View source for MRD:zxt201019

{{fontcolor1|red|Really good report! You did all the tasks very carefully, the report is well structured, references are given plentiful and where appropriate, the discussion is thorough and demonstrates understanding. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:33, 16 May 2020 (BST)}}

{{fontcolor1|red|Well done. 🌟 [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 13:50, 16 May 2020 (BST)}}

== Transition-state theory ==
=== Conventional Transition-state Theory (CTST) ===
While being one of the most important theories in calculating the reaction rate, '''Transition-state theory (TST)''' fails very often the times. There are three main assumptions underpinning this theory:<ref>Laidler, K. J., & King, M. C. (1983). The development of transition-state theory. Journal of Physical Chemistry, 87(15), 2657–2664. https://doi.org/10.1021/j100238a002</ref>

::''1. Once a trajectory have surmounted the transition state, it cannot come back to reform the reactants;''
::''2. Motion along the trajectory can be separated from the other motions;''
::''3. The reaction can be treated classically without considering the quantum effects;''
::''4. The energy distribution of reactants' molecules obeys the Maxwell-Boltzmann distribution.''<br>
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In accordance to the last hypothesis, a further assumption was made, suggesting that there is an equilibrium between the initial and the activated states. Based on these cornerstones, the rate of a bimolecular reaction can be predicted via the following equation, where the <math>q_A</math> and <math>q_B</math> are the partition functions of reactants A and B, while the <math>q_{\ddagger}'</math> is the partition function of the activated species:

{{fontcolor1|red|I would leave out the partition functions here - it doesn't further your explanation, and you would need to define it and everything. Ockham's razor also applies to scientific writing! [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 09:50, 16 May 2020 (BST)}}

::<math>k=\frac{k_B T}{h} \frac{q_{\ddagger}'}{q_A q_B}e^{-\frac{E_0}{k_B T}}</math><br>
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However, as discussed in the section for [[#Reactive and unreactive trajectories|reactive and unreactive trajectories]], this prediction is overestimating the reaction rate. Thus, ''Dynamic approach'' has been advocated rather than the equilibrium approach. '''Transmission coefficient <math>(\kappa)</math>''' has been used to quantify the deviations of the rate calculated from the transition-state theory from the rate in reality.<ref>Sharia, O., & Henkelman, G. (2016). Analytic dynamical corrections to transition state theory. New Journal of Physics, 18(1), 13023. https://doi.org/10.1088/1367-2630/18/1/013023</ref>
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=== Mathematical definition of transition state ===
Using '''Potential energy surface (PES)''' as a function of multi variables, molecular geometry and chemical reaction dynamics can be well-studied. This is a very powerful tool for illustrating the potential energy at each atomic arrangement. 
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Classifying '''critical points''', at which the gradient is either zero or undefined,<ref>Adams, Robert A.; Essex, Christopher (2009). Calculus: A Complete Course. Pearson Prentice Hall. p. 744. ISBN 978-0-321-54928-0</ref> {{fontcolor1|red|What do you mean by undefined? For chemical systems, the gradient should always be definable. It might be different if you, for example, approach zero distance in the dihydrogen system. But that also depends on your model - at some point you will get Helium. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 09:55, 16 May 2020 (BST)}} on the PES is crucial for studying chemical reaction dynamics. They can be found easily by taking first derivative of the function. For the two-variable function, a point is crucial if both partial derivative with respect to r<sub>1</sub> and that with respect to r<sub>2</sub> equal to zero. There are three types of critical points: local minima, local maxima and saddle points. For a general chemical reaction, it would proceed from reactants, through a activation barrier, then to the intermediate or the products. Products or intermediate are corresponding to the local minima on the surface. On the other hand, the highest point on the activation barrier correspond to the transition state, which is defined as a first-order '''saddle point.''' In other words, it is a point which is the local minimum from one direction, and is the local maximum from other directions {{fontcolor1|red|Re-read that sentence - it is the other way round. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 09:57, 16 May 2020 (BST)}} .<ref>Frank Jensen (1999). Introduction to Computational Chemistry. England: John Wiley and Sons Ltd.</ref> 
'''Second derivative test''' can be used to distinguish a saddle point from local minima. After finding all the critical points, taking all the second partial derivatives of ''f'' and put them in the matrix form of: <br>
<br>

