In the gaseous reaction `2A+B Leftrightarrow A_2B, triangleG^@ `= 1200 cal at `227^@C.` What total pressure would be necessary to produce 60% conversion of B into `A_2B` when 2:1 mixture is used?

Answer

To solve the problem, we need to determine the total pressure necessary to produce a 60% conversion of B into A2B in the reaction \(2A + B \leftrightarrow A_2B\) given the standard Gibbs free energy change (\(\Delta G^\circ\)) and the temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- We start with a 2:1 mixture of A and B. Let's denote the initial pressure of A as \(2P\) and the initial pressure of B as \(P\).

2. **Determine the Change in Pressure**:
- Given that 60% of B is converted into A2B, we can calculate the amount of B that reacts:
\[ \text{Amount of B reacted} = 0.6P \]
- Since 2 moles of A react with 1 mole of B, the amount of A that reacts will be:
\[ \text{Amount of A reacted} = 2 \times 0.6P = 1.2P \]

3. **Calculate Equilibrium Pressures**:
- After the reaction, the pressures at equilibrium will be:
- Pressure of A:
\[ 2P - 1.2P = 0.8P \]
- Pressure of B:
\[ P - 0.6P = 0.4P \]
- Pressure of A2B produced:
\[ 0.6P \quad (\text{since 60% of B is converted}) \]

4. **Total Pressure at Equilibrium**:
- The total pressure at equilibrium can be calculated as:
\[ P_{\text{total}} = P_A + P_B + P_{A2B} = 0.8P + 0.4P + 0.6P = 1.8P \]

5. **Relate Gibbs Free Energy to Equilibrium Constant**:
- The relationship between Gibbs free energy and the equilibrium constant \(K_p\) is given by:
\[ \Delta G^\circ = -RT \ln K_p \]
- Rearranging gives:
\[ K_p = e^{-\Delta G^\circ / RT} \]
- We need to convert \(\Delta G^\circ\) from calories to joules (1 cal = 4.184 J):
\[ \Delta G^\circ = 1200 \text{ cal} \times 4.184 \text{ J/cal} = 5020.8 \text{ J} \]

6. **Calculate \(K_p\)**:
- Using \(R = 8.314 \text{ J/(mol K)}\) and \(T = 227 + 273 = 500 \text{ K}\):
\[ K_p = e^{-5020.8 / (8.314 \times 500)} \approx e^{-1.205} \approx 0.3 \]

7. **Set Up the Expression for \(K_p\)**:
- The expression for \(K_p\) in terms of partial pressures is:
\[ K_p = \frac{P_{A2B}}{(P_A)^2 (P_B)} = \frac{0.6P}{(0.8P)^2(0.4P)} \]
- Simplifying gives:
\[ K_p = \frac{0.6P}{0.64P^2 \cdot 0.4P} = \frac{0.6}{0.256P} = \frac{2.34375}{P} \]

8. **Equate and Solve for \(P\)**:
- Setting \(K_p = 0.3\):
\[ 0.3 = \frac{2.34375}{P} \implies P = \frac{2.34375}{0.3} \approx 7.8125 \]

9. **Calculate Total Pressure**:
- Now substituting back to find the total pressure:
\[ P_{\text{total}} = 1.8P = 1.8 \times 7.8125 \approx 14.0625 \text{ atm} \]

### Final Answer:
The total pressure necessary to produce 60% conversion of B into \(A_2B\) is approximately **14.06 atm**.