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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2009 · Paper 2 · Chemistry · Thermodynamics (C)

In a constant volume calorimeter, 3.5 ~g of a gas with molecular...

In a constant volume calorimeter, 3.5 g3.5 \mathrm{~g} of a gas with molecular weight =28=28 was burnt in excess oxygen at 298.0 K298.0 \mathrm{~K}. The temperature of the calorimeter was found to increases from 298.0 K298.0 \mathrm{~K} to 298.45 K298.45 \mathrm{~K} due to the combustion process. Given, that the heat capacity of the calorimeter is 2.52.5 kJK−1\mathrm{kJ} \mathrm{K}^{-1}, the numerical value for the enthalpy of combustion of the gas in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} is

Correct answer

0

Step-by-step solution

The temperature rise is : ΔT=T2−T1=298.45−298=0.45 K\Delta T=T_2-T_1=298.45-298=0.45 \mathrm{~K} This indicates that heat produced from combustion of 3.5 g3.5 \mathrm{~g} of compound rises temperature of calorimeter by 0.45 K0.45 \mathrm{~K}. Heat produced =0.45 K×2.5kJK−1=1.125 kJ=0.45 \mathrm{~K} \times 2.5 \mathrm{k} \mathrm{JK}^{-1}=1.125 \mathrm{~kJ} ⇒\Rightarrow Heat produced from 28 g28 \mathrm{~g} of compound (1.0 mol)=1.1253.5×28=9 kJ(1.0 \mathrm{~mol})=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}
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