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### Unit of Equilibrium Constant K

Posted: Mon Mar 13, 2017 12:20 am
Hello, I have a question regarding one of the questions in quiz 2.

The question is:
Consider the following reaction at equilibrium.
$A\rightleftharpoons B+C$
The equilibrium concentration were measured as follows:
[A] = 0.0100M, [B] = 0.500M, [C] = 2.50M
In a kinetic experiment, the rate constant of the forward reaction k is determined to be $200s^{-1}$. Calculate the rate constant of the reverse reaction.

I calculated K to be
$K = \frac{[products]}{[reactants]} = \frac{[B][C]}{[A]} = \frac{(0.500)(2.50)}{0.0100} = 125$

And since $K = \frac{k(forward)}{k(reverse)}$
$k(reverse) = \frac{k(forward)}{K} = \frac{200s^{-1}}{125} = 1.60s^{-1}$

However, the right answer was supposed to be $1.60s^{-1}M^{-1}$. Equilibrium constant K was supposed to have the unit M.

I am slightly confused by this. I've always thought that K is an expression of the "activities" of the species (as learned in Chem 14A), so it should be a unitless value?
Can someone please clarify when K should have a unit and when it shouldn't?
Thanks in advance!

### Re: Unit of Equilibrium Constant K

Posted: Mon Mar 13, 2017 12:45 am
Hi!
K as an equilibrium constant does not have any units. However, this question asks for k, the rate constant, which does have units. The units are different occasionally depending on the order of the reaction. This is because the rate law's units are always constant, so k units change to accommodate for the order of the reaction.