JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 25)

A sample of n -octane $(1.14 \mathrm{~g})$ was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is $5 \mathrm{~kJ} \mathrm{~K}^{-1}$. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K . The magnitude of the heat of combustion of octane at constant volume is__________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer).
Answer
2500

Explanation

Calculate the moles of octane:

$ \text{Moles of octane} = \frac{1.14 \, \text{g}}{114 \, \text{g/mol}} = 0.01 \, \text{moles} $

Determine the heat evolved during combustion:

The heat capacity of the calorimeter is $ 5 \, \text{kJ/K} $ and the temperature increase is $ 5 \, \text{K} $.

$ \text{Heat evolved} = C \times \Delta T = 5 \, \text{kJ/K} \times 5 \, \text{K} = 25 \, \text{kJ} $

Calculate the magnitude of the heat of combustion:

The heat of combustion per mole of octane is found by dividing the total heat evolved by the moles of octane combusted:

$ \text{Magnitude of Heat of combustion} = \frac{25 \, \text{kJ}}{0.01 \, \text{moles}} = 2500 \, \text{kJ/mol} $

Therefore, the magnitude of the heat of combustion of octane at constant volume is $ 2500 \, \text{kJ/mol} $.

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