In a fuel cell, methanol if used as fuel and oxygen gas is used as an oxidiser. The reaction is
`CH_(3)OH(l) +(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)`
Calculated standard Gibbs free enegry change for the reaction that can be converted into electircal work. If standard enthalpy of combustion for methanol is `-702 kJ mol^(-1)`, calculate the efficiency of converstion of Gibbs energy into useful work.
`Delta_(f)G^(Theta)` for `CO_(2),H_(2)O, CH_(3)OH,O_(2)` is `-394.00, -237.00,-166.00`and `0 kJ mol^(-1)` respectively.

Explanation

`CH_(3)OH_((l))+3//2O_(2(g))toCO_(2(g))+2H_(2)O_((l))`
`Delta_(r)G^(@)=Delta_(f)G_(CO_(2))^(@)+2xxDelta_(f)G_(H_(2)O)^(@)-Delta_(f)G_(CH_(3)OH)^(@)-3//2xxDelta_(f)G_(O_(2))^(@)`
`Delta_(r)G^(@)=-702.35 kJ mol^(-1)`
The efficiency of conversion of Gibbs free energy into usefule work
`=(Delta_(r)G^(@))/(DeltaH^(@))xx100`
or efficiency `=(702.35)/(-726)xx100`
`=-96.7%`
i.e., `96.7%` of Gibbs free energy can be converted into useful work.