Chemistry - Thermochemistry Question with Solution | TestHub
The strain energy (in kJ/mol) of cyclopropane from the following data:
Δf H [C3H6(g)] = 53.0 kJ/mol; Δf H [C(g)] = 715.0 kJ/mol; Δf H [H(g)] = 218.0 kJ/mol
BE (C-C) = 356.0 kJ/mol; BE (C-H) = 408.0 kJ/mol.
Options:
Answer:
Solution:
For an unstrained cyclopropane ring, which contains three C–C bonds and six C–H bonds, we first calculate the energy required to atomize the constituent elements:
3C(s) + 3H2(g) tarrow atomized elements
The energy for atomization is 3(715) + 6(218) = 3453 kJ.
Next, we calculate the energy released when these atomized elements form the bonds in an unstrained cyclopropane ring:
3(C–C bond energy) + 6(C–H bond energy) = 3(356) + 6(408) = 3516 kJ
Therefore, the enthalpy of formation for an unstrained cyclopropane ring, Δ H°f (unstrained), is the energy absorbed during atomization minus the energy released during bond formation:
Δ H°f (unstrained) = 3453 kJ - 3516 kJ = -63 kJ/mol
The strain energy is the difference between the actual enthalpy of formation and the calculated unstrained enthalpy of formation:
Strain energy = Δ H°f (actual) - Δ H°f (unstrained)
Given the actual enthalpy of formation is 53 kJ/mol, the strain energy is:
Strain energy = 53 kJ/mol - (-63 kJ/mol) = 116 kJ/mol
This corresponds to option (b).
