== Characterization of Stratum Corneum Molecular Dynamics by Natural-Abundance 13C Solid-State NMR (Bjorklund, Nowacka, Bouwstra, Sparr, Topgaard (2013) ==

JBChemYear4ProjNotes

2019-09-27T19:02:47Z

Jb4814:

== '''T1 Relaxation (Longitudinal Relaxation)''' ==
* The time taken for a proportion of the nuclei of a sample to align their spins with the magnetic field.
* This proportion is 1-(1/e)
* This is about 63%.
* At a value of 5× T1, the number of nuclei with aligned spins gets very close to its maximum value.
* T1 can be measured by the release of energy. This energy is released as it is energetically favourable to be aligned with the magnetic field.
* The loss of energy is completed by collisions with other molecules, collisions with the lattice surrounding the material or electromagnetic decay.
* It can be called longitudinal relaxation as the nuclei decay to point in the same direction as the magnetic field.
* Relaxation in ALL types of NMR have to be induced. For T1 this is often a case of another magnetic field fluctuating near the Larmor frequency in the transverse plane.
* The applied magnetic field simply chooses the longitude that will direct the lowest energy spins, other local magnetic fields will cause relaxation.
* T1 relaxation is always accompanied by T2 relaxation

== T2 Relaxation ==
* Considered to follow first order kinetics.
* Has a simple exponential decay with time period of T2
* Ie the time required for the transverse magnetization to fall to 1/e (~37%) of it’s starting value.
* Once the radio frequency has been passed through the sample, we have a tiny fraction of the molecules that are now coherently pointing in the transverse direction.
* These coherent spins form a NET magnetisation at 90° to the magnetic field but it’s not perfect.
* As with the Larmour frequency, the NET magnetisation now moves in a circular fashion around the plane, with a frequency that is known as the Larmour frequency
* It is this movement in magnetisation that the receiver picks up and converts into the decay signal.
* The decay happens as nuclei become scrambled and leave the coherent group resulting in a weaker signal.
* The decay can occur in a number of ways.
* The coherent molecules can decay in a similar fashion to T1 decay. Any interaction with its surroundings displaces the spin and knocks that nuclei out. This is known as T1 in T2 relaxation.
* T2 relaxation does not always occur with T1. This can be known as secular contribution to T2.
* A molecule can be located near to a local magnetic field caused by something like an iron clump. This produces it’s own magnetic field and so the frequency now becomes w=g(B(local)+B0)
* This means that eventually the molecule is no longer circling in phase and is pointless. Thus relaxation has occurred.
* Spin spin flip flop occurs randomly and has no effect on T1. The longitudinal and transverse motions swap and so it loses coherence and as a result relaxation occurs.

== Larmour Frequency ==
* In a similar fashion to a clock, or a weight in a string (think mechanics circular motion), a magnetic moment will align with the field.
* As it relaxes into the field, it will initially spin around in a circle.
* From circular motion, frequency = 2(pi)r/T.
* The Larmour frequency is also given by the geomagnetic ratio and the strength of the magnetic field
* W=gB

== Phase Coherence ==
* During the initial relaxation to the initial magnetic field, there is a NET nucleus consensus towards the longitudinal direction in which the magnetic field is pointing.
* This is not coherent though as the other fields do not add up to cancel each other out.
* When the radio frequency wave is turned on, it is at right angles to the magnetic field. This forces only some of the spins to point in the transverse direction. Eg 90° to the magnetic field. The other spins are now completely random and their NET spin cancels out. Therefore, we are left with a few nuclei but with a coherent spin 90° to the magnetic field.

== 1H Relaxation ==
* Consider a nucleus or an impaired electron to be an ideal dipole magnet. That is to have a north pole and south pole . These will have fields that will interact through space known as dipole-dipole interactions.
* Electrons are much stronger at inducing relaxation than protons. Therefore, electron -proton interactions will be much quicker than proton – proton interactions.
* This interaction scales with respect to distance by 1/r^6
* Rotation rates and how close they are to the Larmour frequency result in whether T1 orT2 will predominate.

==== Chemical Shift Anisotropy ====
* Chemical shift results from the differing quantity of protection afforded to the nucleus by the electron clouds. Electrons rotate fast around the nucleus producing a magnetic field that counters the magnetic field in which the molecule is placed. Electron rich centres are afforded more protection than those in electron poor areas.
* Chemical shift anisotropy arises when the molecule is not symmetrical and so results in a different shift depending on the direction of the field compared to the direction of the molecule. This does NOT happen in liquids as the constant tumbling averages this out.
* Magic spinning in solid state NMR can average this out to a non zero but known value in order to increase the resolution of the spectrum.
* This is important when looking at biological systems where the attached water molecules cannot necessarily move when bound to a protein.
* It is also important in something like phosphorus NMR

==== Molecular flow, translation and diffusion ====
* Relaxation can be caused by a change in the local surroundings of a molecule. A change in the gravity, the magnetic field (even if caused by the presence of a magnetized iron clump which increases the local magnetic field or something similar).
* Non uniform magnetic fields cause relaxation. This can be caused by foreign magnetic fields or the such like.

==== Chemical Exchanges ====
* Mostly with protons
* Protons can cause relaxation by simply swapping with other protons in neighbouring molecules. This can be accompanied by a change in structure or nothing at all.
* Relaxation is proportional to the square of the magnetic field.

==== J-Coupling ====

== Brownian Motion ==
The constantly and erratic path formed by molecules in a liquid as a consequence of collisions between molecules and general tumbling.

