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 bond enthalpy of C-C bond (kJ/mol) is
Options:
Answer:
B
Solution:
Atomise n-butane: 4×717.0 + 5×436.0 = 2868 + 2180 = 5048 kJ (5 mol H2 = 10 H atoms). n-Butane has 10 C-H and 3 C-C bonds. -126.0 = 5048 - (10×413.0 + 3 εC-C), so 3 εC-C = 5048 + 126 - 4130 = 1044 and εC-C = 348.0 kJ/mol. This is option (b).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
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