Calculate the concentrations of all gases once equilibrium is reestablished.

Question

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.22-L rigid container are [H₂] = 0.0501 M, [F₂] = 0.0104 M, and [HF] = 0.409 M.

H₂(g) + F₂(g)  2 HF(g)

If 0.207 mol of F₂ is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

[H2]	M
[F2]	M
[HF]

Answer 1

Let us find the Keq for this reaction:

H₂(g) + F₂(g) <==> 2HF(g)

Keq = [HF]² / [H₂][F₂] = (0.409)² / (0.0501)(0.0104)

Keq = 321

Adding 0.207 moles F2 to a 5.22 L container produces is equivalent to 0.207 mol / 5.22 L = 0.0397 M

Adding this to the equilibrium [F2] we get 0.0397 + 0.0104 = 0.0501 M

This will push the reaction to the right toward the product side (Le Chatelier)

H₂(g) + F₂(g) <==> 2HF(g)

0.0501... 0.0501.........0.409....Initial

-x.............. -x...............+2x........Change

0.0501-x....0.0501-x.....2x......Equilibrium

Keq = 321 = (0.0501-x)(0.0501-x) / 2x

The rest is algebra. Solve for x and subtract that value from 0.0501 to get the concentration of H₂ and F₂. Multiply the value of x by 2 to get the concentration of HF.