From the following data, calculate the enthalpy change (in kJ/mol) for the combustion of cyclopropan — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Enthalpy of isomerization $\text{cyclopropane} \to \text{propene}$ is -33 kJ/mol. $\Delta H_f(\text{propene}) - \Delta H_f(\text{cyclopropane}) = -33 \implies 20 - \Delta H_f(\text{cyclopropane}) = -33 \implies \Delta H_f(\text{cyclopropane}) = 53\text{ kJ/mol}$. Combustion of cyclopropane ($\text{C}_3\text{H}_6$): $\Delta H_c = 3(-394) + 3(-286) - 53 = -1182 - 858 - 53 = -2093\text{ kJ/mol}$.