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Question

In a fuel cell, methanol is used as a fuel and O2 is used as an oxidizer. The standard enthalpy of combustion of methanol is 726 kJmol1. The standard free energies of formation of CH3OH(l), CO2(g) and H2O(l) are − 166.3, −394.4 and 237.1 kJmol1 respectively. The efficiency of the fuel cell will be:

A
96.8%
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B
66.2%
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C
41.3%
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D
85.1%
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Solution

The correct option is A 96.8%
Points to remember:
1. ΔrG0=ΣΔfG0productsΣΔfG0reactants
2.Efficiency %=useful workinput×100
ΔrG0=useful work
3. ΔfG0O2(g)=0
The reaction for the combustion of methanol is:
CH3OH(l)+32O2(g)CO2(g)+2H2O(l)
ΔrG0=12ΔfG0CO2+2ΔfG0H2OΔfG0CH3OH32ΔfG0O2
On substituting values ,we get
ΔrG0=702.6kJmol1

efficiency=ΔGΔH×100=96.7%

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