Question:

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction isCH3OH()+32O2(g)CO2(g)+2H2O()CH_{3}OH\left(\ell\right)+\frac{3}{2}O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right)+2H_{2}O\left(\ell\right) At 298K298K standard Gibb?s energies of formation for CH3OH(),H2O()CH_{3}OH\left(\ell\right), H_{2}O\left(\ell\right) and CO2(g)CO_{2}\left(g\right) are 166.2,237.2-166.2, -237.2 and 394.4kJmol1-394.4\, kJ\, mol^{-1} respectively. If standard enthalpy of combustion of methanol is 726kJmol1-726kJ \,mol^{-1} , efficiency of the fuel cell will be

Updated On: Jul 5, 2022
  • 80%80\,\%
  • 87%87\,\%
  • 90%90\,\%
  • 97%97\,\%
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The Correct Option is D

Solution and Explanation

CH3OH()+32O2(g)CO2(g)+2H2O()ΔH=726kJmol1CH_{3}OH\left(\ell\right)+\frac{3}{2}O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right)+2H_{2}O\left(\ell\right) \,\Delta H=-726kJ\,mol^{-1} Also ΔGfoCH3OH()=166.2kJmol1\Delta G^{o}_{f}CH_{3}OH\left(\ell\right)=-166.2\,kJ\,mol^{-1} ΔGfoCO2()=394.4kJmol1\Delta G^{o}_{f}CO_{2}\left(\ell\right)=-394.4\,kJ\,mol^{-1} ΔG=ΔGfo\because \Delta G=\sum\Delta G^{o}_{f} products ΔGfo-\sum \Delta G^{o}_{f} reactants. =394.42(237.2)+166.2=-394.4-2\left(237.2\right)+166.2 =702.6kJmol1=-702.6\,kJ\,mol^{-1} now Efficiency of fuel cell =ΔGΔH×100=\frac{\Delta G}{\Delta H}\times100 =702.6726×100=\frac{702.6}{726}\times100 =97%=97\%
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Concepts Used:

Gibbs Free Energy

The energy associated with a chemical reaction that can be used to do work.It is the sum of its enthalpy plus the product of the temperature and the entropy (S) of the system.

The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. In completely reversible process maximum enthalpy can be obtained.

ΔG=ΔH−TΔS

The Conditions of Equilibrium

If both it’s intensive properties and extensive properties are constant then thermodynamic system is in equilibrium. Extensive properties imply the U, G, A.