In an experiment , 2 moles of HI are taken into an evacuated 10.0 litre container at 720 K. The equilibrium constant equals to 0.0156for the gaseous reaction, `2 HI (g) hArr H_(2) (g) + I_(2)(g).` find equilibrium concentration of ` HI (g) , H_(2) (g) , I_(2)(g).`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial concentration of HI
Given that we have 2 moles of HI in a 10.0 L container, we can calculate the initial concentration of HI using the formula:

\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{2 \text{ moles}}{10.0 \text{ L}} = 0.2 \text{ M} \]

### Step 2: Set up the equilibrium expression
The reaction is:

\[ 2 \text{HI} (g) \rightleftharpoons \text{H}_2 (g) + \text{I}_2 (g) \]

Let \( x \) be the change in concentration of H2 and I2 at equilibrium. Therefore, at equilibrium, we have:

- Concentration of HI = \( 0.2 - 2x \)
- Concentration of H2 = \( x \)
- Concentration of I2 = \( x \)

### Step 3: Write the equilibrium constant expression
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

Substituting the equilibrium concentrations into the expression gives:

\[ K_c = \frac{x \cdot x}{(0.2 - 2x)^2} = \frac{x^2}{(0.2 - 2x)^2} \]

### Step 4: Substitute the value of \( K_c \)
We know that \( K_c = 0.0156 \). Therefore, we can set up the equation:

\[ 0.0156 = \frac{x^2}{(0.2 - 2x)^2} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:

\[ 0.0156 (0.2 - 2x)^2 = x^2 \]

Expanding and rearranging leads to a quadratic equation. First, we can take the square root of both sides:

\[ \sqrt{0.0156} (0.2 - 2x) = x \]

Calculating \( \sqrt{0.0156} \):

\[ \sqrt{0.0156} \approx 0.1249 \]

Now substituting back:

\[ 0.1249 (0.2 - 2x) = x \]

Expanding gives:

\[ 0.02498 - 0.2498x = x \]

Rearranging leads to:

\[ 0.02498 = x + 0.2498x \]
\[ 0.02498 = 1.2498x \]
\[ x = \frac{0.02498}{1.2498} \approx 0.01999 \approx 0.02 \]

### Step 6: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:

- Concentration of H2 = \( x = 0.02 \, \text{M} \)
- Concentration of I2 = \( x = 0.02 \, \text{M} \)
- Concentration of HI = \( 0.2 - 2x = 0.2 - 2(0.02) = 0.2 - 0.04 = 0.16 \, \text{M} \)

### Final Answer:
- Equilibrium concentration of HI = 0.16 M
- Equilibrium concentration of H2 = 0.02 M
- Equilibrium concentration of I2 = 0.02 M