Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110 kJ mol–1, – 393 kJ mol–1, 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
Solution 1
ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfHvalue of reactants.
`triangle_r H = sum triangle_f H("products") - sum triangle_f H("reactants")`
For the given reaction,
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
`triangle_rH = [{triangle_f H(N_2O) + 3triangle_f H(CO_2)}-{triangle_fH(N_2O_4) + 3triangle_fH(CO)}]`
Substituting the values of ΔfH for N2O, CO2, N2O4, and CO from the question, we get:
`triangle_rH = [{81 kJ mol^(-1) + 3(-393)kJ mol^(-1)}-{9.7 kJ mol^(-1) + 3(-110)kJ mol^(-1)}]`
`triangle_rH = -777.7 kJ mol^(-1)`
Hence, the value of ΔrH for the reaction is `- 777.7 kJ mol^(-1)`
Solution 2
Enthalpy of reaction (∆r,H) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kj mol-1
