In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(ℓ)+32O2(g)→CO2(g)+2H2O(ℓ) At 298K standard Gibb’s energies of formation for CH3OH(ℓ),H2O(ℓ) and CO2(g) are −166.2,−237.2 and −394.4kJ mol−1 Respectively. If standard enthalpy of combustion of methanol is −726kJ mol−1 . Efficiency of the fuel cell will be
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a
80%
b
87%
c
90%
d
97%
answer is D.
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Detailed Solution
CH3OH(ℓ)+32O2(g)→CO2(g)+2H2O(ℓ)ΔH=−726kJ mol−1 Also ΔGf0CH3OH(ℓ)=−166.2kJ mol−1 ΔGf0H2O(ℓ)=−237.2kJ mol−1 ΔGf0CO2(ℓ)=−394.4kJ mol−1 ∵ΔG=∑ΔGf0 Products −∑ΔGf0 reactants. =−394.4−2(237.2)+166.2 =−702.6kJ mol−1 Now efficiency of fuel cell =ΔGΔH×100 =702.6726×100 =97%