In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for
`HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)`

Explanation

To calculate \( K_p \) for the dissociation of HI given that 20% of HI is dissociated at equilibrium, we can follow these steps:

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

### Step 2: Define the degree of dissociation
Let the degree of dissociation (α) be 0.2 (since 20% of HI is dissociated).

### Step 3: Set up the initial and equilibrium concentrations
Assuming we start with 1 mole of HI:
- Initial moles of HI = 1
- Initial moles of H2 = 0
- Initial moles of I2 = 0

At equilibrium, the moles will be:
- Moles of HI = \( 1 - \alpha = 1 - 0.2 = 0.8 \)
- Moles of H2 = \( \frac{\alpha}{2} = \frac{0.2}{2} = 0.1 \)
- Moles of I2 = \( \frac{\alpha}{2} = \frac{0.2}{2} = 0.1 \)

### Step 4: Write the expression for \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{H_2})^{1/2} \cdot (P_{I_2})^{1/2}}{P_{HI}} \]
Where \( P \) represents the partial pressures of the gases.

### Step 5: Calculate the partial pressures
Assuming the total volume is \( V \) (which cancels out in the calculation):
- \( P_{HI} = \frac{(1 - \alpha)RT}{V} = \frac{0.8RT}{V} \)
- \( P_{H_2} = \frac{\left(\frac{\alpha}{2}\right)RT}{V} = \frac{0.1RT}{V} \)
- \( P_{I_2} = \frac{\left(\frac{\alpha}{2}\right)RT}{V} = \frac{0.1RT}{V} \)

### Step 6: Substitute the values into the \( K_p \) expression
Now substituting the values into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{0.1RT}{V}\right)^{1/2} \cdot \left(\frac{0.1RT}{V}\right)^{1/2}}{\frac{0.8RT}{V}} \]

### Step 7: Simplify the expression
\[ K_p = \frac{\frac{0.1RT}{V} \cdot \frac{0.1RT}{V}}{\frac{0.8RT}{V}} = \frac{0.01(RT)^2/V^2}{0.8RT/V} = \frac{0.01RT}{0.8V} \]

### Step 8: Calculate \( K_p \)
Since \( R \) and \( T \) are constants, we can simplify further:
\[ K_p = \frac{0.01}{0.8} = 0.0125 \]

Thus, the value of \( K_p \) is:
\[ K_p = 0.125 \]

### Final Answer
The equilibrium constant \( K_p \) for the dissociation of HI is \( 0.125 \).

---