ChemWiki - User contributions [en]

Talk:MRD:jjb215

2017-05-30T17:34:14Z

Bg1512:

Great work! I especially liked your detailed description of the saddle point criterion and the TS critique. Here are some tips to improve even further:

*TS critique: What you are saying is correct, however I found it interesting that you chose to reference the fifth trajectory in your work, when the fourth shows unsuccessful barrier recrossing. Double recrossing like in the 5th trajectory is not accounted for in TS theory, but it makes no difference because the outcome is the same. The fourth trajectory is the one that really produces a different outcome.

*Activation energies: From what I can tell your method seems mostly correct, even though your numerical values are quite off. You located the TS correctly, but its energy in your simulation is different from the one I get. It would have been helpful here to show the settings window next to the graph so I can pinpoint the problem more closely. Your energies for the reactants and products are almost correct. If you run the MEP from very close to the TS you might be able to see what is wrong.

*Experimental verification: Calorimetry shows you that an exo-/endo-thermic reaction occurred, but it can't distinguish between translational and vibrational energy. Can you give another method that allows the distinction between the two modes of movement?

*Polanyi's rules: Your answers are again correct, but I would have liked to see more evidence for/against the rules. One trajectory could be the result of noise or a peculiar outlier. Polanyi's rules are empirical, so you can only confirm them with a large amount of experiments.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 18:32, 30 May 2017 (BST)

Talk:MRD:jjb215

2017-05-30T17:32:58Z

Bg1512: Created page with "Great work! I especially liked your detailed description of the saddle point criterion and the TS critique. Here are some tips to improve even further: *TS critique: What you..."

Great work! I especially liked your detailed description of the saddle point criterion and the TS critique. Here are some tips to improve even further:

*TS critique: What you are saying is correct, however I found it interesting that you chose to reference the fifth trajectory in your work, when the fourth shows unsuccessful barrier recrossing. Double recrossing like in the 5th trajectory is not accounted for in TS theory, but it makes no difference because the outcome is the same. The fourth trajectory is the one that really produces a different outcome.

*Activation energies: From what I can tell your method seems mostly correct, even though your numerical values are quite off. You located the TS correctly, but its energy in your simulation is different from the one I get. It would have been helpful here to show the settings window next to the graph so I can pinpoint the problem more closely. Your energies for the reactants and products are almost correct. If you run the MEP from very close to the TS you might be able to see what is wrong.

*Polanyi's rules: Your answers are again correct, but I would have liked to see more evidence for/against the rules. One trajectory could be the result of noise or a peculiar outlier. Polanyi's rules are empirical, so you can only confirm them with a large amount of experiments.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 18:32, 30 May 2017 (BST)

Talk:MRD:BC physlab

2017-05-30T17:21:41Z

Bg1512: Created page with "Excellent work! I specifically liked your involved treatment of Polanyi's rules! Here are some pointers to help you improve even further: *TS properties: Your answer is corre..."

Excellent work! I specifically liked your involved treatment of Polanyi's rules! Here are some pointers to help you improve even further:

*TS properties: Your answer is correct. Can you now also give a mathematical criterion for the identification of saddle points?

*TS critique: The points you stated are all correct, but I would have liked you to relate your answer to your own findings. Since you observed barrier recrossing, does your work support or contradict TS theory? It might well be that you know the answer but thought it obvious to write down. Make sure that you always highlight your own contribution in a discussion section so lazy readers still understand your input.

*Polanyi's rules: You gave an explicit assessment of each trajectory, so it would be nice to have a summarising paragraph. How many trajectories were in favour of the rules, how many against? Are any of the trajectories less important than others? Can you put forward a modification to the rules that explains the discrepancies?

*Figures: Make sure you number your figures so you can refer to them later in the text.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 18:21, 30 May 2017 (BST)

Talk:MRD:BS3314

2017-05-30T12:32:39Z

Bg1512: Created page with "Good work! Here is some feedback to improve further: *Raw data & methods: In order to convince the reader of your findings, you need to supply the method with which you gener..."

Good work! Here is some feedback to improve further:

*Raw data & methods: In order to convince the reader of your findings, you need to supply the method with which you generated your results so someone else can repeat the experiment. Your first answer for example is correct, but I do not know how you got to it. Even saying trial-and-error is better than just quoting the result. Similarly you have to show all the raw data that you base your results and discussion on (which you didn't do for the activation energy of HHF).

*MEP: Your description is correct. How do the calculations underlying MEP and dynamics differ?

*Proofreading: Please proofread your work before submitting. This way you should be able to avoid mistakes like the formatting error in your reference section.

*Discussion: When presenting your raw data, also make sure to comment on its appearance. Any outstanding features, trends or regularities should be pointed out.
--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 13:32, 30 May 2017 (BST)

Talk:MRD:Clarachong

2017-05-30T12:17:26Z

Bg1512: Created page with "Great work! You gave excellent answers for the TS critique, referring to your own results to reinforce your claims and for the activation energies, clearly reporting raw data,..."

Great work! You gave excellent answers for the TS critique, referring to your own results to reinforce your claims and for the activation energies, clearly reporting raw data, methods and result. Here are some pointers for further improvement:

*TS properties: So, according to your definition of the TS, what is the difference between a TS and a maximum? The PES is a 2D object, so you can define two gradients and three curvatures. The distinguish between between maxima, minima and TSs you need to use curvatures in 2D. See whether you can come up with a new definition.

*Figures: Number your figures so you can refer to them in the text. Have a look at papers for correct formatting of zoom-ins and subplots.

*Polanyi's rules: You stated the rules correctly, but to discuss them you need to draw conclusions from your own data. Do the HHF trajectories you computed match with Polanyi's rules?

*Experimental verification of the release of vibrational energy: You seem to have forgotten this answer unfortunately.

Great work! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 13:17, 30 May 2017 (BST)

Talk:JIR1501069122

2017-05-30T10:47:53Z

Bg1512: Created page with "Good work! Here are some pointers towards improvement: *TS properties: You've only given a criterion that minima and saddle points share. How can you distinguish between them..."

Good work! Here are some pointers towards improvement:

*TS properties: You've only given a criterion that minima and saddle points share. How can you distinguish between them?

*Locating the TS: How did you find it? The answer is correct, but I need to see your method as well (even if it is just trial and error) so I can replicate the experiment and see whether I get the same result.

*MEP: "This is due to the lack of energy change between translational and vibrational bonds." What causes this effect in the underlying calculation? How is the equation to calculate the forces in dynamics modified to obtain the MEP?

*TS critique: How does this critique relate to your own work? Can you reinforce some of these arguments you have given to your own results? What effect does the absence of recrossing in the theory have on its deviation from experimental values? Are QM effects important in your simulation?

*TS location and activation energies: Again, please supply a method and screenshots of the activation energies as proof.

*Experimental evaluation of vibrational energy release: Can you draw the spectrum you expect from a vibrationally excited molecule? What sort of wavelengths are you looking at? (IR, UV, Vis)

*Polanyi's rules: You've state the rules, but can you confirm them with your simulations? Can you disprove them? Any research requires data before you can make any claims, so you need to base your discussion on your own results!
--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 11:47, 30 May 2017 (BST)

Talk:MRD:ll4915

2017-05-22T18:39:17Z

Bg1512: Created page with "Great work! Here is some advice to help you improve even further: *TS classification: You are working with a 2D problem here, so there are two gradients and three curvatures...."

Great work! Here is some advice to help you improve even further:

*TS classification: You are working with a 2D problem here, so there are two gradients and three curvatures. The key phrase is "maximum on the minimum energy path". You described how to identify a maximum, but not how you would find a maximum on the minimum path. This requires the use of 2D calculus.

*TS theory critique: Your answer is correct and I like that you use your own work as a base of your critique rather than stating facts from a text book! Would an experimental reaction rate by higher or lower than the one predicted with TS theory?

*Activation energies: It is good practice to show all your experimental data and methods, well done!

*Experimental verification of vibrational energy release: Which transition in IR would allow you to observe vibrational energy release? Which approximations do you have to invoke and which ones do not work? How would NMR be useful to understand vibrational modes?

*Polanyi's rules: Your reasoning is correct, however I would have liked more references to the trajectories you just calculated. Since there's only one successful reaction, can you come up with more successful reaction conditions based on your knowledge of Polanyi's rules?

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 19:39, 22 May 2017 (BST)

Talk:MRD:01060479

2017-05-22T18:20:46Z

Bg1512: Created page with "Great work! Here are a couple of pointers to improve even further: *TS classification: This is a 2D problem, so there are two gradients and three curvatures. According to you..."

Great work! Here are a couple of pointers to improve even further:

*TS classification: This is a 2D problem, so there are two gradients and three curvatures. According to your definition you can't differentiate between TSs and maxima. Considering that the TS is a minimum in one direction and a maximum in the other, can you come up with a new criterion for saddle points?

*TS theory critique: This is correct, but not the entire truth. You observe barrier recrossing in trajectory 4 of the last exercise, but the simulation you're working with doesn't consider QM. This tells you that there are more factors at play here than just the absence of QM. How would the absence of barrier recrossings impact the validity of the theory?

*Activation energies: How did you calculate the activation energies? You need to supply the reader with your experimental data if you are to convince them of the validity of your results.

*Polanyi's rules: This is correct, but in order to evaluate a theory you need to provide more results. One positive result in each direction could just be the result of chance.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 19:20, 22 May 2017 (BST)

Talk:MRD:OHC15

2017-05-22T18:03:27Z

Bg1512: Created page with "This is excellent, well done! I'm afraid I don't have many suggestions for improvement, but I will highlight the things I particularly liked so you can carry on! *Experimenta..."

This is excellent, well done! I'm afraid I don't have many suggestions for improvement, but I will highlight the things I particularly liked so you can carry on!

*Experimental data & methods: You showed all your data and methods, allowing me to clearly follow your reasoning. I particularly liked that you reasoned in your discussions drawing from evidence of your experimental work. In the case of Polanyi's rules it is important to show a range of results, which you did. Numbering of figures is also good practice so you can refer to them in your main text.

*Vibrational energy release: Could you draw a diagram of the IR transition that would be observable? Which approximations do you have to invoke and which ones can't be used here?

*Introduction: It is more important what the calculation is that you performed compared to how you did it. So instead of talking about "a MATLAB program", mention "molecular reaction dynamics simulation". The exact execution of the simulation (things like MATLAB code, calculation details, parallelism) can be mentioned in the methods section of a paper or even in the supporting information.

*English: Your English is on point, but there are three small flaws I found: Pay attention to singular/plural nouns, don't use definite articles in front of names and use semi-colons when splitting sentences. E.g. "This can be confirmed using the surface plot; the black dot representing our system is situated at the maximum point of the lowest energy trajectory.". In this sentence you switch subject between "This" and "the black dot". You can't separate the halves by a comma; you can only do that when the subject stays the same or when you are entering a subordinate clause. Use a semi-colon instead (this is a very minor point).

Great report! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 19:03, 22 May 2017 (BST)

Talk:MRD:mmn115

2017-05-22T17:41:09Z

Bg1512: Created page with "Great work! I especially liked your clear description of experimental results and how you illustrated all your answers with experimental data. Here are a couple of pointers to..."

Great work! I especially liked your clear description of experimental results and how you illustrated all your answers with experimental data. Here are a couple of pointers to help you improve further:

*TS properties: Since this is a 2D problem, there are two principal gradients and three curvatures (<math>d^2V/dx^2</math>, <math>d^2V/dy^2</math> & <math>d^2V/dxdy</math>). The transition state is not an ordinary maximum, but a maximum in one direction and a minimum in another (''i.e'' "a maximum of minima"). Can you now come up with the right mathematical formulation?

*MEP: Did you realise that you can extend the timeframe of the simulation? If you let the MEP run along enough the simulation will trace out the path along the entire valley. Well done on recognising that the MEP resets the mass!

*TS theory critique: The facts you mentioned are true, but I would have liked you to draw conclusions from your own results as well. What about trajectory #4, does this comply with TS theory?

*Experimental verification of vibrational energy release: Do you know IR spectroscopy well enough to draw out the transitions you are trying to observe and how these differ from the ground state transitions? Which approximations do you need to invoke / which ones aren't sufficient?

*Polanyi's rules: Your answers are correct, but two sets of experimental data aren't enough to convince me of the validity of a theory. For all I know, these two could be just happy accidents.

*English: Your English is grammatically and syntactically correct, but I think a more formal style would be more suitable. Avoid personal pronouns, such as "I" and "we" and use nouns instead. "These results show that....", "To obtain this result, a rough estimate of..."

Great work! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 18:41, 22 May 2017 (BST)

Talk:MRD:pc2715

2017-05-22T16:27:49Z

Bg1512: Created page with "Good work! Here is some advice to improve even further: *TS properties: Your explanation is qualitatively correct. Can you now come up with a mathematical description using s..."

Good work! Here is some advice to improve even further:

*TS properties: Your explanation is qualitatively correct. Can you now come up with a mathematical description using second derivatives?

*MEP: Do you realise that you can extend the timeframe of your simulation? At the moment you can't say anything about the outcome of the reaction because you haven't allowed for enough time. How does the MEP differ from a normal simulation mathematically?

*TS theory critique: The facts you stated are correct, but I would have liked to see more of your own interpretation. How do your own results compare with TS theory (especially trajectory #4 from the previous exercise)?

*F-H-H reaction: Every result you quote here needs to be argued from experimental evidence. That means you need to show screenshots for all the results you obtained. For example, I do not know how you calculated the activation energies; for all I know you could have invented those values. It becomes especially important when you reason about the validity of someone else's theory (like Polanyi's rules) as you will need to convince the reader that you are right in your reasoning. The more clearly you present your findings and show how you arrived at your conclusion from the experimental results you observed, the more likely they are to believe you.

*Release of energy: What is an actual physical experiment that allows us to measure the output of vibrational energy?

Good work! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 17:27, 22 May 2017 (BST)

Talk:MRD:hrc115

2017-05-22T16:09:47Z

Bg1512: Created page with "Good work! I especially liked your detailed search for the TSs, using both symmetry and dynamics arguments. Here are a number of pointers to help you improve even further: *T..."

Good work! I especially liked your detailed search for the TSs, using both symmetry and dynamics arguments. Here are a number of pointers to help you improve even further:

*TS classification: The gradients will be zero at all types of extrema. To distinguish between maxima, minima and transition states you will need to use the second derivatives. Can you come up with a mathematical criterion?

*MEP: How is the dynamics calculation modified to obtain the MEP?

*TS theory critique: Your reasoning is mostly correct. You mentioned that not all reactions cross directly at the TS. What's the percentage of reactions crossing directly on the TS? The TS is a single 0-dimensional point on the surface with an infinitesimally small area. Given you answer to this question, how applicable is TS theory to all the reactions not crossing the TS directly? You did well referencing your own results in the discussion! Make sure to number your figures so I can follow your argument more easily.

*Activation energies: The potential energies you are showing are wildly oscillating. Which value of the potential energy did you take for the activation energy calculation? Activation energies are typically calculated with reference to molecules in their ground state. Where can this atomic geometry be found on your PES?

*Polanyi's rules: I assume you ran out of time for this answer. Please supply more experimental data and comment on whether they confirm or contradict Polanyi's rules.

Great work! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 17:09, 22 May 2017 (BST)

Talk:MRD:milandeleev

2017-05-22T15:46:50Z

Bg1512: Created page with "Good work! I've got a number of pointers for you to improve even further: *TS classification: The problem is set in 2D, not 1D. Therefore two gradients - one partial derivati..."

Good work! I've got a number of pointers for you to improve even further:

*TS classification: The problem is set in 2D, not 1D. Therefore two gradients - one partial derivative for each coordinate - and three curvatures - <math>d^2V/dx^2</math>, <math>d^2V/dy^2</math> & <math>d^2V/dydx</math> - exist. The TS is the defined as the maximum on the minimum energy trajectory, so it is a maximum in one and a minimum in the other direction. Can you come up with a mathematical definition for the TS based on these facts?

*Locating the TS: The answer you got for the H-H-H reaction is correct, but <math>r_1 = r_2</math> only holds for symmetric reactions. Can you come up with a new, more general method? If you start the reaction at the TS, you should obtain zero movement of the nuclei, which is the proof that you have actually found the TS. Given your data you could not be sure that you actually found the TS. In the F-H-H reaction you mentioned that you kept one variable constant. How did you choose to specific value at which you kept the variable constant? I like your use of Hammond's Postulate, but you need to specify your method more clearly.

*MEP: Do you realise that you can extend the timeframe for your simulation? Eventually the products will be formed, even in the MEP. How does the MEP differ from the normal simulation mathematically?

*TS theory critique: Refer to your own results to reinforce your critique. You have observed that there is a reaction which proceeds past the TS, but doesn't from products, which is contradictory to TS theory. Explicitly mention any results that support your arguments.

*Polanyi's rules: Your reasoning is correct, but more experimental data is needed to support your claim. One example of a successful reaction is not enough as it could simply be a happy accident. You can play around with the settings more. When refuting a claim, add your own experiments that prove the opposite. When agreeing with a claim, show that you've tried to negate it and couldn't do so.