== Water T1/T2 relaxation times ==
* Water relaxation times are generally very long. As they tumble so quickly, they have large spaces between each molecule compared to something like collagen.
* Spin spin interactions and energy exchange between molecules are therefore comparatively slow when compared to tightly packed proteins such as collagen where energy exchanges are frequent and spin spin interactions are more common.
*

JBChemYear4ProjNotes

2019-09-27T19:00:05Z

Jb4814:

=== '''T1 Relaxation (Longitudinal Relaxation)''' ===
* The time taken for a proportion of the nuclei of a sample to align their spins with the magnetic field.
* This proportion is 1-(1/e)
* This is about 63%.
* At a value of 5× T1, the number of nuclei with aligned spins gets very close to its maximum value.
* T1 can be measured by the release of energy. This energy is released as it is energetically favourable to be aligned with the magnetic field.
* The loss of energy is completed by collisions with other molecules, collisions with the lattice surrounding the material or electromagnetic decay.
* It can be called longitudinal relaxation as the nuclei decay to point in the same direction as the magnetic field.
* Relaxation in ALL types of NMR have to be induced. For T1 this is often a case of another magnetic field fluctuating near the Larmor frequency in the transverse plane.
* The applied magnetic field simply chooses the longitude that will direct the lowest energy spins, other local magnetic fields will cause relaxation.
* T1 relaxation is always accompanied by T2 relaxation

=== T2 Relaxation ===
* Considered to follow first order kinetics.
* Has a simple exponential decay with time period of T2
* Ie the time required for the transverse magnetization to fall to 1/e (~37%) of it’s starting value.
* Once the radio frequency has been passed through the sample, we have a tiny fraction of the molecules that are now coherently pointing in the transverse direction.
* These coherent spins form a NET magnetisation at 90° to the magnetic field but it’s not perfect.
* As with the Larmour frequency, the NET magnetisation now moves in a circular fashion around the plane, with a frequency that is known as the Larmour frequency
* It is this movement in magnetisation that the receiver picks up and converts into the decay signal.
* The decay happens as nuclei become scrambled and leave the coherent group resulting in a weaker signal.
* The decay can occur in a number of ways.
* The coherent molecules can decay in a similar fashion to T1 decay. Any interaction with its surroundings displaces the spin and knocks that nuclei out. This is known as T1 in T2 relaxation.
* T2 relaxation does not always occur with T1. This can be known as secular contribution to T2.
* A molecule can be located near to a local magnetic field caused by something like an iron clump. This produces it’s own magnetic field and so the frequency now becomes w=g(B(local)+B0)
* This means that eventually the molecule is no longer circling in phase and is pointless. Thus relaxation has occurred.
* Spin spin flip flop occurs randomly and has no effect on T1. The longitudinal and transverse motions swap and so it loses coherence and as a result relaxation occurs.

=== Larmour Frequency ===
* In a similar fashion to a clock, or a weight in a string (think mechanics circular motion), a magnetic moment will align with the field.
* As it relaxes into the field, it will initially spin around in a circle.
* From circular motion, frequency = 2(pi)r/T.
* The Larmour frequency is also given by the geomagnetic ratio and the strength of the magnetic field
* W=gB

=== Phase Coherence ===
* During the initial relaxation to the initial magnetic field, there is a NET nucleus consensus towards the longitudinal direction in which the magnetic field is pointing.
* This is not coherent though as the other fields do not add up to cancel each other out.
* When the radio frequency wave is turned on, it is at right angles to the magnetic field. This forces only some of the spins to point in the transverse direction. Eg 90° to the magnetic field. The other spins are now completely random and their NET spin cancels out. Therefore, we are left with a few nuclei but with a coherent spin 90° to the magnetic field.

=== 1H Relaxation ===
* Consider a nucleus or an impaired electron to be an ideal dipole magnet. That is to have a north pole and south pole . These will have fields that will interact through space known as dipole-dipole interactions.
* Electrons are much stronger at inducing relaxation than protons. Therefore, electron -proton interactions will be much quicker than proton – proton interactions.
* This interaction scales with respect to distance by 1/r^6
* Rotation rates and how close they are to the Larmour frequency result in whether T1 orT2 will predominate.

==== Chemical Shift Anisotropy ====
* Chemical shift results from the differing quantity of protection afforded to the nucleus by the electron clouds. Electrons rotate fast around the nucleus producing a magnetic field that counters the magnetic field in which the molecule is placed. Electron rich centres are afforded more protection than those in electron poor areas.
* Chemical shift anisotropy arises when the molecule is not symmetrical and so results in a different shift depending on the direction of the field compared to the direction of the molecule. This does NOT happen in liquids as the constant tumbling averages this out.
* Magic spinning in solid state NMR can average this out to a non zero but known value in order to increase the resolution of the spectrum.
* This is important when looking at biological systems where the attached water molecules cannot necessarily move when bound to a protein.
* It is also important in something like phosphorus NMR

==== Molecular flow, translation and diffusion ====
* Relaxation can be caused by a change in the local surroundings of a molecule. A change in the gravity, the magnetic field (even if caused by the presence of a magnetized iron clump which increases the local magnetic field or something similar).
* Non uniform magnetic fields cause relaxation. This can be caused by foreign magnetic fields or the such like.

==== Chemical Exchanges ====
* Mostly with protons
* Protons can cause relaxation by simply swapping with other protons in neighbouring molecules. This can be accompanied by a change in structure or nothing at all.
* Relaxation is proportional to the square of the magnetic field.

