Calculate the degree of dissociation of HI at `450^(@)C` if the equilibrium constant for the dissociation reaction is `0.263`.

Explanation

To calculate the degree of dissociation of HI at 450°C given that the equilibrium constant (K) for the dissociation reaction is 0.263, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of hydrogen iodide (HI) can be represented as:
\[ \text{2 HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Step 2: Define the degree of dissociation
Let the degree of dissociation be represented by \( \alpha \). This means that if we start with 1 mole of HI, at equilibrium:
- The amount of HI that dissociates = \( \alpha \)
- The amount of HI remaining = \( 1 - \alpha \)
- The amount of H2 produced = \( \frac{1}{2} \alpha \)
- The amount of I2 produced = \( \frac{1}{2} \alpha \)

### Step 3: Set up the equilibrium concentrations
Assuming the volume of the container is \( V \):
- Concentration of HI at equilibrium = \( \frac{1 - \alpha}{V} \)
- Concentration of H2 at equilibrium = \( \frac{\frac{1}{2} \alpha}{V} = \frac{\alpha}{2V} \)
- Concentration of I2 at equilibrium = \( \frac{\frac{1}{2} \alpha}{V} = \frac{\alpha}{2V} \)

### Step 4: 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 equilibrium concentrations into the expression:
\[ K_c = \frac{\left(\frac{\alpha}{2V}\right) \left(\frac{\alpha}{2V}\right)}{\left(\frac{1 - \alpha}{V}\right)^2} \]

### Step 5: Simplify the expression
This simplifies to:
\[ K_c = \frac{\frac{\alpha^2}{4V^2}}{\frac{(1 - \alpha)^2}{V^2}} = \frac{\alpha^2}{4(1 - \alpha)^2} \]

### Step 6: Substitute the known value of \( K_c \)
Given \( K_c = 0.263 \):
\[ 0.263 = \frac{\alpha^2}{4(1 - \alpha)^2} \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 0.263 \cdot 4(1 - \alpha)^2 = \alpha^2 \]
\[ 1.052(1 - 2\alpha + \alpha^2) = \alpha^2 \]
\[ 1.052 - 2.104\alpha + 1.052\alpha^2 = \alpha^2 \]
\[ 0 = 0.052\alpha^2 + 2.104\alpha - 1.052 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 0.052 \), \( b = 2.104 \), and \( c = -1.052 \).

Calculating the discriminant:
\[ b^2 - 4ac = (2.104)^2 - 4(0.052)(-1.052) \]
\[ = 4.417616 + 0.219424 = 4.63704 \]

Now applying the quadratic formula:
\[ \alpha = \frac{-2.104 \pm \sqrt{4.63704}}{2 \cdot 0.052} \]
\[ = \frac{-2.104 \pm 2.155}{0.104} \]

Calculating the two possible values:
1. \( \alpha = \frac{0.051}{0.104} \approx 0.490 \)
2. \( \alpha = \frac{-4.259}{0.104} \) (not physically meaningful)

### Final Answer
Thus, the degree of dissociation of HI at 450°C is approximately:
\[ \alpha \approx 0.490 \]