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

For the reaction
`CH_(3)OH(l) +(3)/(2)O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)`
`Delta_(r)G^(Theta) = Delta_(f)G^(Theta) (CO_(2)) +2Delta_(f)G^(Theta)(H_(2)O)`
`-{Delta_(f)G^(Theta) (CH_(3)OH) +(3)/(2)Delta_(f)G^(Theta) (O_(2))}`
`=- 394.4 +2 (-237.2) - (-166.2)-0`
`=- 394.4 -474.4 +166.2`
`=- 702.6 kJ mol^(-1)`
Therefore, Gibbs enegry change which can be converted to electrical work `= 702.6 kJ mol^(-1)`
Now efficiency `= (DeltaG)/(DeltaH) = (702.6)/(726.0) xx 100 = 96.78%`