==== J-Coupling ====

=== Brownian Motion ===
The constantly and erratic path formed by molecules in a liquid as a consequence of collisions between molecules and general tumbling.

=== Water T1/T2 relaxation times ===
* Water relaxation times are generally very long. As they tumble so quickly, they have large spaces between each molecule compared to something like collagen.
* Spin spin interactions and energy exchange between molecules are therefore comparatively slow when compared to tightly packed proteins such as collagen where energy exchanges are frequent and spin spin interactions are more common.
*

JBChemYear4ProjNotes

2019-09-27T18:54:59Z

Jb4814: Created page with "T1 Relaxation (Longitudinal Relaxation) The time taken for a proportion of the nuclei of a sample to align their spins with the magnetic field. This proportion is 1-(1/e) T..."

T1 Relaxation (Longitudinal Relaxation)
The time taken for a proportion of the nuclei of a sample to align their spins with the magnetic field.
This proportion is 1-(1/e)
This is about 63%.
At a value of 5× T1, the number of nuclei with aligned spins gets very close to its maximum value.
T1 can be measured by the release of energy. This energy is released as it is energetically favourable to be aligned with the magnetic field.
The loss of energy is completed by collisions with other molecules, collisions with the lattice surrounding the material or electromagnetic decay.
It can be called longitudinal relaxation as the nuclei decay to point in the same direction as the magnetic field.
Relaxation in ALL types of NMR have to be induced. For T1 this is often a case of another magnetic field fluctuating near the Larmor frequency in the transverse plane.
The applied magnetic field simply chooses the longitude that will direct the lowest energy spins, other local magnetic fields will cause relaxation.
T1 relaxation is always accompanied by T2 relaxation

T2 Relaxation
Considered to follow first order kinetics.
Has a simple exponential decay with time period of T2
Ie the time required for the transverse magnetization to fall to 1/e (~37%) of it’s starting value.
Once the radio frequency has been passed through the sample, we have a tiny fraction of the molecules that are now coherently pointing in the transverse direction.
These coherent spins form a NET magnetisation at 90° to the magnetic field but it’s not perfect.
As with the Larmour frequency, the NET magnetisation now moves in a circular fashion around the plane, with a frequency that is known as the Larmour frequency
It is this movement in magnetisation that the receiver picks up and converts into the decay signal.
The decay happens as nuclei become scrambled and leave the coherent group resulting in a weaker signal.
The decay can occur in a number of ways.
The coherent molecules can decay in a similar fashion to T1 decay. Any interaction with its surroundings displaces the spin and knocks that nuclei out. This is known as T1 in T2 relaxation.
T2 relaxation does not always occur with T1. This can be known as secular contribution to T2.
A molecule can be located near to a local magnetic field caused by something like an iron clump. This produces it’s own magnetic field and so the frequency now becomes w=g(B(local)+B0)
This means that eventually the molecule is no longer circling in phase and is pointless. Thus relaxation has occurred.
Spin spin flip flop occurs randomly and has no effect on T1. The longitudinal and transverse motions swap and so it loses coherence and as a result relaxation occurs.

Larmour Frequency
In a similar fashion to a clock, or a weight in a string (think mechanics circular motion), a magnetic moment will align with the field.
As it relaxes into the field, it will initially spin around in a circle.
From circular motion, frequency = 2(pi)r/T.
The Larmour frequency is also given by the geomagnetic ratio and the strength of the magnetic field
W=gB

Phase Coherence
During the initial relaxation to the initial magnetic field, there is a NET nucleus consensus towards the longitudinal direction in which the magnetic field is pointing.
This is not coherent though as the other fields do not add up to cancel each other out.
When the radio frequency wave is turned on, it is at right angles to the magnetic field. This forces only some of the spins to point in the transverse direction. Eg 90° to the magnetic field. The other spins are now completely random and their NET spin cancels out. Therefore, we are left with a few nuclei but with a coherent spin 90° to the magnetic field.

1H Relaxation
Consider a nucleus or an impaired electron to be an ideal dipole magnet. That is to have a north pole and south pole . These will have fields that will interact through space known as dipole-dipole interactions.
Electrons are much stronger at inducing relaxation than protons. Therefore, electron -proton interactions will be much quicker than proton – proton interactions.
This interaction scales with respect to distance by 1/r^6
Rotation rates and how close they are to the Larmour frequency result in whether T1 orT2 will predominate.
Chemical Shift Anisotropy
Chemical shift results from the differing quantity of protection afforded to the nucleus by the electron clouds. Electrons rotate fast around the nucleus producing a magnetic field that counters the magnetic field in which the molecule is placed. Electron rich centres are afforded more protection than those in electron poor areas.
Chemical shift anisotropy arises when the molecule is not symmetrical and so results in a different shift depending on the direction of the field compared to the direction of the molecule. This does NOT happen in liquids as the constant tumbling averages this out.
Magic spinning in solid state NMR can average this out to a non zero but known value in order to increase the resolution of the spectrum.
This is important when looking at biological systems where the attached water molecules cannot necessarily move when bound to a protein.
It is also important in something like phosphorus NMR
Molecular flow, translation and diffusion
Relaxation can be caused by a change in the local surroundings of a molecule. A change in the gravity, the magnetic field (even if caused by the presence of a magnetized iron clump which increases the local magnetic field or something similar).
Non uniform magnetic fields cause relaxation. This can be caused by foreign magnetic fields or the such like.
Chemical Exchanges
Mostly with protons
Protons can cause relaxation by simply swapping with other protons in neighbouring molecules. This can be accompanied by a change in structure or nothing at all.
Relaxation is proportional to the square of the magnetic field.
J-Coupling

Brownian Motion
The constantly and erratic path formed by molecules in a liquid as a consequence of collisions between molecules and general tumbling.

