Question

Question 6.12 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)

Class XI Thermodynamics Page 183

Answer 1

Reaction is ,
N2O4(g) + 3CO(g) -----> N2O(g) + 3CO2(g)

we know,
heat of reaction ( ∆rH°) = ( sum of heat of formation of products ) - ( sum of heat of formation of reactants )

e.g., ∆rH° = [∆fH°(N2O) + 3fH°(CO2) ] - [∆fH°(N2O4) + 3fH°(CO) ]

Given, enthalpies of formation of CO, CO2 , N2O, and N2O4 are -110, -393 , 81 and 9.7 KJ/mol respectively .

now, ∆rH° = [81 + 3(-393)] - [9.7 + 3(-110)]
= -777.7 KJ

Answer 2

Answer:

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