1. Given this NET reaction
HESS LAW CH 5 # 64 10/10/07
Given this NET reaction
And the following mechanism, find enthalpy of reaction ∆H, for the above reaction.
N2O(g) +
NO2 (g)
3 NO (g) 
This reaction was halved to get one N2O as in NET
2 N2O(g) 
2N2(g) +
O2 (g)
N2O(g)
N2(g) +
1/2O2 (g) 
2 NO (g) +
O2 (g) 
2 NO2(g)
NO2 (g) 
NO (g) +
1/2O2 (g)
N2(g) +
O2 (g) 
2 NO (g)
N2(g) +
O2 (g) 
2 NO (g)
This reaction was halved to cancel O2 and get one NO2 as in NET. It was reversed to place NO2 as a reactant!

2. NOTICE REACTION 2 WAS REVERSED AND THE SIGN OF ∆H2 WAS RECIPROCATED
2 N2O(g) 
2N2(g) +
O2 (g)
∆H1 = kJ
∆H2 = kJ
2 NO (g) +
O2 (g) 
2 NO2(g)
∆H3 = kJ
N2(g) +
O2 (g) 
2 NO (g)
NOTICE REACTION 2 WAS REVERSED AND THE SIGN OF ∆H2 WAS RECIPROCATED
N2O(g)
N2(g) +
1/2O2 (g)
∆H1 = ½(-163.2kJ) 
NO2 (g) 
NO (g) +
1/2O2 (g)
∆H2 = ½( kJ)
N2(g) +
O2 (g) 
2 NO (g)
∆H3 = kJ
N2O(g) +
NO2 (g)
3 NO (g)
∆H3 = kJ 