Water T1/T2 relaxation times
Water relaxation times are generally very long. As they tumble so quickly, they have large spaces between each molecule compared to something like collagen.
Spin spin interactions and energy exchange between molecules are therefore comparatively slow when compared to tightly packed proteins such as collagen where energy exchanges are frequent and spin spin interactions are more common.

Rep:LUT1235

2018-06-27T12:59:30Z

Jb4814: /* Introduction */

= '''Transition States Exercises - 3rd Year Computational Lab''' =

=== Introduction ===

On a potential energy surface, a minimum is a point at which

== Diels Alder ==

So done the IRC - looks like its been successful - got to check about the optimisation of th eproducts at the PM6 level. I have optimised it and got a negative frequency of -44 but when using the Hartree-Fock 3-21G method I got all positive frequencies. Is this wrong?

Rep:LUT1235

2018-06-20T15:03:27Z

Jb4814: Created page with "= '''Transition States Exercises - 3rd Year Computational Lab''' = === Introduction === On a potential energy surface, a minimum is a point at which"

= '''Transition States Exercises - 3rd Year Computational Lab''' =

=== Introduction ===

On a potential energy surface, a minimum is a point at which

MRD:LUT1235

2017-05-18T14:52:20Z

Jb4814: /* Exercise 1: H + H2 System */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0<ref>Atkins. P, Paula. J, Physical Chemistry, ''Oxford University Press'', '''2014'''</ref>. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

'''dV(ri)/dri =0''' where dV(ri)/dri is the partial derivative of r with respect to V.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms<ref name=":0">T. L. Cottrell, The Strengths of Chemical Bonds, 2d ed., Butterworth, London, 1958; B. deB. Darwent, National
Standard Reference Data Series, NationalBureau of Standards, no. 31, Washington, 1970; S. W. Benson, J. Chem. Educ.
42:502 (1965); and J. A. Kerr, Chem. Rev. 66:465 (1966).</ref> whilst '''r2''' will be much greater than the 0.94 Angstrom<ref name=":0" /> bond length of HF.

The approximate transition state was found by setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant. During the trials, by starting from a smaller HF bond length than the HF bond length in the transition state, this means that the molecules lay on a gradient that took them towards the HF + H products side of the reaction and so further measurements were required in which the HF bond length was expanded. This can be seen in the graph below, where the HF bond length was still too short for equilibrium.
[[File:H2+F position trial.PNG|none|frame]]
The approximate transition state position was found to be:

'''r1= '''0.745 Angstroms

'''r2= '''1.81 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.3-(-133.9)= 30.6 kcal mol-1 = 128 kJ mol-1

An MEP run reveals that the transition state energy is more accurately found to be -103.75 kcal.
[[File:MEP transition state energy.PNG|none|frame]]

Completing the same calculations for the activation energy gives the following results:

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.75-(-103.9)=0.15 kcal mol-1 = 627.6 J mol-1 (4 s.f)

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.75-(-133.9)= 30.15 kcal mol-1 = 126.1 kJ mol-1 (4 s.f)

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy, this can be seen due to the increase in oscillations up and down the wall of the potential well as the molecule disappears off into the distance. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]'''''6) 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 Rules state that vibrational energy is more efficient in promoting a transition state resembling the products - a late transition state - than translational energy. The opposite is true for a transition state resembling the reactants. In this case translational energy is more likely to promote the transition state.<ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman, 2012 3 (23), 3416-3419, DOI: 10.1021/jz301649w</ref>

Therefore, as the reaction of HF + H has a late transition state, if the value of the translational energy is too high, then the reaction will not work as the products will split up again. This reduces the efficiency of the reaction as, even if the molecules have the required energy to theoretically get over the activation energy and past the transition complex, with too much translational or vibrational energy, the molecule will bounce back to the reactants, meaning it was unsuccessful.

These are empirical rules as they have not been proved and have been bought about to explain the experimental observations.

The following graph shows that when rotational energy is high, and the transition state is late, then the reaction has a greater chance of succeeding.
[[File:High RE Use.PNG|none|frame]]

The following graph shows that when translational energy is high, and the transition state is late, then the reaction has a smaller chance of succeeding.
[[File:High TE Use.PNG|none|frame]]

== References ==
<references />

MRD:LUT1235

2017-05-18T14:51:39Z

Jb4814: /* Physical Modelling Course - Year 2 Chemistry Imperial College London */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0<ref>Atkins. P, Paula. J, Physical Chemistry, ''Oxford University Press'', '''2014'''</ref>. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

dV(ri)/dri =0 where dV(ri)/dri is the partial derivative of r with respect to V.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms<ref name=":0">T. L. Cottrell, The Strengths of Chemical Bonds, 2d ed., Butterworth, London, 1958; B. deB. Darwent, National
Standard Reference Data Series, NationalBureau of Standards, no. 31, Washington, 1970; S. W. Benson, J. Chem. Educ.
42:502 (1965); and J. A. Kerr, Chem. Rev. 66:465 (1966).</ref> whilst '''r2''' will be much greater than the 0.94 Angstrom<ref name=":0" /> bond length of HF.

