The reaction is shown below: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) The enthalpy change for the reaction is calculated as: ΔH°reaction=∑m.ΔH°f( products )−∑n.ΔH°f( reactants ) =[1.ΔH°f(N2O)+3.ΔH°f(CO2)]−[1.ΔH°f(N2O4)+3.ΔH°f(CO)] ΔH°reaction=[1(81)+3(−393)]−[1(−10)+3(−10)] =−1098+40 =−1058kJ∕mol