In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is…

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
  1. $80 \%$
  2. $87 \%$
  3. $90 \%$
  4. $97 \%$

Solution

$ \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} $

Asked in: JEE Main 2009

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