Good work!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 16:46, 22 May 2017 (BST)

Talk:MRD:ZH3615

2017-05-16T14:38:32Z

Bg1512: Created page with "Excellent work! I specifically liked your critique of TS theory by making an argument based on your data and your comment that activation energies can only be calculated in th..."

Excellent work! I specifically liked your critique of TS theory by making an argument based on your data and your comment that activation energies can only be calculated in the limit of infinite internuclear distances. Here is me advice for further improvement:

*TS properties: Your answer is correct for 1D systems, but the PES we're dealing with here is 2D. There are two gradients and three curvatures as you can differentiate the surface with respect to one direction or the other. A saddle point is a minimum in one direction and a maximum in another. Can you come up with a new criterion involving the second derivatives to distinguish it from minima?

*H-H-H TS: Good use of a heuristic to find the coarse region of the TS and then closing in on it with a more expensive calculation!

*MEP: You say the MEP neglects internuclear momenta. How is the calculation of the force modified to achieve this effect?

*Activation energies: Your comment about infinite distances is correct. If you had an analytic function describing the PES, you could mathematically take the limit of the function. Do you have a way of extracting an analytic function from the PES? What about experimental errors?

*Experimental evidence of vibrational energy: Could you name a specific method that allows you to undertake these measurements?

*Polanyi's rules: How do these rules compare to your experimental evidence?

*General writing:
**Abstract: An abstract's goal is to convince someone to read the entire paper. It needs to include a description of the methods used (MATLAB is not a method; that's like saying you calculated something using mathematics.), an explanation of why your work is useful and the main result of your work. At the moment your abstract is a bit short.
**Introduction: This should allow me to read the paper without having to read any other work first. I would have liked to see something about the calculation method employed in obtaining the PES. Any reader who doesn't know the lab script would have trouble figuring out what the contour plots were that you are showing.
**Conclusion: In case the reader didn't understand anything about your paper, you summarise all the information in the conclusion. This needs to include (quantitative) measurements of your most important results. Furthermore, you want to indicate which parts of your experiments align with theory and which ones contradict theory.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:38, 16 May 2017 (BST)

Talk:MRD:jef15

2017-05-16T14:18:39Z

Bg1512:

Great work! Let me give you a couple of pointers to improve even further:

*TS properties: Almost right! The TS is a minimum in one direction and a maximum in the other. Thus, e.g. <math>f_{xx} < 0</math> & <math>f_{yy} > 0</math> (or the other way around). Also, <math>f_{xy}^2 >= 0</math> since it is a square. How do you have to formulate the discriminant to always be negative under these conditions? It is also possible to find saddle points which are not subject to the minimax criterion. Given your new formulation of the discriminant, what is the condition on <math>f_{xy}^2</math> for non-minimax saddle-points? (Feel free to drop me an e-mail if you'd like to discuss.)

*MEP: How is the force calculation in your simulation modified to yield the MEP?

*TS theory: I would have liked you to compare the TS theory predictions to the experimental results you collected. In trajectories 4 & 5 of the previous exercise you saw that TS recrossing can occur, how does this compare to the TS theory predictions? Also, given that you are conducting atomic simulations, do concepts such as temperature have any meaning? (interesting reference by the way)

*Activation energies: Please supply a screenshot of your experimental data, both to validate your method and your results. Considering that the PES tails off in both the reagents and products directions, how did you come up with a criterion of which reagent/product geometry was good enough?

*Polanyi's rules: In order to lead a substantial discussion here, you need to supply experimental data. You have just stated the results of the rules, but did neither reinforce nor contradict them on the basis of your experiments.

Well done!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:14, 16 May 2017 (BST)

Talk:MRD:jef15

2017-05-16T14:14:04Z

Bg1512: Created page with "Great work! Let me give you a couple of pointers to improve even further: *TS properties: Almost right! The TS is a minimum in one direction and a maximum in the other. Thus,..."

Great work! Let me give you a couple of pointers to improve even further:

*TS properties: Almost right! The TS is a minimum in one direction and a maximum in the other. Thus, e.g. <math>f_{xx} < 0</math> & <math>f_{yy} > 0</math> (or the other way around). Also, <math>f_{xy}^2 >= 0</math> since it is a square. How do you have to formulate the discriminant to always be negative under these conditions?

*MEP: How is the force calculation in your simulation modified to yield the MEP?

*TS theory: I would have liked you to compare the TS theory predictions to the experimental results you collected. In trajectories 4 & 5 of the previous exercise you saw that TS recrossing can occur, how does this compare to the TS theory predictions? Also, given that you are conducting atomic simulations, do concepts such as temperature have any meaning? (interesting reference by the way)

*Activation energies: Please supply a screenshot of your experimental data, both to validate your method and your results. Considering that the PES tails off in both the reagents and products directions, how did you come up with a criterion of which reagent/product geometry was good enough?

*Polanyi's rules: In order to lead a substantial discussion here, you need to supply experimental data. You have just stated the results of the rules, but did neither reinforce nor contradict them on the basis of your experiments.

Well done!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:14, 16 May 2017 (BST)

Talk:MRD:cej15

2017-05-16T13:46:11Z

Bg1512: Created page with "Good work! Let me give you a couple of pointers to improve further. I assume you ran out of time since a number of answers are missing and will concentrate on feedback of the..."

Good work! Let me give you a couple of pointers to improve further. I assume you ran out of time since a number of answers are missing and will concentrate on feedback of the questions you have answered.

*TS properties: The graphic you used is correct, but a 1D representation of the PES. We're dealing with a 2D problem here (with two gradients and three curvatures). How can you generalise the definition of a saddle point in 1D to 2D? Think about what the reaction coordinate in your second figure represents on the PES.

*Finding the TS: Your method is only valid for reactions where the products and reactants are the same (which are typically restricted to models, since a chemist usually wants to transform one substance into another). At the TS, reactants and products are balanced. By themselves, they will form neither reactants nor products. Can you use this to come up with a new method?

*Reaction trajectories: Did you realise that you can change the time frame of your calculations? A lot of your trajectories can't be classified as reactive or unreactive because you didn't let enough time elapse.

*MEP: What does the "wiggle" in the black line represent physically? What is the basis of the MEP calculations, ''i.e.'' how is the calculation of the force modified to get the MEP?

*H-H-H trajectories: Your fourth table entry says "reactive", but your description indicates that the products weren't formed. I'll take this as a typo and you mean "unreactive".

*TS theory: How does this relate to your findings? How does TS theory predictions compare to experimental values (not simulations)?

*Exothermicity: You cannot determine whether a reaction is exothermic or endothermic by looking at an arbitrary trajectory. Exothermicity is determined by the geometry of the products vs. that of the reactants. Therefore, the shape of the entire PES tells you about exothermicity, not a specific path on the PES. The specific trajectory only tells you whether a set of initial conditions is reactive, not whether energy is released when the reaction is successful.

*Activation Energies: Are the geometries you obtained for the reactants and products the final geometries? What would happen if you extended the timeframe of your reactions?

*F-H-H trajectories: You need to increase the timeframe. These trajectories are not conclusive; they could potential be successful.--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 14:46, 16 May 2017 (BST)

Talk:MRD:fabianthomaslabphyscomp

2017-05-16T13:18:13Z

Bg1512: Created page with "Good work! Here are a couple of pointers towards improvement: *Experimental Data: You need to substantiate your claims with experimental data. In fact, any discussion you lea..."

Good work! Here are a couple of pointers towards improvement:

*Experimental Data: You need to substantiate your claims with experimental data. In fact, any discussion you lead should be on the basis of the experimental data you collected. Furthermore, you need to describe the means by which you obtained your results so that others can critique their validity. I would have liked to see data for the last four answers. The activation energies you calculated are incorrect, but I do not know whether it was a mistake or whether your entire method is flawed. For the discussion of Polanyi's rules you could show the results of an investigation similar to that of the 5 trajectories in the H-H-H case.

*TS properties: The answer you've given is correct for 1D problems. Here we're dealing with a 2D problem and one can distinguish maxima, minima and saddle points by analysing all three second derivatives (<math>d^2V/dx^2</math>, <math>d^2V/dy^2</math> & <math>d^2V/dxdy</math>). Can you formulate a new rule for saddle points in 2D (hint: they are called minimax points for a reason)?

*MEP: Why is the case that you observe no vibrations in the MEP? What change is made to the calculation of the forces (originally <math>F = m * d^2x/dt^2</math>) during MEP ?

*TS theory: How does TS theory compare to your results? Since your simulations do not contain any QM, QM can't be the only cause for deviations of the theory to your results.

*Polanyi's rules: The discussion of early vs. late TSs is correct, albeit convoluted. You use a lot of relative terms in your discussion. How much is "too high an energy"? To answer this, supply experimental data showing the onsets of certain behaviours, etc.
--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 14:18, 16 May 2017 (BST)

Talk:MRD:lp2015

2017-05-16T12:57:08Z

Bg1512: Created page with "Good work! I especially liked your discussion of TS theory and how you drew from your own experiments to supplement your critique of the theory. Here is some advice to improve..."

Good work! I especially liked your discussion of TS theory and how you drew from your own experiments to supplement your critique of the theory. Here is some advice to improve further:

*TS properties: The maximum of the PES is not the TS! If that were the case, the TS would be at the origin, when both atomic distances are zero and the potential energy is infinitely high. Instead, the TS is the maximum of the reaction coordinate, i.e. a point that's both a minimum and a maximum of the PES. How is that possible? Since we're dealing with a 2D problem, there are also two directions that need to be considered when evaluating extrema. Also I don't understand the caption of your first picture. The TS is a point on the surface, not the entire surface.

*F-H-H bond energies: I'm not sure what the line "F + H2 reaction: +568 kJ/mol - 436 kJ/mol = +132 kJ/mol (Endothermic)" refers to. Are F + H_2 the reactants or the products? If a stronger bond is formed, will that require or release energy? You didn't comment on the connection between bond strength and enthalpy.

*Activation energies: Are the geometries you obtained for reactants/products their final geometries? What would happen if you extended the time frame of your simulation? The geometries you indicated are not minima along both coordinates of the PES, only along one of them.

*Polanyi's rules: In order to comment on the validity of a theory, you need to provide more experimental data. That two trajectories you show could be coincidence. At the very least you could perform the reaction the other way round to confirm that an early barrier has the opposite requirements.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 13:57, 16 May 2017 (BST)

Talk:MRD:hz4315

2017-05-16T12:30:51Z

Bg1512: Created page with "Great work! You enabled me to follow effortlessly and the answers are mostly correct, well done! Here are a couple of tips for further improvement: *TS properties: Your answe..."

Great work! You enabled me to follow effortlessly and the answers are mostly correct, well done! Here are a couple of tips for further improvement:

*TS properties: Your answer is almost correct, but also confusingly expressed. I understood your answer in this way: There are a set of minima on the PES. When travelling along this line of minima, you will encounter a maximum. This is the TS. To calculate its position, first find all the minima by setting the first derivative equal to zero and collecting all the points with a positive curvature. Then, of these minima, find the maximum. What I would have liked to see is a more straightforward explanation employing two types of second derivatives (<math>d^2V/dx^2</math> & <math>d^2V/dy^2</math>) and generally utilising the concept of directionality in the PES. ''I.e.'' the reaction coordinate is the path connecting all the minima in the PES and the TS is the maximum along this path.
*TS theory: You're right in saying that QM (specifically tunnelling) will have an effect on reaction dynamics in reality. However, in this simulation you don't have any QM, but still observe barrier recrossing. There must be other problems with TS theory other than the absence of QM effects.
*Polanyi's rules: Your reasoning is correct, but you have to support your argument with more data. Two examples is not enough to discuss the validity of a theory. TS theory was critiqued earlier in your report by examining five trajectories for example.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 13:30, 16 May 2017 (BST)

Talk:MRD:ar3015

2017-05-16T12:00:53Z

Bg1512: Created page with "Great work! Your wiki is concise and to the point. Your reasoning is sound throughout! Here are a couple of pointers to further improve: *Expression: I sometimes find it diff..."

Great work! Your wiki is concise and to the point. Your reasoning is sound throughout! Here are a couple of pointers to further improve:

*Expression: I sometimes find it difficult to follow your writing. Especially your use of relative pronouns and relative clauses is confusing. Consider the "Determining the activation energies" section for example. Most of your sentence start with "this". While I understand what you are saying, it does require effort to read. I also think that you're using a number of superfluous words (unfortunately this happens often in scientific literature, so I don't blame you for it). Take the sentence "Due to the conservation of energy, as the kinetic energy has increased this means the potential energy must've decreased which is demonstrated in the potential energy graph." for instance. You've used four linking or relative words in one sentence. I would rewrite this as "Due to the conservation of energy, any increase in kinetic energy is balanced by a decrease in potential energy. This is demonstrated in the potential energy graph.". Lastly, a word about the use of "if" (this is nit-picking already). "If we invoke Hammond's postulate, the early transition state indicates an exothermic reaction." An early transition state always indicates an exothermic reaction, whether or not you invoke Hammond's postulate. A neater way of expressing your sentence is: "Within Hammond's postulate, an early transition state indicates an exothermic reaction." Now it is clear that you are talking about something enclosed in the framework of a specific theory.

*Answers to questions:
**Transition State properties: First, remember that we are dealing with a 2D problem. It is true that both derivatives are 0 at minima, maxima and saddle points. Furthermore you're right in saying that the second derivatives allows you to distinguish between the different types of extrema. However, since this is a 2D problem there are three types of second derivatives (<math>d^2V/dx^2</math>, <math>d^2V/dy^2</math> & <math>d^2V/dxdy</math>). Considering that a saddle point is a maximum in one direction and a minimum in the other, can you come up with a criterion for the second derivatives to determine whether a point is a saddle point?
**TS theory: I'm in two minds about this answer. You've done well in identifying a conflict of your own data with the theory, but your invocation of quantum mechanics is out of place. The simulation you ran does not contain QM, so QM can't be responsible for any deviations from the theory you see in your experiments.
**Determining the activation energies: Do you think the geometries you obtained for reactants and products are indeed their final geometries? What would happen if you increased the time frame of your simulations? Your activation energies are slightly too low; can you figure out why?
**Polyani's Rules: Your reasoning is sound based on the evidence you presented. However, basing a judgement on one experiment for each setting is dangerous. I would like you to investigated the four cases (early TS, high vib; early TS, low vib; etc.) more thoroughly so you gain a statistic understanding of how often the theory works and how often it doesn't.

Well done! --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 13:00, 16 May 2017 (BST)

Talk:MRD:01089606

2017-05-11T15:32:36Z

Bg1512:

==Feedback==

Good work! Here are a couple of pointers on how to improve even further:

'''TS properties'''

A TS is a saddle point, which is a combination of a maximum in one and a minimum in the other direction. Can you come up with a mathematical formulation (in 2D) that combines these properties?

'''HHH trajectories'''

In trajectory 4 you reason that not enough energy is present to form the products, but trajectory 3, which contains less energy, is still successful. So clearly the total amount of energy alone cannot be responsible for the success of a reaction. To answer this correctly you need to think about Polanyi's rules. You went into the right direction at the very end of your week talking about early and late TSs. The key is how translational and vibrational energy are distributed, not necessarily their total amount.

'''Activation energies'''

When reporting results (here the activation energies), you need to supply the data (screenshot) as well! You're trying to convince me of a scientific fact and you can only do so by producing evidence. Furthermore, I do not have your method either, so I cannot check the result myself. Unfortunately your result is off by a little bit. You could have compared your values to literature to make sure your value is right.

'''References'''

Please include citations in your work. You shouldn't just list your references at the end of the report, but link each foreign fact exactly to a reference. At the moment I doubt the use of references to enzymatic TSs and you would need to convince me of the opposite by showing me which fact you took from their research.

Well done!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 16:32, 11 May 2017 (BST)

Talk:MRD:01089606

2017-05-11T15:32:27Z

Bg1512: Created page with "==Feedback Good work! Here are a couple of pointers on how to improve even further: '''TS properties''' A TS is a saddle point, which is a combination of a maximum in one a..."

==Feedback

Good work! Here are a couple of pointers on how to improve even further:

'''TS properties'''

A TS is a saddle point, which is a combination of a maximum in one and a minimum in the other direction. Can you come up with a mathematical formulation (in 2D) that combines these properties?

'''HHH trajectories'''

In trajectory 4 you reason that not enough energy is present to form the products, but trajectory 3, which contains less energy, is still successful. So clearly the total amount of energy alone cannot be responsible for the success of a reaction. To answer this correctly you need to think about Polanyi's rules. You went into the right direction at the very end of your week talking about early and late TSs. The key is how translational and vibrational energy are distributed, not necessarily their total amount.

'''Activation energies'''

When reporting results (here the activation energies), you need to supply the data (screenshot) as well! You're trying to convince me of a scientific fact and you can only do so by producing evidence. Furthermore, I do not have your method either, so I cannot check the result myself. Unfortunately your result is off by a little bit. You could have compared your values to literature to make sure your value is right.

