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### Why is K = kfor/krev?

Posted: Tue Mar 07, 2017 8:51 pm
Can someone explain why K = k/krev? In lecture, did Dr. Lavelle cover how to derive or infer this?

### Re: Why is K = kfor/krev?

Posted: Tue Mar 07, 2017 10:13 pm
Check out page 71 on the course reader.

### Re: Why is K = kfor/krev?

Posted: Tue Mar 07, 2017 10:30 pm
We were told that the rate of the forward reaction is equal to the rate of the reverse reaction in a given chemical equation. For example, if you have A + B -> C + D, then kfor[A][B] = krev[C][D]. If you divide the right side by [A][B] and the left side by krev, you get kfor/krev = [C][D]/[A][B], or more simply, kfor/krev = [products]/[reactants]. From chem 14A, we learned that the equilibrium constant K is = [products]/[reactants] and thus K = kfor/krev.

### Re: Why is K = kfor/krev?

Posted: Tue Mar 07, 2017 10:46 pm
Laura Rabichow 1J wrote:We were told that the rate of the forward reaction is equal to the rate of the reverse reaction in a given chemical equation. For example, if you have A + B -> C + D, then kfor[A][B] = krev[C][D]. If you divide the right side by [A][B] and the left side by krev, you get kfor/krev = [C][D]/[A][B], or more simply, kfor/krev = [products]/[reactants]. From chem 14A, we learned that the equilibrium constant K is = [products]/[reactants] and thus K = kfor/krev.

Thanks! Makes sense!