Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryBond EnthalpyEasy2 minQB
ChemistryEasysingle 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≡N bond (kJ/mol) is
Options:
Answer:
C
Solution:
CH3CH2CN is formed from 3C + 5/2H2 + 1/2N2. Atomisation: 3×717.0 + 2.5×436.0 + 0.5×945.0 = 2151 + 1090 + 472.5 = 3713.5 kJ. Bonds: 5 C-H, 2 C-C, 1 C≡N. 51.0 = 3713.5 - (5×413.0 + 2×348.0 + εC≡N) = 3713.5 - 2761 - εC≡N, so εC≡N = 901.5 kJ/mol. This is option (c).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