::<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2f}{\partial (r_1)^2} & \dfrac{\partial^2f}{\partial r_1 \partial r_2} \\[3.3ex]
\dfrac{\partial^2 f}{\partial r_2 \partial r_1}\ & \dfrac{\partial^2 f}{\partial (r_2)^2}
\end{bmatrix}</math>


This symmetric matrix form is known as the '''Hessian matrix (H)'''. Second derivative test is based on its '''determinant:'''<br>
::<math>\mathbf D(x,y) =
f_{xx}(r_1, r_2)f_{yy}(r_1, r_2)-f_{xy}(r_1, r_2)^2
</math>

{{fontcolor1|red|Minor comment: you are using Schwartz' theorem here, if you want to be perfect you could mention that. But I am really struggling to find anything to complain about here! [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 09:59, 16 May 2020 (BST)}}

For a critical point <math>(r_1, r_2)</math>:<br>

If <math>\mathbf D(r_1, r_2)>0</math>, and <math>\mathbf f_{aa}(r_1, r_2)>0</math>, then this is a local minimum;<br>
If <math>\mathbf D(r_1, r_2)>0</math>, and <math>\mathbf f_{aa}(r_1, r_2)<0</math>, then this is a local maximum;<br>
If <math>\mathbf D(r_1, r_2)<0</math>, then this is a saddle point;<br>
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{{fontcolor1|red|The last criterion will give you any saddle point, including higher orders. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:01, 16 May 2020 (BST)}}

== Locating the transition state ==

Transition state is lying at the energy maximum on the potential energy surface, where the first derivatives with respect to both directions equal to zero. Since the force is defined as negative gradient of potential, the transition state must have zero force on atoms in both directions. For the reaction of H + H<sub>2</sub>, the transition state is expected to be symmetrical. In other words, it should be at a point along the diagonal of the contour plot shown below. Calculation showed that when AB=BC=90.775 pm, force along either bond equals to zero. Transition state forms at this point. <br>

{{fontcolor1|red|Well discussed. You could have elaborated more on 'Calculation showed' - what exactly did you do? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:05, 16 May 2020 (BST)}}

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[[File:diagonalcontour0512_xjh.png|300px|thumb|centre|Contour plot with AB = BC = 230 pm]]
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Although the transition state can be located in this case, it is more conventionally to assume that the transition state takes place in a region of thickness. '''Conventional Transition State Theory (CTST)''' states that there is a ''quasiequilibrium'' between the activated complexes and the reactants, and the activated complexes can be found within a very small distance separated by two ''dividing surfaces,'' each at one side of the col.<ref>Laidler, K. J. (Keith J. (1987). Chemical kinetics (3. ed.). Harper & Row.</ref> Therefore, a reasonable estimation can be made: one of the dividing surfaces is at 90.77 pm, and the other is at 90.78 pm, giving the force boundaries along A-B and B-C equal to +0.001 and -0.001 kJmol<sup>-1</sup>pm<sup>-1</sup> respectively. This can be further verified from its plot of intermolecular distances against time, in which the graphs of r<sub>AB</sub> and r<sub>BC</sub> overlap and stay constant as time goes. Any system entering the this narrow region from one end as reactant must come out as product from the other end. <br> 
[[File:rts0512_xjh.png|300px|thumb|centre|Plot of Internuclear Distances vs Time, when AB = BC = 90.77 pm.]]
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== Trajectories ==
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[[File:Mepdynamics_xjh.jpg|600px|thumb|centre|Trajectory under Dynamics (graph on left-hand-side) and MEP (graph on right-hand-side) calculations, when AB = 90.775 pm, BC = 91.775 pm, p<sub>1</sub>=p<sub>2</sub>= 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>.]] <br>
Molecule is picking up momentum while it is rolling down the hill. Thus, its momentum would be different at each step of the measurement. As the MEP approach restore the momentum to zero at each step, the output would be a smooth trajectory along the product channel. While for the Dynamic approach, changes of momentum would be taken into account, giving a shaking trajectory. Therefore, for the purpose of revealing the reaction dynamics, such as the vibrational modes of the molecules during the reaction, Dynamics approach was used rather than using MEP.<br>
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[[File:distance-momenta-time500_xjh.jpg|600px|thumb|centre|Graphs of internuclear distances (graph on left-hand-side) and momenta (graph on right-hand-side) against time, when AB = 91.775 pm, BC = 90.775 pm, p<sub>1</sub> = p<sub>2</sub> = 0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, maximum time = 50 fs.]] <br>
For a large enough time t (50 fs), A-C distance rises dramatically from 183.2 pm to 427.1 pm. Although A-B and B-C are equally separated (91.9 pm), A-B climbs to 352.4 pm at the end, while the distance between B and C becomes closer (73.7 pm). From the momenta's point of view, both momenta of AB and that of BC equal to zero. For the first few femtoseconds, both momenta drop below zero. A-B momentum bounces back after about 6 fs, then keeps climbing until it flattens at about 5.06 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. B-C momentum turns positive at 15 fs. For most of the remaining time, its momentum keeps oscillating around 2.52 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. At 50 fs, it reaches its peak at 3.2 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>. <br>
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== Reactive and unreactive trajectories ==
<div id="Reactive and unreactive trajectories">
A trajectory is reactive if it can successfully proceed from reactant channel to the product channel, otherwise it is unreactive. Previous investigations have shown that for the trajectories with r<sub>1</sub>=74 pm and r<sub>2</sub>=200 pm, if its p<sub>1</sub> is in the range between -3.1 and -1.6 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, and p<sub>2</sub> equals to -5.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>, it can be classified as a reactive trajectory. The relationship between momenta, kinetic energy and the reactivity nature of a reaction deserves further study. With the same coordinates r<sub>1</sub> and r<sub>2</sub>, another set of five trajectories crossing a potential energy surface with a range of momenta were tested as follows: 

