In a fuel celll, methanol is used as a fuel and `O_(2)` is used as oxidizer. The standard enthalpy of combustion of methanol is -726 kJ `mol^(-1)`. The standard free energies of formation of `CH_(3)OH(I), CO_(2)(g) and H_(2)O(I)` are -166.3, -394.4 and -237.1 kJ `mol^(-1)` respectively.
The standard free energy change of the reaction will be

Options

A. `-597.8` kJ `mol^(-1)`
B. `-298.9` kJ `mol^(-1)`
C. `-465.2` kJ `mol^(-1)`
D. `-702.3` kJ `mol^(-1)`

Answer

To calculate the standard free energy change (ΔG°) of the combustion reaction of methanol in a fuel cell, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methanol.
The balanced equation for the combustion of methanol (CH₃OH) with oxygen (O₂) 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 standard free energies of formation (ΔG°f) for each species.
From the problem, we have the following standard free energies of formation:
- ΔG°f (CH₃OH) = -166.3 kJ/mol
- ΔG°f (CO₂) = -394.4 kJ/mol
- ΔG°f (H₂O) = -237.1 kJ/mol
- ΔG°f (O₂) = 0 kJ/mol (since it is in its elemental form)

### Step 3: Use the formula for standard free energy change (ΔG°) of the reaction.
The formula for calculating ΔG° of the reaction is:
\[ \Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_f \text{(products)} - \sum \Delta G^\circ_f \text{(reactants)} \]

### Step 4: Calculate ΔG° for the products and reactants.
**Products:**
- For CO₂: 1 × ΔG°f (CO₂) = 1 × (-394.4 kJ/mol) = -394.4 kJ/mol
- For H₂O: 2 × ΔG°f (H₂O) = 2 × (-237.1 kJ/mol) = -474.2 kJ/mol

Total for products:
\[ \text{Total (products)} = -394.4 + (-474.2) = -868.6 \text{ kJ/mol} \]

**Reactants:**
- For CH₃OH: 1 × ΔG°f (CH₃OH) = 1 × (-166.3 kJ/mol) = -166.3 kJ/mol
- For O₂: 1 × ΔG°f (O₂) = 1 × (0 kJ/mol) = 0 kJ/mol

Total for reactants:
\[ \text{Total (reactants)} = -166.3 + 0 = -166.3 \text{ kJ/mol} \]

### Step 5: Substitute the totals into the ΔG° formula.
Now, substituting the totals into the ΔG° formula:
\[ \Delta G^\circ_{\text{reaction}} = (-868.6 \text{ kJ/mol}) - (-166.3 \text{ kJ/mol}) \]
\[ \Delta G^\circ_{\text{reaction}} = -868.6 + 166.3 = -702.3 \text{ kJ/mol} \]

### Conclusion
The standard free energy change of the reaction is:
\[ \Delta G^\circ_{\text{reaction}} = -702.3 \text{ kJ/mol} \]

### Final Answer
The correct answer is -702.3 kJ/mol.

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