In 5 litre container 1 mole of `H_(2)` and 1 mole of `I_(2)` are taken initially and at equilibrium of HI is `40%`. Then `K_(p)` will be

Options

A. `(9)/(4)`
B. `(16)/(9)`
C. `(1)/(4)`
D. None

Answer

To solve the problem step by step, we will follow the reaction and the given conditions to find the equilibrium constant \( K_p \).

### Step 1: Write the balanced chemical equation
The reaction we are considering is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

### Step 2: Determine initial moles and changes
Initially, we have:
- Moles of \( H_2 = 1 \)
- Moles of \( I_2 = 1 \)
- Moles of \( HI = 0 \)

At equilibrium, we are told that the amount of \( HI \) is 40% of the total moles. Since we have 2 moles of reactants initially (1 mole of \( H_2 \) and 1 mole of \( I_2 \)), at equilibrium:
\[ \text{Moles of } HI = 0.4 \text{ moles} \]

### Step 3: Calculate the change in moles
From the stoichiometry of the reaction, if \( x \) moles of \( H_2 \) and \( I_2 \) react, then \( 2x \) moles of \( HI \) are formed. Given that \( 2x = 0.4 \), we can find \( x \):
\[ x = \frac{0.4}{2} = 0.2 \]

### Step 4: Calculate moles at equilibrium
At equilibrium, the moles will be:
- Moles of \( H_2 = 1 - x = 1 - 0.2 = 0.8 \)
- Moles of \( I_2 = 1 - x = 1 - 0.2 = 0.8 \)
- Moles of \( HI = 2x = 0.4 \)

### Step 5: Calculate concentrations
Since the volume of the container is 5 liters, we can calculate the concentrations:
- Concentration of \( H_2 = \frac{0.8 \text{ moles}}{5 \text{ L}} = 0.16 \text{ M} \)
- Concentration of \( I_2 = \frac{0.8 \text{ moles}}{5 \text{ L}} = 0.16 \text{ M} \)
- Concentration of \( HI = \frac{0.4 \text{ moles}}{5 \text{ L}} = 0.08 \text{ M} \)

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the concentrations:
\[ K_c = \frac{(0.08)^2}{(0.16)(0.16)} = \frac{0.0064}{0.0256} = 0.25 \]

### Step 7: Relate \( K_c \) and \( K_p \)
Since the reaction involves gases, we can relate \( K_c \) and \( K_p \) using the formula:
\[ K_p = K_c (RT)^{\Delta n} \]
Where \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants):
\[ \Delta n = 2 - (1 + 1) = 0 \]
Thus:
\[ K_p = K_c (RT)^0 = K_c \]
So,
\[ K_p = 0.25 \]

### Final Answer
The value of \( K_p \) is:
\[ \boxed{0.25} \]