Question

Consider the following reaction.

A(aq) ⇄ 3B(aq)   Kc=1.79*10^-6 at 500K

If a 2.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

The given reaction in equilibrium is

-------------------- A(aq) <======> 3B(aq); Kc = 1.79*10^-6

Init.conc(M): 2.90, -------------------- 0

Eqm.conc(M):(2.90 - x), ------------- 3x

The equilibrium constant, Kc for the above reaction can be written as

Kc = 1.79*10^-6 = [B(aq)]³ / [A(aq)] = (3x)³ / (2.90 - x) = 27x³ / (2.90 - x)

=>27x³ + 1.79*10^-6 x - 5.191*10^-6 = 0

=> x = 5.77*10^-3 M

Hence concentration of B(aq) at equilibrium = 3x = 3* 5.77*10^-3 M = 0.0173 M (answer)