Question

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

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

Answer 1

Answer: (D)

For the given reaction,

$CH _{3} OH _{(l)}+\frac{3}{2} O _{2(g)} \longrightarrow CO _{2(g)}+2 H _{2} O _{(l)},$

$\Delta H=-726\, kJ\, mol ^{-1}$

also, $\Delta G_{f}^{\circ}\left[ CH _{3} OH _{(l)}\right]=-166.2 \, kJ\, mol ^{-1}$;

$\Delta G_{f}^{\circ}\left[ H _{2} O _{(l)}\right]=-237.2 \, kJ\, mol ^{-1}$

and $\Delta G_{f}^{\circ}\left[ CO _{2(g)}\right]=-394.4 \, kJ\, mol ^{-1}$

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

$=[-394.4+2 \times(-237.2)]-(-166.2)=-702.6\, kJ\, mol ^{-1}$

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

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