`CH_(3)COCH_(3(g))hArrC_(2)H_(6) (g)+CO_((g))` , The initial pressure of `CH_(3)COCH_(3)` is `50 mm`. At equilibrium the mole fraction of `C_(2)H_(6)` is 1/5 then `K_(p)` will be `:-`

Options

A. `3.333` mm
B. `4.166` mm
C. `8.333` mm
D. `16.666` mm

Explanation

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

\[ \text{CH}_3\text{COCH}_3(g) \rightleftharpoons \text{C}_2\text{H}_6(g) + \text{CO}(g) \]

### Step 1: Define Initial Conditions
The initial pressure of \( \text{CH}_3\text{COCH}_3 \) is given as \( 50 \, \text{mm} \). At the start of the reaction, the pressures of \( \text{C}_2\text{H}_6 \) and \( \text{CO} \) are both \( 0 \, \text{mm} \).

### Step 2: Define Changes at Equilibrium
Let \( x \) be the amount of pressure that has reacted (decreased) from \( \text{CH}_3\text{COCH}_3 \). Therefore, at equilibrium:
- Pressure of \( \text{CH}_3\text{COCH}_3 = 50 - x \)
- Pressure of \( \text{C}_2\text{H}_6 = x \)
- Pressure of \( \text{CO} = x \)

### Step 3: Total Pressure at Equilibrium
The total pressure at equilibrium can be expressed as:
\[ P_{\text{total}} = (50 - x) + x + x = 50 + x \]

### Step 4: Use Mole Fraction to Find \( x \)
We are given that the mole fraction of \( \text{C}_2\text{H}_6 \) at equilibrium is \( \frac{1}{5} \). The mole fraction is given by:
\[ \text{Mole fraction of } \text{C}_2\text{H}_6 = \frac{P_{\text{C}_2\text{H}_6}}{P_{\text{total}}} \]
Substituting the values:
\[ \frac{x}{50 + x} = \frac{1}{5} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ 5x = 50 + x \]
Rearranging the equation:
\[ 5x - x = 50 \implies 4x = 50 \implies x = 12.5 \, \text{mm} \]

### Step 6: Calculate Equilibrium Pressures
Now we can find the equilibrium pressures:
- \( P_{\text{C}_2\text{H}_6} = x = 12.5 \, \text{mm} \)
- \( P_{\text{CO}} = x = 12.5 \, \text{mm} \)
- \( P_{\text{CH}_3\text{COCH}_3} = 50 - x = 50 - 12.5 = 37.5 \, \text{mm} \)

### Step 7: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{C}_2\text{H}_6} \cdot P_{\text{CO}}}{P_{\text{CH}_3\text{COCH}_3}} \]
Substituting the values:
\[ K_p = \frac{(12.5)(12.5)}{37.5} \]
Calculating:
\[ K_p = \frac{156.25}{37.5} = 4.1667 \, \text{mm} \]

### Final Answer
Thus, the value of \( K_p \) is approximately \( 4.1667 \, \text{mm} \).

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