For the reaction below, Kc = 9.2 × 10⁻⁵. What is the equilibrium concentration of D if the reaction begins with 0.72 M A? A (aq) + 2 B (s) ⇌ C (s) + 2 D (aq)

Answer 1

Use an ICE table to begin:

A(aq) + 2B(s) <===> C(s) + 2D(aq)

0.72...........................................0.........Initial

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

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

Keq = [D]² / [A] (note that solids are not included in the equilibrium expression)

9.2x10^-5 = (2x)² / (0.72-x)

4x² = (9.2x10^-5) (0.72-x)

Solve for x and the equilibrium concentration of D will be 2 times that value.