At a particular temperature, K = 1.00 102 for the following reaction.

Question

H2(g) + I2(g) 2 HI(g)
In an experiment, 1.23 mol H2, 1.23 mol I2, and 1.23 mol HI are introduced into a 1.00 L container. Calculate the concentrations of all species when equilibrium is reached.

Answer 1

(H2) = 1.23/1L = 1.23M
(I2) = 1.23/1L = 1.23M
(HI) = 1.23/1 = 1.23M
First you want to calculate Q = (HI)^2/(H2)(I2) = (1.23)^2(1.23)(1.23) = 1.00 which means HI is too small and H2 and I2 are too large so the reaction will go to the right.
............H2 + I2 ===> 2HI
initial....1.23..1.23.....1.23
change......-x....-x.....2x
equil.....1.23-x..1.23-x..1.23+2x

Substitute from the ICE chart into K expression and solve for x.

Answer 2

x =.9225
and when you substitute (1.23-.9225) makes H2 and I2=.3075
(1.23+2*.9225)=3.075