concentration of reaction

Question

Consider the reaction.

A(aq)↽−−⇀3B(aq) 𝐾c=4.82×10^−6 at 500 K

If a 2.40 M sample of A is heated to 500 K , what is the concentration of B at equilibrium?

Answer 1

A(aq) <===> 3B(aq) Kc = 4.82x10^-6

2.4 M..............0......Initial

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

2.4-x...............3x....Equilibrium

Kc = 4.82x10^-6 = [B]³ / [A] = (3x)³ / 2.4 - x (if we ignore x in the denominator we can avoid some math)

4.82x10^-6 = 27x³ / 2.4

1.16x10^-5 = 27x³

x³ = 4.28x10^-7

x = 7.5x10^-3 M (note: this is insignificant compared to 2.4 M so our assumption to ignore it was valid)

[B] @ equilibrium = 7.5x10^-3 M