Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH($\mathcal{l}$) + $\frac{3}{2}$O2 (g)$\longrightarrow$CO2(g) + 2H2O(l) At 298 K, standard Gibb’s energies of formation for CH3OH($\mathcal{l}$), H2O($\mathcal{l}$) and CO2 (g) 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. 87%
• B. 90%
• C. 97%
• D. 80%

Answer 1

Answer: (C)

97%

Answer 2

CH3OH($\mathcal{l}$) + $\frac{3}{2}$O2 (g) ¾® CO2(g) + 2H2O($\mathcal{l}$)

$\Delta$Gr = $\Delta$Gf (CO2 ,g) + 2$\Delta$Gf (H2O, ($\mathcal{l}$)) – $\Delta$Gf

(CH3OH, ($\mathcal{l}$)) – $\frac{3}{2}\Delta$Gf (O2 ,g)

= –394.4 + 2 (–237.2) – (–166.2) – 0 = – 394.4 – 474.4 + 166.2 = – 868.8 × 166.2

$\Delta$Gr = –702.6 kJ

% efficiency = $\frac{7.2.6}{726} \times 100$= 97%.