JEE MAIN - Chemistry (2023 - 1st February Evening Shift - No. 20)
$$0.3 \mathrm{~g}$$ of ethane undergoes combustion at $$27^{\circ} \mathrm{C}$$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $$0.5^{\circ} \mathrm{C}$$. The heat evolved during combustion of ethane at constant pressure is ____________ $$\mathrm{kJ} ~\mathrm{mol}{ }^{-1}$$. (Nearest integer)
[Given : The heat capacity of the calorimeter system is $$20 \mathrm{~kJ} \mathrm{~K}^{-1}, \mathrm{R}=8.3 ~\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$$.
Assume ideal gas behaviour.
Atomic mass of $$\mathrm{C}$$ and $$\mathrm{H}$$ are 12 and $$1 \mathrm{~g} \mathrm{~mol}^{-1}$$ respectively]
Answer
1006
Explanation
$\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell)$
$\begin{aligned} & \text { No. of moles of ethane }=\frac{0.3}{30}=0.01 \\\\ & \text { Heat evolved in Bomb calorimeter }=20 \times 0.5 \\\\ & =10 \mathrm{~kJ} \\\\ & \Delta \mathrm{U}=-\frac{10}{0.01}=-1000 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ & =-1000+(-2.5) \times \frac{8.3 \times 300}{1000} \\\\ & =-1000-6.225 \\\\ & =-1006.225 \\\\ & |\Delta \mathrm{H}| \simeq 1006 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \end{aligned}$
$\begin{aligned} & \text { No. of moles of ethane }=\frac{0.3}{30}=0.01 \\\\ & \text { Heat evolved in Bomb calorimeter }=20 \times 0.5 \\\\ & =10 \mathrm{~kJ} \\\\ & \Delta \mathrm{U}=-\frac{10}{0.01}=-1000 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ & =-1000+(-2.5) \times \frac{8.3 \times 300}{1000} \\\\ & =-1000-6.225 \\\\ & =-1006.225 \\\\ & |\Delta \mathrm{H}| \simeq 1006 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \end{aligned}$
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