`A(g)+2B(s) hArr 2C(g)`
Initially 2 mole A (g), 4 mole of B(s) and 1 mole of an inert gas are present in a closed container. After equilibrium has established, total pressure of container becomes 9 atm.If A(g) is 50% consumed at equilibrium, then , calculate `K_p` for the :

Options

A. 9 atm
B. `36/5` atm
C. 12 atm
D. `2/3` atm

Answer

To solve the problem step by step, we will analyze the reaction and the information given:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ A(g) + 2B(s) \rightleftharpoons 2C(g) \]

### Step 2: Identify the initial moles of each component
Initially, we have:
- Moles of \( A(g) = 2 \)
- Moles of \( B(s) = 4 \) (Note: B is a solid and does not affect the equilibrium constant)
- Moles of inert gas = 1

### Step 3: Determine the change in moles at equilibrium
It is given that \( A(g) \) is 50% consumed at equilibrium. Therefore:
- Moles of \( A \) consumed = \( 50\% \) of \( 2 \) moles = \( 1 \) mole
- Moles of \( A \) remaining at equilibrium = \( 2 - 1 = 1 \) mole

From the stoichiometry of the reaction:
- For every \( 1 \) mole of \( A \) consumed, \( 2 \) moles of \( C \) are produced.
- Therefore, moles of \( C \) produced = \( 2 \times 1 = 2 \) moles

### Step 4: Calculate the total moles at equilibrium
At equilibrium, we have:
- Moles of \( A(g) = 1 \)
- Moles of \( C(g) = 2 \)
- Moles of inert gas = 1 (remains unchanged)

Thus, the total moles of gas at equilibrium:
\[ \text{Total moles} = 1 + 2 + 1 = 4 \]

### Step 5: Calculate the partial pressures
The total pressure in the container at equilibrium is given as \( 9 \, \text{atm} \).

To find the partial pressure of \( C(g) \):
- The partial pressure of a gas is given by the formula:
\[ P_i = \left( \frac{n_i}{n_{\text{total}}} \right) \times P_{\text{total}} \]
Where:
- \( n_i \) = moles of the gas
- \( n_{\text{total}} \) = total moles of gas
- \( P_{\text{total}} \) = total pressure

For \( C(g) \):
\[ P_C = \left( \frac{2}{4} \right) \times 9 \, \text{atm} = \frac{1}{2} \times 9 \, \text{atm} = 4.5 \, \text{atm} \]

### Step 6: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_C)^2}{(P_A)} \]

Where:
- \( P_A \) is the partial pressure of \( A(g) \):
\[ P_A = \left( \frac{1}{4} \right) \times 9 \, \text{atm} = 2.25 \, \text{atm} \]

Substituting the values into the equation:
\[ K_p = \frac{(4.5)^2}{2.25} = \frac{20.25}{2.25} = 9 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ K_p = 9 \, \text{atm} \]

---