CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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)

Open in App
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


flag
Suggest Corrections
thumbs-up
6
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon