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7th edition 5I.13

Posted: Sun Jan 13, 2019 7:41 pm
by Alyssa Bryan 3F
Part B of the question says that "if 2.0 mmol F2 was placed into the reaction vessel instead of the chlorine, what would be its equilibrium composition at 1000K."
I was able to solve correctly for part A, but I can't seem to get the correct answer for part B. Can someone explain the steps to solving part B of this question?

Re: 7th edition 5I.13

Posted: Mon Jan 14, 2019 9:25 am
by Hai-Lin Yeh 1J
So for part B, the equation should be: F2(g) -> 2(F) BUT, the equilibrium constant (Kc) is 1.2 x 10^-4
Essentially, your ICE table for part B should be very similar to your ICE table for part A. Looks something like this:
F2 F
I 0.001 0
C -x +x
E 0.001-x x

Kc= [F]^2 /[F2] = (2X)^2/(0.001-X) = 4X^2/0.001-X= 1.2 X 10^-4
Then, if you multiply it out and use the quadratic equation, you should get x= 0.000159 or 1.59 x 10^-4
Thus, [F2] = 8.4 X 10^-4, [F] = 3.2 X 10^-4 ( to 2 significant figures)