Solid A and B are taken in a closed container at a certain temperature. These two solids decompose and following equilibria are established simultaneously
`A(s)hArr X(g) + Y(g) K_(P_(1)) = 250 atm^(2)`
`B(s)hArrY(g) + Z(g) K_(P_(2))` = ?
If the total pressure developed over the solid mixture is 50 atm. Then the value of `K_(P)` for the `2^(nd)` reaction.

Options

A. 375
B. 625
C. 225
D. 250

Answer

To solve the problem, we will analyze the two equilibrium reactions involving solids A and B, and use the given total pressure to find the equilibrium constant \( K_{P} \) for the second reaction.

### Step-by-Step Solution:

1. **Identify the Reactions and Constants**:
- The first reaction is:
\[ A(s) \rightleftharpoons X(g) + Y(g) \]
with \( K_{P1} = 250 \, \text{atm}^2 \).
- The second reaction is:
\[ B(s) \rightleftharpoons Y(g) + Z(g) \]
and we need to find \( K_{P2} \).

2. **Define Pressures**:
- Let the partial pressure of \( X \) be \( P_1 \).
- Let the partial pressure of \( Y \) be \( P_Y \).
- Let the partial pressure of \( Z \) be \( P_2 \).
- Since \( Y \) is produced in both reactions, we can express the total pressure as:
\[ P_{total} = P_1 + P_Y + P_2 \]

3. **Given Total Pressure**:
- The total pressure developed over the solid mixture is given as \( 50 \, \text{atm} \):
\[ P_1 + P_Y + P_2 = 50 \, \text{atm} \] (Equation 1)

4. **Using the First Reaction's Equilibrium Constant**:
- For the first reaction, the equilibrium constant \( K_{P1} \) can be expressed as:
\[ K_{P1} = \frac{P_1 \cdot P_Y}{1} \]
- Given \( K_{P1} = 250 \):
\[ P_1 \cdot P_Y = 250 \] (Equation 2)

5. **Express \( P_Y \) in terms of \( P_1 \)**:
- From Equation 2, we can express \( P_Y \):
\[ P_Y = \frac{250}{P_1} \]

6. **Substituting \( P_Y \) into Equation 1**:
- Substitute \( P_Y \) into Equation 1:
\[ P_1 + \frac{250}{P_1} + P_2 = 50 \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ P_2 = 50 - P_1 - \frac{250}{P_1} \] (Equation 3)

8. **Using the Second Reaction's Equilibrium Constant**:
- For the second reaction, the equilibrium constant \( K_{P2} \) can be expressed as:
\[ K_{P2} = \frac{P_Y \cdot P_2}{1} \]
- Substituting \( P_Y \) and \( P_2 \) from previous equations:
\[ K_{P2} = \frac{\frac{250}{P_1} \cdot \left(50 - P_1 - \frac{250}{P_1}\right)}{1} \]

9. **Calculating \( K_{P2} \)**:
- To find \( K_{P2} \), we need to solve for \( P_1 \) first. We can use numerical methods or trial and error to find \( P_1 \) such that the total pressure equals 50 atm.
- After finding \( P_1 \), substitute back into the equations to find \( P_Y \) and \( P_2 \), and finally calculate \( K_{P2} \).