Question

A mixture of 0.327 M Cl 2 , 0.579 M F 2 , and 0.839 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g ) − ⇀ ↽ − 2 ClF ( g ) K c = 20.0 at 2500 K Calculate the equilibrium concentration of each gas at 2500 K .

Answer 1

Cl2 (g) + F2 (g) ⇌ 2ClF (g)

Kc = [ClF]^2 / [Cl2][F2]

Let x be the change in concentration of ClF, Cl2, and F2 at equilibrium. Then the equilibrium concentrations can be expressed as:

[ClF] = 0.839 M + x

[Cl2] = 0.327 M - x

[F2] = 0.579 M - x

Substituting these expressions into the equilibrium constant expression and solving for x gives:

20.0 = ([0.839 + x]^2) / ([0.327 - x][0.579 - x])

Expanding the numerator and denominator and simplifying, we get:

20.0 = (0.704x^2 + 3.321x + 0.702) / (-0.189x^2 + 0.463x - 0.190)

Multiplying both sides by the denominator and rearranging, we get a quadratic equation:

0.189x^2 - 3.880x + 3.032 = 0

Using the quadratic formula, we find that:

x = 7.68 × 10^-2 M

Substituting this value back into the expressions for the equilibrium concentrations gives:

[ClF] = 0.839 M + 7.68 × 10^-2 M = 0.917 M

[Cl2] = 0.327 M - 7.68 × 10^-2 M = 0.250 M

[F2] = 0.579 M - 7.68 × 10^-2 M = 0.501 M

Therefore, the equilibrium concentrations of ClF, Cl2, and F2 at 2500 K are 0.917 M, 0.250 M, and 0.501 M, respectively.