Question

A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K\(^{-1}\). As a result of combustion, the temperature of the calorimeter increased by 5 K. The magnitude of the heat of combustion at constant volume is ___

Hint:

In bomb calorimeter problems, the heat evolved by the reaction at constant volume is equal to the heat absorbed by the calorimeter (\( q_v = C \Delta T \)). To find the molar heat of combustion, divide the total heat evolved by the number of moles of the substance burnt. Remember to use the correct molar mass of the substance.

Answer 1

Correct Answer: 2500

Mass of n-octane = 1.14 g Heat capacity of the bomb calorimeter (C) = 5 kJ K\(^{-1}\) 
Increase in temperature (\( \Delta T \)) = 5 K The heat evolved during the combustion of n-octane at constant volume (\( q_v \)) is absorbed by the calorimeter, causing the temperature increase. 
Magnitude of heat evolved = \( q_v = C \times \Delta T \) \( q_v = 5 \, \text{kJ K}^{-1} \times 5 \, \text{K} = 25 \, \text{kJ} \) 
This is the heat evolved from the combustion of 1.14 g of n-octane. We need to find the heat of combustion per mole of n-octane. 
The molecular formula of n-octane is C\( _8 \)H\( _{18} \). 
The molar mass of n-octane is: \( (8 \times 12) + (18 \times 1) = 96 + 18 = 114 \, \text{g mol}^{-1} \) Number of moles of n-octane burnt = \( \frac{\text{mass of n-octane}}{\text{molar mass of n-octane}} \) \[ \text{Moles of n-octane} = \frac{1.14 \, \text{g}}{114 \, \text{g mol}^{-1}} = 0.01 \, \text{mol} \] The heat evolved from the combustion of 0.01 mol of n-octane is 25 kJ. The heat of combustion per mole of n-octane (\( \Delta U \)) at constant volume is: \[ \Delta U = \frac{\text{Heat evolved}}{\text{Moles of n-octane}} = \frac{25 \, \text{kJ}}{0.01 \, \text{mol}} = 2500 \, \text{kJ mol}^{-1} \] The magnitude of the heat of combustion at constant volume is 2500 kJ mol\(^{-1}\). 
The nearest integer is 2500.