Azoisopropane decomposes acording to the reaction:
`(CH_(3))_(2)CHN = NCH(CH_(3))_(2)(g) overset(250-290^(@)C)rarr N_(2)(g) + C_(6)H_(14)(g)`
It is found to be a first order reaction. If the initial pressure is `P_(0)` and pressure of the mixture at time `t` is `(P_(t))`, then the rate constant `(k)` would be

Explanation

To find the rate constant \( k \) for the decomposition of azoisopropane, we can follow these steps:

### Step 1: Understand the Reaction
The decomposition of azoisopropane is given by the equation:
\[ (CH_3)_2CHN = NCH(CH_3)_2(g) \rightarrow N_2(g) + C_6H_{14}(g) \]
This reaction is a first-order reaction.

### Step 2: Define Initial and Final Pressures
Let:
- \( P_0 \) = initial pressure of the system (before decomposition)
- \( P_t \) = pressure of the mixture at time \( t \)

### Step 3: Relate Pressure Changes to Decomposition
As the reaction proceeds, some amount of azoisopropane decomposes, leading to the formation of nitrogen and hexane. If \( X \) is the amount of pressure decrease due to the decomposition of azoisopropane, then:
- The pressure of the system at time \( t \) can be expressed as:
\[ P_t = P_0 + X \]
Since \( X \) is the pressure contributed by the products formed (N2 and C6H14), we can rearrange this to find \( X \):
\[ X = P_t - P_0 \]

### Step 4: Express Remaining Pressure
The remaining pressure of the azoisopropane at time \( t \) can be expressed as:
\[ P_0 - X = P_0 - (P_t - P_0) = 2P_0 - P_t \]

### Step 5: Apply the First-Order Rate Law
For a first-order reaction, the rate constant \( k \) can be expressed using the integrated rate law:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_t} \right) \]
In terms of pressure, this becomes:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{2P_0 - P_t} \right) \]

### Final Expression for Rate Constant
Thus, the rate constant \( k \) for the decomposition of azoisopropane can be expressed as:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{2P_0 - P_t} \right) \]

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