In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer and its efficiency is calculated as Δ H ∘ Δ G ∘ × 100 . The reaction is. CH 3 OH(ℓ) + 3/2O 2 (g) → CO 2 (g) + 2H 2 O(ℓ).

Question

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer and its efficiency is calculated as $\frac{Δ G ^{\circ}}{Δ H ^{\circ}} \times 100$ . The reaction is.

CH₃OH(ℓ) + 3/2O₂ (g) → CO₂ (g) + 2H₂O(ℓ).

At 300 K standard Gibb's energies of formation of CH₃OH(ℓ), H₂O(ℓ) and CO₂(g) are –166, 237 and –397 kJmol^-1 respectively. If standard entropy of combustion of methanol 700/3 is J mol^-1, efficiency of the fuel cell will be (approx) :

TestHub · Accepted Answer

CH 3 ​ OH ( ℓ ) + 2 3 ​ O 2 ​ ( g ) → CO 2 ​ ( g ) + 2 H 2 ​ O ( ℓ ) Δ r ​ G ∘ = − 394 − 2 × 236 + 166 = − 700 Δ G ∘ = Δ H ∘ − T Δ S ∘ − 700 = Δ H − 600 × 3 × 1000 700 ​ Δ H ∘ = − 770 kJ / mol %efficiency = − 770 − 700 ​ × 100 = 11 1000 ​ = 90.9%

Chemistry - Thermodynamics - II Question with Solution | TestHub

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