Pe baza urmatoarelor reactii termochimice:
a) NO (g) + O3(g) -> NO2(g) + O2(g) delta H1 = -199kj
b) O3(g) -> 1,5 O2(g) -> delta H2= -142 Kj
c) O2(g) -> 2[O] delta H3 = 495 Kj
Calculati variatia de entalpie a reactiei NO(g) + [O](g) -> NO2(g)

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Răspunsuri

2014-07-13T15:00:08+03:00
Delta H = H1-H2-1/2H3
NO+O3-O3-1/2O2--->NO2+O2-3/2O2-[O]
=>NO+[O]--->NO2
delta H = -199-(-142)-495=-552 kj
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