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At 40 °C, Kc = 3.03×10-2 for the reaction below. H2 + Cl2 ⇌ 2 HCl...

At 40 °C, Kc = 3.03×10-2 for the reaction below. H2 + Cl2 ⇌ 2 HCl In an experiment carried out at this temperature, the initial concentrations were: [H2]o = [Cl2]o = [HCl]o = 1.50 mol L−1 What are the equilibrium concentrations?

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Answer #1

H2      + Cl2    <--------------->   2 HCl

1.50    1.50                           1.50 M

1.50-x       1.50 -x                        1.50 + 2x

Kc = [HCl]^2 / [H2][Cl2]

3.03 x 10^-2 = (1.50 + 2x)^2 / (1.50 - x)^2

0.174 = 1.50 + 2x / 1.50 - x

0.2611 - 0.174x = 1.50 + 2x

1.2389 = - 2.174 x

x = 0.570

equilibrium concentrations :

[H2] = 2.07 M

[Cl2] = 2.07 M

[HCl] = 0.360 M

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