A sample of `HI(g)` is placed in flask at a pressure of `0.2 atm`. At equilibrium. The partial pressure of `HI(g)` is `0.04 atm`. What is `K_(p)` for the given equilibrium?
`2HI(g) hArr H_(2)(g)+I_(2)(g)`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

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

we will follow these steps:

### Step 1: Identify the Initial Conditions
We are given:
- Initial pressure of HI, \( P_{\text{HI}}^{\text{initial}} = 0.2 \, \text{atm} \)
- Equilibrium pressure of HI, \( P_{\text{HI}}^{\text{eq}} = 0.04 \, \text{atm} \)

### Step 2: Set Up the Change in Pressure
Let \( x \) be the change in pressure of HI that dissociates to form H₂ and I₂. The stoichiometry of the reaction indicates that for every 2 moles of HI that dissociate, 1 mole of H₂ and 1 mole of I₂ are formed. Therefore, we can express the changes in pressure as follows:

- At equilibrium:
- Pressure of HI = \( 0.2 - 2x \)
- Pressure of H₂ = \( x \)
- Pressure of I₂ = \( x \)

### Step 3: Write the Equilibrium Expression
From the equilibrium condition, we know that:

\[ P_{\text{HI}}^{\text{eq}} = 0.2 - 2x \]

Given \( P_{\text{HI}}^{\text{eq}} = 0.04 \, \text{atm} \), we can set up the equation:

\[ 0.2 - 2x = 0.04 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:

\[ 2x = 0.2 - 0.04 \]
\[ 2x = 0.16 \]
\[ x = 0.08 \]

### Step 5: Calculate Equilibrium Pressures
Now we can find the equilibrium pressures for H₂ and I₂:

- \( P_{\text{H}_2}^{\text{eq}} = x = 0.08 \, \text{atm} \)
- \( P_{\text{I}_2}^{\text{eq}} = x = 0.08 \, \text{atm} \)

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) is given by:

\[ K_p = \frac{P_{\text{H}_2}^{\text{eq}} \cdot P_{\text{I}_2}^{\text{eq}}}{(P_{\text{HI}}^{\text{eq}})^2} \]

Substituting the equilibrium pressures into the expression:

\[ K_p = \frac{(0.08) \cdot (0.08)}{(0.04)^2} \]

### Step 7: Calculate \( K_p \)
Calculating the values:

\[ K_p = \frac{0.0064}{0.0016} = 4 \]

### Final Answer
The equilibrium constant \( K_p \) for the reaction is:

\[ K_p = 4 \]