Question

A student ran the following reaction in the laboratory at 529 K: CoCl2(g) P CO(g) + Cl2(g) When she introduced 1.68 moles of CoCl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.61*10-2 M. Calculate the equilibrium constant, Ke she obtained for this reaction. Kc =

Answer 1

ANSWER:

Given,

number of moles COCl2 = 1.68 moles

volume of container = 1.0 L

[COCl2]initial = (1.68 mol)/(1.0 L) = 1.68 M

[Cl2]equilibrium = 3.61 x 10-2 M

reaction is :

COCl2 (g) $\rightleftharpoons$ CO (g) + Cl2 (g)

initial 1.68 M 0.0 M 0.0 M

at equili (1.68 - x) M x M x M

So, x = 3.61 x 10-2 M

[COCl2]euilibrium = (1.68 - 3.61 x 10-2 ) M = 1.644 M

[CO]equilibrium = 3.61 x 10-2 M

[Cl2]equilibrium = 3.61 x 10-2 M

Now,

K = Coci COCl2

3.61 x 10-21[3.61 x 10-21 [1.644

Kc = 7.93 x 10-4