Consider a first order gas phase decomposition reaction given below:
`A(g) to B_(g) + C_(g)`
The initial pressure of the system before decomposition of A was `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) - x)`
B. `k = (2.303)/(t) log (p_(i))/(2p_(i) - p_(t))`
C. `k = (2.303)/(t) log (p_(i))/(2p_(i) + p_(i))`
D. `k = (2.303)/(t) log (p_(i))/(p_(i) + x)`

Explanation

To solve the problem step by step, we will analyze the first-order gas phase decomposition reaction \( A(g) \rightarrow B(g) + C(g) \).

### Step 1: Understand the Initial Conditions
At the start of the reaction (time \( t = 0 \)):
- The initial pressure of the system is \( P_i \).
- The concentration of \( A \) is \( P_i \), and the concentrations of \( B \) and \( C \) are both \( 0 \).

### Step 2: Analyze the Change Over Time
After a time \( t' \):
- The total pressure of the system increases by \( x \) units, resulting in a total pressure \( P_t \).
- Therefore, we can express this as:
\[ P_t = P_i + x \]

### Step 3: Relate the Changes in Pressure to the Reaction
Since \( A \) decomposes into \( B \) and \( C \), for every mole of \( A \) that decomposes, one mole of \( B \) and one mole of \( C \) is produced. If \( y \) moles of \( A \) have decomposed, then:
- The pressure of \( A \) at time \( t' \) will be:
\[ P_A = P_i - y \]
- The pressure of \( B \) and \( C \) will each be \( y \), leading to:
\[ P_t = (P_i - y) + y + y = P_i + y \]

### Step 4: Relate the Change in Pressure to \( x \)
From the previous step, we can substitute \( y \) with \( x \):
- Since \( x = y \), we have:
\[ P_t = P_i + x \]

### Step 5: Express \( y \) in Terms of \( P_t \) and \( P_i \)
From the equation \( P_t = P_i + x \), we can rearrange to find:
\[ x = P_t - P_i \]

### Step 6: Substitute \( x \) into the Pressure of \( A \)
Now substituting \( x \) back into the equation for the pressure of \( A \):
\[ P_A = P_i - (P_t - P_i) = 2P_i - P_t \]

### Step 7: Use the First-Order Kinetics Equation
For a first-order reaction, the rate constant \( k \) can be expressed using the equation:
\[ k = \frac{2.303}{t} \log \left( \frac{P_i}{P_A} \right) \]
Substituting \( P_A = 2P_i - P_t \):
\[ k = \frac{2.303}{t} \log \left( \frac{P_i}{2P_i - P_t} \right) \]

### Final Result
Thus, the rate constant \( k \) for the reaction is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{P_i}{2P_i - P_t} \right) \]