Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MitchTwitchita on March 23, 2008, 12:11:50 AM
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Hey Guys, I'm having a really hard time with this one.
Consider the reaction H2(g) + I2(g) ---> 2HI(g). Starting with a concentration of 0.040 M HI, calculate the concentrations of HI, H2, and I2 at equilibrium. The Kc for the reaction is 54.3.
2HI ----> H2 + I2
0.040 M 0 0
-x +x +x
0.040 -x x x
Kc = [H2][I2] / [HI]^2
54.3 = x^2 / (0.040 - x)^2
7.37 = x / (0.040 - x)
0.2948 - 7.37x = x
x = 3.5 x 10^-2
[HI]0.040 - 2 x (3.5 x 10^-2) = 0.03
[H2] = 3.5 x 10^-2
[I2] = 3.5 x 10^-2
I have no idea where I'm going wrong. The answers in the book are: [HI] = 0.031 M [H2] = 4.3 x 10^-3 [I2] = 4.3 x 10^-3
Can somebody please help me out with this one?
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For HI, it should be your equilibrium concentration will be 0.040 - 2x because of the stoichiometry of the reaction. It looks like you definitely know what you are doing, but just made a small oversight.
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Aaaaaah! Thanks again Yggdrasil!