For a gaseous phase reaction `A+2B hArr AB_(2), K_(c)=0.3475 L^(2) "mole"^(-2)` at `200^(@)C`. When `2` moles of B are mixed with one "mole" of A, what total pressure is required to convert `60%` of A in `AB_(2)`?

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript while ensuring clarity and completeness.

### Step 1: Write the Reaction and Initial Conditions
The given reaction is:
\[ A + 2B \rightleftharpoons AB_2 \]

Initially, we have:
- 1 mole of A
- 2 moles of B

### Step 2: Define the Degree of Dissociation
We are told that 60% of A is converted into \( AB_2 \). Thus, the degree of dissociation (α) is:
\[ \alpha = 0.6 \]

### Step 3: Calculate Moles at Equilibrium
At equilibrium:
- Moles of A remaining = \( 1 - \alpha = 1 - 0.6 = 0.4 \)
- Moles of B remaining = \( 2 - 2\alpha = 2 - 2(0.6) = 0.8 \)
- Moles of \( AB_2 \) formed = \( \alpha = 0.6 \)

### Step 4: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of A} + \text{Moles of B} + \text{Moles of } AB_2 \]
\[ = 0.4 + 0.8 + 0.6 = 1.8 \]

### Step 5: Calculate Partial Pressures
Let \( P \) be the total pressure. The partial pressures can be calculated using the mole fractions:
- Partial pressure of A:
\[ P_A = \left( \frac{0.4}{1.8} \right) P \]
- Partial pressure of B:
\[ P_B = \left( \frac{0.8}{1.8} \right) P \]
- Partial pressure of \( AB_2 \):
\[ P_{AB_2} = \left( \frac{0.6}{1.8} \right) P \]

### Step 6: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{AB_2}}{P_A \cdot P_B^2} \]

Substituting the expressions for partial pressures:
\[ K_p = \frac{\left( \frac{0.6}{1.8} P \right)}{\left( \frac{0.4}{1.8} P \right) \left( \frac{0.8}{1.8} P \right)^2} \]

### Step 7: Simplify the Expression
This simplifies to:
\[ K_p = \frac{0.6 P}{\left( \frac{0.4 \cdot 0.8^2}{1.8^3} P^3 \right)} \]
\[ K_p = \frac{0.6 \cdot 1.8^3}{0.4 \cdot 0.8^2} \cdot \frac{1}{P^2} \]

### Step 8: Relate \( K_p \) and \( K_c \)
We know:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]
Where:
- \( K_c = 0.3475 \, \text{L}^2 \text{mol}^{-2} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 200 + 273 = 473 \, \text{K} \)
- \( \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 3 = -2 \)

### Step 9: Calculate \( K_p \)
Substituting the values:
\[ K_p = 0.3475 \cdot (0.0821 \cdot 473)^{-2} \]

### Step 10: Solve for Total Pressure \( P \)
Equate the two expressions for \( K_p \) and solve for \( P \):
\[ 0.3475 \cdot (0.0821 \cdot 473)^{-2} = \frac{0.6 \cdot 1.8^3}{0.4 \cdot 0.8^2} \cdot \frac{1}{P^2} \]

After simplification, you will find:
\[ P \approx 181.5 \, \text{atm} \]

### Final Answer
The total pressure required to convert 60% of A into \( AB_2 \) is approximately:
\[ P \approx 181.5 \, \text{atm} \]