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. `50 mm`
B. `100 mm`
C. `33.3 mm`
D. `75 mm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction is given as:
\[ \text{CH}_3\text{COCH}_3(g) \rightleftharpoons \text{CH}_3\text{CH}_3(g) + \text{CO}(g) \]

### Step 2: Define initial conditions.
The initial pressure of \(\text{CH}_3\text{COCH}_3\) is given as \(150 \, \text{mm}\). The initial pressures of the products \(\text{CH}_3\text{CH}_3\) and \(\text{CO}\) are both \(0 \, \text{mm}\).

### Step 3: Define changes at equilibrium.
Let \(x\) be the change in pressure of \(\text{CH}_3\text{COCH}_3\) that dissociates at equilibrium. Thus, at equilibrium:
- Pressure of \(\text{CH}_3\text{COCH}_3 = 150 - x\)
- Pressure of \(\text{CH}_3\text{CH}_3 = x\)
- Pressure of \(\text{CO} = x\)

### Step 4: Calculate total pressure at equilibrium.
The total pressure at equilibrium is:
\[ P_{\text{total}} = P_{\text{CH}_3\text{COCH}_3} + P_{\text{CH}_3\text{CH}_3} + P_{\text{CO}} = (150 - x) + x + x = 150 + x \]

### Step 5: Use the mole fraction of CO to find x.
The mole fraction of \(\text{CO}\) is given as \(\frac{1}{3}\). The mole fraction is defined as:
\[ \text{Mole fraction of CO} = \frac{P_{\text{CO}}}{P_{\text{total}}} = \frac{x}{150 + x} \]
Setting this equal to \(\frac{1}{3}\):
\[ \frac{x}{150 + x} = \frac{1}{3} \]

### Step 6: Solve for x.
Cross-multiplying gives:
\[ 3x = 150 + x \]
Rearranging gives:
\[ 3x - x = 150 \implies 2x = 150 \implies x = 75 \]

### Step 7: Calculate pressures at equilibrium.
Now we can find the pressures at equilibrium:
- \(P_{\text{CH}_3\text{COCH}_3} = 150 - x = 150 - 75 = 75 \, \text{mm}\)
- \(P_{\text{CH}_3\text{CH}_3} = x = 75 \, \text{mm}\)
- \(P_{\text{CO}} = x = 75 \, \text{mm}\)

### Step 8: Calculate \(K_P\).
The equilibrium constant \(K_P\) is given by:
\[ K_P = \frac{P_{\text{CH}_3\text{CH}_3} \cdot P_{\text{CO}}}{P_{\text{CH}_3\text{COCH}_3}} = \frac{75 \cdot 75}{75} \]
Calculating this gives:
\[ K_P = \frac{5625}{75} = 75 \, \text{mm} \]

### Final Answer:
The value of \(K_P\) is \(75 \, \text{mm}\).

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