The decomposition of azo methane, at certain temperature acccording to the equation `(CH_(3))_(2)N_(2)toC_(2)H_(6)+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 azo methane. The value of rate constant k is :

Options

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

Answer

To solve the problem, we need to determine the rate constant \( k \) for the first-order decomposition of azomethane. Here’s the step-by-step solution:

### Step 1: Understand the Reaction
The decomposition of azomethane is given by the equation:
\[ (CH_3)_2N_2 \rightarrow C_2H_6 + N_2 \]
This is a first-order reaction.

### Step 2: Identify Initial and Final Pressures
- Initial pressure of azomethane, \( P_0 = 200 \, \text{mm Hg} \)
- Total pressure after 40 minutes, \( P_t = 350 \, \text{mm Hg} \)

### Step 3: Determine the Change in Pressure
Let \( P \) be the pressure decrease due to the decomposition of azomethane. The pressure of azomethane at time \( t \) will be:
\[ P_0 - P \]
The total pressure at time \( t \) can be expressed as:
\[ P_t = (P_0 - P) + P + P = P_0 + P \]
Substituting the known values:
\[ 350 = 200 + P \]
From this, we can solve for \( P \):
\[ P = 350 - 200 = 150 \, \text{mm Hg} \]

### Step 4: Calculate Remaining Pressure of Azomethane
The pressure of azomethane remaining after 40 minutes is:
\[ P_0 - P = 200 - 150 = 50 \, \text{mm Hg} \]

### Step 5: Use the First-Order Rate Constant Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right) \]
Where:
- \( t = 40 \, \text{minutes} = 40 \times 60 \, \text{seconds} = 2400 \, \text{seconds} \)
- \( P_0 = 200 \, \text{mm Hg} \)
- \( P_t = 50 \, \text{mm Hg} \)

### Step 6: Substitute Values into the Formula
Now substituting the values into the equation:
\[ k = \frac{2.303}{2400} \log \left( \frac{200}{50} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{200}{50} \right) = \log(4) \approx 0.602 \]
Now substituting this value back into the equation for \( k \):
\[ k = \frac{2.303}{2400} \times 0.602 \]
Calculating \( k \):
\[ k \approx \frac{1.384}{2400} \approx 5.77 \times 10^{-4} \, \text{s}^{-1} \]

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

---