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, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 1: Determine Initial Conditions
Initially, we have:
- 1 mole of \( H_2 \)
- 1 mole of \( I_2 \)
- Volume of the container = 5 L

### Step 2: Calculate Initial Concentrations
The initial concentrations of \( H_2 \) and \( I_2 \) can be calculated as follows:

\[ \text{Concentration of } H_2 = \frac{1 \text{ mole}}{5 \text{ L}} = 0.2 \text{ M} \]
\[ \text{Concentration of } I_2 = \frac{1 \text{ mole}}{5 \text{ L}} = 0.2 \text{ M} \]

### Step 3: Determine Equilibrium Concentrations
At equilibrium, it is given that \( HI \) is 40%. Since \( HI \) is produced from \( H_2 \) and \( I_2 \), we can calculate the moles of \( HI \) at equilibrium.

\[ \text{Moles of } HI = 0.4 \times (1 \text{ mole of } H_2 + 1 \text{ mole of } I_2) = 0.4 \times 2 = 0.8 \text{ moles} \]

### Step 4: Calculate Change in Moles
Let \( x \) be the moles of \( H_2 \) and \( I_2 \) that react to form \( HI \). Since 2 moles of \( HI \) are formed from 1 mole of \( H_2 \) and 1 mole of \( I_2 \):

\[ x = \frac{0.8}{2} = 0.4 \text{ moles of } H_2 \text{ and } I_2 \text{ reacted} \]

### Step 5: Calculate Remaining Moles at Equilibrium
At equilibrium:
- Moles of \( H_2 \) = \( 1 - 0.4 = 0.6 \) moles
- Moles of \( I_2 \) = \( 1 - 0.4 = 0.6 \) moles
- Moles of \( HI \) = \( 0.8 \) moles

### Step 6: Calculate Equilibrium Concentrations
Now, we can calculate the equilibrium concentrations:

\[ \text{Concentration of } H_2 = \frac{0.6 \text{ moles}}{5 \text{ L}} = 0.12 \text{ M} \]
\[ \text{Concentration of } I_2 = \frac{0.6 \text{ moles}}{5 \text{ L}} = 0.12 \text{ M} \]
\[ \text{Concentration of } HI = \frac{0.8 \text{ moles}}{5 \text{ L}} = 0.16 \text{ M} \]

### Step 7: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.16)^2}{(0.12)(0.12)} = \frac{0.0256}{0.0144} \approx 1.7778 \]

### Step 8: Calculate \( K_p \)
Since the change in the number of moles (\( \Delta n \)) is 0 (1 mole of \( H_2 \) + 1 mole of \( I_2 \) gives 2 moles of \( HI \)), we have:

\[ K_p = K_c \cdot R^{\Delta n} = K_c \]

Thus, \( K_p \approx 1.7778 \).

### Final Answer
The value of \( K_p \) is approximately **1.78**.

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