Two solid compounds X and Y dissociates at a certain temperature as follows
`X(s) hArr A(g)+2B(g),K_(p1)=9xx10^(-3)atm^(3)`
`Y(s) hArr 2B(g)+C(g),K_(p2)=4.5xx10^(-3)atm^(3)`
The total pressure of gases over a mixture of X and Y is `:`

Options

A. (a) `4.5`atm
B. (b) `0.45` atm
C. (c) `0.6` atm
D. (d) None of these

Answer

To solve the problem step by step, we will analyze the dissociation of the solid compounds X and Y, and then calculate the total pressure of the gases produced.

### Step 1: Write the dissociation reactions
We have two solid compounds:
1. \( X(s) \rightleftharpoons A(g) + 2B(g) \) with \( K_{p1} = 9 \times 10^{-3} \, \text{atm}^3 \)
2. \( Y(s) \rightleftharpoons 2B(g) + C(g) \) with \( K_{p2} = 4.5 \times 10^{-3} \, \text{atm}^3 \)

### Step 2: Combine the reactions
When we combine the two reactions, we get:
\[ X(s) + Y(s) \rightleftharpoons A(g) + 4B(g) + C(g) \]

### Step 3: Write the expression for the equilibrium constant of the combined reaction
The equilibrium constant \( K_{p} \) for the combined reaction is the product of the individual equilibrium constants:
\[ K_{p} = K_{p1} \times K_{p2} \]
Substituting the values:
\[ K_{p} = (9 \times 10^{-3}) \times (4.5 \times 10^{-3}) \]

### Step 4: Calculate \( K_{p} \)
Calculating the product:
\[ K_{p} = 9 \times 4.5 \times 10^{-6} = 40.5 \times 10^{-6} = 4.05 \times 10^{-5} \, \text{atm}^6 \]

### Step 5: Set up the expression for partial pressures
Let \( p \) be the partial pressure of \( A \) and \( C \), and \( 4p \) be the partial pressure of \( B \):
- Partial pressure of \( A \) = \( p \)
- Partial pressure of \( B \) = \( 4p \)
- Partial pressure of \( C \) = \( p \)

The total pressure \( P \) is given by:
\[ P = p + 4p + p = 6p \]

### Step 6: Write the expression for \( K_{p} \) in terms of \( p \)
Using the expression for \( K_{p} \):
\[ K_{p} = \frac{(p)(4p)^4}{(1)} = 256p^5 \]

### Step 7: Set the equation equal to \( K_{p} \)
Now we set the expression equal to the calculated \( K_{p} \):
\[ 256p^5 = 4.05 \times 10^{-5} \]

### Step 8: Solve for \( p \)
Rearranging gives:
\[ p^5 = \frac{4.05 \times 10^{-5}}{256} \]
Calculating:
\[ p^5 = 1.58 \times 10^{-7} \]
Taking the fifth root:
\[ p \approx (1.58 \times 10^{-7})^{1/5} \approx 0.0735 \, \text{atm} \]

### Step 9: Calculate the total pressure
Now substituting back to find the total pressure:
\[ P = 6p = 6 \times 0.0735 = 0.441 \, \text{atm} \]

### Final Answer
The total pressure of gases over the mixture of X and Y is approximately:
\[ P \approx 0.441 \, \text{atm} \]

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