calculate the concentration of H₂ after equilibrium is reestablished.

Question

For the following reaction at a certain temperature H₂(g) + F₂(g) ⇌ 2HF(g) the equilibrium concentrations in a 10.00-L rigid container are [H₂] = 0.0600 M, [F₂] = 0.0200 M, and [HF] = 0.620 M. If 0.300 moles of F₂ is added to the equilibrium mixture, calculate the concentration of H₂ after equilibrium is reestablished.

Answer 1

Introducing more fluorine gas into a fixed volume system results in a few changes:

1. The concentration of the reactants in the reaction increases.
2. The number of moles of gas increases.

In the reaction, the number of moles of gas on each side of the equation is identical, and so the reaction will not shift any direction in order to compensate for the change in moles of gas. However, the reaction will shift to the right to compensate for the increase in concentration of reactants, according to Le Chatelier's principle.

We have to solve this in steps:

1. Find the equilibrium constant for the reaction, under the conditions of the problem.
2. Identify the new starting concentrations of the reactants.
3. Determine the changes in concentration that will make the final state satisfy equilibrium.

Step 1:

Keq = [HF]² / ( [F₂] [H₂] ) = (0.620 M)² / (0.06 M * 0.02 M ) = 320.33

Step 2:

0.3 mol F₂ was added to a 10L container, so the concentration of F₂ should increase by 0.3 mol / 10L = 0.03 M. Thus, the new concentration is 0.02 M + 0.03M = 0.05 M.

Step 3:

Suppose the reaction proceeds to the right "x times", where we measure x in molar. Then, the new concentrations of all the chemicals will be:

F₂: 0.05 - x

H₂: 0.06 - x

HF: 0.620 + 2x

For these to satisfy equilibrium, we need:

(0.620 + 2x)^2 / ((0.05 - x) (0.06 - x)) = Keq = 320.33

This can be solved via the quadratic equation if it needs to be done by hand, or it can just be input to Wolfram alpha.

It has two solutions, x = {0.10122, 0.01801}. The first solution would result in a negative concentration for hydrogen and fluorine gas, so we can ignore it and choose the second as the real answer (bolded).

Thus, the final concentration of H₂ is:

[H₂] = 0.0600M - x = 0.0600M - 0.01801M = 0.0420M.

Answer 2

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