Why is K = kfor/krev?

[Michael_Johanis_2K]

Can someone explain why K = k/krev? In lecture, did Dr. Lavelle cover how to derive or infer this?

[Andrew Nguyen 3G]

Check out page 71 on the course reader.

[Laura Rabichow 1J]

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.

[hfrazer1B]

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!