Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is$CH_{3}OH\left(\ell\right)+\frac{3}{2}O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right)+2H_{2}O\left(\ell\right)$ At $298K$ standard Gibb’s energies of formation for $CH_{3}OH\left(\ell\right), H_{2}O\left(\ell\right)$ and $CO_{2}\left(g\right)$ are $-166.2, -237.2$ and $-394.4\, kJ\, mol^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726kJ \,mol^{−1}$ , efficiency of the fuel cell will be

• A. $80\,\%$
• B. $87\,\%$
• C. $90\,\%$
• D. $97\,\%$

Answer 1

Answer: (D)

$CH_{3}OH\left(\ell\right)+\frac{3}{2}O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right)+2H_{2}O\left(\ell\right) \,\Delta H=-726kJ\,mol^{-1}$
Also $\Delta G^{o}_{f}CH_{3}OH\left(\ell\right)=-166.2\,kJ\,mol^{-1}$
$\Delta G^{o}_{f}CO_{2}\left(\ell\right)=-394.4\,kJ\,mol^{-1}$
$\because \Delta G=\sum\Delta G^{o}_{f}$ products $-\sum \Delta G^{o}_{f}$ reactants.
$=-394.4-2\left(237.2\right)+166.2$
$=-702.6\,kJ\,mol^{-1}$
now Efficiency of fuel cell $=\frac{\Delta G}{\Delta H}\times100$
$=\frac{702.6}{726}\times100$
$=97\%$