In the reaction `NH_(4)NO_(2)(aq.) rarr N_(2)(g) + 2H_(2)O(l)` the volume of `N_(2)` after `20` min and after a long time is `40 ml` and `70 ml` respectively, The value of rate constant is :

Options

A. `(1//20) In (7//4) "min"^(-1)`
B. `(2.303//1200) log(7//3)sec^(-1)`
C. `(1//20) log (7//3) "min"^(-1)`
D. `(2.303//20) log (11//7) "min"^(-1)`

Explanation

To find the rate constant \( k \) for the reaction \( NH_4NO_2(aq) \rightarrow N_2(g) + 2H_2O(l) \), we can use the first-order rate equation. Here are the steps to solve the problem:

### Step 1: Identify the given information
- Volume of \( N_2 \) after 20 minutes (\( V_T \)) = 40 ml
- Volume of \( N_2 \) after a long time (\( V_{\infty} \)) = 70 ml
- Initial volume of \( N_2 \) (\( V_0 \)) = 0 ml (since no gas is produced at the start)
- Time (\( T \)) = 20 minutes

### Step 2: Use the first-order rate constant formula
The formula for the first-order reaction rate constant \( k \) is given by:

\[ k = \frac{2.303}{T} \log \left( \frac{V_{\infty} - V_0}{V_{\infty} - V_T} \right) \]

### Step 3: Substitute the values into the formula
Substituting the known values into the equation:

- \( V_{\infty} = 70 \, \text{ml} \)
- \( V_0 = 0 \, \text{ml} \)
- \( V_T = 40 \, \text{ml} \)
- \( T = 20 \, \text{minutes} \)

Thus, we have:

\[ k = \frac{2.303}{20} \log \left( \frac{70 - 0}{70 - 40} \right) \]

### Step 4: Simplify the logarithmic expression
Calculating the values inside the logarithm:

\[ \frac{70 - 0}{70 - 40} = \frac{70}{30} = \frac{7}{3} \]

Now, substituting back into the equation for \( k \):

\[ k = \frac{2.303}{20} \log \left( \frac{7}{3} \right) \]

### Step 5: Calculate the logarithm
Using a calculator, we find:

\[ \log \left( \frac{7}{3} \right) \approx 0.5682 \]

### Step 6: Calculate \( k \)
Now, substituting this value back into the equation for \( k \):

\[ k = \frac{2.303}{20} \times 0.5682 \]

Calculating this gives:

\[ k \approx \frac{2.303 \times 0.5682}{20} \approx \frac{1.308}{20} \approx 0.0654 \, \text{min}^{-1} \]

### Final Answer
The value of the rate constant \( k \) is approximately \( 0.0654 \, \text{min}^{-1} \).

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