Question

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g)

given the following reactions and enthalpies of formation:

12N2(g)+O2(g)→NO2(g),   ΔH∘A=33.2 kJ

12N2(g)+12O2(g)→NO(g),  ΔH∘B=90.2 kJ

Answer 1

2NO(g)+O₂(g) → 2NO₂(g) ; H = ?                   --------------------- (1)

Given reactions are

1/2N₂(g)+O₂(g) → NO₂(g),   ΔH^∘_A =33.2 kJ            --------------------------(2)

1/2N₂(g)+1/2O₂(g) → NO(g),  ΔH^∘_B =90.2 kJ          -------------------------(3)

Eqn(1) can be obtained from Eqn(2) & Eqn(3) as follows :

Eqn(1) = [2x Eqn(1)] + [2xreverse of Eqn(2) ]

So H = (2x ΔH^∘_A ] + [2x(-ΔH^∘_B )]

             = (2x33.2 kJ ) + [2x(-90.2 kJ)]

            = -114.0 kJ

Therefore the enthalpy of the reaction is -114.0 kJ