[MitchTwitchita]

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?

[Yggdrasil]

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.

[MitchTwitchita]

Aaaaaah!  Thanks again Yggdrasil!