Question

Consider the following reaction. A(aq) ⇄ 3B(aq)   Kc=1.79*10^-6 at 500K If a 2.90 M sample of...

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?

Homework Answers

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)

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