## Post Assessment Part 2

katherinemurk 2B
Posts: 68
Joined: Wed Nov 15, 2017 3:02 am

### Post Assessment Part 2

There is a question in this post assessment asking about finding the equilibrium constant of reaction 2BrCI <-> Br2 + Cl2. The question says something about 18.3% of the BrCI gas remains at equilibrium. However, no matter how many times I attempt the question I cant get the answer. Can someone explain how to get the answer or is there possibly something wrong with the post assessment values?

Chris Freking 2G
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: Post Assessment Part 2

If 18.3% of the BrCl gas remains at equilibrium and we have the initial concentration of BrCl, it is possible to calculate the amount of BrCl left at equilibrium. 0.183 * (1.84 * 10^-4) = 3.36 * 10^-5 M of BrCl at equilibrium. Then use the ICE box and plug in this value:

2BrCl(g) ⇌ Br2 (g) + Cl2(g)
I 1.84 * 10^-4, 0, 0
C -2x, +x, +x
E 3.36 * 10^-5, x, x

Then solve for x:
(1.84 * 10^-4) - 2x = 3.36 * 10^-5
x = 7.52 * 10^-5

Since x equals the concentration of [Br2] and [Cl2], you can now solve for K by plugging in the equilibrium concentrations of BrCl and Br2 and Cl2 into K=[P]/[R]:

K = [Br2] [Cl2] / [BrCl]^2 = (7.52*10^-5)(7.52*10^-5)/(3.36*10^-5)^2