The approximate transition state was found by setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant. During the trials, by starting from a smaller HF bond length than the HF bond length in the transition state, this means that the molecules lay on a gradient that took them towards the HF + H products side of the reaction and so further measurements were required in which the HF bond length was expanded. This can be seen in the graph below, where the HF bond length was still too short for equilibrium.
[[File:H2+F position trial.PNG|none|frame]]
The approximate transition state position was found to be:

'''r1= '''0.745 Angstroms

'''r2= '''1.81 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.3-(-133.9)= 30.6 kcal mol-1 = 128 kJ mol-1

An MEP run reveals that the transition state energy is more accurately found to be -103.75 kcal.
[[File:MEP transition state energy.PNG|none|frame]]

Completing the same calculations for the activation energy gives the following results:

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.75-(-103.9)=0.15 kcal mol-1 = 627.6 J mol-1 (4 s.f)

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.75-(-133.9)= 30.15 kcal mol-1 = 126.1 kJ mol-1 (4 s.f)

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy, this can be seen due to the increase in oscillations up and down the wall of the potential well as the molecule disappears off into the distance. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]'''''6) 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 Rules state that vibrational energy is more efficient in promoting a transition state resembling the products - a late transition state - than translational energy. The opposite is true for a transition state resembling the reactants. In this case translational energy is more likely to promote the transition state.<ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman, 2012 3 (23), 3416-3419, DOI: 10.1021/jz301649w</ref>

Therefore, as the reaction of HF + H has a late transition state, if the value of the translational energy is too high, then the reaction will not work as the products will split up again. This reduces the efficiency of the reaction as, even if the molecules have the required energy to theoretically get over the activation energy and past the transition complex, with too much translational or vibrational energy, the molecule will bounce back to the reactants, meaning it was unsuccessful.

These are empirical rules as they have not been proved and have been bought about to explain the experimental observations.

The following graph shows that when rotational energy is high, and the transition state is late, then the reaction has a greater chance of succeeding.
[[File:High RE Use.PNG|none|frame]]

The following graph shows that when translational energy is high, and the transition state is late, then the reaction has a smaller chance of succeeding.
[[File:High TE Use.PNG|none|frame]]

== References ==
<references />

File:High TE Use.PNG

2017-05-18T14:50:32Z

Jb4814:

File:High RE Use.PNG

2017-05-18T14:50:22Z

Jb4814:

MRD:LUT1235

2017-05-18T14:42:16Z

Jb4814: /* Exercise 1: H + H2 System */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0<ref>Atkins. P, Paula. J, Physical Chemistry, ''Oxford University Press'', '''2014'''</ref>. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

dV(ri)/dri =0 where dV(ri)/dri is the partial derivative of r with respect to V.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms<ref name=":0">T. L. Cottrell, The Strengths of Chemical Bonds, 2d ed., Butterworth, London, 1958; B. deB. Darwent, National
Standard Reference Data Series, NationalBureau of Standards, no. 31, Washington, 1970; S. W. Benson, J. Chem. Educ.
42:502 (1965); and J. A. Kerr, Chem. Rev. 66:465 (1966).</ref> whilst '''r2''' will be much greater than the 0.94 Angstrom<ref name=":0" /> bond length of HF.

The approximate transition state was found by setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant. During the trials, by starting from a smaller HF bond length than the HF bond length in the transition state, this means that the molecules lay on a gradient that took them towards the HF + H products side of the reaction and so further measurements were required in which the HF bond length was expanded. This can be seen in the graph below, where the HF bond length was still too short for equilibrium.
[[File:H2+F position trial.PNG|none|frame]]
The approximate transition state position was found to be:

'''r1= '''0.745 Angstroms

'''r2= '''1.81 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.3-(-133.9)= 30.6 kcal mol-1 = 128 kJ mol-1

An MEP run reveals that the transition state energy is more accurately found to be -103.75 kcal.
[[File:MEP transition state energy.PNG|none|frame]]

Completing the same calculations for the activation energy gives the following results:

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.75-(-103.9)=0.15 kcal mol-1 = 627.6 J mol-1 (4 s.f)

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.75-(-133.9)= 30.15 kcal mol-1 = 126.1 kJ mol-1 (4 s.f)

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy, this can be seen due to the increase in oscillations up and down the wall of the potential well as the molecule disappears off into the distance. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]'''''6) 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 Rules state that vibrational energy is more efficient in promoting a transition state resembling the products - a late transition state - than translational energy. The opposite is true for a transition state resembling the reactants. In this case translational energy is more likely to promote the transition state.<ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman, 2012 3 (23), 3416-3419, DOI: 10.1021/jz301649w</ref>

Therefore, as the reaction of HF + H has a late transition state, if the value of the translational energy is too high, then the reaction will not work as the products will split up again. This reduces the efficiency of the reaction as, even if the molecules have the required energy to theoretically get over the activation energy and past the transition complex, with too much translational or vibrational energy, the molecule will bounce back to the reactants, meaning it was unsuccessful.

These are empirical rules as they have not been proved and have been bought about to explain the experimental observations.

The following graph shows that when rotational energy is high, and the transition state is late, then the reaction has a greater chance of succeeding.

The following graph shows that when translational energy is high, and the transition state is late, then the reaction has a smaller chance of succeeding.

== References ==
<references />

MRD:LUT1235

2017-05-18T14:40:08Z

Jb4814: /* EXERCISE 2: F - H - H system */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0<ref>Atkins. P, Paula. J, Physical Chemistry, ''Oxford University Press'', '''2014'''</ref>. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

dV(ri)/dri =0 where dV(ri)/dri is the partial derivative of r with respect to V.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.907755 Angstroms

To 3 significant figures this is found to be:

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms<ref name=":0">T. L. Cottrell, The Strengths of Chemical Bonds, 2d ed., Butterworth, London, 1958; B. deB. Darwent, National
Standard Reference Data Series, NationalBureau of Standards, no. 31, Washington, 1970; S. W. Benson, J. Chem. Educ.
42:502 (1965); and J. A. Kerr, Chem. Rev. 66:465 (1966).</ref> whilst '''r2''' will be much greater than the 0.94 Angstrom<ref name=":0" /> bond length of HF.

