[File:Mln_rxndyn_66.PNG|thumb|200px]]
|-
| -1.5 || -1.5 || Yes || [[File:Mln_rxndyn_7.PNG|thumb|200px]] || [[File:Mln_rxndyn_77.PNG|thumb|200px]]
|}

'''''{{fontcolor|red|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''''
[[File:|thumb|300px|left|Internuclear Momenta vs Time; ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, ''pBC'' = 0]]
Consider the Internuclear Momenta vs Time plot produced from the reactive trajectory with initial conditions of ''rAB'' = 2.3, ''rBC'' = 0.74, ''pAB'' = -2.5, and ''pBC'' = 0. From this, and the animation, we can see that, initially, the H2 molecule (BC) has momentum and it is oscillating as it moves towards the F atom (A). The molecule therefore has kinetic energy in the form of vibrational and translational energy, which is converted to potential energy, due to the Coulombic Forces between the atoms, as it approaches the F atom.

The negative momentum increases to zero (a decrease in absolute momentum), so kinetic energy also decreases, as potential energy increases to a maximum. At the potential energy maximum, where the momentum is zero, the system is in the transition state. As the system passes over the potential maximum, this potential energy is converted to kinetic energy of the products, in the form of translational energy of the H atom and vibrational energy of the HF molecule, so kinetic energy and therefore absolute momentum increases. The internuclear momentum of BC (unbound H H) remains approximately constant as the H atom travels away from the HF atom, and the HF atom is oscillating so its momentum is also oscillating. The oscillations of the final AB momentum (oscillations of the HF produced) is greater than the initial oscillations of BC momentum (H2 reactant), because momentum is a function of mass and the total mass of HF is greater than the total mass of HH.

Overall, translational and vibrational (kinetic) energy of the H2 molecule is converted into translational energy of the unbound H atom and vibrational energy of the HF molecule.

This could be confirmed experimentally by measuring the change in temperature, because the reaction is exothermic so there will be a release of energy causing the temperature to increase. This can also be seen in the greater vibrational energy of the products compared to reactants, so heat is generated by higher energy vibrations.

 
'''''{{fontcolor|red|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.}}'''''
[[File:|thumb|300px|left]]

 

==References==
<references/>

File:Mln HF H acten.PNG

2018-05-25T14:35:55Z

Ml5816: Ml5816 uploaded a new version of File:Mln HF H acten.PNG

File:Mln rxndyn 77.PNG

2018-05-25T14:34:30Z

Ml5816:

File:Mln rxndyn 66.PNG

2018-05-25T14:34:22Z

Ml5816: