Given this NET reaction

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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!

NOTICE REACTION 2 WAS REVERSED AND THE SIGN OF ∆H2 WAS RECIPROCATED 2 N2O(g)  2N2(g) + O2 (g) ∆H1 = -163.2kJ ∆H2 = -113.1 kJ 2 NO (g) + O2 (g)  2 NO2(g) ∆H3 = +180.7 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(+113.1 kJ) N2(g) + O2 (g)  2 NO (g) ∆H3 = +180.7 kJ N2O(g) + NO2 (g) 3 NO (g) ∆H3 = +155.7 kJ 