For a first order homogeneous gaseous reaction
`Ato 2B+C`
if the total pressure after time t was `P_(1)` and after long time `(t to infty)` was `P_(infty)` then K in terms of `P_(t_(1))` `P_(infty)` t is

Options

A. `k=(2.303)/(t)log((Poo)/(Poo-P_(t)))`
B. `k=(2.303)/(t)log((2Poo)/(Poo-P_(t)))`
C. `k=(2.303)/(t)log((2Poo)/(3(Poo-P_(t))))`
D. none of these

Answer

To solve the problem step by step, we will analyze the first-order reaction \( A \rightarrow 2B + C \) and derive the expression for the rate constant \( K \) in terms of the pressures \( P_{t_1} \) and \( P_{\infty} \).

### Step 1: Define Initial and Final Conditions
At time \( t = 0 \):
- Let the initial pressure of \( A \) be \( P_A = P_0 \).
- The pressures of \( B \) and \( C \) are both zero.

At time \( t = \infty \):
- All of \( A \) has been converted into products.
- The total pressure \( P_{\infty} \) can be expressed as the sum of the pressures of \( B \) and \( C \):
\[ P_{\infty} = 2P_A + P_A = 3P_A \]

### Step 2: Define Pressure at Time \( t \)
At time \( t = t_1 \):
- Let \( P' \) be the pressure of \( A \) that has reacted.
- The pressure of \( B \) formed is \( 2P' \) and the pressure of \( C \) formed is \( P' \).
- The total pressure at time \( t_1 \) is:
\[ P_{t_1} = P_A - P' + 2P' + P' = P_A + 2P' = P_A + 2(P_A - P_A) = P_A + 2P' \]

### Step 3: Relate \( P_A \) and \( P_{\infty} \)
From the total pressure at \( t = \infty \):
\[ P_A = \frac{P_{\infty}}{3} \]

### Step 4: Express \( P' \) in Terms of \( P_{t_1} \)
We can rearrange the equation for \( P_{t_1} \) to find \( P' \):
\[ P_{t_1} = P_A + 2P' \implies P' = \frac{P_{t_1} - P_A}{2} \]

Substituting \( P_A = \frac{P_{\infty}}{3} \):
\[ P' = \frac{P_{t_1} - \frac{P_{\infty}}{3}}{2} = \frac{3P_{t_1} - P_{\infty}}{6} \]

### Step 5: Substitute \( P' \) into the Expression for \( K \)
The rate constant \( K \) for a first-order reaction is given by:
\[ K = \frac{1}{t} \ln \left( \frac{P_A}{P_A - P'} \right) \]

Substituting \( P_A = \frac{P_{\infty}}{3} \) and \( P' = \frac{3P_{t_1} - P_{\infty}}{6} \):
\[ K = \frac{1}{t} \ln \left( \frac{\frac{P_{\infty}}{3}}{\frac{P_{\infty}}{3} - \frac{3P_{t_1} - P_{\infty}}{6}} \right) \]

### Step 6: Simplify the Expression
Simplifying the denominator:
\[ \frac{P_{\infty}}{3} - \frac{3P_{t_1} - P_{\infty}}{6} = \frac{2P_{\infty}}{6} + \frac{3P_{t_1}}{6} = \frac{2P_{\infty} + 3P_{t_1}}{6} \]
Thus, we have:
\[ K = \frac{1}{t} \ln \left( \frac{\frac{P_{\infty}}{3}}{\frac{2P_{\infty} + 3P_{t_1}}{6}} \right) \]
This simplifies to:
\[ K = \frac{1}{t} \ln \left( \frac{2P_{\infty}}{2P_{\infty} + 3P_{t_1}} \right) \]

### Final Expression
The final expression for the rate constant \( K \) is:
\[ K = \frac{2.303}{t} \log \left( \frac{2P_{\infty}}{2P_{\infty} + 3P_{t_1}} \right) \]