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-C bond (kJ/mol) is
Options:
Answer:
Solution:
To atomize C₃H₈ into gaseous atoms, the energy required is:
(3 × 719.0) + (4 × 435.0) = 2157 + 1740 = 3897 kJ
(Note: 4 mol of H₂ yields 8 H atoms, as each H₂ dissociation produces 2 H atoms).
C₃H₈ contains 8 C-H bonds and 2 C-C bonds. The enthalpy of formation, Δf H, is given by the atomization energy minus the bond energy released.
So, we have:
Δf H = (atomization energy) - (bond energy released)
-85.0 = 3897 - (8 × 414.0 + 2 εC-C)
Now, we can solve for 2 εC-C:
2 εC-C = 3897 + 85 - 3312 = 670
Therefore, the bond energy for a C-C bond, εC-C, is:
εC-C = 670/2 = 335.0 kJ/mol
This corresponds to option (b).
