The decomposition of azo methane, at certain temperature according to the equation
`(CH_(3))_(3)N_(2)rarr C_(2)H_(5)+N_(2)` is a first order reaction.
After 40 minutes from the start, the total pressure developed is found to be `350 mm Hg` in place of initial pressure `200 mm Hg` of axo methane . The value of rate constant `k` is `-`

Options

A. `2.88 xx 10^(-4) sec ^(-1)`
B. `1.25 xx10^(-4)sec^(-1)`
C. `5.77xx10^(-4)sec^(-1)`
D. None of these

Explanation

To solve the problem, we need to calculate the rate constant \( k \) for the decomposition of azomethane, which is a first-order reaction. The steps are as follows:

### Step 1: Identify Initial and Final Pressures
- The initial pressure \( P_0 \) of azomethane is given as \( 200 \, \text{mm Hg} \).
- The total pressure after 40 minutes is \( 350 \, \text{mm Hg} \).

### Step 2: Calculate the Change in Pressure
- The change in pressure \( X \) can be calculated using the formula:
\[ P_{\text{total}} = P_0 + X \]
Rearranging gives:
\[ X = P_{\text{total}} - P_0 = 350 \, \text{mm Hg} - 200 \, \text{mm Hg} = 150 \, \text{mm Hg} \]

### Step 3: Relate Change in Pressure to Reaction Stoichiometry
- From the reaction:
\[ (CH_3)_3N_2 \rightarrow C_2H_5 + N_2 \]
For every mole of azomethane decomposed, 2 moles of products (C2H5 and N2) are formed. Therefore, if \( X \) is the change in pressure due to the decomposition of azomethane, we can express the change in pressure as:
\[ P_{\text{total}} = P_0 + 2X \]
However, since we already calculated \( X \) directly from the total pressure, we can use it directly.

### Step 4: Use the First-Order Rate Constant Formula
- The formula for the rate constant \( k \) for a first-order reaction is:
\[ k = \frac{2.303}{T} \log \left( \frac{P_0}{P_0 - X} \right) \]
Here, \( T \) is the time in seconds. Since \( T = 40 \, \text{minutes} = 40 \times 60 = 2400 \, \text{seconds} \).

### Step 5: Substitute Values into the Formula
- Substitute \( P_0 = 200 \, \text{mm Hg} \) and \( X = 150 \, \text{mm Hg} \):
\[ k = \frac{2.303}{2400} \log \left( \frac{200}{200 - 150} \right) \]
\[ k = \frac{2.303}{2400} \log \left( \frac{200}{50} \right) \]
\[ k = \frac{2.303}{2400} \log (4) \]

### Step 6: Calculate the Logarithm and Final Value
- Calculate \( \log(4) \):
\[ \log(4) = 2 \log(2) \approx 2 \times 0.301 = 0.602 \]
- Now substitute back to find \( k \):
\[ k = \frac{2.303}{2400} \times 0.602 \]
\[ k \approx \frac{1.384}{2400} \approx 5.77 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer
The value of the rate constant \( k \) is approximately:
\[ k \approx 5.77 \times 10^{-4} \, \text{s}^{-1} \]

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