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))rarr CO_2((g))+2H_(2)O_((l))`
At `298K` standard Gibb's energies of formation for `CH_(3)OH(l), H_(2)O(l)` and `CO_(2)(g)` are `-166.2,-237.2` and `-394.4kJ mol^(-1)` respectively. If standard enthalpy of combustion of methanol is `-726kJ mol^(-1)`, efficiency of the fuel cell will be `:`

Options

A. 0.87
B. 0.9
C. 0.97
D. 0.8

Answer

C

Explanation

`CH_(3)OH(l) + (3)/(2) O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l)`
`DeltaG_(r) = DeltaG_(f)(CO_(2),g)+2DeltaG_(f)(H_(2)O, (l))-DeltaG_(r)(CH_(3)OH,(l)) -(3)/(2) DeltaG_(r)(O_(2),g)`
`=-394.4 + 2(-237.2) -(-166.2)-0=-394.4 -474.4 + 166.2 =- 868.8 xx 166.2`
`DeltaG_(r) =- 702.6 KJ`
%efficiency = `(702.6)/(726) xx100=97%.`