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QJEEM23Ch4b9b: JEE Main 2023 Chemistry - Class 11 JEEM23Ch4b9b Solution: of ethane undergoes combustion at in a bomb calorimeter. The…

0.3 g0.3 \mathrm{~g} of ethane undergoes combustion at 27∘C27^{\circ} \mathrm{C} in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5∘C0.5^{\circ} \mathrm{C} . The heat evolved during combustion of ethane at constant pressure is ____________ kJ mol−1\mathrm{kJ} ~\mathrm{mol}{ }^{-1} . (Nearest integer)

[Given : The heat capacity of the calorimeter system is 20 kJ K−1,R=8.3 JK−1 mol−120 \mathrm{~kJ} \mathrm{~K}^{-1}, \mathrm{R}=8.3 ~\mathrm{JK}^{-1} \mathrm{~mol}^{-1} .

Assume ideal gas behaviour.

Atomic mass of C\mathrm{C} and H\mathrm{H} are 12 and 1 g mol−11 \mathrm{~g} \mathrm{~mol}^{-1} respectively]

Step-by-Step Verified Solution

C2H6( g)+72O2( g)→2CO2( g)+3H2O(ℓ)\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)

amp; No. of moles of ethane =0.330=0.01amp; Heat evolved in Bomb calorimeter =20×0.5amp;=10 kJamp;ΔU=−100.01=−1000 kJ mol−1amp;ΔH=ΔU+ΔngRTamp;=−1000+(−2.5)×8.3×3001000amp;=−1000−6.225amp;=−1006.225amp;∣ΔH∣≃1006 kJ mol−1amp;\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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