The vapour pressure of two miscible liquids `(A)` and `(B)` are `300` and `500 mm` of `Hg` respectively. In a flask `10` mole of `(A)` is mixed with `12` mole of (B). However, as soon as `(B)` is added, `(A)` starts polymerising into a completely insoluble solid. The polymerisation follows first-order kinetics. After `100` minute, `0.525` mole of a solute is dissolved whivh arrests the polymerisation completely. The final vapour pressure of the solution is `400 mm` of `Hg`. Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation and ideal behaviour for the final solution.

Explanation

To solve the problem, we need to follow a systematic approach. Let's break it down step by step.

### Step 1: Determine the initial conditions
We have:
- Vapour pressure of liquid A, \( P^0_A = 300 \, \text{mm Hg} \)
- Vapour pressure of liquid B, \( P^0_B = 500 \, \text{mm Hg} \)
- Moles of A, \( n_A = 10 \)
- Moles of B, \( n_B = 12 \)

### Step 2: Calculate the total moles before polymerization
The total moles before polymerization is:
\[ n_{total} = n_A + n_B = 10 + 12 = 22 \, \text{moles} \]

### Step 3: Write the equation for the total vapour pressure before polymerization
Using Raoult's Law, the total vapour pressure \( P_m \) before polymerization can be expressed as:
\[ P_m = \frac{n_A}{n_{total}} P^0_A + \frac{n_B}{n_{total}} P^0_B \]
Substituting the values:
\[ P_m = \frac{10}{22} \times 300 + \frac{12}{22} \times 500 \]

### Step 4: Calculate the total vapour pressure before polymerization
Calculating each term:
\[ P_m = \frac{10 \times 300}{22} + \frac{12 \times 500}{22} = \frac{3000 + 6000}{22} = \frac{9000}{22} \approx 409.09 \, \text{mm Hg} \]

### Step 5: Determine the moles of A left after polymerization
After 100 minutes, we know that 0.525 moles of a solute are dissolved, which arrests the polymerization. The final vapour pressure is given as \( P_f = 400 \, \text{mm Hg} \).

Using the final vapour pressure:
\[ P_f = \frac{n_A'}{n_A' + n_B + 0.525} P^0_A + \frac{n_B}{n_A' + n_B + 0.525} P^0_B \]
Where \( n_A' \) is the moles of A left after polymerization.

### Step 6: Set up the equation for final vapour pressure
Substituting the known values:
\[ 400 = \frac{n_A'}{n_A' + 12 + 0.525} \times 300 + \frac{12}{n_A' + 12 + 0.525} \times 500 \]

### Step 7: Solve for \( n_A' \)
This equation can be solved for \( n_A' \). After some algebraic manipulation, we find:
\[ n_A' \approx 9.9 \, \text{moles} \]

### Step 8: Calculate the rate constant \( k \)
For a first-order reaction, the rate constant \( k \) is given by:
\[ k = \frac{2.303}{t} \log\left(\frac{n_{A, initial}}{n_A'}\right) \]
Substituting the values:
\[ k = \frac{2.303}{100} \log\left(\frac{10}{9.9}\right) \]

### Step 9: Calculate \( k \)
Calculating the logarithm:
\[ \log\left(\frac{10}{9.9}\right) \approx \log(1.0101) \approx 0.00432 \]
Thus,
\[ k = \frac{2.303}{100} \times 0.00432 \approx 0.0000997 \, \text{min}^{-1} \approx 10^{-4} \, \text{min}^{-1} \]

### Final Answer
The rate constant of the polymerization reaction is approximately:
\[ k \approx 10^{-4} \, \text{min}^{-1} \]