In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $C H_{3} O H (ℓ) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (ℓ)$ At 298K standard Gibb’s energies of formation for $C H_{3} O H (ℓ), H_{2} O (ℓ) a n d C O_{2} (g) a r e - 166.2, - 237.2 a n d - 394.4 k J m o l^{- 1}$ Respectively. If standard enthalpy of combustion of methanol is $- 726 k J m o l^{- 1}$ . Efficiency of the fuel cell will be

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80%

87%

90%

97%

answer is D.

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Detailed Solution

$C H_{3} O H (ℓ) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (ℓ) Δ H = - 726 k J m o l^{- 1}$

Also $Δ G_{f}^{0} C H_{3} O H (ℓ) = - 166.2 k J m o l^{- 1}$

$Δ G_{f}^{0} H_{2} O (ℓ) = - 237.2 k J m o l^{- 1}$

$Δ G_{f}^{0} C O_{2} (ℓ) = - 394.4 k J m o l^{- 1}$

$∵ Δ G = \sum Δ G_{f}^{0}$ Products $- \sum Δ G_{f}^{0}$ reactants.

$= - 394.4 - 2 (237.2) + 166.2$

$= - 702.6 k J m o l^{- 1}$

Now efficiency of fuel cell $= \frac{Δ G}{Δ H} \times 100$

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

$= 97 %$

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