In Kolbe's electrolysis sodium propanoate gives `:`

Options

A. `(P_(A))/(P_(B))=2`
B. `(P_(A))/(P_(B))=1`
C. `(P_(A))/(P_(B))=4`
D. `(P_(A))/(P_(B))=3`

Explanation

To solve the question regarding Kolbe's electrolysis of sodium propanoate, let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Electrolysis Process**:
- Kolbe's electrolysis involves the electrolysis of carboxylate salts, in this case, sodium propanoate (CH3CH2COONa).
- When sodium propanoate is electrolyzed, it dissociates into its constituents: propanoate ions (CH3CH2COO⁻) and sodium ions (Na⁺).

2. **Formation of Free Radicals**:
- During the electrolysis, the propanoate ions undergo oxidation at the anode.
- Two moles of propanoate ions can combine to form a dimer, leading to the formation of free radicals.
- The reaction can be represented as:
\[ 2 \text{CH}_3\text{CH}_2\text{COO}^- \rightarrow \text{CH}_3\text{CH}_2\cdots\text{CH}_2\text{COO} + 2 \text{CO}_2 + 2 \text{e}^- \]
- This process generates ethyl radicals (CH3CH2•).

3. **Formation of Products**:
- The ethyl radicals can combine in different ways:
- Two ethyl radicals can combine to form butane (C4H10):
\[ \text{CH}_3\text{CH}_2\cdots\text{CH}_2 \rightarrow \text{C}_4\text{H}_{10} \text{ (butane)} \]
- Ethyl radicals can also react with hydrogen ions (H⁺) to form ethane (C2H6):
\[ \text{CH}_3\text{CH}_2\cdots\text{H} \rightarrow \text{C}_2\text{H}_6 \text{ (ethane)} \]
- Additionally, the loss of hydrogen from ethyl radicals can lead to the formation of ethene (C2H4):
\[ \text{CH}_3\text{CH}_2\cdots \rightarrow \text{C}_2\text{H}_4 \text{ (ethene)} \]

4. **Conclusion**:
- Therefore, the products formed from the Kolbe's electrolysis of sodium propanoate are a mixture of ethane (C2H6), ethene (C2H4), and butane (C4H10).
- The correct answer is a mixture of these three compounds.

### Final Answer:
In Kolbe's electrolysis, sodium propanoate gives a mixture of ethane, ethene, and butane.

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