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In a fuel cell methanol is used as fuel ...

In a fuel cell methanol is used as fuel and oxygen is used as an oxidiser. The reaction is :
`CH_(3)OH(l)+(3)/(2)O_(2)(g)rarr CO_(2)(g)+2H_(2)O(l)`
At 298 K, 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 `-394.4` kJ/mol respectively. If standard enthalpy of combustion of methanol is - 726 kJ/mol, effeciency of the fuel cell will be :

A

`80%`

B

`87%`

C

`90%`

D

`97%`

Text Solution

Verified by Experts

The correct Answer is:
D
`CH_(3)OH (l) + (1)/(3)O_(2) (g) rarr CO_(2) (g) + 2H_(2)O (l)`
`Delta_(f) H = - 726 kJ//mole`
`Delta_(r) G = Sigma Delta_(f) G_("products") - Sigma Delta_(f) G_("ractants")`
`[Delta_(f) G_(CO_(2)) + 2 xx Delta_(f) G_(H_(2)O)] - [Delta_(f) G_(CH_(3)OH) + (3)/(2) Delta_(f) G_(O_(2))]`
`= [- 394.4 + 2 xx (- 237.2)] - [- 166.2 + (3)/(2) (0)]`
`= - 702.6 kJ//mol`
`% efficiency = (Delta_(f) G)/(Delta_(f)H) xx 100 = (- 702.6)/(- 726) xx 100`
= 96.77% = 97%
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