Consider a first order gas phase decompostion reaction gives below
`A(g) to B(g) to C(g)`
The initial pressure of the system before decomposition of A `p_(i)`. After lapse of time 't' total pressure of the system increased by x units and became `p_(t)`. The rate constant K for the reaction is given as....... .

Options

A. `k = (2.303)/(t) log.(p_(i))/(p_(i) -p_(x))`
B. `k = (2.303)/(t) log.(p_(i))/(2p_(i) -p_(t))`
C. `k = (2.303)/(t) log.(p_(i))/(2p_(i) -p_(t))`
D. `k = (2.303)/(t) log.(p_(i))/(2p_(i) -x)`

Explanation

To find the rate constant \( K \) for the first-order gas phase decomposition reaction \( A(g) \to B(g) \to C(g) \), we can follow these steps:

### Step 1: Understand the Reaction and Initial Conditions
- The initial pressure of the system before the decomposition of \( A \) is \( P_i \).
- At time \( t \), the total pressure of the system increases by \( x \) units, resulting in a total pressure \( P_t \).

### Step 2: Relate Initial and Final Pressures
- Since \( A \) decomposes to form \( B \) and \( C \), the change in pressure can be expressed as:
\[ P_t = P_i + x \]
- Rearranging gives:
\[ x = P_t - P_i \]

### Step 3: Apply the First-Order Reaction Formula
- For a first-order reaction, the rate constant \( K \) can be expressed using the formula:
\[ K = \frac{2.303}{t} \log \left( \frac{P_i}{P_i - x} \right) \]
- Here, \( P_i - x \) represents the pressure of \( A \) remaining after some has decomposed.

### Step 4: Substitute for \( x \)
- Substitute \( x \) from Step 2 into the equation:
\[ K = \frac{2.303}{t} \log \left( \frac{P_i}{P_i - (P_t - P_i)} \right) \]
- This simplifies to:
\[ K = \frac{2.303}{t} \log \left( \frac{P_i}{2P_i - P_t} \right) \]

### Step 5: Final Expression for \( K \)
- The final expression for the rate constant \( K \) is:
\[ K = \frac{2.303}{t} \log \left( \frac{P_i}{2P_i - P_t} \right) \]

### Conclusion
- The rate constant \( K \) for the reaction is given by:
\[ K = \frac{2.303}{t} \log \left( \frac{P_i}{2P_i - P_t} \right) \]