Enthalpy- Chemistry

Question

Calculate ΔH^o for the process ½ N₂(g) + ½ O₂(g) → NO(g)

from the following information

N2(g) + 2 O2(g) → 2 NO2(g)	ΔHo = -129.7 kJ/mol
2 NO(g) + O2 → 2 NO2(g)	ΔHo = -417.1 kJ/mol

Answer 1

This is an example of Hess' Law

1/2 N₂(g) + 1/2 O₂(g) <==> NO(g) TARGET EQUATION

eq.1: N₂(g) + 2 O₂(g) → 2 NO₂(g) ... ∆Hº = -129.7 kJ/mol

eq.2: 2 NO(g) + O₂ → 2 NO₂(g) ... ∆Hº = -417.1 kJ/mol

Procedure:

copy eq.1 & divide by 2: 1/2 N₂(g) + O₂(g) → NO₂(g) ... ∆Hº = -64.85 kJ/mol

reverse eq.2 & divide by 2: NO₂(g) -->NO(g) + 1/2 O₂ ... ∆Hº = +208.6 kJ/mol

Add these together and combine/cancel like terms to get the following:

1/2 N₂(g) + O₂(g) → NO₂(g)

NO₂(g) -->NO(g) + 1/2 O₂

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1/2 N₂(g) + O₂(g) + NO₂(g) --> NO₂(g) + NO(g) + 1/2 O₂

1/2 N₂(g) + 1/2 O₂(g) --> NO(g) TARGET EQUATION

∆Hº = -64.85 + 208.6 = 143.8 kJ/mol