{| class="wikitable" border="1"
|+ Trajectory tests
! 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 dynamics !! Illustration of the trajectory 
|-
| -2.56 || -5.1 || -414.280 || Reactive || According to the initial momenta input, A-B distance is shrinking at a faster rate. Reactants cross the col to form oscillating products without bouncing back. || [[File:test1_xjh.png|200px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Reactants bounce back before reaching the transition state, indicating that the translational energy is not efficient enough to drive the reactant molecules over the barrier. Reactants are mainly fuelled by the vibrational energy. || [[File:test2_xjh.png|200px]]
|-
| -3.1 || -5.1 || -413.977 || Reactive || According to the initial momenta input, A-B distance is shrinking at a faster rate. There is a slight fluctuation of the trajectory in the reactant channel. Reactants cross the col to form oscillating products without bouncing back. || [[File:test3_xjh.png|200px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Reactants recross the transition state and re-enter the reactant channel with a much greater vibrational motion. No product formed in this case. || [[File:test4_xjh.png|200px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || Reactants evolve into strongly vibrating products after recrossing the barrier two times. || [[File:test5_xjh.png|200px]]
|}
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Since the negative momentum means that the bond length are decreasing, the more negative momentum a bond has, the faster the rate of two atoms coming together. Instead of saying 'a reaction would be more reactive if its momenta are higher', it is more accurate to say that the total energy becomes less negative when the difference between two momenta gets larger. <br>
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It is important to note that the barrier crossing does not necessarily stand for a reactive trajectory. Recrossing might occur in some cases, turning the direction back to its original side. This recrossing phenomenon reveals the limitations of the transition state theory, which is based on the non-recrossing hypothesis. In reality, the activated species formed from the reactants might not surmount the col at once into the product valley due to some quantum effects, which is also not considered in this classical approach. The portion of the reactants successfully heading into the product channel can be quantified via ''transmission coefficient''.  For the systems having much greater energy than that is required for overcoming the activation barrier, they have much smaller ''transmission coefficients'' than unity, and more excited species are prohibited from passing through. In some cases, even though the reactants do not have enough energy to get to the col, they can go directly into the product side by ''tunnelling'' through the hill. However, this '''tunnelling effect''' matters only when dealing with small masses such as electrons. For atoms, this effect is negligible.<ref>Sansón, J. A., Sánchez, M. L., & Corchado, J. C. (2006). Importance of anharmonicity, recrossing effects, and quantum mechanical tunneling in transition state theory with semiclassical tunneling. A test case: The H2 + CI hydrogen abstraction reaction. Journal of Physical Chemistry A, 110(2), 589–599. https://doi.org/10.1021/jp052849d</ref>
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{{fontcolor1|red|Tunneling is actually also important (or rather 'significant') for light atoms - i.e. hydrogen. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:12, 16 May 2020 (BST)}}

