Chem 12 ice table

Question

At 50 degrees C the reaction:

HBr (g) +FCl (g) <-> HF (g) +BrCl (g) has Keq = 17.8.

2 mol of each reactant is added 1 L flask. Determine concentrations for each gas at equilibrium.

Answer 1

HBr (g) +FCl (g) <-> HF (g) + BrCl (g) ... Keq = 17.8.

2mol......2mol.............0mol.....0mol......Initial

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

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

K = 17.8 = [HF][BrCl] / [HBr][FCl]

17.8 = (x)(x) / (2-x)(2-x)

17.8x² - 71.2x + 71.2 = x²

16.8x² - 71.2x + 71.2 = 0 (use quadratic equation to solve for x)

x = 1.62

Substitute this value of x into the ICE table to get final equilibrium concentrations:

[HBr] = 2 - 1.6 = 0.4 M

[FCl] = 2 - 1.6 = 0.4 M

[HF] = 1.6 M

[BrCl] = 1.6 M