Chemistry - Thermochemistry Question with Solution | TestHub
Use the data (all values are in kJ per mole at 25°C) given below to answer the following:
Enthalpy of formation of CH3CN = +88.0
Enthalpy of formation of C3H8 = -85.0
Enthalpy of sublimation of graphite = 719.0
Enthalpy of dissociation of nitrogen = 948.0
Enthalpy of dissociation of hydrogen = 435.0
Bond enthalpies: C-H = 414.0; C-N = 378.0; N-H = 426.0
The bond enthalpy of C≡N bond (kJ/mol) is
Options:
Answer:
Solution:
For CH₃CN, the formation reaction is:
2C(graphite) + 3/2H2 + 1/2N2 tarrow CH3CN
The atomization energy is calculated as:
2 × 719.0 + 3/2 × 435.0 + 1/2 × 948.0 = 1438 + 652.5 + 474 = 2564.5 kJ
The bonds in CH₃CN are: 3 C-H bonds, 1 C-C bond, and 1 C≡N bond.
Using the given enthalpy of formation, we can set up the equation:
88.0 = 2564.5 - (3 × 414.0 + 335.0 + εC≡N)
Solving for εC≡N:
εC≡N = 2564.5 - 1242 - 335 - 88 = 899.5 kJ/mol
This corresponds to option (c).
