In an experiment 5 moles of HI were enclosed in a 5 litre container . At 717 K equilibrium constant for the gaseous reaction , `2HI (g) hArr `H_2(g) +I_2(g) ` is 0.025 . Calculate the equilibrium concentration of HI `H_2 and I_2` . What is the fraction of HI that decomposes ?

Answer

To solve the problem step by step, let's break it down into manageable parts:

### Step 1: Determine Initial Concentration
We start with 5 moles of HI in a 5-liter container. The initial concentration of HI can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Substituting the values:

\[ \text{Concentration of HI} = \frac{5 \text{ moles}}{5 \text{ L}} = 1 \text{ M} \]

### Step 2: Set Up the Reaction and Change in Concentration
The reaction is given by:

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

Let \( x \) be the change in concentration of H2 and I2 at equilibrium. Since 2 moles of HI produce 1 mole of H2 and 1 mole of I2, the changes in concentration can be expressed as follows:

- Initial concentration of HI = 1 M
- Change in concentration of HI = \( -2x \)
- Equilibrium concentration of HI = \( 1 - 2x \)
- Equilibrium concentration of H2 = \( x \)
- Equilibrium concentration of I2 = \( x \)

### Step 3: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K_c \) is given by:

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

Substituting the equilibrium concentrations:

\[ K_c = \frac{x \cdot x}{(1 - 2x)^2} = \frac{x^2}{(1 - 2x)^2} \]

Given that \( K_c = 0.025 \):

\[ 0.025 = \frac{x^2}{(1 - 2x)^2} \]

### Step 4: Solve for x
To simplify the equation, we can take the square root of both sides:

\[ \sqrt{0.025} = \frac{x}{1 - 2x} \]

Calculating \( \sqrt{0.025} \):

\[ \sqrt{0.025} = 0.158 \]

Now, we have:

\[ 0.158 = \frac{x}{1 - 2x} \]

Cross-multiplying gives:

\[ 0.158(1 - 2x) = x \]

Expanding this:

\[ 0.158 - 0.316x = x \]

Rearranging the terms:

\[ 0.158 = x + 0.316x \]

\[ 0.158 = 1.316x \]

Now, solving for \( x \):

\[ x = \frac{0.158}{1.316} \approx 0.120 \]

### Step 5: Calculate Equilibrium Concentrations
Now that we have \( x \), we can find the equilibrium concentrations:

- Equilibrium concentration of HI:

\[ [\text{HI}] = 1 - 2x = 1 - 2(0.120) = 1 - 0.240 = 0.760 \text{ M} \]

- Equilibrium concentration of H2:

\[ [\text{H}_2] = x = 0.120 \text{ M} \]

- Equilibrium concentration of I2:

\[ [\text{I}_2] = x = 0.120 \text{ M} \]

### Step 6: Calculate the Fraction of HI that Decomposes
The fraction of HI that decomposes can be calculated as:

\[ \text{Fraction decomposed} = \frac{\text{Initial moles of HI} - \text{Equilibrium moles of HI}}{\text{Initial moles of HI}} = \frac{5 - (0.760 \times 5)}{5} \]

Calculating the moles at equilibrium:

\[ \text{Equilibrium moles of HI} = 0.760 \times 5 = 3.8 \text{ moles} \]

Now, substituting back:

\[ \text{Fraction decomposed} = \frac{5 - 3.8}{5} = \frac{1.2}{5} = 0.24 \]

### Step 7: Convert to Percentage
To express the fraction as a percentage:

\[ \text{Percentage decomposed} = 0.24 \times 100 = 24\% \]

### Final Results
- Equilibrium concentrations:
- \([\text{HI}] = 0.760 \text{ M}\)
- \([\text{H}_2] = 0.120 \text{ M}\)
- \([\text{I}_2] = 0.120 \text{ M}\)
- Fraction of HI that decomposes: 0.24 or 24%