For a decomposition of azoisopropane at `270^@C` it was found that at t = 0 , the total pressure was found that at t = 0 , total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction.
`(CH_3)_2CHN = NCH (CH_2)_2rarrN_2+C_6H_(14)`

Options

A. `0.168 "min"^(-1)`
B. `0.173 "min"^(-1)`
C. `0.18 "min"^(-1)`
D. `0.154 "min"^(-1)`

Answer

To solve the problem of calculating the rate constant (K) for the decomposition of azoisopropane at 270°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Pressures:**
- At time \( t = 0 \), the total pressure \( P_0 = 33.15 \, \text{mmHg} \).
- After 3 minutes, the total pressure \( P_t = 46.3 \, \text{mmHg} \).

2. **Calculate the Increase in Pressure:**
- The increase in pressure due to the formation of products is given by:
\[ \Delta P = P_t - P_0 = 46.3 \, \text{mmHg} - 33.15 \, \text{mmHg} = 13.15 \, \text{mmHg} \]

3. **Determine the Pressure of Reactants Remaining:**
- The pressure of the reactants remaining after 3 minutes can be calculated as:
\[ P_{\text{remaining}} = P_0 - \Delta P = 33.15 \, \text{mmHg} - 13.15 \, \text{mmHg} = 20.00 \, \text{mmHg} \]

4. **Use the First Order Reaction Rate Equation:**
- For a first-order reaction, the rate constant \( K \) can be calculated using the formula:
\[ K = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right) \]
- Here, \( t = 3 \, \text{minutes} \), \( P_0 = 33.15 \, \text{mmHg} \), and \( P_t = 20.00 \, \text{mmHg} \).

5. **Substitute the Values into the Equation:**
- Substitute the values into the equation:
\[ K = \frac{2.303}{3} \log \left( \frac{33.15}{20.00} \right) \]

6. **Calculate the Logarithm:**
- Calculate the logarithm:
\[ \log \left( \frac{33.15}{20.00} \right) = \log(1.6575) \approx 0.219 \]

7. **Calculate K:**
- Now substitute the logarithm back into the equation for \( K \):
\[ K = \frac{2.303}{3} \times 0.219 \approx 0.168 \, \text{min}^{-1} \]

### Final Answer:
The value of \( K \) for the decomposition of azoisopropane at \( 270^\circ C \) is approximately \( 0.168 \, \text{min}^{-1} \).