JEE Advanced · Chemistry · 6. Thermodynamics (C)
In a constant volume calorimeter, \(3.5 \mathrm{~g}\) of a gas with molecular weight \(=28\) was burnt in excess oxygen at \(298.0 \mathrm{~K}\). The temperature of the calorimeter was found to increases from \(298.0 \mathrm{~K}\) to \(298.45 \mathrm{~K}\) due to the combustion process. Given, that the heat capacity of the calorimeter is \(2.5\) \(\mathrm{kJ} \mathrm{K}^{-1}\), the numerical value for the enthalpy of combustion of the gas in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
The temperature rise is : \(\Delta T=T_2-T_1=298.45-298=0.45 \mathrm{~K}\)
This indicates that heat produced from combustion of \(3.5 \mathrm{~g}\) of compound rises temperature of calorimeter by \(0.45 \mathrm{~K}\).
Heat produced \(=0.45 \mathrm{~K} \times 2.5 \mathrm{k} \mathrm{JK}^{-1}=1.125 \mathrm{~kJ}\)
\(\Rightarrow\) Heat produced from \(28 \mathrm{~g}\) of compound \((1.0 \mathrm{~mol})=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}\)
This indicates that heat produced from combustion of \(3.5 \mathrm{~g}\) of compound rises temperature of calorimeter by \(0.45 \mathrm{~K}\).
Heat produced \(=0.45 \mathrm{~K} \times 2.5 \mathrm{k} \mathrm{JK}^{-1}=1.125 \mathrm{~kJ}\)
\(\Rightarrow\) Heat produced from \(28 \mathrm{~g}\) of compound \((1.0 \mathrm{~mol})=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}\)
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