ThermochemistryHard

Question

The strain energy (in kJ/mol) of cyclopropane from the following data is

ΔfH [C3H6(g)] = 53.0 kJ/mol;

ΔfH [C(g)] = 715.0 kJ/mol;

ΔfH [H(g)] = 218.0 kJ/mol

B.E. (C−C) = 356.0 kJ/mol;

B.E. (C−H) = 408.0 kJ/mol.

Options

A.770 kJ
B.116 kJ
C.240 kJ
D.346 kJ

Solution

$3C(s) + 3H_{2}(g) \rightarrow C_{3}H_{6}(g);\Delta{H\text{ kJ}}_{\exp}$

$\Delta H_{\text{theo}} = \lbrack 3 \times 715 + 6 \times 218\rbrack - \lbrack 3 \times 356 + 6 \times 408\rbrack = - 63\text{ kJ}$

$\therefore\text{ Strain energy = }\Delta H_{\text{exp}} - \Delta H_{\text{theo}} = 116\text{ kJ}$

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