Single Choice

In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is: $$\mathrm {C}\mathrm{H}_{3}\mathrm{O}\mathrm{H}(l)+\dfrac{3}{2}\mathrm{O}_{2}(\mathrm{g})\rightarrow \mathrm{C}\mathrm{O}_{2}(\mathrm{g})+2\mathrm{H}_{2}\mathrm{O}(l)$$ At $$298\ \mathrm{K}$$, standard Gibb's energies of formation for $$\mathrm{C}\mathrm{H}_{3}\mathrm{O}\mathrm{H}(l),\ \mathrm{H}_{2}\mathrm{O}(l)$$ and $$\mathrm{C}\mathrm{O}_{2}(\mathrm{g})$$ are$$-166.2, - 237.2$$ and $$-394.4\ \mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$ respectively. If standard enthalpy of combustion of methanol is $$-726\ \mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$$, efficiency of the fuel cell will be:

A$$ 80\%$$
B$$ 87\%$$
C$$ 90\%$$
D$$ 97\%$$
Correct Answer

Solution

$$\Delta G=\Delta G_{prod}-\Delta G_{react}=2(-237.2)-394.4+166.2$$

$$=-474.4-394.4+166.2=702.6\ kJ/mol$$

Efficiency $$=\displaystyle\frac{\Delta G}{\Delta H}\times 100$$

$$= \cfrac{702.6}{726}\times 100$$

$$=96.77\approx 97$$%

SIMILAR QUESTIONS

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