ThermodynamicsHard
Question
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is -
CH3OH(l) + 
O2(g) → CO2(g) + 2H2O(l)
At 298K, standard Gibb’s energies of formation for CH3OH)(l), H2O(l) and CO2(g) are -166.2, -237.2 and -394.4 kJ mol-1 respectively. If standard enthalpy of combustion of methanol is -726kJ mol-1, Efficiency of the fuel cell will be
Options
A.80 %
B.87 %
C.90 %
D.97 %
Solution
CH3OH(l) + O2(g) + 2H2O(l) áƒH = - 726kJ mol-1
Also áƒGfo CH3OH(l) = -166.2 kJ mol-1
áƒGfoH2O(l) = -237.2 kJ mol-1
áƒGfoCO2(l) = -394.4 kJ mol-1
∵  áƒG = ΣáƒGfo products -ΣáƒGfo reactants.
= -394.4 -2 (237.2) + 166.2
= -702.6 kJ mol-1
now Efficiency of fuel cell =
× 100
=
× 100
= 97%
O2(g) + 2H2O(l) áƒH = - 726kJ mol-1 Also áƒGfo CH3OH(l) = -166.2 kJ mol-1
áƒGfoH2O(l) = -237.2 kJ mol-1
áƒGfoCO2(l) = -394.4 kJ mol-1
∵  áƒG = ΣáƒGfo products -ΣáƒGfo reactants.
= -394.4 -2 (237.2) + 166.2
= -702.6 kJ mol-1
now Efficiency of fuel cell =
× 100=
× 100= 97%
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