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. `87%`
B. `90%`
C. `97%`
D. `80%`

Answer

To solve the problem, we need to calculate the Gibbs free energy change (ΔG) for the reaction and then use it to find the efficiency of the fuel cell. Here’s the step-by-step solution:

### Step 1: Write the Reaction
The 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 ΔG for the Reaction
The Gibbs free energy change (ΔG) can be calculated using the standard Gibbs energies of formation of the products and reactants:
\[ \Delta G = \sum \Delta G_f^{\circ} \text{(products)} - \sum \Delta G_f^{\circ} \text{(reactants)} \]

Using the given values:
- ΔG_f for \( \text{H}_2\text{O}_{(l)} = -237.2 \, \text{kJ/mol} \)
- ΔG_f for \( \text{CO}_2_{(g)} = -394.4 \, \text{kJ/mol} \)
- ΔG_f for \( \text{CH}_3\text{OH}_{(l)} = -166.2 \, \text{kJ/mol} \)

Calculating for products:
\[ \Delta G_f^{\circ} \text{(products)} = \Delta G_f^{\circ} \text{(CO}_2) + 2 \times \Delta G_f^{\circ} \text{(H}_2\text{O)} \]
\[ = -394.4 + 2 \times (-237.2) \]
\[ = -394.4 - 474.4 = -868.8 \, \text{kJ/mol} \]

Calculating for reactants:
\[ \Delta G_f^{\circ} \text{(reactants)} = \Delta G_f^{\circ} \text{(CH}_3\text{OH)} + \frac{3}{2} \times \Delta G_f^{\circ} \text{(O}_2) \]
Since the standard Gibbs energy of formation for \( \text{O}_2 \) is 0:
\[ = -166.2 + 0 = -166.2 \, \text{kJ/mol} \]

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

### Step 3: Calculate Efficiency of the Fuel Cell
The efficiency of the fuel cell can be calculated using the formula:
\[ \text{Efficiency} = \frac{\Delta G}{\Delta H} \times 100 \]

Given that the standard enthalpy of combustion of methanol (ΔH) is:
\[ \Delta H = -726 \, \text{kJ/mol} \]

Substituting the values:
\[ \text{Efficiency} = \frac{-702.6}{-726} \times 100 \]
\[ = \frac{702.6}{726} \times 100 \]
\[ \approx 96.8\% \]

### Final Answer
The efficiency of the fuel cell is approximately **97%**.

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