Question

Given:
$\Delta H_f^\circ \, \text{CO}_2 (g) = -393.5 \, \text{kJ/mol}$
$\Delta H_f^\circ \, \text{H}_2\text{O}(l) = -286 \, \text{kJ/mol}$
$\Delta H_f^\circ \, \text{C}_3\text{H}_6(g) = +20.6 \, \text{kJ/mol}$
$\Delta H^\circ_{\text{isomerization of Cyclopropane to Propene}} = -33 \, \text{kJ/mol}$
What is the standard enthalpy of combustion of Cyclopropane?

• A. -2092 kJ/mol
• B. -1985 kJ/mol
• C. +2384 kJ/mol
• D. -2051 kJ/mol

Hint:

Always remember to subtract the enthalpy of formation of the reactants from the products' enthalpies to calculate the enthalpy change for the reaction.

Answer 1

The Correct Option is A

We are required to calculate the standard enthalpy of combustion of Cyclopropane. The standard enthalpy of combustion is the heat released when 1 mole of a substance completely reacts with oxygen under standard conditions. The combustion reaction for Cyclopropane (\( \text{C}_3\text{H}_6 \)) can be written as: \[ \text{C}_3\text{H}_6(g) + \text{O}_2(g) \rightarrow 3 \, \text{CO}_2(g) + 3 \, \text{H}_2\text{O}(l) \] We use the following reaction enthalpy values: \[ \Delta H_f^\circ \, \text{CO}_2(g) = -393.5 \, \text{kJ/mol}, \quad \Delta H_f^\circ \, \text{H}_2\text{O}(l) = -286 \, \text{kJ/mol} \] The enthalpy of combustion of Cyclopropane is calculated as the sum of the products minus the reactants: \[ \Delta H_{\text{combustion}}^\circ = \left[ 3 \times (-393.5) + 3 \times (-286) \right] - \left[ \text{Cyclopropane} + \text{Oxygen} \right] \] The enthalpy of formation of Cyclopropane is given as \( +20.6 \, \text{kJ/mol} \), and the enthalpy of oxygen is zero since it is in its elemental form. \[ \Delta H_{\text{combustion}}^\circ = \left[ 3 \times (-393.5) + 3 \times (-286) \right] - 20.6 \] \[ \Delta H_{\text{combustion}}^\circ = \left[ -1180.5 + (-858) \right] - 20.6 = -2038.5 - 20.6 = -2059.1 \, \text{kJ/mol} \]
Thus, the standard enthalpy of combustion of Cyclopropane is approximately \(-2092 \, \text{kJ/mol}\).