For the reaction
`2HI(g)hArr H_(2)(g)+I_(2)(g)`. The value of `K_(c)` is 4. If 2 moles of `H_(2)," 2 moles of "I_(2)` and 2 moles of HI are present in one litre container then moles of `I_(2)` present at equilibrium is :

Options

A. 0.8
B. 3.2
C. 2.4
D. 4.4

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation and identify the initial concentrations.
The reaction given is:
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

Initially, we have:
- Moles of HI = 2
- Moles of H2 = 2
- Moles of I2 = 2

Since the volume of the container is 1 L, the initial concentrations are:
- \([ \text{HI} ]_0 = 2 \, \text{M}\)
- \([ \text{H}_2 ]_0 = 2 \, \text{M}\)
- \([ \text{I}_2 ]_0 = 2 \, \text{M}\)

### Step 2: Set up the expression for the equilibrium constant \( K_c \).
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]
Given that \( K_c = 4 \).

### Step 3: Define the changes in concentrations at equilibrium.
Let \( \alpha \) be the change in moles of H2 and I2 at equilibrium. Therefore:
- At equilibrium:
- Moles of HI = \( 2 - 2\alpha \)
- Moles of H2 = \( 2 + \alpha \)
- Moles of I2 = \( 2 + \alpha \)

### Step 4: Substitute the equilibrium concentrations into the \( K_c \) expression.
Substituting the equilibrium concentrations into the expression for \( K_c \):
\[ 4 = \frac{(2 + \alpha)(2 + \alpha)}{(2 - 2\alpha)^2} \]

### Step 5: Simplify the equation.
This can be rewritten as:
\[ 4 = \frac{(2 + \alpha)^2}{(2 - 2\alpha)^2} \]

Taking the square root of both sides gives:
\[ 2 = \frac{2 + \alpha}{2 - 2\alpha} \]

### Step 6: Cross-multiply and solve for \( \alpha \).
Cross-multiplying gives:
\[ 2(2 - 2\alpha) = 2 + \alpha \]
Expanding this:
\[ 4 - 4\alpha = 2 + \alpha \]

Rearranging the terms:
\[ 4 - 2 = 4\alpha + \alpha \]
\[ 2 = 5\alpha \]

Thus, we find:
\[ \alpha = \frac{2}{5} = 0.4 \]

### Step 7: Calculate the moles of I2 at equilibrium.
Now, we can find the moles of I2 at equilibrium:
\[ \text{Moles of I2 at equilibrium} = 2 + \alpha = 2 + 0.4 = 2.4 \]

### Final Answer:
The moles of I2 present at equilibrium is **2.4 moles**.

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