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When reaction is at standard state at equilibrium, thenA. `80%`B. `87%`C. `90%`D. `97%`

Answer» Correct Answer - B
Precentage efficiency of the fuel cell `=(DeltaG)/(DeltaH)xx100` . The concerned reaction is
`CH_(3)OH(l)+(3)/(2)O_(2)(g)rarrCO_(2)+2H_(2)O(l)`
`DeltaG_(r)[DeltaG_(f)(CO_(2),g)+2DeltaG_(f)(H_(2)O,l)]`
`[-DeltaG_(f)(CH_(3)OH,l)-(3)/(2)DeltaG_(f)(O_(2),g)]`
`=-394.4+2(-237.2)-(-166.2)-0`
`=-394.4-474.4+166.2=-702.6KJmol^(-1)`
Percentage effieiency `=(702.6)/(762)xx100=96.78%=97%`


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