The approximate transition state was found by setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant. During the trials, by starting from a smaller HF bond length than the HF bond length in the transition state, this means that the molecules lay on a gradient that took them towards the HF + H products side of the reaction and so further measurements were required in which the HF bond length was expanded. This can be seen in the graph below, where the HF bond length was still too short for equilibrium.
[[File:H2+F position trial.PNG|none|frame]]
The approximate transition state position was found to be:

'''r1= '''0.745 Angstroms

'''r2= '''1.81069322293 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.3-(-133.9)= 30.6 kcal mol-1 = 128 kJ mol-1

An MEP run reveals that the transition state energy is more accurately found to be -103.75 kcal.
[[File:MEP transition state energy.PNG|none|frame]]

Completing the same calculations for the activation energy gives the following results:

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.75-(-103.9)=0.15 kcal mol-1 = 627.6 J mol-1 (4 s.f)

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.75-(-133.9)= 30.15 kcal mol-1 = 126.1 kJ mol-1 (4 s.f)

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy, this can be seen due to the increase in oscillations up and down the wall of the potential well as the molecule disappears off into the distance. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]'''''6) 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 Rules state that vibrational energy is more efficient in promoting a transition state resembling the products - a late transition state - than translational energy. The opposite is true for a transition state resembling the reactants. In this case translational energy is more likely to promote the transition state.<ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman, 2012 3 (23), 3416-3419, DOI: 10.1021/jz301649w</ref>

Therefore, as the reaction of HF + H has a late transition state, if the value of the translational energy is too high, then the reaction will not work as the products will split up again. This reduces the efficiency of the reaction as, even if the molecules have the required energy to theoretically get over the activation energy and past the transition complex, with too much translational or vibrational energy, the molecule will bounce back to the reactants, meaning it was unsuccessful.

These are empirical rules as they have not been proved and have been bought about to explain the experimental observations.

The following graph shows that when rotational energy is high, and the transition state is late, then the reaction has a greater chance of succeeding.

The following graph shows that when translational energy is high, and the transition state is late, then the reaction has a smaller chance of succeeding.

== References ==
<references />

MRD:LUT1235

2017-05-18T14:07:10Z

Jb4814: /* EXERCISE 2: F - H - H system */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0<ref>Atkins. P, Paula. J, Physical Chemistry, ''Oxford University Press'', '''2014'''</ref>. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

dV(ri)/dri =0 where dV(ri)/dri is the partial derivative of r with respect to V.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.907755 Angstroms

To 3 significant figures this is found to be:

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms<ref name=":0">T. L. Cottrell, The Strengths of Chemical Bonds, 2d ed., Butterworth, London, 1958; B. deB. Darwent, National
Standard Reference Data Series, NationalBureau of Standards, no. 31, Washington, 1970; S. W. Benson, J. Chem. Educ.
42:502 (1965); and J. A. Kerr, Chem. Rev. 66:465 (1966).</ref> whilst '''r2''' will be much greater than the 0.94 Angstrom<ref name=":0" /> bond length of HF.

The approximate transition state was found by setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant. During the trials, by starting from a smaller HF bond length than the HF bond length in the transition state, this means that the molecules lay on a gradient that took them towards the HF + H products side of the reaction and so further measurements were required in which the HF bond length was expanded. This can be seen in the graph below, where the HF bond length was still too short for equilibrium.
[[File:H2+F position trial.PNG|none|frame]]
The approximate transition state position was found to be:

'''r1= '''0.745 Angstroms

'''r2= '''1.81069322293 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.3-(-133.9)= 30.6 kcal mol-1 = 128 kJ mol-1

An MEP run reveals that the transition state energy is more accurately found to be -103.75 kcal.
[[File:MEP transition state energy.PNG|none|frame]]

Completing the same calculations for the activation energy gives the following results:

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.75-(-103.9)=0.15 kcal mol-1 = 627.6 J mol-1 (4 s.f)

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-103.75-(-133.9)= 30.15 kcal mol-1 = 126.1 kJ mol-1 (4 s.f)

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy in the HF molecule. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]'''''6) 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 Rules state that vibrational energy is more efficient in promoting a transition state resembling the products - a late transition state - than translational energy. The opposite is true for a transition state resembling the reactants. In this case translational energy is more likely to promote the transition state.<ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman, 2012 3 (23), 3416-3419, DOI: 10.1021/jz301649w</ref>

Therefore, as the reaction of HF + H has a late transition state, if the value of the translational energy is too high, then the reaction will not work as the products will split up again. This reduces the efficiency of the reaction as, even if the molecules have the required energy to theoretically get over the activation energy and past the transition complex, with too much translational or vibrational energy, the molecule will bounce back to the reactants, meaning it was unsuccessful.

These are empirical rules as they have not been proved and have been bought about to explain the experimental observations.

== References ==
<references />

2017-05-18T12:33:40Z

Jb4814: /* Physical Modelling Course - Year 2 Chemistry Imperial College London */

= '''Physical Modelling Course - Year 2 Chemistry Imperial College London''' =

== Exercise 1: H + H2 System ==
'''''1)'' ''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at a minimum and at a transition structure is 0. The transition structure can be distinguished as this is the point where maximum potential energy exists on the energy line that links the reactants to the products. The double partial derivative of this point will be negative as it is a maximum on the potential energy line. The transition state can be further distinguished by the fact the the gradient either side of this maximum is steepest along the line between reactants and products.

