Two solids AB and `CB_2` are simultaneously heated in a closed vessel to attain equilibrium.
`AB(s)hArrA(g)+B(g)`..(1)
`CB_2(s)hArrC(s)+2B(g)`….(2)
`K_P` for (1) and (2) reaction are `4.8xx10^(-6)` and `5.76xx10^(-6)`.If partial pressure of A at equilibrium is `2xx10^(-x)`, find x.

Explanation

To solve the problem, we will analyze the two equilibrium reactions and use the given equilibrium constants to find the value of \( x \) in the expression for the partial pressure of \( A \).

### Step 1: Write the equilibrium expressions for both reactions.

For the first reaction:
\[ AB(s) \rightleftharpoons A(g) + B(g) \]
The equilibrium constant \( K_{P1} \) is given by:
\[ K_{P1} = \frac{P_A \cdot P_B}{1} = P_A \cdot P_B \]
where \( P_A \) is the partial pressure of \( A \) and \( P_B \) is the partial pressure of \( B \).

For the second reaction:
\[ CB_2(s) \rightleftharpoons C(s) + 2B(g) \]
The equilibrium constant \( K_{P2} \) is given by:
\[ K_{P2} = \frac{P_B^2}{1} = P_B^2 \]

### Step 2: Substitute the values of \( K_{P1} \) and \( K_{P2} \).

We know:
\[ K_{P1} = 4.8 \times 10^{-6} \]
\[ K_{P2} = 5.76 \times 10^{-6} \]

### Step 3: Set up the equations based on the equilibrium constants.

From the first equilibrium:
\[ P_A \cdot P_B = 4.8 \times 10^{-6} \quad \text{(1)} \]

From the second equilibrium:
\[ P_B^2 = 5.76 \times 10^{-6} \quad \text{(2)} \]

### Step 4: Solve for \( P_B \) from equation (2).

Taking the square root of both sides:
\[ P_B = \sqrt{5.76 \times 10^{-6}} = 2.4 \times 10^{-3} \]

### Step 5: Substitute \( P_B \) back into equation (1) to find \( P_A \).

Substituting \( P_B \) into equation (1):
\[ P_A \cdot (2.4 \times 10^{-3}) = 4.8 \times 10^{-6} \]
\[ P_A = \frac{4.8 \times 10^{-6}}{2.4 \times 10^{-3}} = 2 \times 10^{-3} \]

### Step 6: Relate \( P_A \) to the expression \( 2 \times 10^{-x} \).

We have:
\[ P_A = 2 \times 10^{-3} \]
This can be expressed as:
\[ P_A = 2 \times 10^{-x} \]
From this, we can see that:
\[ -x = -3 \implies x = 3 \]

### Final Answer:
The value of \( x \) is \( 3 \).

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