Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryEnthalpy of FormationMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
Study the following thermochemical equations:
(a) N2O(g) + 3H2(g) → H2O(l) + N2H4(l); ΔH = -75.56 kcal
(b) 4N2(g) + 3H2O(l) → 2NH3(g) + 3N2O(g); ΔH = +241.35 kcal
(c) N2H4(l) + H2O(l) → 2NH3(g) + 1/2 O2(g); ΔH = +34.18 kcal
(d) H2O(l) → H2(g) + 1/2 O2(g); ΔH = +68.32 kcal
The enthalpy of formation (in kcal/mol) of liquid N2H4 is
Options:
Answer:
C
Solution:
As solved in Q1, the equations h = o - 7.24 and 3o + h = 70.53 yield o = 19.44 and h = +12.20 kcal/mol. The answer is (c).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Formation
⏱ 2mℹ️ Source: QB
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