The minima can be distinguished as its double partial derivative will be positive.

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

The best estimate given for the transition state position is given by the position where the vibrations are of a negligible proportion and the final momenta of the hydrogen molecules are 0.

r'''ts'''=0.907755 Angstroms

To 3 significant figures this is found to be:

r'''ts'''=0.908 Angstroms

This was found for a trajectory gave the bond lengths to be constant with time as the following screenshot shows:
[[File:Molecular Dynamics Rts H2.png|frame|The graph of inter-nuclear distance against time with the initial momentum set to 0 and the radii set to values (0.907755 Angstrom) where the vibration would be negligible
|none|767x767px]]

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

The mep has no vibrations after the reaction has taken place. As shown by this surface plot, the molecules move in a straight line along the bottom of the valley once the reaction has taken place. This is due to the fact that the energy resets to 0 upon each calculation and so when the bond lengths reach the equilibrium distance the energy is set to 0 before the next calculation and so no further oscillations will take place. Therefore the momentum of the molecule (p=mv) is also 0 and therefore the molecule can only take the route of minimum energy and not deviate.
[[File:Molecular Dynamics H2 Reaction Path MEP JB.png|none|frame|764x764px|The trajectory of the hydrogen atoms in the minimum energy path setting. The path follows the bottom of the valley closely with no vibrations. ]]

On the other hand, the dynamic trajectory includes some vibrations after the reaction has taken place. This is due to the fact that the programme takes into consideration bond vibrations and so the energy is not 0 and so can deviate from the minimum line, albeit very slightly.
[[File:Molecular Dynamics H2 Reaction Path dynamics JB.png|none|thumb|1008x1008px|The trajectory of the hydrogen atoms in the dynamic setting. The path of the hydrogen atoms does not follow the valley of the surface potential plot as closely as the mep trajectory as the initial conditions were not set exactly to the equilibrium bond length. This means that vibrations will continue after the reaction has taken place.]]

'''''4) What do you observe?'''''

We observe that the momentum of the particles has changed and that their new momentum maps exactly onto the momentum of the particle on the opposite side of the molecule. As the graphs show, the momenta and the internuclear distance behaves exactly the same. This is due to the symmetry of the molecule. The graphs therefore are identical other than the labeling of atoms A, B and C.
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear momenta v t JB 2.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 1.PNG|none|frame]]
[[File:Molecular Dynamics H2 Reaction Path dynamics internuclear dist v t r1+d JB 2.PNG|none|frame]]

'''''5) Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''
{| class="wikitable"
!p1
!p2
!Successful or Not?
!Trajectory Route
!What happens along the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 1 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 2 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which is vibrating initially. Atom C continues to get closer but its relative speed to the molecule of hydrogen decreases as it converts its kinetic energy to potential energy as it draws closer to the molecule. Its nearest approach is X Angstroms before it has no further kinetic energy with which to approach and for a new bond. Then it is repelled by the molecule of hydrogen and its potential energy begins to be turned to kinetic energy and it accelerates away from the molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Successful
|[[File:Yes or No - conditions 3 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has small initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close but as B and C begin to oscillate the speed of A also oscillates such that the relative speed AC remains constant. The resulting hydrogen molecule (BC) continues to oscillate after the reaction has gone to completion.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unsuccessful
|[[File:Yes or No - conditions 4 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen (A-B) at a relative velocity that is quite high. The A-B molecule has negligible vibrations initially. Atom C starts large oscillations around B, pushing atom A further out which is where the reaction co-ordinate appears to cross the barrier, which it does. However, atom A eventually starts oscillating around B as well and these vibrations are large enough to cross the barrier for a second time and slowly push the still oscillating atom C further away from the central atom meaning that we end up with the initial reactants
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Successful
|[[File:Yes or No - conditions 5 - JB.PNG|thumb]]
|Hydrogen atom C approaches the molecule of hydrogen A-B which has minimal initial vibrations. Once atom C gets within X Angstroms, atom A begins to move away from atom B with increasing speed as C gets increasingly close pushing the reaction co-ordinate across the barrier. As atom C begins to oscillate, atom A initially has a lot of vibrational energy and moves back in and begins to oscillate, pushing the reaction back across the barrier. However, C is still oscillating with large vibrational energy and pushes the reaction once again across the barrier and this time atom C does not have sufficient kinetic energy to have to react, moving further and further away whilst still oscillating.
|}
'''''6) 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?'''''

The main assumptions of Transition state theory are that there are reactants that occupy one energy level and products which occupy a different energy level and these two plateaus are joined by a curve that passes through a maximum. This maximum is known as the transition state, where the energy of the arrangement of the molecules is the highest and therefore the arrangement is the least stable. The gap between the reactant energy level and the energy of the transition state is known as the activation energy. This model assumes that if we give a group of molecules enough energy (in excess of the activation energy), then the molecules will always rearrange into the products.

The limitation of this model is in the fact that as we apply more energy to the system then the results get more chaotic and the results of which become more uncertain - something that is not accommodated in the transition state model. Giving the molecules a lot of energy means that the reaction can cross the activation barrier several times and be effected by several different factors including solvent and other reaction molecules.

Therefore the rate of reaction predicted by the Transition State Theory will be larger than the experimental rate of reaction as the experimental rate will be affected by molecules bouncing over the activation barrier several times.

