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. 0.87
B. 0.9
C. 0.97
D. 0.8

Explanation

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

### Step 1: Write down the reaction
The reaction for the combustion of methanol 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: Identify the given data
- Standard Gibbs energies of formation:
- \( \Delta G_f^\circ (\text{CH}_3\text{OH}_{(l)}) = -166.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{H}_2\text{O}_{(l)}) = -237.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{CO}_2_{(g)}) = -394.4 \, \text{kJ/mol} \)
- Standard enthalpy of combustion of methanol:
- \( \Delta H = -726 \, \text{kJ/mol} \)

### Step 3: Calculate the standard Gibbs free energy change (\( \Delta G \))
Using the formula:
\[ \Delta G = \Delta G_f^\circ (\text{products}) - \Delta G_f^\circ (\text{reactants}) \]

For the products:
- \( \Delta G_f^\circ (\text{CO}_2) = -394.4 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{H}_2\text{O}) = -237.2 \, \text{kJ/mol} \) (for 2 moles)

Thus,
\[ \Delta G_f^\circ (\text{products}) = -394.4 + 2 \times (-237.2) = -394.4 - 474.4 = -868.8 \, \text{kJ/mol} \]

For the reactants:
- \( \Delta G_f^\circ (\text{CH}_3\text{OH}) = -166.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{O}_2) = 0 \, \text{kJ/mol} \) (as it is in its standard state)

Thus,
\[ \Delta G_f^\circ (\text{reactants}) = -166.2 + \frac{3}{2} \times 0 = -166.2 \, \text{kJ/mol} \]

Now, substituting into the equation for \( \Delta G \):
\[ \Delta G = -868.8 - (-166.2) = -868.8 + 166.2 = -702.6 \, \text{kJ/mol} \]

### Step 4: Calculate the efficiency of the fuel cell
The efficiency (\( \eta \)) can be calculated using the formula:
\[ \eta = \frac{\Delta G}{\Delta H} \times 100 \]

Substituting the values:
\[ \eta = \frac{-702.6}{-726} \times 100 \]

Calculating this gives:
\[ \eta \approx 96.8\% \]

### Conclusion
The efficiency of the fuel cell is approximately \( 96.8\% \).

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