For the reaction below, Kc = 4.60 × 10⁻⁶. What is the equilibrium concentration of B if the reaction begins with 0.800 M A? A (g) ⇌ 3 B (g)

Answer 1

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

0.8............0..........Initial

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

0.8-x.......3x.........Equilibrium

Kc = [B]³ / [A]

4.60x10^-6 = (3x)3 / 0.8-x

4.60x10^-6 = 27x³ / 0.8-x

3.68x10^-6 - 4.60x10^-6x = 27x³

27x³ + 4.60x10^-6x - 3.68x10^-6 = 0

I'm not going to do the math, but solve this equation for x and then multiply that value by 3 to get the equilibrium concentration of B