'''References'''

Please include citations in your work. You shouldn't just list your references at the end of the report, but link each foreign fact exactly to a reference. At the moment I doubt the use of references to enzymatic TSs and you would need to convince me of the opposite by showing me which fact you took from their research.

Well done!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 16:32, 11 May 2017 (BST)

Talk:MRD:YanZhangPhyComp

2017-05-11T15:09:39Z

Bg1512:

Talk:MRD:YanZhangPhyComp

2017-05-11T14:59:58Z

Bg1512: Created page with "== Feedback == Great work! Your answers to the questions were solid and I could follow your argumentation all the way through. Your writing style is concise and uses a pleasan..."

MRD:YanZhangPhyComp

2017-05-11T14:58:11Z

Bg1512:

='''Molecular Reaction Dynamics: Applications to Triatomic systems'''=

==Introduction==

In this lab, the reaction inbetween a triatomic system is investigated where an single atom A collides with a diatomic molecule BC, forming a new diatomic AB and an atom of C, where it is assumed that the motion of the atoms obey classical mechanics.

The types of atoms, the internuclear distances between the atoms and their internuclear momentum are varied and its reactivity(whether the reaction occurs or not), the molecular dynamic trajectories and minimum energy paths are monitored on the potential energy surface using MATLAB.

Also,as for a reaction to occur, the particles having sufficient energy to overcome the activation energy barrier is not the only pre-requisite, the vibrational energy also has to be in the right time and in the right mode. Therefore, in the reaction, the ratio of the internuclear momenta is varied to investigate the correct combinations for the reaction to proceed.

== Reaction 1: H atom + H2 molecule system ==

In this reaction, one H atom(HA) collides with one H2 molecule(HB-HC) to form a different H2 molecule(HA-HB) and detaching HC.

===Dynamics of the transition state and the minimum===

In this triatomic system, the reaction proceed via transition state to the products, in which the transition state links reactants and products together and is defined by the maximum on the reaction path.

Both gradients are equal to 0 at minimum and at a transition structure. Therefore the '''total value of gradient is 0'''.

However, for the minimum, it will always be at the lowest point on the potential energy surface, therefore, its second derivative will always be positive.
For transition state, as the reaction proceeds via the minimum energy path, the whole trajectory will be minimum in the potential energy surface, so the transition state will have a positive value of second derivative in this aspect; but within the trajectory of the reaction, the reaction has maximum energy and will have a negative value of the second derivative within the path.
Therefore, the minimum and the transition state could be distinguished by their '''second derivatives'''.

===Determination of position of Transition State===

[[File:Min interd.PNG|thumb|none|500px|The internuclear distance vs time graph at transition point.]]

As the three atoms in the system are all identical, the potential energy surface will be symmetrical as well as the reaction path. Therefore, the transition state will occur at the point where two internuclear distances are equal'''(rAB=rBC)'''.

Also, given 0 momentum for all three atoms, the system will remain on transition state so therefore, the internuclear distances will be '''constant''' instead of fluctuating values throughout time.

Therefore, from the internuclear distance vs time plot,the best estimate of the transition state position is determined to be around '''0.908 Å''' for both internuclear distances

===MEP and Dynamic paths===

MEP represents minimum energy path, in which the path simply follows the lowest energy reaction path in the potential energy surface. It does not account for any vibrational energy and will be follow the valley in the surface plot. Also, the atoms will have 0 momentum throughout the MEP because the velocity of the atoms is always reset to 0 at each time step.

The initial conditions are set as below:

rAB=rts=0.908 Å

rBC=rts+0.01=0.918 Å

pAB=pBC=0

The reaction trajectories are run with the initial conditions in both Dynamic mode and MEP mode and the surface plots are shown below :

[[File:Hydrogenandh mep1.png|thumb|500px|none|The surface plot under MEP mode.]]

[[File:Hydrogenandh dyn1.png|thumb|500px|none|The surface plot under dynamic mode.]]

As the MEP does not account for neither vibrations inbetween the molecules, the above graph gives a smooth line, while the dynamic mode accounts for the vibrations of the molecules and appears as a oscillating curve in the potential energy surface.

The intermolecular distance vs time and intermolecular momenta vs time graphs for both modes are shown below:

[[File:h2andh interdis1.png|thumb|500px|none|The intermolecular distance vs time graph under dynamic mode.]]

[[File:h2andh mep interd.png|thumb|500px|none|The intermolecular momenta vs time graph under MEP mode.]]

[[File:h2andh momentum.png|thumb|500px|none|The intermolecular momenta vs time graph under dynamic mode.]]

[[File:H2andh mep momentum.png|thumb|500px|none|The intermolecular momenta vs time graph under MEP mode.]]

Then the values of the two internuclear distances are exchanged with momentum kept constant, which results in the trajectory going in the direction of the reactants, the internuclear distance vs time and internuclear momenta vs time graphs will have identical shapes only with AB and BC inverted as shown below:

[[File:h2andh dyn3.png|thumb|500px|none|The surface plot with values rAB and rBC exchanged under dynamic mode.]]

[[File:h2andh interdis3.png|thumb|500px|none|The intermolecular distance vs time graph with values rAB and rBC exchanged under dynamic mode.]]

[[File:h2andh momentum1.png|thumb|500px|none|The intermolecular momenta vs time graph with values of rAB and rBC exchanged under dynamic mode.]]

Then, the initial conditions are set as the final conditions as above with the signs of momenta inverted, and the resulting path goes from the products back to the transition state:

[[File:h2andh dyn4.png|thumb|500px|none|The surface plot with initial conditions changed under dynamic mode.]]

[[File:h2andh reverse dis.png|thumb|500px|none|The intermolecular distance vs time graph with initial conditions changed under dynamic mode.]]

[[File:h2andh reverse momentum.png|thumb|500px|none|The intermolecular momenta vs time graph with initial conditions changed under dynamic mode.]]

===Reactive and Unreactive Trajectories===

The trajectory is reactive under the following conditions as according to the above calculations:

rAB = 2.0 Å

rBC = 0.74 Å

-1.5 < pBC < -0.8 and pAB = -2.5

As these conditions are reactive, the conditions with higher momentum should have more than sufficient energy to cross the activation barrier and should also be reactive, however, it is not the case as there could be recrossing of the transition state and reformation of the reactants according to the following trajectories.

{| class="wikitable" border="1"
|+ Reactivity with following momenta combination:
! pBC !! pAB !! Reactivity
|-
| -1.25 || -2.5 || Reactive
|-
| -1.5 || -2.0 || Uneactive
|-
| -1.5 || -2.5 || Reactive
|-
| -2.5 || -5.0 || Unreactive
|-
| -2.5 || -5.2 || Reactive
|}

'''Trajectory 1'''
[[File:Traj1.PNG|thumb|500px|none|Trajectory 1]]
[[File:Traj1 dis.PNG|thumb|500px|none|The system crosses the transition state and forms the product.]]

In trajectory 1, enough energy is provided by the initial momentum of the particles to overcome activation barrier, so the the H atom approaches and form a new H-H bond with one H atom in the hydrogen molecule.

'''Trajectory 2'''
[[File:Traj2.PNG|thumb|500px|none|Trajectory 2]]

[[File:Traj2 dis.PNG|thumb|500px|none|The system reverts back to the reactant before reaching transition state due to lack of energy.]]

In trajectory 2, not enough energy is provided by the momentum to overcome the activation barrier, so the H atom approaches the transition state but reverts back before reaching the point.

'''Trajectory 3'''
[[File:Traj3.PNG|thumb|500px|none|Trajectory 3]]

[[File:Traj3 dis.PNG|thumb|500px|none|The system crosses the transition state and forms the product.]]

In trajectory 3, similar cases happen with that in trajectory 1, enough energy is provided to overcome the activation barrier to form the product.

'''Trajectory 4'''
[[File:Traj4.PNG|thumb|500px|none|Trajectory 4]]

[[File:Traj4 dis.PNG|thumb|500px|none|The system crosses the transition state, but then recrosses and reverts back in the pathway to form reactants again.]]

In trajectory 4, the system has enough energy to cross the transition state, but then it falls back and the reactants are reformed.

'''Trajectory 5'''
[[File:Traj5.PNG|thumb|500px|none|Trajectory 5]]

[[File:Traj5 dis.PNG|thumb|500px|none|The system recrosses the transition state, but forms the product in the last.]]

In trajectory 5, the system recrosses the transition state, but it has enough energy to cross again and form the final product.

===Transition State Theory===

In the Transition State Theory, it was assumed that at transition state, the reaction could only proceed in the direction of the products and could not go back to form the reactants.

However, in reality, for reactive trajectories, there is a possiblity for the transition state to recross and revert back to the reactant and then reform the products, in which the trajectory takes another path rather than predicted minimum energy path. Therefore,the experimental values of reaction rates are always slower than that predicted by the Transition State Theory.

== Reaction 2: F-H-H system ==

In this reaction, three atoms are present, in which they could react via two ways:

1. F atom + H2 molecule reacts to give HF

2. H atom + HF molecule reacts to give H2 molecule and F atom

===F+H2 Reaction===

In this reaction, F atom collides with the H2 molecule to give HF molecule and H atom. As the H-F bond has a large difference of electronegativity, its ionic character increases so the [[bond strength of H-F will be much larger than that of H-H]], so larger amount of energy will be released during H-F bond formation than the activation energy, and the reaction will be '''exothermic'''.

According to Hammond's Postulate, the transition state will resemble one of the reactant/product which has a closer energy. Therefore, in this reaction, the [[transition state complex will resemble]] that of the '''starting material''' with a relatively small value of activation energy.

Because the transition state resemble the reactants both in energy and in structure, it will be located near the position of the reactants on the potential energy curve.

As illustrated above, the transition state will be located at the position where if no momentum is given, it will remain at the position and the internuclear position will remain constant.

[[File:fhhacte inter.png|thumb|500px|none|The internuclear distance remains constant at transition state with 0 momentum.]]

In this reaction, the transition state is located at the position where the '''internuclear distance of F and H AB=1.81 Å''' and the '''internuclear distance of H-H BC=0.745 Å'''.

[[File:fhhacte interd.png|thumb|500px|none|The transition state is presented as a point on the potential energy curve.]]

The activation energy is determined by the energy difference of the transition state and the reactants, which in this case was very small of around 0.2 kcal/mol.

[[File:acte ts.png|thumb|500px|none|The energy of the transition state.]]

[[File:acte reactant.png|thumb|500px|none|The energy of the reactants.]]

===H+HF Reaction===

In this reaction, according to Hammond's Postulate, as the [[H-F bond strength is much higher than that of H-H bond]] formed, the product will be less stable than the reactants and the reaction will be '''endothermic'''. Therefore, the '''transition state will resemble the product''' and be located near the position of products on the potential energy curve.

The transition state is located at the position where the internuclear distance of '''H and H''' '''AB=1.81 Å''' and the internuclear distance of '''H-F BC=0.7465 Å'''.

[[File:hfhacte interd.png|thumb|500px|none|The internuclear distance remains constant at transition state with 0 momentum.]]

The activation energy is determined by the energy difference between reactants and transition state, which has the value of around 3.1 kcal/mol.

[[File:hfhacte ts.png|thumb|500px|none|The energy of the transition state.]]

[[File:hfhacte reactant.png|thumb|500px|none|The energy of the reactants.]]

===Reaction Dynamics===

====F + H2 Reaction====

Firstly, one of the reactive trajectory was determined with the following conditions:

rFH=1.9 Å

rHH=0.745 Å

pFH=-1.9 kg m s-1

pHH=0.1 kg m s-1

where its internuclear distance vs time graph is shown below:

[[File:momenta2.png|thumb|500px|none|The internuclear momenta of the particles vs time graph of this set of conditions.]]

We could see that the reaction occurs to form HF molecule and H atom according to following observations:

1. approaching of the F atom(negative momentum) to the vibrating H2 molecule(shown by oscillating values of internuclear momentum)

2. the collision of two and then the dissociation of H2 molecule(shown by 1. increasing extent of oscillation of B-C internuclear momentum indicating lengthening of the bond. 2. constant B-C internuclear momenta afterwards indicating the dissociation of two H atoms and the steady movement of the H atom away from the HF molecule.) and the formation and vibration of the HF molecule(oscillation of A-B internuclear momentum).

the dynamic surface plot of the reaction is shown below:

[[File:momenta1.png|thumb|500px|none|The reaction trajectory of this set of conditions on PES.]]

And the graphs of potential energy vs time and kinetic energy vs time are shown below:

[[File:Pevst.png|thumb|500px|none|The graph of potential energy of the system against time.]]

[[File:Kevst.png|thumb|500px|none|The graph of kinetic energy of the system against time.]]

In the reaction, as the energy is conserved, the energy of the particles would be converted between potential and kinetic energies. At the start of the reaction,there are minor vibrations of the reactants shown by the small extent of fluctuations of energy with time shown in the beginning in both energy vs time graphs, where the energy is converted between two types but is largely potential energy and only has small amount of kinetic energy.

As the reaction is exothermic, the potential energy of the reactants will be much higher than that of the products, therefore there will be release of excess potential energy, which is converted into kinetic energy of the product, which is shown in the decrease in the potential energy and the increase in the kinetic energy of the system shown in the graph above. In addition, after the transition state, the molecule starts vibrate in a larger extent, and it could be observed in the large oscillations of the energy curves after transition state.

To confirm that experimentally, IR spectroscopy could be used as it monitors the vibration movement of the moleucules.

Then the value of internuclear momentum of H-H varied between -3-3 kg m s-1, and the two internuclear distances and the internuclear momentum of F-H are kept constant with values of:

rFH=1.81 Å

rHH=0.74 Å

pFH=-0.5 kg m s-1

The values of are and the reactivity of these conditions are shown below:

{| class="wikitable" border=1
! pHH !! -3.0 !! -2.5 !! -2.0 !! -1.5 !! -1.0 !! -0.5 !! 0.0 !! 0.5 !! 1.0 !! 1.5 !! 2.0 !! 2.5 !! 3.0
|-
| Reactivity || Unreactive || Unreactive || Reactive || Unreactive || Unreactive || Unreactive || Reactive || Unreactive || Unreactive || Unreactive || Unreactive || Reactive || Unreactive
|}

The internuclear distance vs time graphs are shown below:

[[File:m-3-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-3.]]

[[File:m-2.5-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-2.5.]]

[[File:m-2-reactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-2.]]

[[File:m-1.5-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-1.5.]]

[[File:m-1-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-1.]]

[[File:m-0.5-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=-0.5.]]

[[File:m0-reactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=0.]]

[[File:m0.5-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=0.5.]]

[[File:m1-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=1.]]

[[File:m1.5-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=1.5.]]

[[File:m2-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=2.]]

[[File:m2.5-reactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=2.5.]]

[[File:m3-unreactive-interd.png|thumb|500px|none|The graph of internuclear distance vs time with pHH=3.]]

As the activation energy is quite small in this exothermic reaction, significant energy is put into the internuclear momentum of the atoms compared that with the activation energy, therefore, it is predicted that the trajectory should always be reactive. However, this is not the case for these sets of reaction, as only certain values of pHH leads to reactive trajectory.

Then the same initial internuclear distances are used, with internuclear momentum changed to:

pFH=-0.8 kg m s-1

pHH=0.1 kg m s-1

This set of conditions gives a reactive trajectory as shown below:

[[File:Fh2 dynamics surf.png|thumb|500px|none|The surface plot of this set of conditions.]]

[[File:Fh2 dynamics interd.png|thumb|500px|none|The internuclear distance vs time graph of this set of conditions.]]

In this reaction, it could be observed that in the reactive trajectory,the reactants have low vibrational kinetic energy which react in a exothermic reaction to give product with high vibrational kinetic energy.

====H + HF Reaction====

The initial conditions are set to the values below so that the the reactants have low vibrational kinetic energy:

rHH=2.0 Å

rHF=0.91 Å

pHH=-2.2 kg m s-1

pHF=-2.0 kg m s-1

The trajectory is unreactive as shown below:

[[File:low vib hf.png|thumb|500px|none|The reaction trajectory of this set of conditions.]]

Then, the momentum of incoming hydrogen atom is reduced and the internuclear momentum of the HF molecule are increased and several sets of internuclear momenta are tested to give the reactive combinations below:

{| class="wikitable" border=1
! pHH kg m s-1 !! pHF kg m s-1
|-
| -0.9 || 8.5
|-
| -0.9 || 10.0
|-
| -1.0 || 10.0
|-
| -2.0 || 8.5
|}

The surface plots are shown as below:

[[File:-0.9+8.5.png|thumb|500px|none|The reaction trajectory of this set of conditions.]]

[[File:-0.9+10.png|thumb|500px|none|The reaction trajectory of this set of conditions.]]

[[File:-1+10.png|thumb|500px|none|The reaction trajectory of this set of conditions.]]

[[File:-2+8.5.png|thumb|500px|none|The reaction trajectory of this set of conditions.]]

