JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 25)
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} $.
Comments (0)
