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, we need to determine the equilibrium concentrations of HI, H2, and I2, as well as the fraction of HI that decomposes. Let's break down the solution step by step.

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

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of HI = 5 moles
- Moles of H2 = 0 moles
- Moles of I2 = 0 moles

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the number of moles of H2 and I2 formed at equilibrium. According to the stoichiometry of the reaction:
- Moles of HI at equilibrium = \( 5 - 2x \)
- Moles of H2 at equilibrium = \( x \)
- Moles of I2 at equilibrium = \( x \)

### Step 4: Calculate the equilibrium concentrations
The volume of the container is 5 liters, so we can calculate the molar concentrations:
- Concentration of HI = \( \frac{5 - 2x}{5} \)
- Concentration of H2 = \( \frac{x}{5} \)
- Concentration of I2 = \( \frac{x}{5} \)

### Step 5: Write the expression for the equilibrium constant
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 concentrations we found:
\[ K_c = \frac{\left(\frac{x}{5}\right)\left(\frac{x}{5}\right)}{\left(\frac{5 - 2x}{5}\right)^2} \]

### Step 6: Substitute the given value of \( K_c \)
We know that \( K_c = 0.025 \):
\[ 0.025 = \frac{\left(\frac{x}{5}\right)^2}{\left(\frac{5 - 2x}{5}\right)^2} \]
This simplifies to:
\[ 0.025 = \frac{x^2}{(5 - 2x)^2} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 0.025(5 - 2x)^2 = x^2 \]
Expanding and rearranging leads to a quadratic equation.

### Step 8: Calculate the value of \( x \)
After solving the quadratic equation, we find that:
\[ x = 0.6 \text{ moles} \]

### Step 9: Calculate the equilibrium concentrations
Now substituting \( x \) back into the expressions for concentrations:
- Concentration of HI = \( \frac{5 - 2(0.6)}{5} = \frac{3.8}{5} = 0.76 \, \text{mol/L} \)
- Concentration of H2 = \( \frac{0.6}{5} = 0.12 \, \text{mol/L} \)
- Concentration of I2 = \( \frac{0.6}{5} = 0.12 \, \text{mol/L} \)

### Step 10: Calculate the fraction of HI that decomposes
The fraction of HI that decomposes is given by:
\[ \text{Fraction decomposed} = \frac{2x}{\text{Initial moles of HI}} = \frac{2(0.6)}{5} = \frac{1.2}{5} = 0.24 \]

### Final Results
- Equilibrium concentration of HI = 0.76 mol/L
- Equilibrium concentration of H2 = 0.12 mol/L
- Equilibrium concentration of I2 = 0.12 mol/L
- Fraction of HI that decomposes = 0.24