Polanyi's empirical rules states that the vibrational energy is more efficient in promoting a late-barrier reaction than the translational energy.

As a result, as H+HF reaction is an endothermic and so a late-barrier reaction, it should obey the rules.

As the vibrational energy is low in the above trajectory, it does not react as according to the rules.

The following four examples, all with high intermolecular momenta and high vibrational energy as shown in the potential energy surface plots, all proceeds to form the product. Compared with the example above, they have large values of pHF and low values of HH, therefore, the ratio between the translational and vibrational energy is quite high.

Also, for the early-barrier reaction of F+H2 reaction, it reacts with a rather small ratio of two momenta(reactive trajectories have the momentum ratio of :2:1,1.67:1,and 2.08:1), which is different from that in H+HF reaction.

Therefore, the reacion trajectories obey the Polanyi's rules.

Talk:MRD:01058157

2017-05-11T14:56:46Z

Bg1512: Created page with "== Feedback == Good work! Your explanations of TS theory and research around experimental verification of results were thorough. Here are nevertheless some pointers for furthe..."

== Feedback ==
Good work! Your explanations of TS theory and research around experimental verification of results were thorough. Here are nevertheless some pointers for further improvement:
* Answers to questions:
** Saddle point: You correctly identified that both first derivatives are zero. However, I'm confused as to why you are only speaking about a single curvature. Are you talking about the mixed derivative (''d2f/dxdy)''? In a 2D PES there are three different second derivatives. Consider that the TS is a minimax point. Does that help you come up with a mathematical definition of the saddle point?
** Energy distribution between vibrational and translational modes: You haven't shown any data. You always have to base your discussion on experimental data (or calculations in this case; a computational project does not relieve you from presenting data just because it can be trivial to generate) so other can judge and compare their data to yours.
** Eyring Theory: It is based on thermodynamical (statistical) arguments. Does it still apply here?
* Introduction:
** Aim to include an introduction to all your academic work. The gold standard is to allow a user who hasn't seen the lab script get up to speed. This means you need to motivate your work (why should I care?) and explain key concepts
* Conclusion:
** Similarly, provide a conclusion at the end. If your reader hasn't taken in any of your report, they should still be able to get the key facts from your conclusion. This means reporting key data, any unexpected results and comparisons with literature.
* Figures:
** Please label and number your figures/graphs. Your report made it clear which figure you were referring to at each point, but when you are writing more complicated (multi-column!) texts, you will need to number your figures and refer to them in the text by their number. --[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:56, 11 May 2017 (BST)

MRD:01058157

2017-05-11T14:54:03Z

Bg1512:

== Transition States on a Potential Energy Surface for Hydrogen Exchange Reaction ==

=== Locating the transition state ===
The total gradient of the potential energy surface is zero at both the minima and the transition state - saddle points. These points on the potential energy surface can be distinguished by considering the second derivative of the potential energy surface: the transition state is indicated by a negative curvature and the minima by a positive curvature.

The transition state for the reaction H + H2 → H2 + H was located by setting the initial momentum for both species to 0 and equating the internuclear distances (r1 and r2) between the outer and central hydrogen, as when r1=r2=rts. No oscillation is observed as the reactants remain on the transition state ridge. The transition state position occurs at (0.908, 0.908).

[[File:HHH_rts.PNG|468x468px]]

=== Comparing Reaction Trajectories ===
The simulation also examines the minimum energy path (mep), which treats the reacting particles as having infinitely slow motion. Hence the reaction trajectory follows the lowest path in the potential well displaying no vibrational tendencies. Here, the trajectory was studied with an initial momentum of 0 and one of the atoms slightly displaced from the transition state radius (+0.01). This was contrasted with dynamic trajectory for the same condition, which incorporates lateral movement along the potential energy surface.

[[File:spe.png|574x574px|border]] [[File:Mepe.png|550x550px]]

The side-by-side comparison heightens the contrast between the two reaction trajectories. As the minimum energy paths follows most stable trajectory from reactants to product the gradient of the path is 0. It begins at the the transition state and falls slowly down the potential surface. In comparison, in the dynamic state, the HF species has vibrational energy which displaces it back to the reactants from the starting transition state due to interconverting between the energies, allowing it to collapse more rapidly. However, it also follows the general minimum energy path, only with additional vibrational motion.

=== Reactive and Unreactive Trajectories ===

==== p1= -1.25, p= -2.5 Reaction Successful ====
This reaction trajectory leads to the successful formation of the product. It follows the valley of the valley of the potential energy surface and displays very little vibrational energy.

[[File:-1.25,-2.5_sp.PNG|481x481px]] [[File:-1.25,-2.5_dt.PNG|452x452px]]

==== p1= -1.5, p= -2.0 Reaction Unsuccessful ====
The H and H2 do not have sufficient translational energy to react. The atoms come together but collapse back down their initial state.

[[File:-1.5,-2 dt.PNG|473x473px]] [[File:-1.5,-2 sp.PNG|450x450px]]

==== p1= -1.5, p= -2.5 Reaction Successful ====
In a similar case for the first conditions, the reaction takes places but this is not surprising due to the subtle change in the initial kinetic energy.

[[File:-1.5,-2.5 sp.PNG|476x476px]] [[File:-1.5,-2.5 dt.PNG|460x460px]]

==== p1= -2.5, p= -5.0 Reaction Unsuccessful ====
These initial distance and momentum conditions lead to an interesting reaction in which the product is formed in transition state but the reactants separate and return to their side of the potential valley. The barrier recrossing is due to the extension of the bond as it breaks after formation.

[[File:-2.5, -5 sp.PNG|468x468px]] [[File:-2.5, -5 dt.PNG|451x451px]]

==== p1= -2.5, p= -5.2 Reaction Successful ====
The hydrogen exchange occurs but the product has very high vibrational energy. This could be a consequence of the system possessing very high energy. It crosses and recrosses until it has the correct ratio of energies.

[[File:-2.5, -5.2 sp.PNG|478x478px]] [[File:-2.5, -5.2 dt.PNG|468x468px]]

=== Main Assumptions in Transition State Theory ===
The governing equation of transition state theory is Eyring's equation

[[File:wdmneq.png|269x269px]]

The main assumptions in transition state theory are
* All steps preceding the formation of the activated complex are not considered
* Atoms are considered as hard spheres
* All translational energy can be ascribed to kinetic energy and momentum
* Quantum mechanical effects including tunnelling are ignored
* If the system is at equilibrium the transition state is at equilibrium and is unaffected by removing product
* One reaction pathway
* Limited recrossing of the barrier
* Obeys Boltzmann distribution
The combined effect of these assumptions is that the observed rate is higher than that predicted by theory <ref>K. J. Laidler and M. C. King, ''J. Phys. Chem.'', 1983, '''87''', 2657–2664.</ref>.

== Potential Energy Surface and Reaction Dynamics of an F - H - H system ==

=== Inspection of the F + H2 → HF + H Potential Energy Surface ===

The F + H2 → HF reaction is exothermic because the product is more energetically stable than the reactants. Conversely the reverse reaction is endothermic because the reactants are more energetically stable than the products. Intuitively this is expected as is the H-F bond is significantly stronger than H-H due to the strong polarisation of the bond.

The transition state was located by visual inspection of a contour plot for the saddle point. It is an early transition state in the exothermic direction (similar to the reactants) and late in the endothermic direction (similar to the products). The activation energy for F + H2 → HF + H is very small and so applying Hammond's postulate, in which the transition state resembles reactant or product it closer to in energy allows it determination. The transition state was found at (1.814, 0.745) .

F + H2 → HF + H Activation Energy= 0.6 kcal-1

H + HF → H2 + F Activation Energy= 30.4 kcal-1

[[File:Pse2.png|569x569px]] [[File:ide.png|569x569px]]

=== Experimental Techniques for Analysing the F + H2 system ===

In the reaction trajectory for F + H2 → HF + H , after reaching the transition state the energy is released into the vibrational mode. The HF molecule displays significant vibrational oscillations as evidenced by the oscillating internuclear momenta against time.

Hence using vibrationally resolved spectroscopy, the reaction dynamics can be investigated. Tizniti et al simulated the conditions of deep space and analysed the reaction using pulsed laser photolysis–pulsed laser-induced fluorescence <ref>M. Tizniti, S. D. Le Picard, F. Lique, C. Berteloite, A. Canosa, M. H. Alexander and I. R. Sims, ''Nat Chem'', 2014, '''6''', 141–145.</ref> . This would allow for probing of vibrational modes, which could be rationalised using the Frank Condon principle. The FH2- ion is readily characterised and as it is close in energy and semblance to the transition state. Qui et al were able to analyse the reaction from a full quantum mechanical perspective using an advanced crossed molecular beam scattering technique in conjunction with the time of flight measurements of Rydberg tagged hydrogen radicals <ref>M. Qiu, Z. Ren, L. Che, D. Dai, S. A. Harich, X. Wang, X. Yang, C. Xu, D. Xie, M. Gustafsson, R. T. Skodje, Z. Sun and D. H. Zhang, ''Science (80-. ).'', 2006, '''311''', 1440 LP-1443.</ref>.

=== Energy Distribution between Vibrational and Translational Modes ===

==== F + H2 → HF + H ====
For the collision of a fluorine radical into a hydrogen molecule, translational energy is considerably more important than vibrational energy. The early transition state means that momentum is more effective when supplied along the bond axis. It is an early barrier exothermic reaction. As the energy released by H-H repulsion is supplied before the H-F bond reaches its equilibrium position, it causes the bond to vary in length - vibrational energy <ref>J. C. Polanyi and J. L. Schreiber, ''Faraday Discuss. Chem. Soc.'', 1977, '''62''', 267–290.</ref>.

==== H + HF → H2 + F ====
In this direction, vibrarional energy is more important as the momentum supplied to H-F elongates and weakensthe strong bond between hydrogen and fluorine. It is a late barrier endothermic reaction. Consequently, the energy supplied from the formation of the H-H bond is in the form of translational energy.

<references />

Talk:MRD:zw4415

2017-05-11T14:50:55Z

Bg1512:

Talk:MRD:zw4415

2017-05-11T14:45:51Z

Bg1512: Created page with "== Feedback == Good work! Your structure and argumentation is clear and I could follow your train of thought throughout. I especially liked the formatting of multiple graphs i..."

MRD:zw4415

2017-05-11T14:44:45Z

Bg1512:

=='''Introduction'''==
In this computational lab, reaction dynamics of triatomic systems was investigated using MATLAB R2016a.
Varies of reaction conditions was experimented by changing the parameters within the program.
Transition states of triatomic system (HHH) and (HHF) are determined. Trajectories of reaction together with the intermolecular momentum vs time graph was used to determine whether the system is reactive or unreactive.
The same method could also be applied to other triatomic systems to determine their transition state and reaction trajectory.

== '''Exercise 1: H + H2 potential energy surface''' ==

=== Determination of the Transition State geometry ===
At transition structure, the gradient ∂V (r1)/∂r1=0 and ∂V (r2) /∂r2=0.

At minimum 1 (reactants), the gradient ∂V (r1)/∂r1=0 but ∂V (r2) /∂r2 is less than 0.

At minimum 2 (products), the gradient ∂V (r1)/∂r1=0 but ∂V (r2) /∂r2 is less than 0. 

[[File:zw4415_1.png|thumb|300px|none|Figure 1. Internuclear Distances vs Times at transition state]]

rts= 0.90775, tried out a range of distances. When r1 and r2 are both 0.90775 and p1=p2=0 the internuclear distance doesn’t change with time, which means it is
at the transition state.

If one starts a trajectory exactly at transition state, with no initial momentum, it will remain there forever.

=== Trajectories from Transition State ===
Distance of A-B increase over time while B-C distance decrease over time and eventually B-C bonded.

According to the internuclear momentum vs time graph calculated by Dynamics method, we can tell that after formation of bond B-C, the translational energy converts to vibrational energy, and hence bond B-C oscillates.

However in MEP calculation, there is no intermolecular momentum. From the contour plot, the B-C will not vibrate over time.

In dynamics method at large t ( t = 50s ):

The average internuclear momentum between B and C is 1.241.

The average internuclear momentum between A and B is 2.481.

The average internuclear momentum between A and C is 0.

The average internuclear distance between B and C is 0.7432.

The average internuclear distance between A and B is 186.5.

{| class="wikitable"
|MEP
|[[File:zw4415_MEP1.png|thumb|300px|none|Figure 2. Contour plot in MEP calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|[[File:zw4415_MEP_momenta_time.png|thumb|300px|none|Figure 3. Intermolecular momentum vs time in MEP calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|[[File:zw4415_MEP_dis_time.png|thumb|300px|none|Figure 4. Intermolecular distance vs time in MEP calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|-
|Dynamics
|[[File:zw4415_Dynamics1.png|thumb|300px|none|Figure 5. Contour plot in Dynamics calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|[[File:zw4415_Dynamics_momenta_time.png|thumb|300px|none|Figure 6. Intermolecular momentum vs time in Dynamics calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|[[File:zw4415_Dynamics_dis_time.png|thumb|300px|none|Figure 7. Intermolecular distance vs time in Dynamics calculation (r1=0.90775, r2=0.91775, p1=p2=0)]]
|}

If we start from the the initial positions correspond to the final positions of the trajectory we just calculated, with the same momenta values as the final momenta before but with signs reversed.

We could obtain a following trajectory. It is very similar to the trajectory with the initial value r1=0.90775, r2=0.91775, p1=p2=0.

[[File:zw4415_inverse_momenta.png|thumb|300px|none|Figure 8. Contour plot in Dynamics calculation (r1=186.5, r2=0.7432, p1=2.481, p2=1.241)]]

=== Reactive and unreactive trajectories ===

{| class="wikitable"
!p1
!p2
!trajectory
!momentum
!description
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:zw4415_react1.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.25,p2=-2.5)]]
|[[File:zw4415_react1_m.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.25,p2=-2.5)]]
|Reactive trajectory. The trajectory passes though the transition state, which indicates that it has sufficient energy to overcome the activation energy.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:zw4415_react2.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.0)]]
|[[File:zw4415_react2_m.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.0)]]
|Unreactive trajectory. The trajectory path reverse back to A-B side. The B-C momentum is not oscillating at all, so no B-C bond forms.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:zw4415_react3.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.5)]]
|[[File:zw4415_react3_m.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.0)]]
|Reactive trajectory. The trajectory passes though the transition state, which indicates that it has sufficient energy to overcome the activation energy.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:zw4415_react4.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-2.5, p2=-5.0)]]
|[[File:zw4415_react4_m.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.0)]]
|Unreactive trajectory, The trajectory path reverse back to A-B side. The momentum plot indicates that B-C was formed at the beginning but then breaks due to the reformation of A-B.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:zw4415_react5.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-2.5, p2=-5.2)]]
|[[File:zw4415_react5_m.png|thumb|300px|none|(r1=0.74, r2=2.0, p1=-1.5, p2=-2.0)]]
|Reactive trajectory. According to the internuclear momentum vs time graph, we can predict that B-C will eventually be oscillating.
|}

According to transition state theory, when the system reaches a transition state configuration, it automatically assume that the system proceed to the product system. It might be due to the fact that the transition state is formed from the initial reactant state, so when it reach transition state, it have a inertia to having a momentum in positive direction. This assumption theory is true in most of the cases. However in the last 2 systems, both system have sufficient energy to overcome the activation energy barrier. However B-C bond only forms in the last system. It can not be used to predict the reaction rate as it does not take into account the rate where the transition state reverse back to reactants.

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
Reaction from F + H2 to H + HF is exothermic.

This is related to the bond strength between H-F is stronger than H-H. Therefore energy released when the reaction taken place.
{| class="wikitable"
![[File:zw4415_HHF_trans.png|thumb|300px|none|Approximate position of the transition state]]
![[File:zw4415_HHF_trans_en.png|thumb|300px|none|Energy of the transition state in surface plot]]
|}

The transition state was estimated to be at r1=1.8107 and r2=0.745 with initial momenta equals to 0.

Activation energy is the energy barrier between the minimum point on the trajectory and the transition state.
{| class="wikitable"
![[File:zw4415_HHF_min1_en.png|thumb|300px|none|Energy of the minimum point 1 on trajectory in surface plot]]
![[File:zw4415_HHF_min2_en.png|thumb|300px|none|Energy of the minimum point 2 on trajectory in surface plot]]
|}

For energy barrier between the H-H bond and transition state:

Thus the Ea could be calculated as Ea=-103.3-(-103.9)= 0.6 Kcal/mol

For energy barrier between the transition state and the H-F bond:

Thus the Ea could be calculated as Ea=-103.3-(-133.9)= 30.6 Kcal/mol

=== Reaction dynamics ===

{| class="wikitable"
![[File:zw4415_RD_1.png|thumb|300px|none|Contour plot (r1=2, r2=0.745, p1=-2, p2=0)]]
![[File:zw4415_RD_2.png|thumb|300px|none|Intermolecular momentum vs time plot (r1=2, r2=0.745, p1=-2, p2=0)]]
|}

Initially the B-C bond which corresponding to H-H is vibrating. Then when F atom approaches, the H-F bond forms and the excess energy was transferred to the vibration energy between H and F atom. Thus at large time t, the inter molecular momentum between B-C flattened and A-B momentum is vigorously oscillating.

