View source for MRD:hz7718

==Exercise 1: H + H<sub>2</sub> system==

Grade: 3/5 - '''''<u>Some good features to this report. Some of your numerical estimates were a little off, and you failed to complete the final excercise. Other individual comments are provided throughout. </u>'''''

===Transition state===

<nowiki> </nowiki>Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero. 

For finding transition state on graph, the slopes of both r<sub>1</sub> and r<sub>2</sub> directions should be zero, which means δV/δr<sub>1</sub> = 0 and δV/δr<sub>2</sub> = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (δ<sup>2</sup>V/δr<sub>1</sub><sup>2</sup>)*(δ<sup>2</sup>V/δr<sub>2</sub><sup>2</sup>) - δ<sup>2</sup>V/(δr<sub>1</sub>*δr<sub>2</sub>) < 0.

A minimum stationary point means in both directions, it is the minimum point with zero gradients (δ<sup>2</sup>V/δx<sup>2</sup> > 0 and (δ<sup>2</sup>V/δr<sub>1</sub><sup>2</sup>)*(δ<sup>2</sup>V/δr<sub>2</sub><sup>2</sup>) -δ<sup>2</sup>V/(δr<sub>1</sub>*δr<sub>2</sub>) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.

'''''<u>Where do these directions lie on your PES though?</u>'''''

{| class="wikitable" border="1"
!saddle point on a graph !! Minimum stationary point plot !! Maximum stationary point plot
|-
| [[file:hz7718_py1.png|500px|thumb|left]] || [[file:hz7718_py2.png|500px|thumb|right]]   || [[file:hz7718_py3.png|500px|thumb|right]]
|}

===locating the transition state===
When the system is at transition state, the potential energies of both p<sub>1</sub> and p<sub>2</sub> are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r<sub>1</sub> and r<sub>2</sub> will keep constant. On the graph, r<sub>1</sub> and r<sub>2</sub> equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r<sub>ts</sub>=90.7pm.

[[file:hz7718_py4.png|300px]]   


'''<u>''A good estimate for the TS, well done!''</u>'''
===Trajectories from r<sub>1</sub>=r<sub>ts</sub>+δ, r<sub>1</sub>=r<sub>ts</sub>===
In MEP type graph, the trajectory follows the valley floor to H<sub>1</sub> + H<sub>2</sub>-H<sub>3</sub>. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r<sub>2</sub> will fluctuate.

'''''<u>What causes these momenta to be zero though? How does the code enforce this property?</u>'''''

{| class="wikitable" border="1"
!MEP graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0!!  Dynamic graph, r<sub>1</sub>=91.7pm, r<sub>2</sub>=90.7pm, p<sub>1</sub>=p<sub>2</sub>=0
|-
| [[file:hz7718_py5.png|300px|thumb|left]] || [[file:hz7718_py6.png|300px|thumb|right]]  
|}

===Reactive and unreactive trajectories===
{| class="wikitable" border="1"
! p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> !! E<sub>tot</sub>/ kJ.mol<sup>-1</sup> !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.28 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. When the system passes the transition state region, new bond H<sub>B</sub>- H<sub>C</sub> will form and H<sub>A</sub>-H<sub>B</sub> bond will break. Then H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. Before H<sub>B</sub>-H<sub>C</sub> forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H<sub>A</sub>-H<sub>B</sub> is in translational energy. After H<sub>B</sub>-H<sub>C</sub> bond forming, the trajectory is wavy because most kinetic energy of H<sub>B</sub>-H<sub>C</sub> is in vibrational energy. ||  [[file:hz7718_py7.png|300px|thumb|right]]
|-
| -3.1 || -4.1 || -420.08 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other at the beginning. But they do not pass the transition state region, H<sub>A</sub>-H<sub>B</sub> bond does not break and H<sub>B</sub>-H<sub>C</sub> bond does not form. Then H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule move away to each other. H<sub>A</sub>-H<sub>B</sub> bond keeps vibrating due to the kinetic energy. ||  [[file:hz7718_py8.png|300px|thumb|right]]
|-
| -3.1 || -5.1 || -413.98 || YES || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H<sub>A</sub>-H<sub>B</sub> bond starts to break, H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom and H<sub>B</sub>-H<sub>C</sub> molecule move away to each other. And H<sub>B</sub>-H<sub>C</sub> bond keeps vibrating.  ||   [[file:hz7718_py9.png|300px|thumb|right]]
|- 
| -5.1 || -10.1 || -357.28 || NO || H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule approach to each other, when they passes the transition state region, H<sub>A</sub>-H<sub>B</sub> bond starts to break and H<sub>B</sub>-H<sub>C</sub> bond starts to form. After that, H<sub>A</sub> atom moves away from H<sub>B</sub>-H<sub>C</sub> molecule but H<sub>B</sub>-H<sub>C</sub> molecule vibrates strongly. Then H<sub>B</sub>-H<sub>C</sub> bond breaks and H<sub>A</sub> and H<sub>B</sub> atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H<sub>A</sub>-H<sub>B</sub> molecule and H<sub>C</sub> atom move away to each other. ||   [[file:hz7718_py10.png|300px|thumb|right]]
|-
| -5.1 || -10.6 || -349.48 || YES || The system passes the transition region for three times: At first, H<sub>C</sub> atom and H<sub>A</sub>-H<sub>B</sub> molecule get closer, H<sub>B</sub>-H<sub>C</sub> bond forms and H<sub>A</sub>-H<sub>B</sub> bond breaks. Then H<sub>A</sub> atom returns to H<sub>B</sub>-H<sub>C</sub> molecule immediately and the system reverts to reactants. After that, H<sub>B</sub>-H<sub>C</sub> bond forms again and H<sub>A</sub>-H<sub>B</sub> bond breaks again. In the end, vibrating H<sub>B</sub>-H<sub>C</sub> molecule and H<sub>A</sub> atom move away to each.  || [[file:hz7718_py11.png|300px|thumb|right]]
|}
From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.


