Transcript PRON_5.64_.ppt

HESS LAW CH 5 # 64 10/10/07
N2O(g) + NO2 (g)  3 NO (g) Given this NET reaction
And the following mechanism, find enthalpy of reaction ∆H, for the
This reaction was
above reaction.
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 N2O(g)  2N2(g) + O2 (g)
2 NO (g) + O2 (g)  2 NO2(g)
∆H1 = -163.2kJ
∆H2 = -113.1 kJ
N2(g) + O2 (g)  2 NO (g)
∆H3 = +180.7 kJ
NOTICE REACTION 2 WAS
REVERSED AND THE SIGN
OF ∆H2 WAS
RECIPROCATED
N2O(g)  N2(g) + 1/2O2 (g)
NO2 (g) NO (g) + 1/2O2 (g)
N2(g) + O2 (g)  2 NO (g)
N2O(g) + NO2 (g)  3 NO (g)
∆H1 = ½(-163.2kJ)
∆H2 = ½(+113.1 kJ)
∆H3 = +180.7 kJ
∆H3 = +155.7 kJ