Asked by sstt
Dinitrogen tetraoxide (N2O4) is known to be used, along with hydrogen, as rocket fuel by NASA since 1950s. It is prepared as follows:
N2 (g) + 2O2 (g) N2O4 (g)
Calculated, using the following data, what would be the enthalpy of formation of N2O4?.
(i) 2NO2 (g) N2 (g) + 2O2 (g) H = –118.7 kJ
(ii) 2NO2 (g) N2O4 (g) H = –101.9 kJ
All Answers 1
Answered by
GPT-5 mini
AI
Use Hess's law.
From (i): 2 NO2 -> N2 + 2 O2, ΔH = -118.7 kJ
0 (N2) + 0 (O2) - 2 ΔHf(NO2) = -118.7 kJ → ΔHf(NO2) = +59.35 kJ/mol
From (ii): 2 NO2 -> N2O4, ΔH = -101.9 kJ
ΔHf(N2O4) - 2 ΔHf(NO2) = -101.9 kJ
ΔHf(N2O4) = -101.9 kJ + 2(59.35 kJ) = +16.8 kJ/mol
Answer: ΔHf°(N2O4, g) = +16.8 kJ/mol.
From (i): 2 NO2 -> N2 + 2 O2, ΔH = -118.7 kJ
0 (N2) + 0 (O2) - 2 ΔHf(NO2) = -118.7 kJ → ΔHf(NO2) = +59.35 kJ/mol
From (ii): 2 NO2 -> N2O4, ΔH = -101.9 kJ
ΔHf(N2O4) - 2 ΔHf(NO2) = -101.9 kJ
ΔHf(N2O4) = -101.9 kJ + 2(59.35 kJ) = +16.8 kJ/mol
Answer: ΔHf°(N2O4, g) = +16.8 kJ/mol.
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