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)`

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction is given as:
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Partial pressure of HI = 0.2 atm
- Partial pressure of H₂ = 0 atm
- Partial pressure of I₂ = 0 atm

### Step 3: Define the change in pressure
Let \( x \) be the change in pressure of H₂ and I₂ at equilibrium. Since 2 moles of HI produce 1 mole of H₂ and 1 mole of I₂, the change in pressure of HI will be \( 2x \).

### Step 4: Write the equilibrium pressures
At equilibrium, the pressures will be:
- \( P_{\text{HI}} = 0.2 - 2x \)
- \( P_{\text{H}_2} = x \)
- \( P_{\text{I}_2} = x \)

### Step 5: Use the given equilibrium pressure of HI
We are given that the equilibrium pressure of HI is 0.04 atm:
\[ 0.2 - 2x = 0.04 \]

### Step 6: Solve for x
Rearranging the equation:
\[ 2x = 0.2 - 0.04 \]
\[ 2x = 0.16 \]
\[ x = 0.08 \, \text{atm} \]

### Step 7: Calculate the equilibrium pressures of H₂ and I₂
Since \( x = 0.08 \):
- \( P_{\text{H}_2} = x = 0.08 \, \text{atm} \)
- \( P_{\text{I}_2} = x = 0.08 \, \text{atm} \)

### Step 8: Write the expression for \( K_p \)
The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{P_{\text{H}_2} \cdot P_{\text{I}_2}}{(P_{\text{HI}})^2} \]

### Step 9: Substitute the equilibrium pressures into the \( K_p \) expression
Substituting the values we found:
\[ K_p = \frac{(0.08)(0.08)}{(0.04)^2} \]

### Step 10: Calculate \( K_p \)
Calculating the values:
\[ K_p = \frac{0.0064}{0.0016} = 4 \]

### Conclusion
Thus, the value of \( K_p \) for the given equilibrium is:
\[ K_p = 4 \]

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