From the F + H2 side where rHH = 0.74 and rFH =2,5 , momentum pFH was set to -0.5 .

Several values of pHH in the range from -3 to 3 was tried out in this system.

{| class="wikitable"
|[[File:zw4415_mom_1.png|thumb|200px|none| pHH = 1]]
|[[File:zw4415_mom_2.png|thumb|200px|none| pHH = 2]]
|[[File:zw4415_mom_2.74.png|thumb|200px|none| pHH = 2.74]]
|[[File:zw4415_mom_2.76.png|thumb|200px|none| pHH = 2.76]]
|[[File:zw4415_mom_3.png|thumb|200px|none| pHH = 3]]
|-
|[[File:zw4415_mom_n1.png|thumb|200px|none| pHH = -1]]
|[[File:zw4415_mom_n2.png|thumb|200px|none| pHH = -2]]
|[[File:zw4415_mom_n2.74.png|thumb|200px|none| pHH = -2.74]]
|[[File:zw4415_mom_n2.76.png|thumb|200px|none| pHH = -2.8]]
|[[File:zw4415_mom_n3.png|thumb|200px|none| pHH = -3]]
|}

Observation: When pHH was taken from -2.74 to 2.76, it shows a unreactive trajectory. Higher or below that, the trajectory becomes reactive.

----
[[File:zw4415_mom_inc.png|thumb|300px|none| pFH= -0.8, pHH = 0.1]]

If the momentum of FH was set slightly higher at -0.8, and pHH was reduced to 0.1 which is very small. We can tell from the trajectory that the system is still reactive.

----

For system starting form H + HF, the entry point should be (r1=1.8107, r2=2.5)

[[File:zw4415_0_5.png|thumb|300px|none| r1=1.8107, r2=2.5, p1=0, p2=-5]]

The initial conditions was set at p1=0 and p2=5. It can be seen from the contour plot that the H-F bond was initially broke by the high momentum between H and H. However after H-F bond breaks, F could back attack the H-H molecule and reforms the H-F. Thus the total system is in oscillation.

A reactive trajectory was obtained by setting the momentum H-H to -0.5 and gradually increasing momentum H - F.

It can be observed that when pHF goes below 1.45, the system becomes unreactive.
{| class="wikitable"
![[File:zw4415_phf_1.65.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.65]]
![[File:zw4415_phf_1.60.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.60]]
![[File:zw4415_phf_1.55.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.55]]
![[File:zw4415_phf_1.50.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.50]]
![[File:zw4415_phf_1.45.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.45]]
![[File:zw4415_phf_1.40.png|thumb|200px|none| r1=1.8107, r2=2.5, p1=-0.5 , p2=-1.40]]
|}

=== Conclusion ===
In this case the pHF corresponds to the vibration energy and pHH corresponds to the transition energy.
From the above analysis, when transitional energy increases, the system is likely to become reactive.

When the transition state is closer to the reactant , the momentum thus energy needed for both species are lower.
When the transition state is closer to the product, higher transitional energy is required to overcome the energy barrier.

Talk:MRD:fp1615 molecular reaction dynamics

2017-05-11T14:43:18Z

Bg1512:

'''Feedback '''

Excellent work! Basically all the questions were answered correctly. Your writing style is on point, the structure is clear and I could follow your argumentation throughout. Here are a couple of pointers on how to improve further:

'''Answers to questions'''

'''Properties of the TS''' I see you invested considerable effort in this answer, but unfortunately it is not quite right. I think you switched the < and > around in your equations. Let me explain: A TS is a maximum in one direction and a minimum in the other. So you're right in saying the partial derivatives are both 0. Let's assume x is the direction along which we find a maximum and the minimum is along y. Then <math>f_{xx} < 0</math> & <math>f_{yy} > 0</math>. You know this from 1D calculus: If you find an extreme point that's curved negatively, the points on either side of the extreme must have lower functional values and you have a maximum (and the other way round for minima). This gives us <math>f_{xx}f_{yy} < 0</math>. Now <math>f_{xy}^2 > 0</math> since it is a square and <math>f_{xx}f_{yy} - f_{xy}^2 < 0</math> for a saddle point. You may have noticed that this derivation of the discriminant actually assumes the minimax property of the TS, so it is actually easier to test for the saddle point by using this assumption rather than using the discriminant itself. There are however saddle points that aren't minimax points, then <math>f_{xx}f_{yy} > 0</math> and <math>f_{xy}^2 > f_{xx}f_{yy}</math>. These don't look like saddles, but more like terasses.

Unfortunately there are two types of TS theory out there. The one you quoted is the Eyring theory. It is mainly based on thermodynamic arguments, which themselves are statistical (that's why we can talk about equilibria). Do you think this is an appropriate description of the system at hand?

'''Experimental verification of release of vibrational energy''' What specific aspect of IR allows you to infer the occupation of vibrational levels? Could you draw this out on a diagram? What about the role of harmonic and anharmonic oscillators?

'''Grammar'''

You used articles in front of capitalised nouns, such as "The Hammond's postulate". Omit the "the" before such nouns.

'''Introduction'''

In the introduction you state that every successful reaction has to go over the TS, but later on you show the opposite. It would be good to indicate in the introduction that you will contradict this widely accepted fact and later on, when you actually disprove it, remind people of the original statement in the introduction. This emphasize your input in the work and your critical evaluation of the literature.
You could have also disproven this point directly in the introduction. The TS (as a mathematical construct) is a 0-dimensional point and hitting it would require an atomic geometry with infinite precision, which is not possible.

'''Results'''

The graph for the endothermic direction of the F-H-H reaction looks almost like a step function. This is an unanticipated result to me and I would have liked to see a comment from you. When calculating the TS energy, you do actually know what the energy of the TS is. The potential energy of the atomic configurations is independent of how you got to them, so the potential energy in the perfectly symmetrical TS calculation can be used, even though it would never go on to react. You may also be able to get a better estimate of the reactant potential energy by fitting the potential energy curve with an analytical function and then taking the limit of time approaching infinity.--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:09, 11 May 2017 (BST)

Talk:MRD:fp1615 molecular reaction dynamics

2017-05-11T14:39:15Z

Bg1512:

'''Feedback '''

Excellent work! Basically all the questions were answered correctly. Your writing style is on point, the structure is clear and I could follow your argumentation throughout. Here are a couple of pointers on how to improve further:

'''Answers to questions'''

'''Properties of the TS''' I see you invested considerable effort in this answer, but unfortunately it is not quite right. I think you switched the < and > around in your equations. Let me explain: A TS is a maximum in one direction and a minimum in the other. So you're right in saying the partial derivatives are both 0. Let's assume x is the direction along which we find a maximum and the minimum is along y. Then <math>f_{xx} < 0</math> & <math>f_{yy} > 0</math>. You know this from 1D calculus: If you find an extreme point that's curved negatively, the points on either side of the extreme must have lower functional values and you have a maximum (and the other way round for minima). This gives us <math>f_{xx}f_{yy} < 0</math>. Now <math>f_{xy}^2 > 0</math> since it is a square and <math>f_{xx}f_{yy} - f_{xy}^2 < 0</math> for a saddle point. You may have noticed that this derivation of the discriminant actually assumes the minimax property of the TS, so it is actually easier to test for the saddle point by using this assumption rather than using the discriminant itself. There are however saddle points that aren't minimax points, then <math>f_{xx}f_{yy} > 0</math> and <math>f_{xy}^2 > f_{xx}f_{yy}</math>. These don't look like saddles, but more like terasses.

Unfortunately there are two types of TS theory out there. The one you quoted is the Eyring theory. It is mainly based on thermodynamic arguments, which themselves are statistical (that's why we can talk about equilibria). Do you think this is an appropriate description of the system at hand?

'''Experimental verification of release of vibrational energy''' IR spectroscopy only shows you which vibrational modes are available to the molecule, not which ones are currently occupied. You need some method of monitoring the vibrational energy without exciting the molecule into a vibrationally excited state (which is what IR spectroscopy would do).

'''Grammar'''

You used articles in front of capitalised nouns, such as "The Hammond's postulate". Omit the "the" before such nouns.

'''Introduction'''

In the introduction you state that every successful reaction has to go over the TS, but later on you show the opposite. It would be good to indicate in the introduction that you will contradict this widely accepted fact and later on, when you actually disprove it, remind people of the original statement in the introduction. This emphasize your input in the work and your critical evaluation of the literature.
You could have also disproven this point directly in the introduction. The TS (as a mathematical construct) is a 0-dimensional point and hitting it would require an atomic geometry with infinite precision, which is not possible.

'''Results'''

The graph for the endothermic direction of the F-H-H reaction looks almost like a step function. This is an unanticipated result to me and I would have liked to see a comment from you. When calculating the TS energy, you do actually know what the energy of the TS is. The potential energy of the atomic configurations is independent of how you got to them, so the potential energy in the perfectly symmetrical TS calculation can be used, even though it would never go on to react. You may also be able to get a better estimate of the reactant potential energy by fitting the potential energy curve with an analytical function and then taking the limit of time approaching infinity.--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:09, 11 May 2017 (BST)

Talk:MRD:fp1615 molecular reaction dynamics

2017-05-11T14:36:03Z

Bg1512:

'''Feedback '''

Excellent work! Basically all the questions were answered correctly. Your writing style is on point, the structure is clear and I could follow your argumentation throughout. Here are a couple of pointers on how to improve further:

'''Answers to questions'''

'''Properties of the TS''' I see you invested considerable effort in this answer, but unfortunately it is not quite right. I think you switched the < and > around in your equations. Let me explain: A TS is a maximum in one direction and a minimum in the other. So you're right in saying the partial derivatives are both 0. Let's assume x is the direction along which we find a maximum and the minimum is along y. Then <math>f_{xx} < 0</math> & <math>f_{yy} > 0</math>. You know this from 1D calculus: If you find an extreme point that's curved negatively, the points on either side of the extreme must have lower functional values and you have a maximum (and the other way round for minima). This gives us <math>f_{xx}f_{yy} < 0</math>. Now <math>f_{xy}^2 > 0</math> since it is a square and <math>f_{xx}f_{yy} - f_{xy}^2 < 0</math> for a saddle point. You may have noticed that this derivation of the discriminant actually assumes the minimax property of the TS, so it is actually easier to test for the saddle point by using this assumption rather than using the discriminant itself. There are however saddle points that aren't minimax points, then <math>f_{xx}f_{yy} > 0</math> and <math>f_{xy}^2 > f_{xx}f_{yy}</math>. These don't look like saddles, but more like terasses.

'''Experimental verification of release of vibrational energy''' IR spectroscopy only shows you which vibrational modes are available to the molecule, not which ones are currently occupied. You need some method of monitoring the vibrational energy without exciting the molecule into a vibrationally excited state (which is what IR spectroscopy would do).

'''Grammar'''

You used articles in front of capitalised nouns, such as "The Hammond's postulate". Omit the "the" before such nouns.

'''Introduction'''

In the introduction you state that every successful reaction has to go over the TS, but later on you show the opposite. It would be good to indicate in the introduction that you will contradict this widely accepted fact and later on, when you actually disprove it, remind people of the original statement in the introduction. This emphasize your input in the work and your critical evaluation of the literature.
You could have also disproven this point directly in the introduction. The TS (as a mathematical construct) is a 0-dimensional point and hitting it would require an atomic geometry with infinite precision, which is not possible.

'''Results'''

The graph for the endothermic direction of the F-H-H reaction looks almost like a step function. This is an unanticipated result to me and I would have liked to see a comment from you. When calculating the TS energy, you do actually know what the energy of the TS is. The potential energy of the atomic configurations is independent of how you got to them, so the potential energy in the perfectly symmetrical TS calculation can be used, even though it would never go on to react. You may also be able to get a better estimate of the reactant potential energy by fitting the potential energy curve with an analytical function and then taking the limit of time approaching infinity.--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:09, 11 May 2017 (BST)

Talk:MRD:fp1615 molecular reaction dynamics

2017-05-11T14:09:40Z

Bg1512: Created page with "'''Feedback ''' Excellent work! Basically all the questions were answered correctly. Your writing style is on point, the structure is clear and I could follow your argumentat..."

MRD:fp1615 molecular reaction dynamics

2017-05-11T14:08:11Z

Bg1512:

== Introduction ==
In this investigation, the software MatLab was used to calculate molecular dynamics trajectories in order to analyze the reactivity of triatomic systems in which an atom and a diatomic molecule collide. Reactions were simulated by describing the relative motion of the atoms, which can be approximated using the laws of classical mechanics as their mass is relatively large when compared to electrons.

[[File:fp1615_schematic1.PNG|thumb|centre|'''Figure 1.''' Schematic of the triatomic system analysed in this report.[1] Before the collision r2 is approximately constant (it only oscillates slightly due to vibrational energy); r1 decreases steadily as m1 approaches m2 leading to the collision. After the reaction, r1 is now constant and r2 instead increases steadily as m3 moves away from the diatomic molecule that has just formed. |600px]]

Interatomic interactions, which affect the relative motion of the atoms and depend on their relative positions, can be expressed as a ''potential energy surface V(r1,r2...), ''a plot of the potential energy as a function of the relative positions coordinates ri, of all the atoms taking part in the reaction.[1] In an atom-diatom collision A + BC, the potential energy is a function of the three interatomic distances (or two distances and the angle between them or one distance and two angles). In order to simplify the analysis, the three atoms are constrained to be collinear (A-B-C) so that there are only two independent interatomic distances (AB and BC), with AC being equal the sum of the others.[2] The potential energy surfaces that are going to be examined are a plot of the potential as a function of these two coordinates and can be visualized in 3-dimensions using the software MatLab. The actual paths of the atoms during the encounter depends on the sum of kinetic and potential energy, but it is possible to obtain an idea of the paths available to the system by identifying paths that correspond to the least potential energy: the minimum energy path that leads from the reactants to the products is a trajectory defined as the ''reaction path ''or ''MEP''.[2] On the other hand, a more realistic description of a reactive collision is the ''dynamics reaction path, ''which also takes into account atomic motion.[2]

=== The Transition State (TS) ===
The transition state is defined as the highest point along the minimal energy path, the saddle point: all the reactive trajectories must pass through it, while the ones that do not go directly over the saddle point do not result in a reaction.[2] This happens because the energy increases if at the saddle point we deviate sideways from the reaction coordinate, so all the trajectories that do not pass through it require a significantly higher total energy.[3]

At the transition structure, the value of the gradient of the potential energy surface is equal to zero, meaning that all the partial derivatives are equal to zero (∂V('''ri''')/∂'''ri'''=ƒri=0). However, also at a minimum the total gradient of the potential energy surface and thus all the partial derivatives are equal to zero. A minimum can be distinguished from a saddle point using the ''second derivative test discriminant'': if ƒr1r1ƒr2r2 - ƒ2r1r2>0 at (a,b) then (a,b) is a saddle point (transition state), whereas if ƒr1r1ƒr2r2 - ƒ2r1r2<0 at (a,b) then (a,b) is either a maximum or a minimum.[4] In particular, a minimum is characterized by the fact that the second partial derivatives with respect to both coordinates are positive (ƒr1r1>0 and ƒr2r2>0) at (a,b).[4]

[[File:fp1615_potentialenergy.PNG|thumb|centre|'''Figure 2.''' Potential energy surface for the A + BC --> AB + C reaction in the collinear configuration. The dashed line represents the reaction coordinate and X is the saddle point of the plot, which is the highest point along the minimal energy path and corresponds to the transition state.[2] |500px]]

== H - H - H System ==
The first system that is going to be analyzed is composed of three hydrogen atoms and the collision being studied is the one between a single H atom and a H2 molecule. It is important to state that for a given potential surface the positions r1(t),r2(t) and the momenta p1(t), p2(t) at time t are determined by the conditions at t=0 r1(0),r2(0) and p1(0), p2(0). Furthermore, a negative value of p corresponds to a velocity that decreases the interatomic distance.

This investigation will therefore examine the following collinear collision:'''H + H2 → H2 + H'''

<nowiki> </nowiki>[A=Hydrogen, B=Hydrogen, C=Hydrogen so the above reaction can also be written as BC + A → AB + C.]

[[File:fp1615_schematic2.PNG|thumb|centre|'''Figure 3.''' A schematic of the triatomic system being studied.[1] Before the collision, r1=rBC=0.74 is the H-H bond length and r2=rAB. After the collision a new bond is formed r2=rBC=0.74.  |600px]]

=== Locating the Transition State in the H-H-H System ===
The potential energy surface of H + H2 is symmetric, so the transition state must have r1=r2: following this, the transition state was located by testing different initial conditions with rAB=rBC and pAB=pBC=0. The force acting on a given interatomic coordinate ri is the variation of momentum with time (dp/dt) and depends on the derivative of the potential energy surface with respect to that coordinate (dp/dt=-∂V('''rAB,rBC''')/∂'''ri'''): as explained above, at transition state the total gradient of the potential energy surface has to be equal to zero, so it follows that also the change in momentum has to be equal to zero.[1] Hence the best estimate of the transition state position is '''rts=0.908 Å''', which was found by inspecting the ''internuclear distance vs time ''plot (Figure 4) for the different initial conditions tested: the position chosen was the one in which the internuclear distances appear as a straight line, indicating that the atoms have no momentum and are thus still; if the system is not perturbed the atoms will not move closer together nor will they move away from each other.

