In a closed cylinder of capacity 24.6 L the following reaction occurs at `27^@C` ,
`A_2(s) hArr B_2(s) +2C(g)`
At equilibrium 1 g of `B_2(s)` (molar mass = 50 g `mol^(-1)` ) is present . The equilibrium constant `K_p` for the equilibrium in `atm^2` unit is (R=0.082 L atm `K^(-1) mol^(-1)` )

Options

A. `1.6xx10^(-2)`
B. `1.6xx10^(-5)`
C. `1.6xx10^(-3)`
D. `1.6xx10^(-4)`

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction given:

\[ A_2(s) \rightleftharpoons B_2(s) + 2C(g) \]

### Step 1: Determine the moles of \( B_2 \)
We know that at equilibrium, 1 g of \( B_2 \) is present, and the molar mass of \( B_2 \) is 50 g/mol.

To find the number of moles of \( B_2 \):

\[ \text{Moles of } B_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{50 \text{ g/mol}} = 0.02 \text{ mol} \]

### Step 2: Determine the moles of \( C \)
From the stoichiometry of the reaction, for every mole of \( B_2 \) produced, 2 moles of \( C \) are produced. Therefore, if 0.02 moles of \( B_2 \) are formed, the moles of \( C \) will be:

\[ \text{Moles of } C = 2 \times \text{moles of } B_2 = 2 \times 0.02 = 0.04 \text{ mol} \]

### Step 3: Calculate the pressure of \( C \)
Using the ideal gas law \( PV = nRT \), we can rearrange it to find pressure \( P \):

\[ P = \frac{nRT}{V} \]

Where:
- \( n = 0.04 \) mol (moles of \( C \))
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 27^\circ C = 300 \, K \) (convert Celsius to Kelvin)
- \( V = 24.6 \, L \)

Substituting the values:

\[ P = \frac{0.04 \times 0.082 \times 300}{24.6} \]

Calculating this gives:

\[ P = \frac{0.04 \times 24.6}{24.6} = 0.04 \, \text{atm} \]

### Step 4: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by the expression:

\[ K_p = P_C^2 \]

Since we only have the pressure of \( C \):

\[ K_p = (0.04)^2 = 0.0016 \, \text{atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is:

\[ K_p = 1.6 \times 10^{-3} \, \text{atm}^2 \]