F2 (g) + Cl2 (g) = 2 ClF (g) Kc = 272.3

Question

If you put 3.000 M F2 and 3.000 M Cl2 into a container at these conditions and let it go to equilibrium, what are the equilibrium concentrations of F2 and ClF?

Answer 1

Use an ICE table:

F₂(g) + Cl₂(g) ==> 2ClF(g) Kc= 272.3

3.000......3.000............0.......Initial

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

3.000-x....3.000-x........2x.....Equilibrium

Kc = [ClF]² / [F₂][Cl₂]

272.3 = (2x)² / (3-x)(3-x)

272.3 = 4x² / (x² - 6x + 9)

4x² = 272.3x² - 1634x + 2451

268x² - 1634x + 2451 = 0

x = 2.664

At equilibrium,

[F₂] = 3.000 - 2.664 = 0.336 M

[ClF] = 2 x 2.664 = 5.328 M