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NVAEasyJEE 2025Internal Energy & Enthalpy

JEE Chemistry 2025 Question with Solution

A sample of n-octane (1.14 g1.14 \, \text{g}) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K−15 \, \text{kJ K}^{-1}. As a result of combustion, the temperature of the calorimeter increased by 5 K5 \, \text{K}. The magnitude of the heat of combustion at constant volume is _____ kJ mol−1\text{kJ mol}^{-1} (nearest integer).

Answer

Correct answer:2500

Step-by-step solution

Standard Method

Given: mass of n-octane = 1.14 g1.14 \, \text{g}, heat capacity of bomb calorimeter = 5 kJ K−15 \, \text{kJ K}^{-1}, temperature rise = 5 K5 \, \text{K}.

Find: magnitude of heat of combustion at constant volume in kJ mol−1\text{kJ mol}^{-1}.

In a bomb calorimeter, the heat evolved by the reaction at constant volume is absorbed by the calorimeter.

qv=CΔTq_v = C\Delta T

Substituting the given values:

qv=5 kJ K−1×5 K=25 kJq_v = 5 \, \text{kJ K}^{-1} \times 5 \, \text{K} = 25 \, \text{kJ}

This heat corresponds to combustion of 1.14 g1.14 \, \text{g} of n-octane.

The molecular formula of n-octane is C8H18\mathrm{C_8H_{18}}.

Its molar mass is:

(8×12)+(18×1)=114 g mol−1(8 \times 12) + (18 \times 1) = 114 \, \text{g mol}^{-1}

Number of moles burnt:

n=1.14 g114 g mol−1=0.01 moln = \frac{1.14 \, \text{g}}{114 \, \text{g mol}^{-1}} = 0.01 \, \text{mol}

Therefore, heat of combustion per mole is:

ΔU=25 kJ0.01 mol=2500 kJ mol−1\Delta U = \frac{25 \, \text{kJ}}{0.01 \, \text{mol}} = 2500 \, \text{kJ mol}^{-1}

Therefore, the magnitude of the heat of combustion at constant volume is 2500 kJ mol−12500 \, \text{kJ mol}^{-1}. Hence the numerical answer is 2500.

The solution explicitly states the correct answer as 2500.

Direct Calorimeter Ratio

Given: calorimeter heat absorbed = CΔTC\Delta T and sample mass = 1.14 g1.14 \, \text{g}.

Find: molar heat of combustion.

First compute the heat released by the given sample:

q=5×5=25 kJq = 5 \times 5 = 25 \, \text{kJ}

Now note that 1.14 g1.14 \, \text{g} is exactly 1.14114=0.01\frac{1.14}{114} = 0.01 mole of n-octane.

So multiplying the sample heat by 100100 gives heat per mole:

25×100=2500 kJ mol−125 \times 100 = 2500 \, \text{kJ mol}^{-1}

Therefore, the required numerical value is 2500.

Common mistakes

  • Using the mass 1.14 g1.14 \, \text{g} directly as if it were moles is incorrect because heat of combustion is required per mole. First convert mass to moles using the molar mass of n-octane.

  • Forgetting that the bomb calorimeter works at constant volume leads to confusion between different heat quantities. Here the required quantity is obtained from the calorimeter relation qv=CΔTq_v = C\Delta T.

  • Calculating the molar mass of C8H18\mathrm{C_8H_{18}} incorrectly gives a wrong final answer. The correct molar mass is 114 g mol−1114 \, \text{g mol}^{-1}, not 112112 or 118118.

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