In reaction:
`CH_(3)COCH_(3)(g)hArrCH_(3)CH_(3)(g)+CO(g),`
if the initial pressure of `CH_(3)COCH_(3)(g)` is `150 mm` and at equilibrium the mole fraction of `CO(g)` is `1/3`, then the value `K_(P)` is

Options

A. 100 mm
B. 50 mm
C. 25 mm
D. 150 mm

Answer

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

\[ \text{CH}_3\text{COCH}_3(g) \rightleftharpoons \text{CH}_3\text{CH}_3(g) + \text{CO}(g) \]

### Step 1: Understand the given information
- Initial pressure of \( \text{CH}_3\text{COCH}_3 \) is \( 150 \, \text{mm} \).
- At equilibrium, the mole fraction of \( \text{CO} \) is \( \frac{1}{3} \).

### Step 2: Relate mole fraction to moles
The mole fraction of \( \text{CO} \) being \( \frac{1}{3} \) means that out of a total of 3 moles, 1 mole is \( \text{CO} \). Thus, we can infer that:
- Moles of \( \text{CO} = 1 \)
- Moles of \( \text{CH}_3\text{CH}_3 = 1 \)
- Moles of \( \text{CH}_3\text{COCH}_3 = 1 \)

### Step 3: Calculate total moles at equilibrium
Since the mole fraction of \( \text{CO} \) is \( \frac{1}{3} \), the total moles at equilibrium can be expressed as:
\[ \text{Total moles} = \text{Moles of } \text{CO} + \text{Moles of } \text{CH}_3\text{CH}_3 + \text{Moles of } \text{CH}_3\text{COCH}_3 = 1 + 1 + 1 = 3 \]

### Step 4: Calculate partial pressures
Using the initial pressure of \( \text{CH}_3\text{COCH}_3 \):
- The partial pressure of \( \text{CO} \) can be calculated as:
\[ P_{\text{CO}} = \text{Mole fraction of } \text{CO} \times \text{Total pressure} = \frac{1}{3} \times 150 \, \text{mm} = 50 \, \text{mm} \]

- Since the reaction produces equal moles of \( \text{CH}_3\text{CH}_3 \) and \( \text{CO} \), the partial pressure of \( \text{CH}_3\text{CH}_3 \) is also:
\[ P_{\text{CH}_3\text{CH}_3} = 50 \, \text{mm} \]

- The remaining pressure for \( \text{CH}_3\text{COCH}_3 \) at equilibrium will also be:
\[ P_{\text{CH}_3\text{COCH}_3} = 150 \, \text{mm} - 50 \, \text{mm} - 50 \, \text{mm} = 50 \, \text{mm} \]

### Step 5: Calculate \( K_p \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{P_{\text{CH}_3\text{CH}_3} \times P_{\text{CO}}}{P_{\text{CH}_3\text{COCH}_3}} \]
Substituting the values:
\[ K_p = \frac{(50 \, \text{mm}) \times (50 \, \text{mm})}{50 \, \text{mm}} = 50 \, \text{mm} \]

### Final Answer
Thus, the value of \( K_p \) is \( 50 \, \text{mm} \).

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