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

Question

H2(g) + I2(g) equilibrium reaction arrow 2 HI(g)
In an experiment, 1.13 mol H2, 1.13 mol I2, and 1.13 mol HI are introduced into a 1.00 L container. Calculate the concentrations of all species when equilibrium is reached.

Answer 1

In 1L these mols give molar concentrations of
1.13 M H2, 1.13 M I2, and 1.13 M HI
............... H2(g) + I2(g) <==> 2 HI(g)
I...............1.13........1.13...........1.13
First determine which direction the reaction moves in order to reach equilibrium.
Qc = [(HI)^2/(H2)(I2)] = [(1.13)^2/(1.13)(1.13)] = 1
Since Kc = 100, then Qc < Kc which means that Q should be larger to make it equal to Kc at equilibrium. How can it get larger. By moving to the left to make the numerator larger and the denominator smaller so it will shift to the left. Now, reading from right to left we have
............... H2(g) + I2(g) <==> 2 HI(g)
I...............1.13........1.13...........1.13
C...............+x............+x...............-x
E.............1.13+x......1.13+x......1.13-x

Now substitute the E line into the Kc expression and solve for x then evaluate each of the values in the E line to give you the concentrations in mols/L. Obviously you will need to solve a quadratic equation. Post your work if you run into trouble.