Enthalpies of formation of CO(g),CO2(g),N2O(g) and N2O4(g)are–11 0kJ 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)
ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.
ΔrH=∑ΔfH (products) −∑ΔfH (rectants)
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:
ΔfH=[{81 kJ mol−1+3(−393)kJ mol−1}−{9.7 kJ 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