2H2 + O2(g) ---> 2H2(l) ΔH=-571,5 KJ
N2O5(g) + H2O(l) ------> 2HNO3(l) ΔH=-76,6 KJ
1/2 N2 (g) + 3/2O1 + 1/2 H2(g) ----> HNO3(l) ΔH= -174,1 KJ
Calcule ΔH para a reação
2N2(g)+5O2(g) ---> 2N2O5
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H= -571,5 + (-76,6) + (-174,1) = -882,2
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Verified answer
2H2 + O2(g) ---> 2H2(l) ΔH=-571,5 KJ
N2O5(g) + H2O(l) ------> 2HNO3(l) ΔH=-76,6 KJ
1/2 N2 (g) + 3/2O1 + 1/2 H2(g) ----> HNO3(l) ΔH= -174,1 KJ
________________________________________________
2N2(g)+5O2(g) ---> 2N2O5
H= -571,5 + (-76,6) + (-174,1) = -882,2