Question

For the reaction: 2H2O(g) ⇌ 2H2(g) + O2(g) KC = 2.40 x 10-3 at a given temperature. At equilibrium, it is found that [H2O] = 0.294 M and [H2] = 1.80 x 10-2 M. If the reaction is run in a 5.50 L container, how many moles of O2 are present at equilibrium?

Answer 1

Given that

At equilibrium, [H2O] = 0.294 M and [H2] = 1.80 x 10^-2 M

volume of the container = 5.5 L

Hence, at equilibrium

moles of H2O = molarity x volume = 0.294 M x 5.5 L = 1.617 moles

[H2O] = 1.617 moles

moles of H2 = molarity x volume = 1.80 x 10^-2 M x 5.5 L = 0.099 moles

[H2] = 0.099 moles

Given that Kc = 2.40 x 10^-3

2H2O(g) ⇌ 2H2(g) + O2(g)

Kc = [H2]² [O2] / [H2O]²

[O2] =  Kc  [H2O]²/ [H2]²

= (2.40 x 10^-3) (1.617)² / (0.099)²

      = 0.64 moles

[O2] = 0.64 moles

Therefore,

0.64 moles of O2 are present at equilibrium.