Question

Consider the following reaction. A(aq) <--> 2B(aq) Kc = 4.69 x 10^-6 at 500K If a 1.80 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

The Kc expression

Kc = [B]^2 / [A]

initially

[B] = 0

[A] = 1.8

in equilibrium, due to stoichioemtry

[B] = 0 + 2x

[A] = 1.8 - x

now, substitute in Kc

Kc = [B]^2 / [A]

4.69*10^-6 = (2x)^2 / (1.8-x)

(4.69*10^-6)(1.8) - (4.69*10^-6)x = 4x^2

4x^2 + (4.69*10^-6)x - 0.000008442 = 0

solve for x

x = 0.001452

[B] = 0 + 2*0.001452 = 0.002904

[A] = 1.8 - 0.001452 = 1.798548

now, proove:

Kc = [B]^2 / [A] = (0.002904^2)/(1.798548) = 0.0000046889 = 4.6889*10^-6; which is pretty near to KC shown above