Question:

In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is: CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l). At 298 K, standard Gibb's energies of formation for CH3OH(l), H2O(l), and CO2(g) are -166.2, -237.2, and -394.4 kJ/mol respectively. If standard enthalpy of combustion of methanol is -726 kJ/mol, efficiency of the fuel cell will be:

ΔG = ΔGprod - ΔGreact = 2(-237.2) - 394.4 + 166.2 = -474.4 - 394.4 + 166.2 = -702.6 kJ/mol

The efficiency of the fuel cell is given by:

Efficiency = (ΔG / ΔH) * 100

where ΔG is the change in Gibbs free energy and ΔH is the change in enthalpy.

Given:
ΔG = -702.6 kJ/mol
ΔH = -726 kJ/mol

Efficiency = (-702.6 / -726) * 100 = 96.75%

Therefore, the efficiency of the fuel cell is approximately 97%.