For a first order gas phase reaction :
`A_((g)) to 2B_((g)) +C_((g))`
`P_(0)` be initial pressure of A and `P_(t)` the total pressure at time 't'. Integrated rate equation is :

Options

A. `(2.303)/(t)log((P_(0))/(P_(0)-P_(t)))`
B. `(2.303)/(t)log((2P_(0))/(3P_(0)-P_(t)))`
C. `(2.303)/(t)log((P_(0))/(2P_(0)-P_(t)))`
D. `(2.303)/(t)log((2P_(0))/(2P_(0)-P_(t)))`

Explanation

To solve the problem, we need to derive the integrated rate equation for the first-order gas-phase reaction given by:

\[ A_{(g)} \rightarrow 2B_{(g)} + C_{(g)} \]

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let \( P_0 \) be the initial pressure of gas A.
- Initially, the pressures of B and C are zero since the reaction has not started.

2. **Change in Pressure**:
- Let \( P \) be the change in pressure of A at time \( t \).
- Therefore, the pressure of A at time \( t \) will be \( P_0 - P \).
- The pressure of B will increase by \( 2P \) (since 2 moles of B are produced for every mole of A consumed).
- The pressure of C will increase by \( P \).

3. **Total Pressure at Time \( t \)**:
- The total pressure \( P_t \) at time \( t \) can be expressed as:
\[ P_t = (P_0 - P) + 2P + P = P_0 + 2P \]

4. **Rearranging for \( P \)**:
- From the equation \( P_t = P_0 + 2P \), we can solve for \( P \):
\[ 2P = P_t - P_0 \implies P = \frac{P_t - P_0}{2} \]

5. **Substituting into the First-Order Rate Equation**:
- For a first-order reaction, the integrated rate equation is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{P_0 - P} \right) \]
- Substitute \( P \) into the equation:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{P_0 - \frac{P_t - P_0}{2}} \right) \]

6. **Simplifying the Denominator**:
- The denominator becomes:
\[ P_0 - \frac{P_t - P_0}{2} = P_0 - \frac{P_t}{2} + \frac{P_0}{2} = \frac{3P_0 - P_t}{2} \]
- Thus, the equation for \( k \) becomes:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{\frac{3P_0 - P_t}{2}} \right) \]

7. **Final Form**:
- This can be rewritten as:
\[ k = \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right) \]

### Conclusion:
The integrated rate equation for the given first-order gas-phase reaction is:
\[ k = \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right) \]
Thus, the correct option is option 2:
\[ \frac{2.303}{t} \log \left( \frac{2P_0}{3P_0 - P_t} \right) \]