A sample of n-octane $(1.14 \mathrm{~g})$ was completely burnt in excess of oxygen in a bomb calorimeter,…

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).

Solution

$\text { Mole of octane }=\frac{1.14}{114}=0.01 \mathrm{~mole}$
Heat evolved $=\mathrm{C} \times \Delta \mathrm{T}$
$\begin{aligned}
& =5 \times 5 \mathrm{~kJ} \\
& =25 \mathrm{~kJ}
\end{aligned}$
$\therefore$ Magnitude of Heat of combustion $=\frac{25}{0.01}=2500$ $\mathrm{kJ} /$ mole

Asked in: JEE Main 2025 (03 Apr Shift 2)

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