PSPM 2019/2020 Ethanol, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \) is produced from fermentation of sugarcane and widely used as an alternative fuel to replace petrol. A 1.00 g sample of ethanol was burnt in a bomb calorimeter, which has total heat capacity of \( 11.0 \mathrm{~kJ}{ }^{\circ} \mathrm{C}^{-1} \). The temperature of the calorimeter and its contents increased from \( \frac{25.0^{\circ} \mathrm{C}}{\mathrm{T}} \) to \( \frac{27.7^{\circ} \mathrm{C} \text {. }}{\mathrm{Tf}} \) (i) Write a balanced chemical equation for the reaction that takes place in the (ii) Calculate the enthalpy of combustion of ethanol per mole.

To tackle this question, let’s start with the combustion of ethanol. (i) The balanced chemical equation for the combustion of ethanol in oxygen is: \[ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + 3 \mathrm{H}_{2}\mathrm{O} \] This indicates that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water. (ii) To calculate the enthalpy of combustion per mole of ethanol, we first determine the heat absorbed by the calorimeter: \[ q = C \times \Delta T = 11.0 \, \mathrm{kJ/^{\circ}C} \times (27.7 - 25.0) \, \mathrm{^{\circ}C} = 11.0 \, \mathrm{kJ/^{\circ}C} \times 2.7 \, \mathrm{^{\circ}C} = 29.7 \, \mathrm{kJ} \] Since 1.00 g of ethanol (molar mass = 46.07 g/mol) was burned, the number of moles of ethanol is: \[ \text{moles of ethanol} = \frac{1.00 \, \mathrm{g}}{46.07 \, \mathrm{g/mol}} \approx 0.0217 \, \text{mol} \] Now, we can ascertain the enthalpy of combustion per mole: \[ \Delta H = \frac{-q}{\text{moles of ethanol}} = \frac{-29.7 \, \mathrm{kJ}}{0.0217 \, \mathrm{mol}} \approx -1366.4 \, \mathrm{kJ/mol} \] Thus, the enthalpy of combustion of ethanol is approximately \(-1366.4 \, \mathrm{kJ/mol}\).