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%`

Answer

B

Explanation

The reaction involved is `CH_(3)OH(l) + (3)/(2) O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) Delta H= -726kJ mol^(-1)`
Now, `Delta G= Delta G^(@) (CO_(2)) + 2Delta G^(@) (H_(2)O) - Delta G^(@) (CH_(3)OH)`
`=-394.4 -2 xx (237.2) + 166.2 = -702.6kJ mol^(-1)`
Efficiency of fuel cell is given by `(Delta G)/(Detla H) xx 100= (702.6)/(726)xx 100= 97%`