[[File:fp1615_ts3d.PNG|thumb|centre|'''Figure 4.'''Three-dimensional view of the potential energy surface of the H-H-H system. The point indicated corresponds to the transition state located.|600px]]

[[File:fp1615_tscontour.PNG|thumb|centre|'''Figure 5.'''Potential energy contour plot of the H-H-H system with its transition state.|600px]]
[[File:fp1615_tsestimate.PNG|thumb|centre|'''Figure 6.''' Internuclear Distance vs Time plot for rAB(0)=rBC(0)=0.908 and pAB(0)=pBC(0)=0.|500px]]

=== Trajectories from rBC = rts+δ, rAB = rts ===
Once the transition state has been located, it is possible to analyze what happens when the system is slightly displaced from it. In order to do so, it is necessary to set the initial conditions as rBC = rts+0.01Å, r2AB = rts, which are rBC = 0.918 Å, rAB = 0.908 Å based on the previous findings. In addition to the ''dynamics energy path'', also the ''minimum energy path (MEP), ''a trajectory in which the velocity always resets to zero at each step and that thus corresponds to infinitely slow motion, will be calculated. Although the ''MEP'' can be useful to characterize a chemical reaction, it does not describe a realistic motion because the atoms being studied are gaseous and possess a mass, so their motion is inertial.

[[File:fp1615_DYNAMICShydrogens1.PNG|thumb|centre|'''Figure 7.''' Potential energy surface plot of the H-H-H system with the trajectory calculated with the ''dynamics'' settings at initial interatomic distance of the diatomic slightly displaced from the transition state.|600px]]

[[File:fp1615_MEPhydrogens2.PNG|thumb|centre|'''Figure 8.''' Potential energy surface plot of the H-H-H system with the trajectory calculated with the ''MEP'' settings at initial interatomic distance of the diatomic slightly displaced from the transition state.|600px]]

[[File:fp1615_kinetic1.PNG|thumb|centre|'''Figure 9.''' ''Kinetic energy vs time ''plot for the ''dynamics reaction path.''|600px]]
[[File:fp1615_kinetic2b.PNG|thumb|centre|'''Figure 10. '''''Kinetic energy vs time ''plot for the ''MEP reaction path.''|600px]]

In Figures 7 and 8, it is possible to observe both the trajectory calculated according to ''dynamics'' and the one that corresponds to the ''minimum energy path ''(note that the number of steps has been increased to obtain a complete trajectory)''. ''In both cases, the trajectories follow the valley floor to the products H and H2 (AB + C); however, the latter differs from the first one because it is a straight line without any oscillation. As mentioned before, the ''MEP ''trajectory does not provide a realistic account as it does not take into account molecular and atomic motion: the ''kinetic energy vs time'' plot corresponding to the ''MEP'' is in fact a straight line at Ek=0 indicating that the atoms do not move (Figure 10). The wavy black line in the other calculated trajectory is in fact given by molecular and atomic vibrations and it is a more realistic account of what actually happens in a real collision, in which de destabilisation of the system due to the motion of the atoms plays a major role. An analysis of the ''kinetic energy vs time ''plot for the ''dynamics reaction path'' reveals that as the system is displaced from its transition state the kinetic energy increases quickly to then transform into vibrational energy of the new diatomic and reach a constant value around which it keeps oscillating (Figure 9)

[[File:fp1615_momentahydrogens1.PNG|thumb|centre|'''Figure 11.''' ''Momenta vs time ''plot for the H + H2 collision calculated with the dynamics setting at at initial rBC interatomic distance of the diatomic slightly displaced from the transition state.|600px]]

[[File:fp1615_internucleardistancehydrogens1.PNG|thumb|centre|'''Figure 12.''' ''Internuclear distance vs time ''plot for the H + H2 collision calculated with the dynamics setting at initial rBC interatomic distance of the diatomic slightly displaced from the transition state.|600px]]

At t=1 the collision has occurred and the reaction has taken place. From the graphs above, it is possible to observe that at this time rAB=0.74 Å, the H2 bond distance, while the atom C has moved away from the system and rBC=2.5 Å. Due to the fact that at large t atom C is moving away from the product AB at constant velocity, also pBC=2.48 remains constant and is positive indicating that the interatomic distance is increasing. On the other hand, pAB oscillates between 1.55 and 0.90 with a mean value of 1.1225: the oscillation observed is caused by the molecular vibrations of the newly formed diatomic molecule AB.

If the above calculated final positions are set to be the initial positions and the values of momenta are inverted, the trajectory goes back to transition state and, if the system is provided with slightly more energy, it then overcomes the barrier and proceeds to the products. In fact, as the valley is discended the potential energy is converted to kinetic energy, and if this kinetic energy is supplied to the system in the opposite direction, the system then follows a trajectory that goes up to the transition state gaining again potential energy.

[[File:fp1615_invertedfinalconditions2.PNG|thumb|centre|'''Figure 13. '''Trajectory calculated by taking the final positions of the reaction above calculated in Figure 8 and inverting the final momenta calculated in Figure 7. The energy supplied to the system is just enough to reach the transition state. |600px]]

[[File:fp1615_invertedfinalconditions.PNG|thumb|centre|'''Figure 14. '''Trajectory calculated by taking the final positions of the reaction above calculated in Figure 8 and using as initial momenta the inverse of a value slightly bigger than the one calculated in Figure 7. In this case, the energy supplied to the system is enough to reach the transition state and proceed to the products. |600px]]

If ''dynamics'' and ''MEP'' trajectories are simulated analogously as before, but rAB is displaced from equilibrium (rAB=rTS+0.01 Å) instead of rBC (always keeping the initial momentum equal to zero), the trajectories will behave in the exact opposite way: the system will go back to the reactants (A+ BC) instead of proceeding to the products as in the previous case.

[[File:fp1615_DYNAMICShydrogensinverted.PNG|thumb|centre|'''Figure 15.'''Potential energy surface plot of the H-H-H system with the trajectory calculated with the ''dynamics'' settings at initial rAB slightly displaced from the transition state.|600px]]

[[File:fp1615_MEPhydrogensinverted.PNG|thumb|centre|'''Figure 16.''' Potential energy surface plot of the H-H-H system with the trajectory calculated with the ''MEP'' settings at initial rAB slightly displaced from the transition state.|600px]]

[[File:fp1615_momentahydrogens2.PNG|thumb|centre|'''Figure 17.''' ''Momenta vs time ''plot for the H + H2 collision calculated with the dynamics setting at initial rAB slightly displaced from the transition state.|600px]]

[[File:fp1615_internucleardistancehydrogens2.PNG|thumb|centre|'''Figure 18. '''''Internuclear distance vs time ''plot for the H + H2 collision calculated with the dynamics setting at initial rAB slightly displaced from the transition state.|600px]]

=== Reaction Dynamics: Reactivity of Different Trajectories ===
The shape of the potential energy surface can be investigated experimentally by changing the relative speed of approach of the molecules and the degree of vibrational excitation and observing whether a reaction occurs and the products are in a vibrationally excited state.[3] For the initial positions rBC=0.74 and rAC=2.0, trajectories were calculated with a set of momenta combinations to analyse how these affect the reactivity of the collisions.

'''Table 1.''' Different initial pBC and pAB at rBC=0.74 and rAB=2.0 to analyse how reactivity varies with initial momenta at fixed initial distances.

{| class="wikitable" border="1"
!
!pBC!!pAB!!Reactivity!!Description of the Collision
|-
|a
|<nowiki>-1.25</nowiki>||-2.5||Reactive||A approaches the diatomic BC resulting in the formation of the AB bond and in C moving away from the newly formed diatomic.
|-
|b
|<nowiki>-1.5</nowiki>|| -2.0 ||Unreactive||A approaches the diatomic BC, but before the collision its momentum changes sign and A moves away from the diatomic.
|-
|c
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive||A approaches the diatomic BC resulting in the formation of the AB bond and in C moving away from the newly formed diatomic.
|-
|d
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive||A approaches very quickly the diatomic BC and seems to have formed an AB bond with C moving away, but then C comes back and the system returns to the initial A+BC configuration.
|-
|e
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive||A approaches the diatomic BC and a symmetric ABC vibration follows for a while before C starts to move away from the newly formed bond AB.
|}

[[File:fp1615_hydrogens_reactivity1.PNG|thumb|centre|'''Figure 19.''' Conditions ''(a)'': the trajectory passes through the transition state and then follows the valley floor to H + H-H (C + AB).|600px]][[File:fp1615_hydrogens_reactivity2.PNG|thumb|centre|'''Figure 20.''' Conditions ''(b)'': the system does not have enough energy to climb to the saddle point, so at some point the momentum goes to zero and inverts its sign so that the system goes back down the valley to the reactants. |600px]]
[[File:fp1615_hydrogens_reactivity3.PNG|thumb|centre|'''Figure 21.''' Conditions ''(c)'': the trajectory passes through the transition state and the follows the valley floor to H + H-H (C + AB). Compared to ''condition (a)'', the H-H bond of the reactant molecule possesses more vibrational energy and thus the product vibrates more too.|600px]]
[[File:fp1615_hydrogens_reactivity4.PNG|thumb|centre|'''Figure 22.''' Conditions ''(d)'': the system crosses the transition state and the AB product bond forms for a while, but then the system reverts back to its reactants A + BC (''barrier recrossing'').|600px]]
[[File:fp1615_hydrogens_reactivity5.PNG|thumb|centre|'''Figure 23.''' Conditions ''(e)'': the system is provided with a great amount of energy that causes various vibrations around the transition state before descending the potential valley to the products.|600px]]The above experimental data show that the assumption that if the system is provided with high enough overall energy than the collision will be reactive is not valid. In fact, reactivity of a certain trajectory depends on the ratio of the momenta provided at initial conditions and not on their overall value.

==== Transition State Theory ====
''Transition state theory (TST) '' describes the reaction rates of elementary chemical reactions, assuming that there is a particular ''quasi-equilibrium ''between the reactants and the activated complex at transition state.[5] This theory is based on the assumption that atomic nuclei behave according to classical mechanics and that atomic motion on the potential energy hypersurface can be accurately described by classical mechanics laws[5]. Furthermore, ''TST'' assumes that at transition state the reaction coordinate degree of freedom can be separated from the rest and that there is no dynamic coupling between it and all the other degrees of freedom.[6] These assumptions are however a limitation of this theory that, especially in the case of a light particle such as a hydrogen atom, does not account for a realistic description of the reaction dynamics.

According to ''TST'', any trajectory that reaches the transition state with infinitesimal kinetic energy in the direction of the products will go down the valley and proceed to the products.[6] This theory therefore predicts that when the barrier has been passed it is not possible to go back to the reactants side: however, experimental evidence (condition d in the above section) suggests that this prediction is wrong and that barrier recrossing can happen.[6]

The ''transition state theory'' also assumes that, in order to travel successfully from reactants to products, a molecule must pass through the exact transition state of the potential surface.[6] By looking at the reactive trajectories calculated in the previous section, it is possible to notice that none of them passes through the exact saddle point and that they all pass through a variety of locations around it, with no single location being dominant. A statistical perspective allows to understand that is very unlikely for a system to have the exact same energy as the activation energy, but is far more likely that the trajectory is slightly displaced resulting in the system to reach an energy higher than the value needed for the reaction to proceed.

In addition, ''TST'' does not take into account the fact that, according to quantum mechanics, particles can tunnel across barriers of finite energies so that there is the possibility that molecules react even if they do not collide with enough energy to climb the saddle point of the potential surface and cross the barrier (note that tunnelling increases with decreasing barrier height).[6]

In conclusion, ''TST'' could be used to gain a qualitative understanding of the system, but its limitations result in the fact that there are many conditions at which its predictions are not valid as shown in this investigation.

== F - H - H System ==

'''H2 + F → HF + H'''

Fluorine=A, Hydrogen=B, Hydrogen=C so the reaction can also be written as BC + A → AB + C.

H2 + F → HF + H is an exothermic reaction and its inverse is instead an endothermic reaction, indicating that the H-F bond is stronger than the H-H bond because more energy is released when the H-F bond (product) is formed than it takes to break H2 (reactant). It is a very exoergic reaction, so the barrier occurs at an early stage and is very small; immediately after the barrier the potential energy instead drops rather steeply: since the exoergicity is released early along the reaction path, it is readily transformed into the vibrational motion of the newly formed bond: this is a typical example of an ''attractive potential energy surface''.[2][3] Considering this a reversible reaction, it follows that the opposite reactionis characterised by a ''repulsive potential energy surface''.[2]

=== Locating the Transition State in the F-H-H System ===
The Hammond's postulate states that in an exothermic reaction, the transition state is closer in energy to the reactants than to the products: knowing that F + H2 is an exothermic reaction, this postulate is used to locate the transition state.

The potential energy surface of this system is not symmetric as the previously examined one was, so at transition state rBC≠rAB. Knowing the approximate position of the saddle point (near the reactants), the exact transition state position can be found following the same procedure as in the H-H-H system, which consisted in finding the initial conditions that would result in the ''internuclear distance vs time'' plot being a straight line (momentum equal to 0). In this way, it was found that the approximate transition state of the system is at '''rAB= 1.812 Å and rBC= 0.745 Å.'''
[[File:fp1615_tsfhh1.PNG|thumb|centre|'''Figure 24.'''Position of the transition state (saddle point) on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_tsfhh.PNG|thumb|centre|'''Figure 25.''' Internuclear Distance vs Time plot for rAB(0)=1.812 Å and rBC(0)=0.745 Å and pAB(0)=pBC(0)=0.|600px]]

=== Activation Energy ===
An inspection of the potential surface energy of the system also allows the determination of the activation energies for the reaction and its inverse. ''MEP'' trajectories at zero initial momentum were calculated slightly displacing the transition state rAB first in the direction of the reactants and then in the direction of products. Using these ''mimimum energy paths'' the most precise estimate of the activation energy possible was obtained by plotting a ''potential energy vs time'' graph.[[File:fp1615_ACTIVATIONFINAL2.PNG|thumb|centre|'''Figure 26.'''Potential energy surface plot of the F-H-H system with the trajectory calculated with the ''MEP'' settings at initial positions slightly displaced from the transition state. In this case the trajectory goes down the valley to the reactants H2 + F, so the activation energy that is calculated from this plot is the one of the exothermic reaction. |600px]]

[[File:fp1615_ACTIVATIONFINAL1.PNG|thumb|centre|'''Figure 27.'''''Potential energy vs time'' plot for the formation of the reactants (H2 + F) starting from initial conditions slightly displaced from transition state.|600px]]In the above pictures rAB was slightly displaced from transition state towards the reactants, so as it can be observed the trajectory starts from around the TS and goes down the valley to the reactants (H2 + F). At t=0 the potential energy is approximately equal to the potential energy at transition state (the displacement is very minor), then it goes down as the trajectory goes down the valley in the potential energy surface and it finally reaches a constant value, which is the potential energy of the reactants. The energy of the transition state is very close to the one of the reactants, so the gradient is very small and it would take infinite steps in a MEP simulation to actually reach a constant value. However, the potential energy value at t=1500 can be considered a good estimate. The difference in these two potential energies is the difference in energy between the reactants and the transition state, which is defined as the '''activation energy of the exothermic reaction H2 + F → HF + H''' and is equal to '''0.244 kcal mol-1'''.[[File:fp1615_ACTIVATIONFINAL3.PNG|thumb|centre|'''Figure 28.'''Potential energy surface plot of the F-H-H system with the trajectory calculated with the ''MEP'' settings at initial positions slightly displaced from the transition state. In this case the trajectory goes down the valley to the products, so the activation energy that is calculated from this plot is the one of the endothermic reaction.|600px]]

[[File:fp1615_ACTIVATIONFINAL4.PNG|thumb|centre|'''Figure 29.''' ''Potential energy vs time'' plot for the formation of the reactants (H2 + F) starting from initial conditions slightly displaced from transition state. |600px]]In a similar way, the above pictures show the ''MEP'' trajectory calculated when the transition state rAB is slightly displaced towards the products. It can be observed that the system is displaced from transition state to the products HF + H, so the difference in potential energy calculated from the ''potential energy vs time'' plot is defined as the '''activation energy for the endothermic reaction HF + H → H2 + F''' and is equal to '''30.2 kcal mol-1'''. It is possible to notice that the change in potential energy is much steeper than the one from TS to reactants due to the fact that there is much more difference between the energy of the transition state and the energy of the products, which are far more stable than the reactants, resulting in the valley of the potential energy surface going down much faster and in the constant potential energy value for the products being reached much sooner.

