When reaction is at standard state at equilibrium, then

1.	When reaction is at standard state at equilibrium, then
Answer» `80%` `87%` `90%` `97%` Solution :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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