===Transition State Theory===
In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. <sup>1</sup>

'''''<u>A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!</u>'''''

==Exercise 2: F-H-H system==
===PES inspection===
{| class="wikitable" border="1"
! potential energy surface graph !! Internuclear distance vs time plot 
|-
| [[file:hz7718_py12.png|300px|thumb|left]] || [[file:hz7718_py13.png|300px|thumb|right]]  
|}
The potential energy surface graph show that the system H<sub>2</sub> and F has higher potential energy than HF and H. It means H<sub>2</sub> + F to HF + H reaction is exothermic and HF + H to H<sub>2</sub> + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H. 

So, from H<sub>2</sub> + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H<sub>2</sub> + F, the system releases energy to the environment.

When it comes to the transition state location, it can be analyzed by Hammond's postulate. <sup>2</sup> For example, in the reaction H<sub>2</sub> + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F<sub>A</sub> and H<sub>B</sub> is very large and the distance between H<sub>B</sub> and H<sub>C</sub> is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.

In the Internuclear distance vs time plot, F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero. 

Then we can know the black point in potential energy surface graph is the transition state(where F<sub>A</sub>-H<sub>B</sub>=180pm, H<sub>B</sub>-H<sub>C</sub>=74pm and pH<sub>1</sub>=pH<sub>2</sub>=0).

'''''<u>A sensible estimation. How did you confirm that this was a saddle point, and not a potnetial energy minimum though?</u>'''''

===Activation EnergyFor finding the actvition energy, the steps are extended to 3500. When F<sub>A</sub>-H<sub>B</sub> is 180pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H<sub>2</sub> + F.===
{| class="wikitable" border="1"
! Contour Plot 1 !! Graph 1
|-
| [[file:hz7718_py14.png|400px|thumb|left]] || [[file:hz7718_py15.png|300px|thumb|right]]  
|}

When F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, the Contour plot 2 shows that the transition state finally forms H<sub>2</sub> + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H<sub>2</sub> + F to HF + H.

'''''<u>A sensible estimate of the activation energy for FH+H. your estimate for F+H2 is a bit too low, likely because you've not let the reaction go to completion. </u>'''''
{| class="wikitable" border="1"
! Contour Plot 2 !! Graph 2
|-
| [[file:hz7718_py16.png|400px|thumb|left]] || [[file:hz7718_py17.png|300px|thumb|right]]  
|}

===Reaction dynamics===
In reaction H<sub>2</sub> + F to HF + H, F<sub>A</sub>-H<sub>B</sub> is 184pm and H<sub>B</sub>-H<sub>C</sub> is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H<sub>C</sub> molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H<sub>B</sub><sub>A</sub> momentum fluctuates strongly but H<sub>c</sub>-H<sub>B</sub> momentum keeps a relatviely low value.

The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy.  

'''''<u>No, this is incorrect. Simply looking at the peak of an IR spectrum is not enough? How would we detect vibrationally excited states in IR spectroscopy?</u>''''' 
{| class="wikitable" border="1"
! Contour Plot  !! Momentum vs Time
|-
| [[file:hz7718_py18.png|400px|thumb|left]] || [[file:hz7718_py19.png|400px|thumb|right]]  
|}

===Translational energy VS Vibrational Energy===
Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.<sup>3</sup> 

Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.<sup>3</sup>

'''''<u>YOu need to provide some examples (i.e. trajectories for the F+h2 system) here to back up your assertion.</u>'''''
{| class="wikitable" border="1"
! Graph 1  !! Graph 2
|-
| [[file:hz7718_py20.png|400px|thumb|left]] || [[file:hz7718_py21.png|400px|thumb|right]]  
|}

==Reference==
1.Veser, Götz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.

2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334–338. doi:10.1021/ja01607a027.

3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680