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For the reaction below at a certain temperature, it is found that the equilibrium concentrations in...

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.05-L rigid container are [H2] = 0.0523 M, [F2] = 0.0121 M, and [HF] = 0.450 M.

H2(g) + F2(g) equilibrium reaction arrow 2 HF(g)

If 0.203 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

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Answer #1

Concentration equilibrium constant, Kc = [HF]eq2 / [H2]eq[F2]eq

Kc = (0.450 M)2 / [(0.0523 M) * (0.0121 M)2]

Kc = 320

moles F2 added = 0.203 mol

concentration F2 added = (moles F2 added) / (volume of container)

concentration F2 added = (0.203 mol) / (5.05 L)

concentration F2 added = 0.0402 M

new concentration F2 = (0.0121 M) + (0.0402 M)

new concentration F2 = 0.0523 M

ICE table H2 (g) F2 (g) \rightleftharpoons 2 HF (g)
Initial conc. 0.0523 M 0.0523 M 0.450 M
Change -x -x +2x
Equilibrium conc. 0.0523 M - x 0.0523 M - x 0.450 M + 2x

Kc = [HF]eq2 / [H2]eq[F2]eq

320 = (0.450 M + 2x)2 / [(0.0523 M - x) * (0.0523 M - x)]

Solving for x, x = 0.0244 M

equilibrium concentration H2 = 0.0523 M - x

equilibrium concentration H2 = 0.0523 M - 0.0244 M

equilibrium concentration H2 = 00279 M

equilibrium concentration F2 = 0.0523 M - x

equilibrium concentration F2 = 0.0523 M - 0.0244 M

equilibrium concentration F2 = 00279 M

equilibrium concentration HF = 0.450 M + 2x

equilibrium concentration HF = 0.450 M + 2 * (0.0244 M)

equilibrium concentration HF = 0.499 M

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