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

Answer

D

Explanation

`CH_(3)OH_((l)) +3/2 O_(2(g)) to CO_(2(g)) + 2H_(2)O_((l))`
`DeltaH=-726 KJ "mol"^(-1)`
Also `DeltaG_(f)^(@) CH_(3)OH_((l)) =-1662 KJ"mol"^(-1)`
`DeltaG_(f)^(@) CO_(2(g)) =-394.4 KJ "mol"^(-1)`
`because DeltaG=sum DeltaG_(f)^(@)` products `- sum DeltaG_(f)^(@)` reactants
`=-394.4 -2(237.2)+166.2`
`=-702.6 KJ "mol"^(-1)`
Now Efficiency of fuel cell `=(DeltaG)/(DeltaH)xx100`
`=(702.6)/(726)xx100=97%`