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Calculate the entropy change for
\(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow 3 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \text {, using }\)
the following data :
\(\begin{array}{lllll}
\text { Substance } & \mathrm{CH}_4(\mathrm{~g}) & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \mathrm{H}_2(\mathrm{~g}) & \mathrm{CO}(\mathrm{g}) \\
\mathrm{S}^{\circ} / \mathrm{JK}^{-1} & 186.2 & 188.7 & 130.6 & 197.6
\end{array}\)
\(\mathrm{mole}^{-1}\)
The entropy change is:
Options:
\(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow 3 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \text {, using }\)
the following data :
\(\begin{array}{lllll}
\text { Substance } & \mathrm{CH}_4(\mathrm{~g}) & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \mathrm{H}_2(\mathrm{~g}) & \mathrm{CO}(\mathrm{g}) \\
\mathrm{S}^{\circ} / \mathrm{JK}^{-1} & 186.2 & 188.7 & 130.6 & 197.6
\end{array}\)
\(\mathrm{mole}^{-1}\)
The entropy change is:
Solution:
1370 Upvotes
Verified Answer
The correct answer is:
\(+214.5 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}\)
\(\begin{aligned}
& \Delta \mathrm{S}^{\circ}=\Sigma \mathrm{S}^{\circ} \text { (products) }-\Sigma \mathrm{S}^{\circ}(\text { reactants }) \\
& =\left[3 \times \mathrm{S}^{\circ}\left(\mathrm{H}_2\right)+\mathrm{S}^{\circ}(\mathrm{CO})\right]-\left[\mathrm{S}^{\circ}\left(\mathrm{CH}_4\right) +\mathrm{S}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right]
\end{aligned}\)
\(\begin{aligned}
& =[3 \times 130.6+197.6]-[186.2+188.7] \\
& =[391.8+197.6]-[374.9] \\
& =214.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}\)
& \Delta \mathrm{S}^{\circ}=\Sigma \mathrm{S}^{\circ} \text { (products) }-\Sigma \mathrm{S}^{\circ}(\text { reactants }) \\
& =\left[3 \times \mathrm{S}^{\circ}\left(\mathrm{H}_2\right)+\mathrm{S}^{\circ}(\mathrm{CO})\right]-\left[\mathrm{S}^{\circ}\left(\mathrm{CH}_4\right) +\mathrm{S}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right]
\end{aligned}\)
\(\begin{aligned}
& =[3 \times 130.6+197.6]-[186.2+188.7] \\
& =[391.8+197.6]-[374.9] \\
& =214.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}\)
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