Hess Law Chemistry

Question

Calculate the enthalpy for: NO (g) + 0.5O2 (g) → NO2 (g),

Given: 0.5N2 (g) + 0.5O2 (g) → NO (g) ΔrH = 92.5 kJ mol-1

0.5N2 (g) + O2 (g) → NO2 (g) ΔrH = 33.9 kJ mol-1

Answer 1

NO (g) + 0.5 O2 (g) → NO2 (g) ... TARGET EQUATION

Given:

(1) 0.5 N2 (g) + 0.5 O2 (g) → NO (g) ΔrH = 92.5 kJ mol-1

(2) 0.5 N2 (g) + O2 (g) → NO2 (g) ΔrH = 33.9 kJ mol-1

Procedure:

reverse (1): NO ==> 0.5 N2 + 0.5 O2 ... ∆H = -92.5 kJ

copy (2): 0.5 N2 (g) + O2 (g) ==> NO2 (g) ... ΔrH = 33.9 kJ

Add them together:

NO + 0.5 N2 + O2 ==> 0.5 N2 + O.5 O2 + NO2

Cancel like terms to get ...

NO + + 0.5 O2 ==> NO2 ... TARGET EQUATION

∆H = -92.5 + 33.9 = -58.6 kJ