Question

Given the following reaction:

H2(g) + I2(g) <---> 2HI(g) Kc = 57.85

1.50 mol of each reactant was placed in a 2.00-L flask.

At equilibrium what is the concentration of HI?
At equilibrium what is the concentration of H2 and I2?

Answer 1

The equilibrium concentration of H2 and I2 is 0.165 M. The equilibrium concentration of HI is 1.17 M.

What is Kc?

The term Kc refers to the equilibrium constant for a aqueous phase reaction. The concentrations of each specie is;

H2 = 1.5/2 = 0.75 M

I2 = 1.5/2  = 0.75 M

Setting up the ICE table;

        H2(g) +     I2(g) <---> 2HI(g)

I      0.75          0.75             0

C     -x                -x                 +2x

E    0.75 - x     0.75 - x           2x

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

57.85 = (2x)^2/(0.75 - x)^2

57.85(0.56 - 1.5x + x^2) = 4x^2

32.3 - 86.7x + 57.85x^2 = 4x^2

4x^2 - 57.85x^2 + 86.7x - 32.3 = 0

-53.85x^2 + 86.7x - 32.3 = 0

x = 0.585 M because x can not be greater than the initial concentration.

At equilibrium;

[HI] = 2x = 2(0.585) = 1.17 M

[H2] =[I2] = 0.75 M  - 0.585 M = 0.165 M

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