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Question

Enthalpies of formation of CO(g),CO2(g),N2O(g) and N2O4(g)are11 0kJ mol1,393 kJ mol1,81 kJ mol1 and 9.7 kJ mol1 respectively. Find the value of ΔrH for the reaction:
N2O4(g)+3CO(g)N2O(g)+3CO2(g)

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Solution

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 mol1+3(393)kJ mol1}{9.7 kJ mol1+3(110) kJ mol1}]ΔrH=777.7 kJ mol1
Hence, the value of ΔrH for the reaction is 777.7 kJ mol1


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