Consider a general, single-step reaction of the type A + B ∆ C. S...

Question

Consider a general, single-step reaction of the type A + B ∆ C. Show that the equilibrium constant is equal to the ratio of the rate constants for the forward and reverse reactions, Kc = kf>kr.

Accepted Answer

hi everyone for this problem it reads for the following certain single step reaction represented by the following derive a relationship between the equilibrium constant and the rate constants of the forward and reversed reaction. So we want to derive a relationship. Okay, and to do this, let's first define what our equilibrium constant represents. So our equilibrium constant K. C. Represents products over react ints. And specifically this is the concentration of products over reacting. So looking at our single step reaction, we see that our products are R. And S. So it's going to be the concentration of our times the concentration of X. So the concentration of products over our reactant, P and Q. So the concentration of P. The concentration of Q. So this is what our K. C. Expression represents in terms of the rates the rate forward is equal to. So we'll write this as the rate constant of the forward is going to be if we're looking for the forward direction it's going to b times the concentration of p times the concentration of Q. And for the rate reverse this is going to equal the equilibrium constant for the reverse is going to be the concentration of our times the concentration of s. Okay, so we're going to set the rate forward equal to the rate reverse. Okay, so rate forward equals rate reverse. We're going to set them equal to each other. So when we set them equal to each other, what we're going to get is equilibrium constant of the forward equals the equilibrium constant of the reverse. Okay, so let's go ahead and isolate or bring to one side. The the equilibrium constant of the forward over the reverse. So we want to focus on these two variables. Okay, so what we're gonna do is we're going to rewrite this where we have the rate constant of the forward. Over the rate constant of the reverse is going to equal the concentration of our times the concentration of S over the concentration of p times the concentration of Q. Okay, so what this means then is R. K. C. Our equilibrium constant is going to equal what we just wrote. So the concentration of our times the concentration of S over the concentration of p times the concentration of Q. So this is going to equal the rate of the forward over the rate of the reverse. Alright, so in terms of the relationship we can say the relationship is going to be, the equilibrium constant is equal to the rate constant of the forward. Over the rate constant of the reverse. So this is going to be our final answer here. Okay, that is the relationship between the equilibrium constant and the rate constants. That is it for this problem. I hope this was helpful

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