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I understand that there are many different catalytic mechanisms that suggest different ways of accounting for the effect of a catalyst on reaction rate. My question is:

Given a termolecular elementary (it is known it is an elementary reaction) reaction of the form: $$\ce{A + B + C <=> D + E + C}$$ Does the rate of reaction simplify down (according to collision theory) to: $$r = [\ce{C}]\bigr(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\bigr)$$

Edit: To clarify my question, I have a gas phase elementary reaction of the form: $$\ce{A + B + C -> D + E + C}$$ I have another of the form: $$\ce{D + E + C -> A + B + C}$$ Whether they are exactly the same mechanism is irrelevant. My question relates to how quantitatively does a collision partner's concentration in a gas mixture affect reaction rate. i.e. is forward rate defined by (assuming that I know the reaction acts as an elementary reaction in that there is no intermediate): $$r_\mathrm{fwd} = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]$$ and reverse rate defined by (reverse is also elementary reaction - not concerned if same mechanism just also acts in the rate law as a single collision event): $$r_\mathrm{rev} = k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$ Therefore, the total rate, not assuming they're the same reaction just two elementary reactions that go in opposite directions and we've called on forward and one reverse, the total rate forward, taking into account our other reaction, is: $$r = r_\mathrm{fwd} - r_\mathrm{rev}$$ $$r = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}] - k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$ $$r = [C]\left(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\right)$$ Therefore quantitatively a collision partner does not change the overall rates direction but rather enhances it in both directions. Is this correct?

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    $\begingroup$ You are correct. What you call a collision partner is a specific example of the general concept of catalysis. "C" is a catalyst in your scheme. Catalysts don't change thermodynamic equilibria, just the speed at which it is approached (i.e. kinetics). Accelerating the forward rate but not the reverse rate of an equilibrium reaction would alter the equilibrium point, and is not possible for any true catalyst. $\endgroup$ – Curt F. Jul 6 '15 at 3:49
  • $\begingroup$ Thank you very much for all your help, if you wouldn't mind just putting that exact paragraph into an answer I'll be sure to accept it for future viewers. $\endgroup$ – J-S Jul 6 '15 at 4:59
  • $\begingroup$ I have downvoted your question for several reasons. The main reason is, that you are still describing an equilibrium as an elementary reaction. Additionally I believe your thinking is still not correct, but I won't and cannot go into detail here. Instead of altering your question to the point where my answer becomes irrelevant you could have asked a new question with a clarification. $\endgroup$ – Martin - マーチン Jul 6 '15 at 13:35
  • $\begingroup$ I understand that there were errors in my question, but I feel as though even without the clarification whilst correct your answer failed to address my issue. Whilst you could have pointed out this issue, I feel as though the main query was clear enough. Also since your answer was not actually answer put rather pointing out an error in the question, I feel it may have been inappropriate to put it in the answers and maybe would have been better in the comments. I do appreciate that my question wasn't the best and understand the down-vote and I thank you for the time put into your answer :) $\endgroup$ – J-S Jul 6 '15 at 23:29
  • $\begingroup$ Also I called it a set of elementary reactions to get across the point that I knew the effect of the concentration of other reactants in both directions. The equilibrium sign was just for convenience. $\endgroup$ – J-S Jul 6 '15 at 23:31
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You are correct. What you call a collision partner is a specific example of the general concept of catalysis. "C" is a catalyst in your scheme. Catalysts don't change thermodynamic equilibria, just the speed at which it is approached (i.e. kinetics). Accelerating the forward rate but not the reverse rate of an equilibrium reaction would alter the equilibrium point, and is not possible for any true catalyst.

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If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an elementary reaction. For example, $$\ce{A + B -> C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -> A + B}$$ is a different elementary reaction. The rates of both reactions \begin{align} r&=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\ r&=k_\mathrm{rev}[\ce{C}][\ce{D}]\\ \end{align} would determine the location of the equilibrium \begin{align} \ce{A + B &<=> C + D}& K&=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}. \end{align}

For a true termolecular reaction, i.e. three species react in one concerted transition state, $$\ce{A + B + C -> D + E + F},$$ the rates are always given as \begin{align} r&=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\ r&=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\ \end{align} You cannot simplify a rate law of an elementary reaction.$%edit$

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    $\begingroup$ But if F = C, then you could just to algebra to simplify the rate law to be equivalent to what user2353082 wrote in the question. Surely algebraic manipulation is still allowed, even for a rate law of an elementary reaction? $\endgroup$ – Curt F. Jul 5 '15 at 14:27
  • $\begingroup$ @Curt No you cannot. An elementary reaction is only dependent on itself, it has only one direction, i.e. one transition state, therefore there is only one possible rate law. If F = C, then these are still two completely different elementary reactions. You can determine the the location of the equilibrium (and its rate law) in the suggested way, but this is no longer an elementary reaction. $\endgroup$ – Martin - マーチン Jul 5 '15 at 17:22
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    $\begingroup$ Your answer so far is very good, but is answering something the question didn't ask, which is, "what is the strict definition of elementary reaction". I think it's possible that the OP meant to say "Consider this system of two coupled elementary reaction_s_..." and really wants a confirmation of their algebra that the final rate law is correct. The algebra is correct. As an aside, I'm not really sure what use the elementary reaction concept is, if it is immune to algebra and does not know about microscopic reversibility either. $\endgroup$ – Curt F. Jul 6 '15 at 3:17
  • $\begingroup$ @CurtF. You are correct, this was my question. I have tried to include some edited clarification. So are you saying that my thinking is indeed correct? $\endgroup$ – J-S Jul 6 '15 at 3:39
  • $\begingroup$ @CurtF. The question clearly did as for an termolecular elementary reaction. I clarified, that there is only one possible rate law for this kind of reaction and that this thinking in this case is wrong. If the OP means something different than what he asks, this is not my fault. $\endgroup$ – Martin - マーチン Jul 6 '15 at 13:28

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