Consider the reaction. A(aq)↽−−⇀3B(aq) 𝐾c=3.14×10−6 at 500. If a 3.20 M sample of A is heated to 500 K , what is the concentration of B at equilibrium?

Answer 1

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

3.20.............0...........Initial

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

3.2-x............3x.........Equilibrium

Kc = 3.14x10^-6 = [B]³ / [A] = (3x)³ / 3.20-x (assume x is small relative to 3.20 and ignore it)

3.14x10^-6 = 27x³ / 3.20

1.00x10^-5 = 27x³

x³ = 3.70x10^-7

x = 7.18x10^-3 M

[B] @ equilibrium = 3x = 3 x 7.18x10^-3 M = 2.15x10^-2 M