# If a reactant appears on both sides of an elementary equation, can it be cancelled?

Me and a friend were debating the following. For the elementary reaction

$$\ce{A + B -> A + C}$$

my friend says that it is unimolecular because $$\ce{A}$$ does not "participate" in the reaction. So, he says that the rate law is

$$\text{rate} = k[\ce{B}].$$

But, I think that that since it is an elementary reaction step, you cannot cancel $$\ce{A}$$ from both sides. I think that it is a bimolecular reaction with

$$\text{rate} = k[\ce{A}][\ce{B}].$$

Who is correct?

• In a real experiment, if one starts with A(aq) + B(aq) and the reaction forms more water, one may, or may not, be actually able to remove the starting water. The issue is does A + B create water, moving the reaction forward, upon heating the dry salts. If so, canceling out the water may be appropriate, otherwise, not. Dec 23, 2019 at 3:08
• This is either not an elementary reaction step, or "A" should be excluded from the equation.
– Karl
Dec 23, 2019 at 9:16
• @Karl It can be an elementary step. For example in reaction of chiral bromoalkane with bromide. Dec 23, 2019 at 14:40

If the elementary step is indeed written

$$\ce{A + B -> A + C}$$

and assuming that you haven't just written in $$\ce{A}$$ for fun, i.e. $$\ce{A}$$ is actually a participant in the step, then yes, this would be considered bimolecular. This sort of step, featuring the same chemical species on both sides of the reaction, does pop up in (for example) the Lindemann mechanism.

With regard to kinetics, however, it is worth noting that steps of this kind can sometimes indicate a reaction that is catalytic with respect to $$\ce{A}$$. If that is the case, then the concentration of $$\ce{A}$$ is essentially constant and you would obtain pseudo-first-order kinetics.

• You wrote that this "can sometimes [emphasis mine] indicate a reaction that is catalytic with respect to A". But if A is, as depicted above, both a participant in, and yet unchanged by, the reaction, then wouldn't it always be considered a catalyst? I.e., are there any reactions in which A is both a participant and unchanged, yet for which it would not be considered a catalyst? Dec 23, 2019 at 3:01
• @theorist I use that wording on purpose because I don't want to rule out the possibility that A is further consumed in a separate step of the mechanism. I can't come up with an example right now, but there may well be something out there I don't know. Dec 23, 2019 at 3:05
• In that case it seems it would at least be catalytic with respect to that particular elementary reaction. The only exception I can think of is if A is added to slow down the reaction, in which case I suppose one might say that it must be either catalytic or anticatalytic. Though until just now I've never thought about anticatalysts in elementary reactions Dec 23, 2019 at 3:12
• @theorist You guys are discussing on the basis of a wrong premise in the original question. Catalytic reactions can never be single step! This is not an elementar reaction, if A not in there just for fun.
– Karl
Dec 23, 2019 at 9:26
• @DrPepper I’m not sure why you say it would be unimolecular - I wrote bimolecular in my answer. Dec 23, 2019 at 15:27

This is common in biochemistry where A is a protein (an enzyme). Enzymes that alter one substrate only, producing a single product, are called isomerases.

Michaelis-Menten kinetics applies. The reaction speed is proportional to the concentration of the enzyme:

$$v = k [\ce E] \frac{[\ce S]}{K + [\ce S]}$$

here [E] is the enzyme concentration (would be A) and [S] is the substrate concentration (would be B). k and K are the reaction-specific constants. Hence no, [E] cannot be "abstracted out". The reaction speed would be close to zero regardless of the amount of the substrate.

This obviously assumes that A and B form the complex first. If A is an unrelated substance at the opposite edge of the Universe than B now converting into C, it probably should not be considered as part that conversion.

• I didn't use M-M kinetics as an example because the question is asking about an elementary step, not an overall reaction. Even though the overall reaction $$\ce{E + S -> E + P}$$ resembles the original form given in the question, the individual steps $$\ce{E + S <-> ES <-> E + P}$$ aren't quite the same as what OP is asking about, I think. Dec 23, 2019 at 21:20