Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryBond EnthalpyMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
Use the data (all values are in kJ per mole at 25°C) given below to answer the following:
Enthalpy of formation of CH3CH2CN = +51.0
Enthalpy of formation of C4H10 (n-butane) = -126.0
Enthalpy of sublimation of graphite = 717.0
Enthalpy of dissociation of nitrogen = 945.0
Enthalpy of dissociation of hydrogen = 436.0
Bond enthalpies: C-H = 413.0; C-N = 305.0; N-H = 390.0
The enthalpy of hydrogenation of CH3CH2CN to CH3CH2CH2NH2 (kJ/mol) is
Options:
Answer:
A
Solution:
CH3CH2CN + 2H2 → CH3CH2CH2NH2. Broken: C≡N 901.5 + 2 H-H (2×436.0 = 872) = 1773.5 kJ. Formed: C-N 305.0 + 2 N-H (2×390.0 = 780) + 2 C-H (2×413.0 = 826) = 1911 kJ. ΔH = 1773.5 - 1911 = -137.5 kJ/mol. This is option (a).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
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