For the reaction below, Kc = 4.60 × 10⁻⁶. What is the equilibrium concentration of B if the reaction begins with 0.250 M A? A (g) ⇌ 3 B (g)

Answer 1

A(g) <===> 3B(g)

0.25.............0........Initial

-x...............+3x......Change

0.25-x.........3x.......Equilibrium

Kc = [B]3 / [A]

4.6x10^-6 = (3x)³ / 0.25-x = 27x³ / 0.25-x

Solve for x and multiply by 3 to get the equilibrium concentration of B