01356570:physicalcomplab

2019-05-24T08:20:28Z

Hk4117:

For a three atom reaction, the potential surface is four dimensional. In the discussions below, the angle between the bonds is fixed to 180o, making the surface collinear to decrease the dimensionality to three. A potential energy profile for the three atom reaction is then plotted as a function of the internuclear distances, the reaction path in order to investigate a reaction of the hydrogen molecule with hydrogen atom and with a fluorine atom.

=Exercise 1: H + H2=

==Transition State==

'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

Transition state of a reaction results in a maximum point on a potential energy surface vs reaction coordinate diagram. This is the highest energy state that the reactants have to go through before turning into the products. All of the kinetic energy is converted to potential energy at that state, resulting in a 0 gradient at the point. The first step in calculation is finding the values for position, r that gives 0 when the first derivative of potential energy, V(r) is taken with respect to the Cartesian coordinates. The calculated r values are then introduced into the second derivative of V(r). A negative value of second derivative means that the r value corresponds to a maximum whereas a positive value shows that the r value is where the minimum of the potential energy is located on the graph. As a result, the negative value will indicate the location of transition state at that point.

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

The reaction between the (HA atom and the HB-HC molecule occurs on a symmetric surface. This means that at the transition state the internuclear distances between the consecutive atoms will be the same; rAB=rBC.

In order to locate the saddle point r values ranging between 0.78 and 1 A were used for experimentation. Since the equilibrium bond length for a Hydrogen molecule is 0.74 A, and since an elongation of the bond is expected at the transition state, a value bigger than that was chosen. The limit was chosen such that the molecule and the atom were close enough such that H atom HA atom and the HB would result in a bond with a length of 0.74 A after crossing over the transition state. After experimenting with the values, it was found that the maximum energy occurred at rAB=rBC= 0.90774 A. All the trials were done by equating the two momenta parameters to 0. When all of the kinetic energy is converted to potential energy and the species are sitting on a maximum with no momentum the species stay there resulting in no reaction trajectories.

[[File:H2transitionstate location.PNG]]
''Figure 1: The location of x shows the position of transition state. There is no trajectory drawn to any of the wells shown in the diagram which means the species are not anywhere that is downhill but right at the maximum point.''

[[File:H2internuclearplotfortransition.PNG]]
''Figure 2: The Internuclear Distances vs Time graph above also confirms that the species are stationary since there is no change in the distances as the time progresses.''

[[File:H2transitionstaeblackdot.PNG]]
''Figure 3 : The location of black dot shows the location of the transition state on a potential energy surface diagram''

==MEP and Dynamics Calculations==

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

The trajectories below were calculated such that the distance between the the atoms which used to be a molecule was increased to 0.91774 and the distance where a new bond will form was kept at the transition state position, 0.90774 with 0 momenta. This means that the elongation of the bond increased and the transition state is crossed over. The reaction path is downhill and so the products are expected to be formed.

[[File:Dynamicscontour.PNG]] [[File:Dynamicsinternuclear.PNG]] Figure 4: Calculation with the Dynamics algorithm

[[File:Mepcontour.PNG]] [[File:Mepinternucleardistance.PNG]] Figure 5: Calculation with the MEP algorithm

MEP (minimum energy path) and dynamics were the two algorithms used to observe the trajectory of the reaction. In MEP calculation of the trajectory, it could be seen that the product has no change in its internuclear distance as time progresses whereas vibration is observed for the latter calculation. This means that in MEP calculation, energy is not conserved, it does not take into account the involvement of kinetic energy after bond formation. The extra energy released after bond formation goes into rotational and vibrational energy. Dynamics algorithm instead uses the dynamic information and shows that the excess energy is converted into vibrational such that the equilibrium bond length goes through oscillation.

[[File:Dynamicsmomentum.PNG]] [[File:Momentummep.PNG]] Figure 6: Momenta from Dynamics and MEP calculations respectively

Even though the species are not given initial momenta, since they are positioned to start from a downhill position, conversion of potential energy to kinetic allows the movement of the species to the most stable position, the energy level of the products, giving rise to changes in momentum. Since MEP calculation ignores the conservation of energy, it assumes that the product will have no energy to vibrate and be stationary.

==Reactive and Unreactive Trajectories==

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Internuclear Distance vs Time Graph !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Momentum1.PNG]] || [[File:Table1.PNG]] || The collision between the HA atom and the HB-HC molecule results in a successful reaction. It could be seen that the product molecule has more vibrational energy than the reactant molecule since the trjectory after crossing the transition state has a bigger frequency and amplitude of vibration.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Momentum2moreaccurate.PNG]] || [[File:Table2.PNG]] || Although the molecule and the atom collide, the collision does not result in a successful reaction. The kinetic energy of the reactants is not enough to get over the activation energy barrier.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Momentum3.PNG]] || [[File:Table3.PNG]] || The trajectory in that case shows that products are produced. In comparison to the first trajectory the reactants have more vibrational energy because of having a slight increase towards negative in the momentum of the stating hydrogen molecule. This resulted in a less defined vibrations in the final molecule when compared to the first trajectory. This could be explained such that the primary hydrogen molecule had less available energy when crossing the barrier which resulted in a smaller energy change and so less energy to be conserved by vibrational and rotational modes.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Momentum4.PNG]] || [[File:Table4.PNG]] || This reaction has enough energy that the reactants reach the transition state and the products are produced. The resulting product has so much energy that, the bond length elongates back again and then breaks reforming the reactants. This case evokes the idea of an anharmonic oscillator. The products produced have a lot of energy that the vibration results the bond length reaching to the levels that the attraction between the two nuclei is not enough to keep the atoms together. The released energy after breaking the bond is again enough to recross the transition state and produce the vibrationally excited HB-HC molecule.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Momentum5.PNG]] || [[File:Table5.PNG]] || This trajectory results in a successful collision in comparison to the case above. The difference is such that more kinetic energy is put into the system. Increasing the pHAHB resulted in a collision that formed a bond which then quickly turned back. The second collision then gave the vibrationally excited products successfully. It could be seen that the vibrations have high amplitude and frequency.
|}

