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In a fuel cell, methanol if used as fuel and oxygen gas is used as an oxidiser. The reaction is `CH_(3)OH(l) +(3)/(2)O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l)` Calculated standard Gibbs free enegry change for the reaction that can be converted into electircal work. If standard enthalpy of combustion for methanol is `-702 kJ mol^(-1)`, calculate the efficiency of converstion of Gibbs energy into useful work. `Delta_(f)G^(Theta)` for `CO_(2),H_(2)O, CH_(3)OH,O_(2)` is `-394.00, -237.00,-166.00`and `0 kJ mol^(-1)` respectively. |
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Answer» The reaction for combustion of methanol is : `CH_(3)OH(l)+(3)/(2)O_(2)(g)toCO_(2)(g)+2H_(2)O(l)` `DeltaG_("reaction")^(@)=[DeltaG_(f)^(@)CO_(2)(g)+2DeltaG_(f)^(@)H_(2)O(l)]-[DeltaG_(f)^(@)CH_(3)OH(l)+(3)/(2)DeltaG_(f)^(@)O_(2)(g)]` `=[-394.36+2(-237.13)]-[-166.27+0]` Efficiency of conversion of Gibbs free energy into useful work `=(DeltaG_("reaction")^(@)xx100)/(DeltaH_("reaction")^(@))` `=(-702.35xx100)/(-726)=96.7%` |
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