In the reaction, `H_2 +I_2 hArr 2HI`. In a 2 litre flask 0.4 moles of each `H_2 and I_2` are taken. At equilibrium 0.5 moles of HI are formed. What will be the value of equilibrium constant, `K_c`

Options

A. 20.2
B. 25.4
C. 0.284
D. 11.1

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( H_2 + I_2 \rightleftharpoons 2HI \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ H_2 + I_2 \rightleftharpoons 2HI \]

### Step 2: Determine initial moles and concentrations
We are given:
- Initial moles of \( H_2 = 0.4 \) moles
- Initial moles of \( I_2 = 0.4 \) moles
- Volume of the flask = 2 L

To find the initial concentrations:
\[ \text{Concentration of } H_2 = \frac{0.4 \text{ moles}}{2 \text{ L}} = 0.2 \, \text{M} \]
\[ \text{Concentration of } I_2 = \frac{0.4 \text{ moles}}{2 \text{ L}} = 0.2 \, \text{M} \]

### Step 3: Determine moles at equilibrium
At equilibrium, we are given that 0.5 moles of \( HI \) are formed.

From the balanced equation, for every 1 mole of \( H_2 \) and \( I_2 \) that reacts, 2 moles of \( HI \) are produced. Therefore, the change in moles of \( H_2 \) and \( I_2 \) will be:
\[ \text{Change in moles of } H_2 = -0.25 \, \text{moles} \quad (\text{since } 0.5 \text{ moles of } HI \text{ formed}) \]
\[ \text{Change in moles of } I_2 = -0.25 \, \text{moles} \]

### Step 4: Calculate equilibrium moles
Now we can calculate the moles of each substance at equilibrium:
- Moles of \( H_2 \) at equilibrium:
\[ 0.4 - 0.25 = 0.15 \, \text{moles} \]
- Moles of \( I_2 \) at equilibrium:
\[ 0.4 - 0.25 = 0.15 \, \text{moles} \]
- Moles of \( HI \) at equilibrium:
\[ 0.5 \, \text{moles} \]

### Step 5: Calculate equilibrium concentrations
Now we convert these moles into concentrations:
\[ \text{Concentration of } H_2 = \frac{0.15 \text{ moles}}{2 \text{ L}} = 0.075 \, \text{M} \]
\[ \text{Concentration of } I_2 = \frac{0.15 \text{ moles}}{2 \text{ L}} = 0.075 \, \text{M} \]
\[ \text{Concentration of } HI = \frac{0.5 \text{ moles}}{2 \text{ L}} = 0.25 \, \text{M} \]

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}} \]
For our reaction:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.25)^2}{(0.075)(0.075)} \]
\[ K_c = \frac{0.0625}{0.005625} \approx 11.11 \]

### Final Answer
The value of the equilibrium constant \( K_c \) is approximately \( 11.11 \).

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