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In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell)$ At $298 \mathrm{~K}$ standard Gibb's energies of formation for $\mathrm{CH}_3 \mathrm{OH}(\ell), \mathrm{H}_2 \mathrm{O}(\ell)$ and $\mathrm{CO}_2(\mathrm{~g})$ are $-166.2,-237.2$ and $-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$, efficiency of the fuel cell will be
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Verified Answer
The correct answer is:
$97 \%$
$97 \%$
$$
\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell) \quad \Delta \mathrm{H}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Also $\Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CH}_3 \mathrm{OH}(\ell)=-166.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{H}_2 \mathrm{O}(\ell)=-237.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CO}_2(\ell)=-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \because \Delta \mathrm{G}=\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { products }-\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { reactants. } \\
& =-394.4-2(237.2)+166.2 \\
& =-702.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \text { now Efficiency of fuel cell }=\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}} \times 100 \\
& =\frac{702.6}{726} \times 100 \\
& =97 \%
\end{aligned}
$$
\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell) \quad \Delta \mathrm{H}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Also $\Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CH}_3 \mathrm{OH}(\ell)=-166.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{H}_2 \mathrm{O}(\ell)=-237.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CO}_2(\ell)=-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \because \Delta \mathrm{G}=\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { products }-\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { reactants. } \\
& =-394.4-2(237.2)+166.2 \\
& =-702.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \text { now Efficiency of fuel cell }=\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}} \times 100 \\
& =\frac{702.6}{726} \times 100 \\
& =97 \%
\end{aligned}
$$
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