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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 3 April, Evening Shift — Question 17

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−15 \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 _____\_\_\_\_\_ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} (nearest integer).

Answer: 2500

Numerical answer — enter this value.

Step-by-step solution

Q=CΔT=5×5=25 kJ\mathrm{Q}=\mathrm{C} \Delta \mathrm{T}=5 \times 5=25 \mathrm{~kJ}

QV=251.14×114=2500 kJ mol−1\mathrm{Q}_{\mathrm{V}}=\frac{25}{1.14} \times 114=2500 \mathrm{~kJ} \mathrm{~mol}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes

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