JEE MAIN - Chemistry (2009 - No. 19)
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OH(l) + 3/2O2 $$\to$$ 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 -726 kJ mol−1, efficiency of the fuel cell will be
CH3OH(l) + 3/2O2 $$\to$$ 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 -726 kJ mol−1, efficiency of the fuel cell will be
87%
90%
97%
80%
Explanation
$$C{H_3}OH\left( l \right) + {3 \over 2}{O_2}\left( g \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$$
$$\Delta {G_r} = \Delta {G_f}\left( {C{O_2},g} \right) + 2\Delta {G_f}\left( {{H_2}O,l} \right) - $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G_f}\left( {C{H_3}OH,l} \right) - {3 \over 2}\Delta {G_f}\left( {{O_2},g} \right)$$
$$ = - 394.4 + 2\left( { - 237.2} \right) - \left( { - 166.2} \right) - 0$$
$$ = - 394.4 - 474.4 + 166.2$$
$$ = - 702.6\,kJ$$
$$\% \,\,$$ efficiency $$\,\, = {{702.6} \over {726}} \times 100$$ $$ = 97\% $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$$
$$\Delta {G_r} = \Delta {G_f}\left( {C{O_2},g} \right) + 2\Delta {G_f}\left( {{H_2}O,l} \right) - $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {G_f}\left( {C{H_3}OH,l} \right) - {3 \over 2}\Delta {G_f}\left( {{O_2},g} \right)$$
$$ = - 394.4 + 2\left( { - 237.2} \right) - \left( { - 166.2} \right) - 0$$
$$ = - 394.4 - 474.4 + 166.2$$
$$ = - 702.6\,kJ$$
$$\% \,\,$$ efficiency $$\,\, = {{702.6} \over {726}} \times 100$$ $$ = 97\% $$
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