== F-H-H systems ==
So far the discussion has been confined to the symmetric H-H-H system. Another matter of importance is the asymmetric exchange reaction in the F-H-H system. In Hammond's postulate,<ref>Hammond, G. S. (1955). A Correlation of Reaction Rates. Journal of the American Chemical Society, 77(2), 334–338. https://doi.org/10.1021/ja01607a027</ref> a transition state could be either early (in an exothermic reaction) or late (in an endothermic reaction). Recognising a reaction is exothermic or endothermic has been regarded as the most important part in the study of reaction dynamics. <br>
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This problem can be addressed with ''infrared chemiluminiscence (IRCL),''<ref>Newman, R. (1952). Emission spectrum of HCl in the near infrared. In The Journal of Chemical Physics (Vol. 20, Issue 4, pp. 749–750). https://doi.org/10.1063/1.1700540</ref><ref>Polanyi, J. C. (1963). Infrared chemiluminescence. Journal of Quantitative Spectroscopy and Radiative Transfer, 3(4), 471–496. https://doi.org/10.1016/0022-4073(63)90026-8</ref> a special spectroscopic technique using the changes in the intensity distribution after the infrared relaxation to probe the energy deposition in vibrational excitation. In addition to the band due to the excitation from the ground state to the first excited state, there would be some overtones as well. However, after the infrared relaxation, the first and second excited vibrational states would not be populated as much as before. In contrast, ground state becomes more and more populated for the excitation to the first excited state. Overall, it would be expected to see a more intense emission absorption band while decreased intensity in the others.<br>
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{{fontcolor1|red|Good discussion. You might also be interested in an [https://en.wikipedia.org/wiki/Hydrogen_fluoride_laser 'application']. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:15, 16 May 2020 (BST)}}

For exothermic reactions, energy released would be absorbed by the product molecules. The absorbed energy would distribute among vibrational <math>(E_{vib})</math>, rotational <math>(E_{rot})</math>, translational <math>(E_{trans})</math> and electronic <math>(E_{ele})</math> energy levels. Energy distribution can be greatly affected by the position of the transition state, which in turns depends on the relative masses of the atoms. <math>F+H_2</math> and <math>H+HF</math> are two of the most representative examples within the F-H-H system, corresponding to exothermic and endothermic reactions respectively. <br>
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{{fontcolor1|red|The mass is not determining the position of the transition state, at least not directly. Think of neon and chlorine. Neon has approximately the same mass as fluorine, whereas chlorine is much heavier. However, we would expect the contour plot (including the transition state) for the H-H-F system to look more like H-H-Cl than like H-H-Ne. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:19, 16 May 2020 (BST)}}

=== F+H<sub>2</sub> ===
::<math>F+H_2 {\longrightarrow} HF+H</math>
Involving breakage of a weak H-H bond and formation of a strong H-F bond, this is an exothermic reaction with an early transition state and a mixed energy release. Since this is an asymmetric exchange, transition-state coordinate cannot be located via the diagonal approach any more. Instead, it can be deduced using its stationary-point nature. In other words, it is at a point at which the species has no tendency to move when there is no energy input. After attempting s series of combinations of A-B and B-C distances with zero p<sub>1</sub> and p<sub>2</sub>, it was found that the transition state occurs when AB=181 pm and BC=74.5 pm, as demonstrated in the contour plot below. Energy of this transition state was shown to be -433.981 kJ.mol<sup>-1</sup>.<br>
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[[File:FH2TS_xjh.jpg|600px|thumb|centre|Locating the transition state for the F+H<sub>2</sub> reaction]]
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Knowing the energies of reactants and transition state allows us to calculate the activation energy. According to ''Hammond's postulate'', which states that the structure of an early transition state would resemble that of the reactants, increasing the F-H distance from 181 pm would push the reaction all the way back to the reactants' level. Table shown below lists all the energy results from simulations. As expected, they finally came to a constant value (-435.057 kJ.mol<sup>-1</sup>) corresponding to the energy of the reactants. Subtracting it from the transition-state energy yields the activation energy, in this case, it is +1.076 kJ.mol<sup>-1</sup>.<br>

