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

The reaction for combustion of methanol is:
`CH_(3)OH(l) +(3)/(2)rarr CO_(2)(g) +2H_(2)O(l)`
`Delta_(reaction) G^(Theta) =[Delta_(f)G^(Theta)CO_(2)(g) +2Delta_(f)G^(Theta)H_(2)O(l)]`
`-[Delta_(f)G^(Theta)CH_(3)OH(l)+(3)/(2)Delta_(f)G^(Theta) O_(2)(g)]`
`=[-394.0 +2(-237.00)-[-166.00+0]`
`=- 702.00 kJ mol^(-1)`
Efficiency of conversion of Gibbs free energy into useful work
`=(Delta_(reaction)G^(Theta)xx100)/(Delta_(reaction)H^(Theta)) = (-702.00xx100)/(-702) = 100%`