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 reaction, the temperature of the calorimeter is increased by

Correct answer: 0 Mole of octane =(1.14)/(114)=0.01 ~mole Heat evolved =C × Δ T aligned & =5 × 5 ~kJ & =25 ~kJ aligned of Heat of combustion = (25)/(0.01)= 2500 kJ / mole

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2025 · 3 April · Evening Shift · Chemistry · Thermodynamics (C)

A sample of n-octane (1.14 ~g) was completely burnt in excess of...

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).

Correct answer

Step-by-step solution

\text{ Mole

octane }=\frac{1.14}{114}=0.01 \mathrm{~mole}

Heat evolved

=\mathrm{C} \times \Delta \mathrm{T}

\begin{aligned}<br />& =5 \times 5 \mathrm{~kJ} \\<br>& =25 \mathrm{~kJ}<br />\end{aligned}

\thereforeMagnitude

of Heat of combustion =

\frac{25}{0.01}=

2500

\mathrm{kJ}

mole

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