== EXERCISE 2: F - H - H system ==
'''''1) Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

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

The reaction plot of H2 + F to form HF +H shows that the energy of the products is lower than the energy of the reactants. This suggests that the reaction is exothermic. The energy of the products is shown in the graph below coming out of the page towards us.

The reverse reaction of HF+ H to form H2 + F is the exact reverse of this reaction and is therefore endothermic.

[[File:Molecular Dynamics H2 + F energy plot.PNG|none|frame]]

The reaction of H+HF to form HF+H is symmetrical and as the products and reactants are identical, they cannot be distinguished unless they are in a lattice or they are chemically labelled. Therefore, the reaction is neither exothermic nor endothermic.

This fact that a reaction is exothermic or endothermic is a reflection on the stability and the bond strength of the reactants and the products. The side of the reaction with lower energy (whether this is the reactants in the case of an endothermic reaction or the products in the case of an exothermic reaction) is the more stable side. This means that the HF bond formed in the case of the above exothermic reaction is stronger than the H-H bond that is present in the reactants. This means that, taking the transition state (point of highest energy on the minimum energy path from reactants to products) as the point of reference, more energy is released when the stronger HF bond is formed compared to the energy released when the H-H bond is formed.

From Hammonds Postulate which suggests that ''<nowiki/>'If two states appear in consecutive steps of a reaction and involve only a small difference in energy, then the difference between the two states involves a small rearrangement of molecules'.''

This suggests that the transition state appears more like the side of higher energy than the side with lower energy. In this case, The transition state will appear more similar to the reactants than the products and therefore the covalent bond between the hydrogen atoms will probably still exist and therefore '''r1= '''0.745 Angstroms whilst '''r2''' will be much greater than the 0.94 Angstrom bond length of HF.

By setting the momenta to 0 and setting '''r1= '''0.745 Angstroms, and then increasing '''r2 '''until the internuclear distance remains constant, it was found that the approximate transition state was when:

'''r1= '''0.745 Angstroms

'''r2= '''1.81069322293 Angstroms

Which to 3 significant figures:

'''r2= '''1.81 Angstroms

This is shown by the internuclear distance graph below:

[[File:H2+F position.PNG|none|frame]]

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

The activation energy of this reaction is the energy difference between the energy of the reactants and the energy of the transition state. The energy of each state can be seen as the z co-ordinate of each of the positions in the graph.

This can be measured by using the co-ordinate picker on the surface plot as can be seen when used for the transition state energy point on the graph below

[[File:H2+F transition state energy.PNG|none|frame]]
Therefore the energy of the transition state is -103.3 kcal mol-1.

Similar calculations find that for the HF + H system, the energy of the system is found to be -133.9 kcal mol-1 whilst the energy of the H2 + F system is found to be -103.9 kcal mol-1.

i) H2 + F --> HF + H

The activation energy of this reaction is found to be:

-103.3-(-103.9)=0.6 kcal mol-1 = 2510 J mol-1

ii) HF + H -->H2 + F

The activation energy of this reaction is found to be:

-133.9-(-103.3)= -30.6 kcal mol-1 = -128 kJ mol-1

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

The reaction conditions that allow for a successful reaction between F and H2 can be seen in the surface plot below and are as follows:

'''rHH:''' 0.74 Angstroms

'''rHF:''' 2.30 Angstroms

'''pHH'''= -0.5 Ns

'''pHF'''= -1 Ns
[[File:Succesful F+H2.PNG|none|frame]]

As can be seen, the momentum of the fluorine atom with respect to the the hydrogen molecule has to be high enough in order that the fluorine atom can approach close enough that it can react with the hydrogen molecule. As previously discussed, the reaction is endothermic, the potential energy decreases. From the conservation of energy, this means that some form of kinetic energy must be increased in order to counter the decrease in potential energy. From the graph, this appears to be an increase in vibrational energy in the HF molecule. The bond length has large oscillations which requires kinetic energy. Furthermore, this energy could be dissipated as rotational energy and in the momentum of the products.

This could be confirmed experimentally by calorimetry. This measures the heat released by the reaction - this heat is in effect the kinetic energy of the particles being passed on to other molecules as they bump into each other without reacting.

'''''4) Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''''

It is noticed that when |3|>'''PHH'''>|2|, the reaction goes through stages when the momentum is enough so that the system is reactive but also goes through stages where the system is not reactive as the vibrational energy of the H-F bond that is formed is so great that it breaks itself once formed and the H2 molecule is reformed.

When '''PHH<'''|2|, no matter what the value, the reaction is unsuccessful. This is due to the fact that the momentum of the fluorine is not relatively large enough to approach close enough to the molecule to react or even influence the H2 bond length, despite the fact that the oscillations between the to hydrogen atoms are still quite large. This can be seen in the following graphs.
[[File:H2+F PHH -2.0.PNG|none|frame]]
[[File:H2+F PHH -2.2.PNG|none|frame]]
[[File:H2+F PHH -2.6.PNG|none|frame]]
[[File:H2+F PHH -2.8.PNG|none|frame]]
[[File:H2+F PHH -3.PNG|none|frame]]'''''5) For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''''

We observe that the system is now set up to be successful but it takes some time for the final result to be known. This is due to the fact that the reaction is proposed to pass through the transition complex several times as the potential and vibrational energy is such that the molecules of HF and HH are continually breaking apart and reforming as can be seen in the graph below.
[[File:H2+F PHH 0.1 PFH 0.8.PNG|none|frame]]