## Self-Test 11.11A in textbook 6th Edition

David Sarkissian 1K
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### Self-Test 11.11A in textbook 6th Edition

Bromine monochloride, BrCl, decomposes into bromine and chlorine and reaches the equilibrium $2BrCl(g)\leftrightharpoons Br^{_{2}}+Cl_{2}$, for which K = 32 at 500. K. If initially pure BrCl is present at a concentration of 3.30 mbar, what is its partial pressure in the mixture at equilibrium?

The problem above is a self test under Example 11.8 of the same type of Ice Table problem. It tells us the answer should end up being 0.3 mbar, however after trying to solve it I keep getting different answers, wrong answers. Is there a trick that I'm missing when it comes to how you found mbars? I dont believe you need to convert anything, however I keep getting massive numbers that end up cancelling in the weirdest way, so it might just be me. I'm just curious if anyone has gotten the right answer attempting to solve this problem. Thanks.

David S
Posts: 54
Joined: Fri Sep 28, 2018 12:15 am

### Re: Self-Test 11.11A in textbook 6th Edition

I got 0.3 and here is how:

1) $K=\frac{(P_{Br_{2}})(P_{Cl_{2}})}{(P_{BrCl})^{2}}$

2) After setting up ice table and finding expressions for equilibrium pressures in terms of x, you get: $K = \frac{x^{2}}{(3.30-2x)^{2}}$

3) multiply both sides by denominator and then simplify the equation by foiling out the (3.30-2x)^2 to get: $32(4x^{2}-13.2x+10.89)=x^{2}$

4) multiply in the 32 up front and subtract x^2 from both sides to get: $127x^{2}-422.4x+348.48 =0$

5) set up quadratic equation, solve for x and you get two values, 1.81 and 1.52. Since 3.30-2x = equilibrium pressure of BrCl, we choose x=1.52 (because 2*1.81 > 3.30, and you cannot use up more reactant than you started with) and we get 3.30-2(1.52) = 0.30 when rounded.