The decomposition of a gaseous substance (A) to yield gaseous products (B), (C) follows First order kinetics. If initially only (A) is present and 10 minutes after the start of the reaction the pressure of (A) is 200 mm Hg and that of overall mixture is 300 mm Hg, then rate constant for `2ArarrB+3C` is :

Options

A. `((1)/(600))"ln" 1.25"sec"^(-1)`
B. ((2.303)/(10))"log"1.5"min"^(-1)`
C. `((1)/(10))"ln" 1.25"sec"^(-1)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the initial conditions
Let the initial pressure of substance A be \( P \). According to the problem, after 10 minutes, the pressure of A is 200 mm Hg, and the total pressure of the mixture is 300 mm Hg.

### Step 2: Set up the equations
From the information given:
- The pressure of A after 10 minutes is \( P - 2x = 200 \) mm Hg (since 2 moles of A decompose).
- The total pressure after 10 minutes is \( P - 2x + x + 3x = 300 \) mm Hg.

### Step 3: Solve the equations
1. From the first equation, we have:
\[ P - 2x = 200 \quad \text{(1)} \]
2. From the second equation, we can simplify:
\[ P + 2x = 300 \quad \text{(2)} \]

### Step 4: Add the equations
Adding equations (1) and (2):
\[ (P - 2x) + (P + 2x) = 200 + 300 \]
This simplifies to:
\[ 2P = 500 \implies P = 250 \, \text{mm Hg} \]

### Step 5: Substitute P back to find x
Substituting \( P = 250 \) into equation (1):
\[ 250 - 2x = 200 \]
Solving for \( x \):
\[ 2x = 250 - 200 = 50 \implies x = 25 \, \text{mm Hg} \]

### Step 6: Calculate the rate constant (k)
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A_0}{A_0 - x} \right) \]
Where:
- \( A_0 = P = 250 \, \text{mm Hg} \)
- \( A = P - 2x = 200 \, \text{mm Hg} \)
- \( t = 10 \, \text{minutes} = 600 \, \text{seconds} \)

Substituting the values:
\[ k = \frac{2.303}{600} \log \left( \frac{250}{250 - 25} \right) = \frac{2.303}{600} \log \left( \frac{250}{225} \right) \]

### Step 7: Calculate the logarithm
Calculating the logarithm:
\[ \frac{250}{225} = \frac{10}{9} \implies \log \left( \frac{10}{9} \right) \approx 0.045757 \]
Now substituting this back into the equation for \( k \):
\[ k = \frac{2.303}{600} \times 0.045757 \approx 0.000175 \, \text{sec}^{-1} \]

### Final Answer
Thus, the rate constant \( k \) for the reaction is approximately \( 0.000175 \, \text{sec}^{-1} \).

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