==Transition State Theory==

'''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 transition state theory employs the occurrence of that maximum state in between the reactants and products in order to equate the expression for thermal rate constant. The primary assumptions of the theory are listed below:

1-Born-Oppenheimer approximation(treating electronic and nuclear motion separately)

2-Distribution of reactants amongst states follows Maxwell-Boltzmann laws.

3-Once the transition state is crossed towards the products, molecules will not roll back to reform the reactants.

4-The motion available to the species in the transition state is translation only.

5-Irrespective of the reversibility of the reaction the states of the species in transition state are distributed in accordance with Maxwell-Boltzman distribution.

In reaction 4 presented above, although the transition state was crossed over and the HA-B distance got smaller than they were at the transition state, the mechanism reverted, producing the reactants back again. The third assumption in that case fails. Theory instead suggests that, the greater kinetic energy should have resulted in a greater rate constant, giving the products even faster. Conservation of energy thus brings about a highly excited vibrational motion. This relates to anharmonic oscillator which tells how bigger displacements might result in the bond between the two atoms to be broken. The energy released by bond breakage also helps get back to the transition state and roll to the reactants side.

=Exercise 2: F + H2 and H + HF=
==Potential Energy Surfaces==

'''By inspecting the potential energy surfaces, 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 PES graphs below show the two local minima, the former representing the level where reactants are found, H2 and F, and the latter which is much lower in energy standing for the products, H and HF.

[[File:Products min.PNG]] [[File:Reactants min.PNG]] [[File:Enth.PNG]]

Since the products are more stable and so have a more negative energy than the reactants, it was expected to have a release of energy going from the reactants to the products. This is confirmed by the enthalpy calculation below which results in a negative value proving that it is an exothermic reaction. It is the strength of the H-F bond that makes the product a lot more stable than the reactants. The huge electronegativity difference between H and F atoms give a highly ionic bond with very strong attractions resulting in a very strong bond.

'''Enthalpy Calculation'''
<pre>
deltaH=Hproducts-Hreactants

=-130.599-(103.783)=-26.82 kcal/mol
</pre>

'''Transition State'''

The transition state seems not to appear in the graph. Since it is not possible to capture a species in a reaction in its transition state, Hammond's postulate uses the configuration of the local minimum closest to the transition state to determine the possible arrangement of the transition state. The local minima are either the products or reactants or intermediates. They could all be captured in a reaction mechanism and the one that has the most similar energy to transition state is said to resemble it.

The transition state occurs at 1.81009 and 0.74489. The equilibrium bond length for the H-H bond is 0.74. This means that the transition state is expected to appear very close in energy to the reactants. The minimal elongation of the bond is also supported by the energy of the transition state which was found by setting momenta 0 and placing the internuclear distances where they do not follow a trajectory and are stable. The table below shows the activation energies from the information:

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

<pre>
For H2+F reaction:

Transition State Energy-Reactants Energy= -104.01-(-103.783)=0.23 kcal/mol

For HF+H reaction=
Transition Energy-Reactant Energy= -130.599-(-104.01)=-26.59 kcal/mol
</pre>

==Reaction Dynamics==

===Polanyi's Rules and Hammond's Postulate===

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

Increasing the kinetic energy of the hydrogen molecule resulted in a reaction with bigger vibrational energy to some extent but after a point the kinetic energy was too high that -3 collision took place but bond formation did not happen.

no reaction when the momentum was positive. kinetic energy increases irrespective of the sign of the momentum

goes back to reactants?

Infrared Chemiluminescence could be used to confirm that the produced HF molecule is vibrationally excited. Once the products are produced during a reaction, an electromagnetic radiation in the IR region is emitted which confirms that the products are vibrationally excited. The technique is primarily employed for the reactions that produce a Hydrogen-Halogen molecule in excited states.

inversion of momentum procedure

'''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.'''

It is the translational energy that is available for use in conversion to products whereas the kinetic energy that comes from vibration and rotation could not be used for this purpose. Once the available energy surpasses the products are produced.

reaction compleex is very close to F2 arti H. Most of the heat released during the reaction is taken up by the HF molecules. The channelling of this energy to the vibration of the olecule could be seen by the trajectory below. As the distance between HC and HB increased and the H-F distance decreased, vibrational excition appeared in the teajectory.

The product energy states are distributed among vibrational, rotational and translational modes.

Fo rotational and vibrational energies, the levels are quantised

how energy is consumed and disposed in a biomolecular reaction.

an early transition

The rules in combination support the idea that an exothermic reaction has an early transition state and that an endothermic reaction has a late transition state. The state should resemble the reactants for the former case whereas the products for the endothermic case. For an efficient reaction, if the transition state occurs at an early state

the vibrational energy of the reactant bein channeled to the translational energy

=References=

2019-05-23T21:51:19Z

Hk4117: