Complete combustion gaseous ethane |MODERNPHYSICSJEEMAIN2023

JEE MAIN 2023

25-1-23 S2

Question

28.0 L of $\mathrm{CO}_2$ is produced on complete combustion of 16.8 L gaseous mixture of ethane and methane at $25^{\circ} \mathrm{C}$ and 1 atm . Heat evolved during the combustion process is $\_\_\_\_$ kJ. Given : $\Delta \mathrm{Hc}\left(\mathrm{CH}_4\right)=-900 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{Hc}_{\mathrm{C}}\left(\mathrm{C}_2 \mathrm{H}_4\right)=-1400 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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Solution

Sol. Let, Volume of $\mathrm{C}_2 \mathrm{H}_4$ is $x$ litre $$ \mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} $$ Initial x Final - $$ 2 x $$ $$ \mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} $$ Initial $(16.8-\mathrm{x})$ Final - $$ (16.8-x) $$ Total volume of $\mathrm{CO}_2=2 \mathrm{x}+16.8-\mathrm{x}$ $$ \begin{array}{ll} \Rightarrow \quad & 28=16.8+x \\ & x=11.2 \mathrm{~L} \\ { }^{\mathrm{n}} \mathrm{CH}_4= & \frac{P V}{R T}=\frac{1 \times 5.6}{0.082 \times 298}=0.229 \mathrm{~mole} \end{array} $$ $\begin{aligned} & { }^{\mathrm{n}} \mathrm{C}_2 \mathrm{H}_2=\frac{11.2}{0.082 \times 298}=0.458 \mathrm{~mole} \\ & \begin{aligned} \therefore \text { Heat evolved } & =0.229 \times 900+0.458 \times 1400 \\ & =206.1+641.2=847.3 \mathrm{~kJ}\end{aligned}\end{aligned}$

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JEE Main

Chemistry

Easy

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