A 5.00L reaction vessel is filled with 1.00mol of H2, 1.00mol of I2, and 2.50mol of HI. If equilibrium constant for the reaction is 129 at 500k, what are the equilibrium concentrations of all species?

H2(g) + I2(g) <-> 2HI(g) Kc = 129

3 answers

(H2) = 1.00/5.00 = 0.2M
(I2) = 1.00/5.00 = 0.2M
(HI) = 2.50/5.00 = 0.5M

.........H2 + I2 ==> 2HI
I........0.2...0.2...0.5
C........-x....-x....+2x
E.......0.2-x..0.2-x..0.5+2x

Kc = (0.5+2x)^2/(0.2-x)(0.2-x)

Note: In the C line, how do we know it is -x. -x, and +2x instead of +x +x, and -2x;i.e., how do we which way the reaction will occur. You run a Qrxn = (HI)^2/(H2)(I2)= (0.5)^2/(0.2)^2 = about 6. Compare with K = 129 and that means the numerator is too small and denominator is too large so the reaction must go to the right.

What does x equal out too?

h2, i2 = 11.4
hi = 22.8