[daylight]

25.0 g of HI(g) is injected into a 4.00 L reaction vessel that contains 20.0 g of I2(g). When the system comes to equilibrium at 400ºC, what will be the total pressure inside the reaction vessel? 2HI(g) ↔ H2(g) + I2(g), Kc = 0.0156 at 400ºC
a. 2.70 atm
b. 13.0 atm
c. 2.43 atm
d. 0.815 atm
e. 3.24 atm

Ok, I have had some help with this, and even that person can't find an answer that matches....

Kc = [H2][I2]/[HI]^2 = 0.0156

Moles I2 = 20.00 g/(253.81 g/mole) = 0.0788 moles
[I2] = 0.0788 moles/4 L = 0.01970 M

Moles HI = 25.0 g/(127.91 g/mole) = 0.195 moles
[HI] initial = 0.195 moles/4 liters = 0.04886 M
[H2] = x
[I2]eq = 0.01970 + x
[HI]eq = 0.04886 M - 2x

Kc = 0.0156 = [H2][I2]/[HI]^2 = x*(0.01970 + x)/(0.04886 - 2x)^2
(3.7242 x 10^-5) - (3.0489 x 10^-3)x + 0.0624x^2 = 0.0197x + x^2
0.9376x^2 + 0.01665x - (3.7424 x 10^-5) = 0
x = 0.0002209 = 2.209E-04

Therefore,
[HI]eq = 0.0489 M - (2*2.209E-04) = 4.8420E-02
[I2]eq = 0.0197 + 2.209E-04 = 1.9921E-02
[H2] = 2.2090E-04


Total concentration of gases = 0.04842 + 0.01992 + 0.00022 = 0.06856 moles/L
PV = nRT, P = (n/V)*RT = (0.0686)*0.0821*(400+273.15) = 3.79 atm
which is close to 3.24, but I am wary of being that far off....

[Borek]

Seems to me like there is no matching answer, and 3.79 looks OK.

Note, that this question is much simpler then you are making it.

Ignore iodine presence for now.

You start with 0.1954 moles of HI.

Does number of moles of gas change in the reaction?

Does number of moles of gas depend on the equilibrium position?

Do you need equilibrium for anything?

Add iodine. Anything have changed?

You have 0.1954 moles of gas from the HI equilibrium and 0.0788 moles gaseous iodine.

PV=nRT is all you need.