The values calculated are approximations because the maximum potential energy is not taken as the exact transition state, but is taken from positions slightly displaced from it. At the same time, for the calculation of the activation energy of the exothermic reaction, the potential energy value of the reactants is just an estimate. The values calculated can however, be considered accurate to a good approximation and allow to gain an understanding of the energetics of the reaction.

=== Reaction Dynamics: Reactivity of Different Trajectories ===
As previously observed in this report, it is wrong to assume that higher values of momenta necessarily correspond to the calculated trajectory to be reactive. In this section, a set of initial conditions that results in a reactive trajectory was identified and can be observed in the following figures: rAB= 2.5 Å, rBC= 0.74 Å (H2 bond length), pAB=-1.63, pBC=-0.55. For the H-H-H system a mathematical approach to the calculation of the initial momenta that lead to a reactive collision is more straightforward because all the three atoms possess the same mass. However, F has a mass that is around 19 times bigger than the one of hydrogen and this has to be taken into account. The ''internuclear momenta vs time'' plot in Figure 31 shows that before the reaction the diatomic BC possesses vibrational energy and the atom A possesses translational energy as it approaches the diatomic. After the collision the reaction energy is released as translational energy for the atom C that moves away from the system and vibrational energy for the AB molecule formed in the reaction, which is higher than the initial BC vibrational energy. In fact, the momenta pAC at large t oscillates around a constant value illustrating that the newly formed molecule AB is vibrating. This could be confirmed experimentally by vibrational spectroscopy (i.e. IR analysis).[[File:fp1615_reactiveconditions1.PNG|thumb|centre|'''Figure 30.''' An example of a reactive trajectory for the reaction H2 + F --> HF + H on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_reactiveconditions2.PNG|thumb|centre|'''Figure 31.''' ''Internuclear momenta vs time ''plot of the reactive trajectory illustrated above.|600px]]In order to further explore reactive trajectories in this system, different initial values of pBC ranging from -3 to 3 were tested for rBC=0.74 Å, rAB=2.5 Å and pAB=-0.5. At these conditions, much more energy than the activation energy is supplied to the system on the H2 vibration.''' '''As it is possible to observe in the Figures 32-44, the trajectories are reactive only for large values of pAB such as -3, 2.5 and 3. The larger the momentum, the more the initial diatomic molecule H2 vibrates. As it can be deduced from the following experimental data, if these vibrations are strong enough they eventually perturb the motion of the fluorine atom to an extent that it results in the collision and thus the reaction to occur.

[[File:fp1615_momentum-3.PNG|thumb|centre|'''Figure 32.''' Reactive trajectory calculated with initial pBC=-3, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]
[[File:fp1615_momentum-25.PNG|thumb|centre|'''Figure 33.''' Unreactive trajectory calculated with initial pBC=-2.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum-2.PNG|thumb|centre|'''Figure 34.''' Unreactive trajectory calculated with initial pBC=-2, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum-15.PNG|thumb|centre|'''Figure 35.''' Unreactive trajectory calculated with initial pBC=-1.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum-1.PNG|thumb|centre|'''Figure 36.''' Unreactive trajectory calculated with initial pBC=-1, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum-05.PNG|thumb|centre|'''Figure 37.''' Unreactive trajectory calculated with initial pBC=-0.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum0.PNG|thumb|centre|'''Figure 38.''' Unreactive trajectory calculated with initial pBC=0, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum05.PNG|thumb|centre|'''Figure 39.''' Unreactive trajectory calculated with initial pBC=0.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum1.PNG|thumb|centre|'''Figure 40.''' Unreactive trajectory calculated with initial pBC=1, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum15.PNG|thumb|centre|'''Figure 41.''' Unreactive trajectory calculated with initial pBC=1.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum2.PNG|thumb|centre|'''Figure 42.''' Unreactive trajectory calculated with initial pBC=2, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum25.PNG|thumb|centre|'''Figure 43.''' Reactive trajectory calculated with initial pBC=2.5, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

[[File:fp1615_momentum3.PNG|thumb|centre|'''Figure 44.''' Reactive trajectory calculated with initial pBC=3, pAB=-0.5, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]For the same initial positions, the initial momentum pAB was increased to -0.8 and the overall energy of the system was considerably reduced by reducing pBC to 0.1. The diatomic H2 vibrates much less, but the increased momentum pAB allows the fluorine atom to approach the diatomic with more energy eventually resulting in a reactive collision. This shows how what matters in reaction dynamics is not only the overall energy of the system, but also how it is distributed among the particles that are part of it. In fact, in the previous calculations many conditions that resulted in an overall bigger energy of the system generated unreactive trajectories, while this collision is reactive with a much reduced overall energy. [[File:fp1615_momenta0801.PNG|thumb|centre|'''Figure 45. '''Reactive trajectory calculated with initial pBC=0.1, pAB=-0.8, rBC=0.74 Å and rAB=2.5 Å on the potential energy surface of the F-H-H system.|600px]]

=== Polanyi's Empirical Rules ===
The Polanyi's empirical rules state that vibrational energy is more efficient in promoting a late-barrier reaction than translational energy and that at the same time the latter is more efficient in promoting an early-barrier reaction.[7] 

<nowiki> </nowiki>We can think about the momentum of the colliding atom to be translational energy and the momentum of the diatomic to result in molecular vibrations. It follows that the trajectories for the forward reaction (early-transition state) illustrated in Figures 32-45 seem to agree with the prediction: in Figure 45, only a very small vibrational energy was necessary for the collision to be reactive because there was enough transitional energy (larger pAB), which is very efficient in promoting an early-barrier reaction. At the same time, in Figures 32-44 a much bigger vibrational energy was required in the reactants to proceed to the products because the translational energy supplied was lower. Therefore it is the translational energy that dominates the reactivity of an early-barrier reaction: in fact, small variations in it affect the reactivity much more than bigger variations in vibrational energy.

An analysis of the backward endothermic reaction (late-barrier) can further confirm this theory: vibrations will in fact be more efficient than translation in promoting this reaction. Figure 46 shows how a reaction with very low initial vibrational energy and relatively high initial transitional energy will not go on to react.
[[File:fp1615_endo4.PNG|thumb|centre|'''Figure 46.''' Unreactive trajectory for the endothermic reaction with arbitrarily high initial translation energy
and very low initial vibrational energy.|600px]]

At the same time, Figure 47 illustrates a reactive trajectory of a system with high initial vibrational energy and low initial translational energy, confirming the rule's prediction. However, the initial momenta combinations that result in a reactive trajectory are not as simple as having an high vibrational energy and a low transitional energy, but are very few and hard to identify. More experimental data would be needed to further explore different combinations of initial momenta and confirm this rule. It is however safe to say that providing the system with energy higher than the activation energy is not enough to obtain reactive collisions, but relative initial momenta values are crucial in determining their reactivity.
[[File:fp1615_endo2.PNG|thumb|centre|'''Figure 47.''' Reactive trajectory for the endothermic reaction with low initial translation energy
and very high initial vibrational energy.|600px]]
== References ==
[1] Imperial College London, ''Molecular Reaction Dynamics: Applications to Triatomic Systems ''(Lab Script May 2017).

[2] R. D. Levine, ''Molecular Reaction Dynamics'', Cambridge University Press, UK, 1st edition, 2005, ch. 5, pp 148-175.

[3] P. Atkins and J. de Paula, ''Atkins' Physical Chemistry'', Oxford University Press, UK, 10th edition, 2014, ch. 21, pp 879-936.

[4] G. B. Thomas and R. L. Finney, ''Calculus and Analytic Geometry, ''Addison-Wesley,'' ''USA,'' ''8th edition,1992, pp. 881-891.

[5] H. Eyring, ''J. Chem. Phys''., '''1935''', 361(2), 107–115.

[6] R. Masel, ''Principles of Adsorption and Reactions on Solid Surfaces'', Wiley, New York, 1996.

[7] Z. Zhang, Y. Zhou and D. H. Zhang ,''J. Phys. Chem. Lett.'', '''2012''', 3 (23), 3416–3419.

Talk:MRD:mhz15

2017-05-11T14:06:36Z

Bg1512:

'''Feedback'''

Good work. Here's some feedback to improve further:

'''Missing answers'''

Unfortunately I couldn't identify any answers concerning the properties of TSs and the critique of TS theory, Polanyi's rules & experimental evidence for the formation of vibrationally excited HF molecules.

'''Presentation of results'''

All your results must be backed up by pictures of your simulations. You are trying to convince me that you have done an experiment and come up with a novel idea! You can only do this on the basis of facts. For example I would have liked to see pictures for how you have calculated the activation energies. Also make sure that you include the left half of the MATLAB window in your screenshots so I can actually see your settings.

'''Activation energies'''

I'm not sure how you found the TS in the HHF reaction, you just show the picture of the finished "product", ''i.e.'' the proof that you've found the TS, not the method. When you show the TS in the surface plot you actually use different coordinates than the ones you reported just three sentences earlier. The geometry in your PES (and I assume also the one you used for the activation energy calculations) is not the TS, which means that your activation energy is a lot lower than it should be. A comparison to literature might have given you a hint that something's awry!

'''FHH Trajectories'''

You stated that your trajectories originated from a minimum. If you look at the endpoints of your trajectories you can see that none of them are located in a well. If started at a minimum the well should have been dark blue at the starting point.

'''Figures'''

Please number your figures so you can refer to them in your text. If you're addressing data in your text (which will happen quite often) you need to make sure the reader knows which graph you're talking about.--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:06, 11 May 2017 (BST)

Talk:MRD:sk4915

2017-05-11T14:05:41Z

Bg1512: Created page with "'''Feedback''' Thank you for the effort! I assume you ran out of time when writing up as a number of answers are missing, but please have a look at this feedback: '''Exother..."

'''Feedback'''

Thank you for the effort! I assume you ran out of time when writing up as a number of answers are missing, but please have a look at this feedback:

'''Exothermicity'''
Please read this paragraph carefully, as I fear you have a significant misunderstanding about thermodynamics here:
"The reaction F + H2 is exothermic which means that some of the kinetic energy used to cause the reaction is stored in the newly formed F-H bond, which is much stronger than the previous H-H bond." In an exothermic reaction energy is released from the molecule to the surroundings (''i.e.'' the expelled atom). Thus, the energy of the F-H bond must be lower than that of the H-H bond. In other words, the energy of the H-H bond is used to form the F-H bond and '''additionally''' give off heat. A strong bond requires a lot of energy to break up; that means it has a very low energy (very large negative number). A weak bond on the other hand requires little energy to break, having an energy more close to zero (small negative number). When a weak bond breaks and a strong one forms instead, less energy is stored in the strong bond than in the weak one and the excess is given off as energy to the environment. If you have a trouble with this concept, feel free to contact me to discuss!

'''Transition State properties'''

This is a 2D problem, meaning that there is not only one gradient, but two! A Transition State is located at a saddle point, which is an extreme point with respect to both axes. So while it is true that both gradients are zero, this also holds for 2D-maxima and 2D-minima. To distinguish the three from one another, consider their second derivatives. The alternative name "minimax" for a saddle point might be a good hint!

'''HHH trajectories'''

In the fourth, unsuccessful, trajectory you claim that the lone atom would not possess enough energy. Trajectory 4 is however initialised with more energy than trajectory 3, which itself is successful. Therefore, the sum of energies cannot be the cause for reactivity by itself. Can you come up with a different explanation? Consider running some more experiments to get more data.

'''TS theory'''

While it is true that QM is neglected in the TS theory, there are many more shortcomings of the theory. One is connected to trajectory 4 above. If you look closely at the PES, you can see that geometry crosses over into the product territory, but then reverts back. What could cause this?

'''FHH system'''

Please include figures with all your answers. Any scientific claim you make has to be backed up by data! You need to convince your reader that you're telling the truth and the only way to do so is to present the raw data you have collected in your experiments! For example your reported TS geometry is significantly off. I as a reader have two options: Either I believe you, necessarily doubting established results, or I reject your claim. If you really have something worth doubting prior knowledge, you need to support my shift in perspective with even better new knowledge!--[[User:Bg1512|Bg1512]] ([[User talk:Bg1512|talk]]) 15:05, 11 May 2017 (BST)

2015-02-01T18:03:26Z

Bg1512:

=2D Ising Model - Third year CMP experiment=
''by Bjorn Gugu; Experiment conducted from 19/01/2015 - 30/01/2015''

==Abstract==

This report discusses a computational approach to the calculation of the heat capacity of the ferromagnetic phase transition in a 2D Ising Model. A Monte Carlo simulation was performed to calculate the average spin and magnetisation of various periodic lattices over temperature. The heat capacities were obtained and the critical temperature of the phase transition was found in relation to the lattice size. The answer of this simulation, <math>T_{C} = 2.267K</math> shows strong agreement with the literature value of 2.269K<ref name="lit value1">Preis, Tobias, et al. "GPU accelerated Monte Carlo simulation of the 2D and 3D Ising model." Journal of Computational Physics 228.12 (2009): 4468-4477.</ref>.

==Methods==

All code used in this experiment was written in Python. References values were provided by an external simulation written in C++. The simulation is based on a Monte Carlo algorithm for importance sampling of thermodynamic states.

==Discussion==

===The Ising Model===

'''Show that the lowest possible energy for the Ising model is <math>E = -DNJ</math>, where D is the number of dimensions and N is the total number of spins. What is the multiplicity of this state? Calculate its entropy.'''

The total energy of a system of D dimensions and N atoms is given by <math>-\frac{J}{2}\sum_{i}^{N}\sum_{j\in adj}{s_i s_j}</math>, where J is the strength of an individual interaction. The lowest energy state is expected to be the one in which all spins are parallel. Therefore, <math>s_i s_j = 1</math> for all interactions. A linear system assigns two direct neighbours to each atom. As the dimensions are independent of one another, every degree of freedom adds another 2 neighbours to each atom. Therefore, the amount of direct neighbours of each atom is <math>2D</math> and <math>\sum_{j \in adj}{s_i s_j} = 2D</math>. This is the number of interactions per atom. In order to evaluate the total interactions, simply the sum overall atoms must be taken. The atomic interactions are constant, as previously established, therefore <math>\sum_i^N\sum_{j \in adj} s_i s_j = 2ND</math> and <math>E = -NJD</math>.

There is only one possible arrangement of spins that yields this configuration, so the number of microstates is 1. The entropy of such a system is given by <math>S = K_B lnW</math>, where <math>K_B</math> is Boltzmann's constant and W the number of microstates. Given that <math>W = 1</math>, <math>S = 0</math>.

'''Imagine that the system is in the lowest energy configuration. To move to a different state, one of the spins must spontaneously change direction ("flip"). What is the change in energy if this happens (D=3, N=1000)? How much entropy does the system gain by doing so?'''

If one atom flips spin, according to the calculation from Task 1, an atom interacts with 2D other atoms (after accounting for double-counting of interactions). The energy of each interaction is +J. The total energy however changes by 2J per interaction, as the new interaction is not only positive, but also removes a former negative interaction. The total change in energy is then 2DJ. For a system (N = 1000, D = 3) this change amounts to 6J. The number of microstates is N, as each atom is equally likely to flip. The change in entropy is therefore <math>S_1 - S_0 = K_B lnN - 0 = K_B lnN = 9.540*10^{-23}</math>.

'''Calculate the magnetisation of the 1D and 2D lattices in figure 1. What magnetisation would you expect to observe for an Ising lattice with D=3, N=1000 at absolute zero?'''

The total magnetisation of a system is given by <math>M = \sum_{i}^{N} {s_i}</math>. For the three example systems indicated, the total magnetisations are M1 and M2, respectively. <math>M_1 = M_2 = 1</math>, as one more up-spin entity is present than down-spin entities. For a system (D = 3, N = 1000, T = 0K) all spins are expected to align in parallel and the total magnetisation is simply the sum of all atoms: <math>|M| = 1000</math>

===Calculating the energy and magnetisation===

'''Complete the functions energy() and magnetisation(), which should return the energy of the lattice and the total magnetisation, respectively. In the energy() function you may assume that J=1.0 at all times (in fact, we are working in reduced units in which <math>J=k_{B}</math>, but there will be more information about this in later sections). Do not worry about the efficiency of the code at the moment — we will address the speed in a later part of the experiment.'''

In order to calculate the energy of the system, the above equation has to be implemented into code. This can be done by looping over all atoms, calculating the interactions each atom exhibits with its neighbours. Additionally, the periodicity of the lattice has to be accounted for. While the randomly defined lattice in the initialisation of the IsingLattice class is two-dimensional, no direct connection between atoms on the "borders" of the lattice are given. Therefore a new property "periodic_lattice" was defined, which is an ndarray of size(n_rows+2, n_cols+2). The core section of this ndarray is the lattice defined as before, but the first and last rows and columns are copied from the respective lines in the original lattice (''Figure 1''). This pseudo-periodic lattice allows looping over all interactions of the atoms in the core section without the necessity of introducing exception clauses. The magnetisation of the lattice can simply be found by adding the values of each matrix entry, looping over both dimensions.

