Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $C{H}_{3}O{H}_{(l)}+\frac{3}{2}{O}_{2(g)}\to C{O}_{2(g)}+2H_2{O}_{(l)}$
At $298K$ standard Gibbs' energies of formation for $C{H}_{3}O{H}_{(l)},H_2{O}_{(l)}$ and $C{O}_{2(g)}$ are $-166.2,-237.2$ and $-394.4kJmo{l}^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726kJmo{l}^{-1},$ efficiency of the fuel cell will be

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

Answer 1

Answer: (D)

For the given reaction,

$C{H}_{3}O{H}_{(l)}+\frac{3}{2}{O}_{2(g)}⟶C{O}_{2(g)}+2H_2{O}_{(l)},$

$\Delta H=-726kJmo{l}^{-1}$

also, $\Delta G_f^{\circ }\left[C{H}_{3}O{H}_{(l)}\right]=-166.2kJmo{l}^{-1}$;

$\Delta G_f^{\circ }\left[H_2{O}_{(l)}\right]=-237.2kJmo{l}^{-1}$

and $\Delta G_f^{\circ }\left[C{O}_{2(g)}\right]=-394.4kJmo{l}^{-1}$

Now, $\Delta G^{\circ }\text{reaction }=\sum \Delta G_{f\text{ products }}^{\circ }-\sum \Delta G_{f\text{ reactants }}^{\circ }$

$=[-394.4+2\times (-237.2)]-(-166.2)=-702.6kJmo{l}^{-1}$

$\%$ Efficiency $=\frac{\Delta G}{\Delta H}\times 100=\frac{-702.6}{-726}\times 100=96.77\%$

$\therefore$ Efficiency $\approx 97\%$