## Why is K = kfor/krev?

$K = \frac{k_{forward}}{k_{reverse}}$

Michael_Johanis_2K
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### Why is K = kfor/krev?

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

Andrew Nguyen 3G
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

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

Check out page 71 on the course reader.

Laura Rabichow 1J
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Joined: Fri Jun 17, 2016 11:28 am

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

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
Posts: 19
Joined: Wed Sep 21, 2016 2:57 pm

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

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!