In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is
${CH}_{3} OH (l) + \frac{3}{2} O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (l)$
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

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97%

answer is D.

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Detailed Solution

Percentage efficiency of the fuel cell = $\frac{∆ G}{∆ H} \times 100 .$ The concerned reaction is

${CH}_{3} OH (l) + \frac{3}{2} O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (l)$

$∆ G_{r} = [∆ G_{f} ({CO}_{2}, g) + 2 ∆ G_{f} (H_{2} O, l)] - [∆ G_{r} ({CH}_{3} OH, l) + \frac{3}{2} ∆ G_{f} (O, g)] = - 394.4 + 2 (- 237.2) - (- 166.2) - 0 = - 394.4 - 474.4 + 166.2 = - 702.6 kJ {mol}^{- 1} Percentage efficiency = \frac{702.6}{726} \times 100 = 96.78 % \approx 97 %$