[[File:Periodic Lattice.jpg|thumb|right|Figure 1: An illustration of the lattice transformation used to emulate periodic boundary conditions.]]

'''Run the ILcheck.py script from the IPython Qt console using the command'''

The values calculated by the above functions were compared to a number of previously calculated systems. It can be seen that the functions above provide exactly correct results (''Figure 2'').

[[File:ILcheck picture.png|thumb|right|Figure 2: A test of the energy() and magnetisation() functions. The calculated values were compared to ab-initio solutions.]]

===Introduction to Monte Monte Carlo simulations===

'''How many configurations are available to a system with 100 spins? To evaluate these expressions, we have to calculate the energy and magnetisation for each of these configurations, then perform the sum. Let's be very, very, generous, and say that we can analyse <math>1\times 10^{9}</math> configurations per second with our computer. How long will it take to evaluate a single value of <math><M>_{T}</math>?'''

The number of energy states given for a system of N particles with n available states is given by <math>W = n^N</math>. In this system, n=2, N=100, therefore <math>W = 2^{100}</math> ≈ <math>1.27 * 10^{30}</math>. Estimating an evaluation of <math>10^9</math> states per second, a calculation of all microstates of the above systems would take approximately 1021 seconds - roughly 1014 years.

'''Implement a single cycle of the above algorithm in the montecarlocycle(T) function. This function should return the energy of your lattice and the magnetisation at the end of the cycle. You may assume that the energy returned by your energy() function is in units of <math>k_{B}</math>! Complete the statistics() function. This should return the following quantities whenever it is called: <math><E></math>,<math><E^{2}></math>,<math><M></math>,<math><M^{2}></math>, and the number of Monte Carlo steps that have elapsed.'''

An implementation of a Monte Carlo simulation essentially requires a method of comparing the probabilities of occurrence of two experimental outcomes. In this case, the ensemble average of the energy and the magnetisation of a system will be calculated. To assess the occurrence of a fluctuation in the system, a random spin flip of one spin, the energy of the lattice before and after the flip is calculated with the energy() function. The flip itself can simply be simulated by changing the sign of the given lattice entry. The algorithm for the acceptance and rejection of states was implemented via two nested if-structures. If a state is accepted, it is added to the list of accepted states, from which the average properties will be calculated via the statistics() function. The exponential factor is not explicitly entered into the average calculation, as it is already implemented in the acceptance-rejection algorithm. Instead of simply adding the newly calculated energies to a running average, lists for all four properties were calculated to be able to relate temperatures to each data point retrospectively.

The statistics() function computes the averages of the four properties energy, energy squared, magnetisation and magnetisation squared from the states stored by the montecarlostep(T) function. The entries in the respective lists are read, summed and divided by the number of Monte Carlo cycles.

''' If <math>T<T_{C}</math>, do you expect a spontaneous magnetisation (i.e. do you expect <math><M> \neq 0</math>)? When the state of the simulation appears to stop changing (when you have reached an equilibrium state), use the controls to export the output to PNG and attach this to your report. You should also include the output from your statistics() function.'''

The Curie temperature TC is the temperature at which entropic contributions balance the energetic stability of well ordered systems. Therefore, at temperatures lower than TC, entropy cannot outweigh the energetic preference for well ordered systems. Therefore, the alignment of spins will be non-random and <M> will be non-zero. At T=0K only the lowest energy state would be occupied. At temperatures higher than absolute zero intermediate states between perfect uniform alignment and completely random alignment may be found. These manifest in the formation of regions of uniform spin that show only few disclinations. The formation of such intermediate states depends heavily on the (randomly chosen) starting conditions.

The ILanim.py function runs the simulation at T=1.0K. The system reaches the predicted equilibrium state of energy/spin = -2 and magnetisation/spin = +1 after approximately 600 Monte Carlo cycles. This corresponds to a fully aligned system. The equilibrium proves very stable, little fluctuation scan be observed (''Figure 3''). This is due to the low temperature at which this simulation was conducted. It is assumed that higher temperatures lead to stronger fluctuations.

[[File:ILanim picture 12.png|thumb|right|Figure 3: An initial Monte Carlo simulation of an 8x8 Lattice at 1.0K.]]

===Accelerating the code===

'''Use the script ILtimetrial.py to record how long your current version of IsingLattice.py takes to perform 2000 Monte Carlo steps. This will vary, depending on what else the computer happens to be doing, so perform repeats and report the error in your average!'''

The ILtimetrial.py function assesses the performance of the montecarlostep(T) function. The time trial was performed 10 times on an otherwise superficially idle computer (no other programs operated by user), yielding an average computational time of 148s (rms 0.7s). Considering that over 100000 Monte Carlo steps will be calculated for multiple lattices this calculation speed is not feasible.

'''Look at the documentation for the NumPy sum function. You should be able to modify your magnetisation() function so that it uses this to evaluate M. The energy is a little trickier. Familiarise yourself with the NumPy roll and multiply functions, and use these to replace your energy double loop (you will need to call roll and multiply twice!).'''

Simply using the numpy.sum function instead of two nested for-loops allows for a faster calculation of the magnetisation of the system. The calculation of the energy however is more involved. The numpy function multiply allows the multiplication of each element in an array with its counterpart in another array. By using the roll function, each entry in the lattice array can be shifted by 1 column or row in either direction, representing both an interaction with a neighbouring element as well as implementing periodic boundary conditions. The total energy can be calculated by <math>-\frac{J}{2}\sum_{i}^{N}\sum_{j\in adj}{s_i s_j}</math>, where the inner sum <math>\sum_{j\in adj}{s_i s_j}</math> represents the interactions each atom shows with its neighbours. By creating a matrix in which each element represents the sum of the interactions of each entry's counterpart in the real lattice, this inner sum is validly expressed. The outer sum is then simply the sum over all elements of this auxiliary matrix.

'''Use the script ILtimetrial.py to record how long your new version of IsingLattice.py takes to perform 2000 Monte Carlo steps. This will vary, depending on what else the computer happens to be doing, so perform repeats and report the error in your average!'''

The updated code exhibits an average computational time of 0.616s, which is about 240 times faster than the original code. The rms was lowered to 0.00939, which is orders of magnitude lower than the average value, indicating a valid experiment.

===The effect of temperature===

'''The script ILfinalframe.py runs for a given number of cycles at a given temperature, then plots a depiction of the final lattice state as well as graphs of the energy and magnetisation as a function of cycle number. This is much quicker than animating every frame! Experiment with different temperature and lattice sizes. How many cycles are typically needed for the system to go from its random starting position to the equilibrium state? Modify your statistics() and montecarlostep() functions so that the first N cycles of the simulation are ignored when calculating the averages. You should state in your report what period you chose to ignore, and include graphs from ILfinalframe.py to illustrate your motivation in choosing this figure.'''

Empirically it can be seen that the time needed to achieve equilibration is less than 100 times the amount of spins in the system (''Figure 4''). The temperature does not seem to affect the time needed to achieve equilibrium, but the added noise creates a need for longer averaging periods (''Figure 5''). Considering this, the statistics() function now calculates the averages only from data points starting after n_cols*n_rows*100 steps. This interval admits a confidence interval should a different set of initial conditions demand a longer equilibration period. Additionally, the function now resets the entries of all four properties and the counter of Monte Carlo steps to enable the correct calculation of averages in future calculations.

[[File:Final frame temp 1.png|thumb|right|Figure 4: The output of the ILfinalframe.py script for lattices of sizes 8x8 (a), 16x16(b) and 32x32(c) at 1.0 K. Equilibration is achieved in each case.]]

[[File:Final frame temp varied.png|thumb|right|Figure 5: The output of the ILfinalframe.py script for an 8x8 lattice at temperatures 1.0K (a), 2.0K(b) and 32x32(c).]]

'''Use ILtemperaturerange.py to plot the average energy and magnetisation for each temperature, with error bars, for an 8\times 8 lattice. Use your initution and results from the script ILfinalframe.py to estimate how many cycles each simulation should be. The temperature range 0.25 to 5.0 is sufficient. Use as many temperature points as you feel necessary to illustrate the trend, but do not use a temperature spacing larger than 0.5. T NumPy function savetxt() stores your array of output data on disk — you will need it later. Save the file as 8x8.dat so that you know which lattice size it came from.'''

As argued in task 1, the number of cycles needed to reach equilibration is N*100, where N is the number of spins in the system. This allows a fast computation (within minutes) while maintaining a high confidence interval. The variation of the critical temperature is given by the expression <math>T_{C, L} = \frac{A}{L} + T_{C, \inf}</math>. A simple plot of <math>T_{C, L}</math> vs. 1/L reveals that the predicted critical temperature variation will vary between 2 - 3 K. Therefore, 0.1K was chosen as an appropriate temperature step size. The total number of cycles was chosen in such a way that at least twice the amount of redundant cycles was calculated to allow for sufficient averaging steps once equilibration is achieved. The script ''ILplotaverage.py'' reads the values generated by ''ILtemperaturerange.py'' and plots the average of energy and magnetisation per spin over three data sets against temperature (''Figure 6''). The errorbars indicate the standard deviation.

[[File:ILplot 8x8.png|thumb|right|Figure 6: The energy and magnetisation per spin against temperature of an 8x8 Ising lattice.]]

The energy per spin rises from the energy of the minimum configuration (-2) to a value closely below zero. The magnetisation starts from an initial value of +2 (perfectly ordered lattice below TC) and falls to 0 (disordered lattice over TC). In the steep transition between the initial and final value, great noise is observed, which is typical of a critical region.

===The effect of system size===

'''Repeat the final task of the previous section for the following lattice sizes: 2x2, 4x4, 8x8, 16x16, 32x32. Make sure that you name each data file that your produce after the corresponding lattice size! Write a Python script to make a plot showing the energy per spin versus temperature for each of your lattice sizes. Hint: the NumPy loadtxt function is the reverse of the savetxt function, and can be used to read your previously saved files into the script. Repeat this for the magnetisation. As before, use the plot controls to save your a PNG image of your plot and attach this to the report. How big a lattice do you think is big enough to capture the long range fluctuations?'''

For each lattice size three simulations were conducted to allow for fluctuations introduced by varying initial conditions. The scripts ''ILplot_average2.py'' & ''ILplot_average3.py''' read three data sets generated for each lattice size, averages their values and calculates the energy/ magnetisation per spin. The data for all lattice sizes are plotted on the same graph (''Figure 7''). Error bars were omitted for legibility.

[[File:Energy plots.png|thumb|right|Figure 7: A plot of the energies per spin against temperature for a number of lattice sizes.]]

[[File:Mag_plots.png|thumb|right|Figure 8: The magnetisation per spin vs. temperature for a number of lattice sizes.]]

The plots produced show similar features across all lattice sizes: As seen before, the energy rises from the minimum of -2 per spin to a maximum value close to zero. The final value is proportional to the system size.

===Determining the heat capacity===

'''Write a Python script to make a plot showing the heat capacity versus temperature for each of your lattice sizes from the previous section. You may need to do some research to recall the connection between the variance of a variable, Var[X], the mean of its square <math><X^{2}></math>, and its squared mean <math><X>^{2}</math>. You may find that the data around the peak is very noisy — this is normal, and is a result of being in the critical region. As before, use the plot controls to save your a PNG image of your plot and attach this to the report.'''

The heat capacity of the system is given by:

<math>C = \frac{Var(E)}{K_{B}T^{2}} = \frac{<E^{2}> - <E>^{2}}{K_{B}T^{2}}</math>

In order to calculate the heat capacity, values for the energy and energy squared vs. temperature are read from files generated beforehand (''ILheatcapacity.py''). As three data sets are present, averages are calculated. The Boltzmann constant can be omitted from the calculation as the energies are set in reduced units. The plots of heat capacity vs. temperature show curves with a well defined maximum (''Figure 9''). The maximum increases in value at higher temperatures and also shift to slightly lower temperatures. Due to the noise present at the critical region, a fit will need to prepared for data of higher resolution. It can also be seen that the function onset is different from the final value, which is typical for a second order phase transitions, such as the one present.

[[File:ILheatcapacity.png|thumb|right|Figure 9: The variation of heat capacity with temperature for various lattice sizes.]]

===Locating the Curie Temperature===

'''A C++ program has been used to run some much longer simulations than would be possible on the college computers in Python. You can view its source code here if you are interested. Each file contains six columns: T,E,E^{2},M,M^{2},C (the final five quantities are per spin), and you can read them with the NumPy loadtxt function as before. For each lattice size, plot the C++ data against your data. For one lattice size, save a PNG of this comparison and add it to your report — add a legend to the graph to label which is which. To do this, you will need to pass the label="..." keyword to the plot function, then call the legend() function of the axis object.'''

The script ''ILreference.py'' plots reference values together with the simulated values calculated above on the same graph for comparison (''Figure 10, 11, 12''). Attention should be paid to the different amount of data points in the two curves, which were accounted for by two different sets of temperature values. It is evident that the calculated data correlate highly with the reference data. The strongest deviations can be found in the region of the phase transition, but large fluctuations are expected in this region. The accuracy of the values could be adjusted by increasing the number of redundant steps in the Monte Carlo simulation as well as the number of averaging steps.

[[File:Ref 4x4 Energy.png|thumb|right|Figure 10 :A comparison between simulated and reference value for the energy per spin of a 4x4 Lattice.]]

[[File:Ref 4x4 Magnetisation.png|thumb|right|Figure 11: A comparison between simulated and reference value for the magnetisation per spin of a 4x4 Lattice.]]

[[File:Ref 4x4 Capacity.png|thumb|right|Figure 12: A comparison between simulated and reference value for the heat capacity per spin of a 4x4 Lattice.]]

'''Write a script to read the data from a particular file, and plot C vs T, as well as a fitted polynomial. Try changing the degree of the polynomial to improve the fit — in general, it might be difficult to get a good fit! Attach a PNG of an example fit to your report.'''

The script ''ILfit1.py'' plots both the heat capacity as well as a polynomial fit against temperature. The degree of the fit can be chosen by the user. A range of fit degrees were tested. However, even a polynomial of degree 30 could not provide a satisfactory fit (''Figure 13'').

[[File:Fit1 30.png|thumb|right|Figure 13: A polynomial fit for the heat capacity of the 32x32 Lattice against temperature.]]

'''Modify your script from the previous section. You should still plot the whole temperature range, but fit the polynomial only to the peak of the heat capacity! You should find it easier to get a good fit when restricted to this region.'''

''ILfit_updated.py'' reduces the range of temperatures over which the polynomial proposed in ''ILfit1.py'' is fitted. As it is of highest importance to accurately sample the region around the maximum in heat capacity, the fitting range was set to start at half the value of the maximum and end after the function has passed through half the maximum value once more - for different functional forms of heat capacity this would need to be adjusted to a more general algorithm. This value can be adjusted to different ranges to provide optimal fits. A fourth degree fit was found to be satisfactory (''Figure 14'').

[[File:Fit2 4.png|thumb|right|Figure 14: A polynomial fit of restricted range for the heat capacity of the 32x32 Lattice against temperature.]]

'''Find the temperature at which the maximum in C occurs for each datafile that you were given. Make a text file containing two colums: the lattice side length (2,4,8, etc.), and the temperature at which C is a maximum. This is your estimate of T_{C} for that side length. Make a plot that uses the scaling relation given above to determine <math>T_{C,\infty }</math>. By doing a little research online, you should be able to find the theoretical exact Curie temperature for the infinite 2D Ising lattice. How does your value compare to this? Are you surprised by how good/bad the agreement is? Attach a PNG of this final graph to your report, and discuss briefly what you think the major sources of error are in your estimate.'''

By finding the maximum in the fitted curve, the maximum in heat capacity and the critical temperature at which it occurs can be approximated. The script mentioned above writes the critical temperature at every lattice size to the output file ''Heatcapacity_maxima.txt''. According to the relation

<math>T_{C, L} = \frac{A}{L} + T_{C, \infty}</math>

the critical temperature of an infinite 2D Ising Lattice can be extrapolated from a plot of TC, L vs. <math>\frac{1}{L}</math>. The constant A can be obtained from the gradient of the linear fit, whereas the intercept with the y-axis equals the critical temperature of an infinite lattice (''Figure 15''). Even though the function shows a poor correlation with a linear fit, the obtained value of <math>T_{C, \infty} = 2.267K</math> is very close to the literature values of 2.269K<ref name = "lit value1" /> and 2.534K<ref name="lit value2">Onsager, Lars. "Crystal statistics. I. A two-dimensional model with an order-disorder transition." Physical Review 65.3-4 (1944): 117.</ref>.

[[File:Curie Temp.png|thumb|right|Figure 15: The variation of the critical temperature with lattice size.]]

==References==

<references/>

File:ILanim picture 12.png

2015-02-01T18:03:11Z

Bg1512: An initial Monte Carlo simulation of an 8x8 Lattice at 1.0K.

An initial Monte Carlo simulation of an 8x8 Lattice at 1.0K.