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A fuel cell operates with methanol as the fuel and oxygen as the oxidant. The overall reaction is CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l) At 298 K, the standard Gibbs energies of formation of CH3OH(l), H2O(l), and CO2(g) are −166.2, −237.2, and −394.4 kJ mol−1, respectively. If the standard enthalpy of combustion of methanol is −726 kJ mol−1, what is the efficiency of the fuel cell?

1. 87%
2. 90%
3. 97%
4. 80%

Correct answer: 97%

Solution

delta G = [-394.4 + 2(-237.2)] - (-166.2) = -702.6 kJ. Efficiency = delta G / delta H = 702.6/726 = 0.968, i.e. about 97%.

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