{| class="wikitable" border="1" align="center"
|+ Finding the energy of reactants
! F-H distance / pm !! H-H distance / pm !! E<sub>tot</sub> / kJ.mol<sup>-1</sup>
|-
| 181 (Transition-state) || 74.5 (Transition-state) || -433.981 (Transition-state)
|-
| 200 || 74.5 || -434.133
|-
| 300 || 74.5 || -434.934
|-
| 400 || 74.5 || -435.044
|-
| 500 || 74.5 || -435.056
|-
| 600 || 74.5 || -435.057
|-
| 700 || 74.5 || -435.057
|-
| 800 || 74.5 || -435.057
|}
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For exothermic reactions, destinations of the released energy is worthy of notice. The following graphs show the results from a set of 9 simulations of this reaction with increasingly negative momentum <math>p_{HH}</math>. Although we are pumping much more energy than its activation energy into the system, reaction was not able to occur. Instead, vibrational motions became more and more significant as the energy increased, implying that most of the energy income was spent on doing vibrations. Although how the incoming energy is distributed among different modes partly depends on the relative atomic masses, the most intrinsic reason is derived from the '''Polanyi's rules,''' which would be further discussed in the [[#Polanyls rules|last section]]. <br>
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[[File:9FH2_xjh.png|600px|thumb|centre|Contour plots for evaluating energy distribution. p<sub>FH</sub>=-1.0 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> for all cases, while the p<sub>HH</sub> changes from -6.1 to 6.1 g.mol<sup>-1</sup>.pm.fs<sup>-1</sup> as annotated in the graph]]
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=== H+HF ===
::<math>H+HF {\longrightarrow} H_2+F</math>
In contrast to the last reaction, it is now the strong H-F bond going to break, forming a much weaker H-H bond. Thus, this must be an endothermic reaction. Using the same method of finding the transition-state coordinates and the activation energy as in the F+H<sub>2</sub> simulation, it was found that the activation energy of this reaction only costs 0.015 kJ.mol<sup>-1</sup>. Its transition state occurs when two H atoms are 400 pm apart while the H-F bond length is 92 pm. That is, the reaction is going to happen even when the incoming hydrogen atom is still very far apart. 
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{{fontcolor1|red|The activation energy for this reaction should be something about 127 kJ/mol. I haven't come across that in this model so far, but you might have found another transition state, i.e. the one for complex formation. I'm not sure about that though. Regarding this experiment: bear in mind that the TS for both directions of the reaction should be the same. In more complicated cases there might be more than one TS connecting reactants and products, but all of them can be crossed in both directions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:28, 16 May 2020 (BST)}}

== Polanyi's empirical rules ==
<div id="Polanyls rules">
From the cases discussed above, their results show that investing large amount of kinetic energy in the system does not necessarily lead to the reaction. Only certain type(s) of energy can drive the reaction effectively. This can be well-explained by the Polanyi's empirical rules, which was first discovered by John Polanyi in 1986.<ref>Evans, M. G., & Polanyi, M. (1936). Further considerations on the thermodynamics of chemical equilibria and reaction rates. In Transactions of the Faraday Society (Vol. 32, Issue 0, pp. 1333–1360). The Royal Society of Chemistry. https://doi.org/10.1039/TF9363201333</ref> Simply put, it tells how the barrier position determines different demand for energy and where the excess energy goes. For the exothermic reactions, their transition states generally happen early during the reaction.<ref>Yeston, J. (2012). Stretching the Polanyi Rules. Science, 338(6113), 1397–1397. https://doi.org/10.1126/science.338.6113.1397-b</ref> Most of the energy driving this type of reactions comes from the translational energy of the reactants, and the energy released would be preferentially absorbed by the products as their vibrational energy. Conversely, vibrational energy would contributes most in endothermic reactions and its excess energy would be deposited into the translational energy of the products. That is the reason why raising the kinetic energy only promoted the vibrational motions of the reactants for the reaction between F and H<sub>2</sub>. <br>
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In spite of being rare, endothermic reactions with an early transition state would be mainly determined by the amount of translational energy. In other words, only the molecules having high enough translational energy can complete the reaction effectively, even though they have relatively low vibration energy. On the other hand, for those having a late transition state, vibrations are much more effective in converting the reactants to the products. More commonly, transition states in endothermic reactions belong to the latter type. 
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== Conclusion ==
In brief, as a classical approach for rate calculations, ''Conventional Transition State Theory (CTST)'' has limited applicability because it omits the recrossing effects. In other words, the rate calculation based on the CTST would overestimate the actual reaction rate. The modern techniques enable us to simulate various triatomic exchange reactions dynamically. It was found that different forms of energy would play different roles in different types of reactions, which is in consistent with the Polanyi's empirical rules.<br>
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{{fontcolor1|red|Well done. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 10:31, 16 May 2020 (BST)}}

=References=