In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is `:`
`CH_(3)OH_((l))+(3)/(2)O_(2(g))rarr CO_2((g))+2H_(2)O_((l))`
At `298K` standard Gibb's energies of formation for `CH_(3)OH(l), H_(2)O(l)` and `CO_(2)(g)` are `-166.2,-237.2` and `-394.4kJ mol^(-1)` respectively. If standard enthalpy of combustion of methanol is `-726kJ mol^(-1)`, efficiency of the fuel cell will be `:`

Options

A. 0.8
B. 0.87
C. 0.9
D. 0.97

Explanation

To find the efficiency of the fuel cell using methanol and oxygen, we will follow these steps:

### Step 1: Write the Reaction
The balanced chemical reaction for the fuel cell is:
\[ \text{CH}_3\text{OH}_{(l)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2 \text{H}_2\text{O}_{(l)} \]

### Step 2: Calculate the Standard Gibbs Free Energy Change (ΔG°)
The standard Gibbs free energy change for the reaction can be calculated using the formula:
\[ \Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_f \text{(products)} - \sum \Delta G^\circ_f \text{(reactants)} \]

Where:
- \(\Delta G^\circ_f\) of \(\text{CO}_2(g) = -394.4 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f\) of \(\text{H}_2\text{O}(l) = -237.2 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f\) of \(\text{CH}_3\text{OH}(l) = -166.2 \, \text{kJ/mol}\)
- \(\Delta G^\circ_f\) of \(\text{O}_2(g) = 0 \, \text{kJ/mol}\) (as it is in its standard state)

Now substituting the values:
\[ \Delta G^\circ_{\text{reaction}} = \left[ (-394.4) + 2 \times (-237.2) \right] - \left[ (-166.2) + 0 \right] \]
\[ = (-394.4 - 474.4 + 166.2) \, \text{kJ/mol} \]
\[ = -702.6 \, \text{kJ/mol} \]

### Step 3: Use the Standard Enthalpy of Combustion (ΔH°)
The standard enthalpy of combustion of methanol is given as:
\[ \Delta H^\circ = -726 \, \text{kJ/mol} \]

### Step 4: Calculate the Efficiency of the Fuel Cell
The efficiency (\( \eta \)) of the fuel cell can be calculated using the formula:
\[ \eta = \frac{\Delta G^\circ}{\Delta H^\circ} \]

Substituting the values we calculated:
\[ \eta = \frac{-702.6 \, \text{kJ/mol}}{-726 \, \text{kJ/mol}} \]
\[ = 0.966 \approx 0.97 \]

### Conclusion
The efficiency of the fuel cell is approximately \(0.97\) or \(97\%\).

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