Question 12
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.
ΔrH = 
For the given reaction,
N2O4(g)+ 3CO(g) → N2O(g) + 3CO2(g)
ΔrH = [{ΔfH (N2O) + 3ΔfH(CO2)} - {ΔfH(N2O4) + 3ΔfH(CO)}]
Substituting the values of ΔfH for N2O, CO2, N2O4,and CO from the question, we get:
ΔrH = [{81 KJ mol-1 + 3(-393) KJ mol-1} - {9.7KJ mol-1 + 3(-110)KJ mol-1}]
ΔrH = -777.7 KJ mol-1
Hence, the value of ΔrH for the reaction is -777.7 KJ mol-1.
