Where do catalysts come into rate laws? For example in this question: EDIT: Ah shit I can't post pictures yet. I'd be inclined to say the rate law is: Rate = k[B ][C] but I don't know whether catalysts are supposed to show up in overall rate equations or not. As for coming up with a 2 step mechanism, wouldn't this work: i) B + C --> BC ii) AB + BC --> A + B[SUB]2[/SUB] + C ? they wouldn't make this question this simple though so I'm guessing I'm missing something here.
My recollection is that because catalysts show up in both sides of the reaction, they don't show up in the rate equation, because they effectively multiply the rate equation by one. Another way to think of it is, to use your first example, for the reaction: bB + cC → dD Then the rate equation is: r = k[sup]b[/sup][C][sup]c[/sup] However, it is my recollection that that itself is derived from this monstrosity: For the equation aA + bB → pP + qQ. \(r = - \frac{1}{a} \frac{d[A]}{dt} = - \frac{1}{b} \frac{d[B]}{dt} = - \frac{1}{p} \frac{d[P]}{dt} = - \frac{1}{q} \frac{d[Q]}{dt}\) The reason why this is important to understand is that if we then introduce a catalyst into the reaction such that it becomes aA + bB + cC → pP + qQ + cC (if you want to think of it in that way), then, for an ideal catalyst, the concentration of the catalyst is a constant, and so doesn't appear in the equation, which brings us back to our original generic reaction: r = k[sup]b[/sup][C][sup]c[/sup] Where any effects of the catalyst would be included in the constant k. Does that make sense?
