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)andCO_(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 :

In a fuel cell methanol is used as fuel and oxygen is used as an oxidi

1.	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)andCO_(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 :
Answer» `80%` `87%` `90%` `97%` Solution :`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%