The strain energy (in kJ/mol) of cyclopropane from the following data:<br>; ; <br>BE (C–C) = 356.0 k — Thermodynamics and Thermochemistry Chemistry Question
Question
The strain energy (in kJ/mol) of cyclopropane from the following data:<br>$\Delta_f H [\text{C}_3\text{H}_6\text{(g)}] = 53.0\text{ kJ/mol}$; $\Delta_f H [\text{C(g)}] = 715.0\text{ kJ/mol}$; $\Delta_f H [\text{H(g)}] = 218.0\text{ kJ/mol}$<br>BE (C–C) = 356.0 kJ/mol; BE (C–H) = 408.0 kJ/mol.
Answer: B
💡 Solution & Explanation
Theoretical atomization energy of $\text{C}_3\text{H}_6$ (3 C-C, 6 C-H) $= 3(356) + 6(408) = 1068 + 2448 = 3516\text{ kJ/mol}$. Experimental atomization energy $= 3\Delta H_f(\text{C,g}) + 6\Delta H_f(\text{H,g}) - \Delta H_f(\text{C}_3\text{H}_6\text{,g}) = 3(715) + 6(218) - 53 = 2145 + 1308 - 53 = 3400\text{ kJ/mol}$. Strain energy = Theoretical - Experimental $= 3516 - 3400 = 116\text{ kJ/mol}$.
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