Question

If a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is $C{H}_{3}O{H}_{\left(l\right)}+\frac{3}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+2H_2{O}^{}\left(g\right)$
At $298K$ standard Gibb’s energies of formation for $C{H}_{3}OH\left(l\right),H_{2}O\left(l\right),\text{and}C{O}_2\left(g\right)$ are $–166.2,–237.2$ and $-394.4{\text{kJ mol}}^{-1},$ respectively. If the standard enthalpy of combustion of methanol is $–726{\text{kJ mol}}^{-1}$, the efficiency of the fuel cell will be:

• A. $80\%$
• B. $87\%$
• C. $90\%$
• D. $97\%$

Answer 1

The correct option is D $97\%$

The given equation is:
$C{H}_{3}OH\left(l\right)+3/2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_2{O}^{}\left(l\right)$
Given that the enthalpy change of this reaction,
$\Delta H=-726\text{kJ/mol}$
The Gibb’s free energy of the reaction, $\Delta G_R$ is
$\Delta G_R=\Delta G_{f\left(Product\right)}-\Delta G_{f\left(reactant\right)}$
Given that, $\Delta G_{C{O}_2}=-394.4\text{kJ/mol}$
$\Delta G_{H_{2}O}=-237.2\text{kJ/mol}$
$\Delta G_{C{H}_{2}OH}=-166.2\text{kJ/mol}$ and
$\Delta G_{O_2}=0$
$\therefore \Delta G_R=[-394.4+(2\times (-237.2)\left)\right]$
$–[–166.2+3/2(0\left)\right]=–702.6\text{kJ/mol}$
Efficiency $=\frac{\Delta G_R}{\Delta H}\times 100=\frac{-702.6}{-726}\times 100$
$=96.77\approx 97\%$
$\therefore$The correct answer is (D).