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), ΔfH∘A=33.2 kJ mol−1
12N2(g)+12O2(g)→NO(g), ΔfH∘B=90.2 kJ mol−1

Answer 1

We write the enthalpy change equation:

2ΔH_formation(NO) = ΔH_reaction + 2ΔH_formation(NO₂)
2(90.2) = ΔHr + 2(33.2)
ΔHr = 114 kJ/mol

The reaction releases 114 kJ/mol