Consider the reaction: 2 AB(g) ↔ A2(g) + B2(g), Kc = 5.69 The reaction vessel initially contained 0.713M AB(g) and no A2 or B2. Determine the concentration of A2 when equilibrium is reached.

Answer 1

Use an ICE table:

2AB(g) <===> A₂(g) + B₂(g)

0.713................0............0.............Initital

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

0.713-2x...........x............x.............Equilibrium

Write the equilibrium expression, and solve:

Kc = [A₂][B₂] / [AB]²

5.69 = (x)(x) / (0.713 - 2x)²

5.69 = x² / 4x² -2.85x + 1.43

21.8x² - 16.2x + 8.14 = 0

x = 0.372 M = [A]

(be sure to check ALL of the math)