In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is `CH _(3) OH (l) + (3)/(2) O_(2) (g) to CO_(2) (g) + 2H _(2) O(l)`
At 298 K, standard Gibbs energies of formation for `CH _(3) OH (l), H _(2) O (l)` and `CO _(2) (g) ` are `-166.2,-237.2` and `-394.4 kJ mol^(-1)` respectively. If standard enthalpy of combustion of methanol is `-726 kJ mol^(-1)` efficiency of the fuel cell will be:

Options

A. `90%`
B. `97%`
C. `80%`
D. `87%`

Explanation

`CH _(3) OH (1) + (3)/(2) O _(2) (g) to CO _(2) + 2 H _(2) O(l) , Delta G ^(@) (CO _(2)) + 2 Delta G ^(@) (H_(2) O ) - Delta G ^(@) (CH _(3) OH)`
`=- 3944 - 2 (237.2) + 166.2 =- 702 =- 702 . 6 kJ mol ^(-1)`
Efficiency of fuel cell `(Delta G ^(@))/( Delta H ^(@)) = (-702 . 6)/( - 726) xx 100 = 97 %`