For a homogeneous gaseous reaction `A rarr B + C + D` , the initial pressure was `P_0` white pressure after time 't' was P. if `(P gt P_0)` The expression for the constant K is

Options

A. `K=2.303/tlog.((2P_0)/(3P_0-P))`
B. `K=2.303/tlog.((3P_0)/(2P_0-P))`
C. `K=2.303/tlog.((P_0)/(P_0-P))`
D. `K=2.303/tlog.((P_0)/(4P_0-P))`

Answer

To derive the expression for the rate constant \( K \) for the reaction \( A \rightarrow B + C + D \), we will follow these steps:

### Step 1: Understand the Reaction
The reaction involves one mole of reactant \( A \) producing one mole each of products \( B \), \( C \), and \( D \). Therefore, for every mole of \( A \) that reacts, the total number of moles of gas increases.

### Step 2: Initial and Final Pressure
Let the initial pressure of the system be \( P_0 \). After a time \( t \), the pressure is \( P \). Since the reaction produces three moles of products from one mole of reactant, we can express the change in pressure.

### Step 3: Change in Pressure
If \( x \) is the change in pressure due to the consumption of \( A \), then:
- The pressure of \( A \) at time \( t \) will be \( P_0 - x \).
- The total pressure at time \( t \) will be the sum of the remaining pressure of \( A \) and the pressures of the products \( B \), \( C \), and \( D \).

Since each mole of \( A \) produces three moles of products, we have:
\[ P = (P_0 - x) + 3x = P_0 + 2x \]

### Step 4: Relate \( x \) to Initial and Final Pressure
From the equation above, we can express \( x \) in terms of \( P \) and \( P_0 \):
\[ P - P_0 = 2x \implies x = \frac{P - P_0}{2} \]

### Step 5: Substitute \( x \) Back
Now, we can substitute \( x \) back into the expression for the pressure of \( A \):
\[ P_A = P_0 - x = P_0 - \frac{P - P_0}{2} = \frac{P_0 + P}{2} \]

### Step 6: Apply the Rate Law
For a first-order reaction, the rate constant \( K \) can be expressed using the logarithmic form:
\[ K = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right) \]

### Step 7: Substitute Concentrations with Pressures
Since pressure can be used as a measure of concentration for gases, we substitute:
\[ K = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} \right) = \frac{2.303}{t} \log \left( \frac{P_0}{\frac{P_0 + P}{2}} \right) \]

### Step 8: Simplify the Expression
This can be simplified further:
\[ K = \frac{2.303}{t} \log \left( \frac{2P_0}{P_0 + P} \right) \]

### Final Expression
Thus, the expression for the rate constant \( K \) is:
\[ K = \frac{2.303}{t} \log \left( \frac{2P_0}{P_0 + P} \right) \]

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