Benzene diazonium chloride (A) decoposes into chlorobenzene (B) and `N_(2)` (g) in first order reaction. Volumes of `N_(2)` collected after 5 min and at the complete decomposition of A are 10 mL. and 50mL respectively. The are constant for the reaction is

Options

A. `0.446"min"^(-1)`
B. `0.0446"min"^(-1)`
C. `0.223 "min"^(-1)`
D. `0.112 "min"(-1)`

Explanation

To find the rate constant for the decomposition of benzene diazonium chloride into chlorobenzene and nitrogen gas, we will follow these steps:

### Step 1: Understand the Reaction
The reaction can be summarized as:
\[ \text{C}_6\text{H}_5\text{N}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \]
This is a first-order reaction.

### Step 2: Gather Initial and Final Volumes
From the problem:
- Initial volume of benzene diazonium chloride (A) = 50 mL (at \(t = 0\))
- Volume of \(N_2\) collected after 5 minutes = 10 mL
- Volume of \(N_2\) at complete decomposition = 50 mL

### Step 3: Calculate Remaining Volume of A after 5 Minutes
After 5 minutes, the volume of benzene diazonium chloride remaining can be calculated as:
\[ \text{Volume of A remaining} = \text{Initial volume} - \text{Volume of } N_2 \text{ collected} \]
\[ \text{Volume of A remaining} = 50 \, \text{mL} - 10 \, \text{mL} = 40 \, \text{mL} \]

### Step 4: Use the First-Order Rate Equation
For a first-order reaction, the rate constant \(k\) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
where:
- \(A_0\) = initial volume of A = 50 mL
- \(A_t\) = volume of A at time \(t\) = 40 mL
- \(t\) = time = 5 minutes

### Step 5: Substitute Values into the Equation
Substituting the values into the equation:
\[ k = \frac{2.303}{5} \log \left( \frac{50}{40} \right) \]

### Step 6: Calculate the Logarithm
Calculating the logarithm:
\[ \log \left( \frac{50}{40} \right) = \log(1.25) \]
Using logarithm values:
\[ \log(1.25) \approx 0.0969 \]

### Step 7: Calculate the Rate Constant
Now substituting back into the equation:
\[ k = \frac{2.303}{5} \times 0.0969 \]
\[ k \approx \frac{2.303 \times 0.0969}{5} \]
\[ k \approx \frac{0.223}{5} \]
\[ k \approx 0.0446 \, \text{min}^{-1} \]

### Final Answer
The rate constant \(k\) for the reaction is approximately \(0.0446 \, \text{min}^{-1}\).

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