Aqueous `AB_(2)` decomposes according to the first order reaction:
`AB_(2)(aq) rarr A(g) + 2B(l)`
After `20 min` the volume of `A(g) ` colledcted during such a reaction is `20 mL`, and that collected after a very long time is `40 mL`. The rate constant is :

Options

A. `A. 3.45 xx 10^(-3) min^(-1)`
B. `B. 3.45 xx 10^(-2) min^(-1)`
C. `C. 1.435 xx 10^(-2) min^(-1)`
D. `D. 6.93 min^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ AB_2(aq) \rightarrow A(g) + 2B(l) \]
This is a first-order reaction.

### Step 2: Identify Initial and Final Conditions
- At \( t = 0 \):
- Concentration of \( AB_2 \) = \( A \) (initial concentration)
- Concentration of \( A \) = 0
- Concentration of \( B \) = 0

- At \( t = 20 \) minutes:
- Volume of \( A \) collected = 20 mL
- Let \( X \) be the amount of \( AB_2 \) decomposed.
- Therefore, concentration of \( A \) = \( X \) (which corresponds to 20 mL)
- Concentration of \( AB_2 \) = \( A - X \)

- At \( t = \infty \):
- Volume of \( A \) collected = 40 mL
- Therefore, concentration of \( A \) = \( A \) (which corresponds to 40 mL)
- Concentration of \( AB_2 \) = 0

### Step 3: Relate Volumes to Concentrations
From the ideal gas law, at constant temperature and pressure, the volume is directly proportional to the number of moles. Thus:
- \( A \) (final concentration of \( A \)) is proportional to 40 mL.
- \( X \) (amount decomposed) is proportional to 20 mL.

### Step 4: Set Up the First-Order Rate Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - X} \right) \]
Where:
- \( t = 20 \) minutes
- \( A = 40 \) mL
- \( X = 20 \) mL

### Step 5: Substitute Values into the Formula
Substituting the values into the equation:
\[ k = \frac{2.303}{20} \log \left( \frac{40}{40 - 20} \right) \]
\[ k = \frac{2.303}{20} \log \left( \frac{40}{20} \right) \]
\[ k = \frac{2.303}{20} \log(2) \]

### Step 6: Calculate the Logarithm
The value of \( \log(2) \) is approximately \( 0.3010 \):
\[ k = \frac{2.303}{20} \times 0.3010 \]

### Step 7: Perform the Final Calculation
Calculating \( k \):
\[ k = \frac{2.303 \times 0.3010}{20} \]
\[ k = \frac{0.693}{20} = 0.03465 \text{ per minute} \]
This can also be expressed as:
\[ k \approx 3.465 \times 10^{-2} \text{ per minute} \]

### Final Answer
The rate constant \( k \) is approximately \( 3.45 \times 10^{-2} \) per minute.

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