Chapter 8: Unit 6. Calculating Equilibrium Constant

Calculation Examples of the Equilibrium Constant

Example 1:

Consider the following equilibrium reactions:
a) 2 NO2 (g) ⇔ N2O4 (g) Keq = 2.2
b) Cu2+(aq) + 2 Ag(s) ⇔Cu(s) + 2Ag+ (aq) Keq = 1 x 10-15
c) Pb2+ (aq) + 2 Cl- (aq) ⇔PbCl2(s) Keq = 6.3 x 104
d) SO2(g) + O2 (g) ⇔SO3 (g) Keq = 110

Which equilibrium favors products to the greatest extent?
Which equilibrium favors reactants to the greatest extent?

Answer:

Higher Keq value favors the products and lower Keq value favors the reactants.

Choice C has the largest value of Keq value and hence the equilibrium reaction of Choice C favors the product.

c) Pb2+ (aq) + 2 Cl- (aq) ⇔PbCl2(s) Keq = 6.3 x 104

Choice B has the lowest value of Keq value and hence the equilibrium reaction of Choice B favors the reactants.

b) Cu2+(aq) + 2 Ag(s) ⇔Cu(s) + 2 Ag+ (aq) Keq = 1 x 10-15

Example 2:

SO3(g) + H2O(g) ⇔H2SO4(g)
At equilibrium [SO3] = 0.400M [H2O] = 0.480M [H2SO4] = 0.600M
Calculate the value of the equilibrium constant.
Answer:

Keq = [ H2SO4(g) ] / [ SO3(g) ] x [ H2O(g) ] = [ 0.600M ] / [ 0.400M ] x [ 0.480M } = 3.13

Example 3:

At equilibrium at 100oC, a 2.0L flask contains:
0.075 mol of PCl5 0.050 mol of H2O 0.750 mol 0f HCl 0.500 mol of POCl3
Calculate the Keq for the reaction:
PCl5 (s) + H2O (g) ⇔2 HCl (g) + POCl3 (g)
Keq = { [ HCl (g) ]2 [ POCl3 (g) ] } / { [ H2O (g) ] }

PCl5 (s) is solid and hence it is discarded from Keq expression.

Keq = { [ HCl (g) ]2 [ POCl3 (g) ] } / { [ H2O (g) ] } = {[0.750 mol/2.0 L]2 x [0.500 mol/2.0L]}/{[ 0.050
Mol/ 2.0 L]} = 1.40625 = 1.4

Example 4:

Keq= 798 at 25oC for the reaction: 2 SO2 (g) + O2 (g)⇔ 2 SO3 (g).
In a particular mixture at equilibrium, [SO2]= 4.20 M and [SO3]=11.0M. Calculate the equilibrium [O2] in
this mixture at 25oC.

Keq = 798 = {[SO3 (g)]2} / {[SO2 (g)]}2x[O2 (g)]} = {[11.0 M]}2 / {[4.20 M]2 x [O2]}
Isolating [O2]
[O2] = {[11.0 M]}2 / {[4.20 M]2 x [798] = 0.00860 M