Bodenstein carried out the determination of equilibrium constants of phosgene equilibrium by introducing `CO and CI_2` at known pressure in a reaction bulb and measuring the equillibrium pressure from the attached monometer. In one experiment, CO at 342 mm and `CI_2` at 351.4 mm were introduced . The equilibrium pressure was found to be 439.5mm . At equilibrium if partial pressure of `COCI_2` be ` x ` mm at `127^@C`.
The volume of reacting vessel decreased by 3 times , `K_p` changed by

Options

A. a) 3 times increased
B. b) Remains constant
C. c) `1/3` times decreased
D. d) `1/9` times increased

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction we are dealing with is:
\[ \text{CO} + \text{Cl}_2 \rightleftharpoons \text{COCl}_2 \]

### Step 2: Identify initial pressures
At the start (t = 0), the pressures of CO and Cl₂ are:
- \( P_{\text{CO}} = 342 \, \text{mmHg} \)
- \( P_{\text{Cl}_2} = 351.4 \, \text{mmHg} \)
- \( P_{\text{COCl}_2} = 0 \, \text{mmHg} \)

### Step 3: Identify equilibrium pressure
At equilibrium, the total pressure is given as:
\[ P_{\text{total}} = 439.5 \, \text{mmHg} \]

### Step 4: Define the change in pressure
Let \( x \) be the partial pressure of COCl₂ at equilibrium. Therefore, the partial pressures at equilibrium will be:
- \( P_{\text{CO}} = 342 - x \)
- \( P_{\text{Cl}_2} = 351.4 - x \)
- \( P_{\text{COCl}_2} = x \)

### Step 5: Set up the equation for total pressure
The total pressure at equilibrium can be expressed as:
\[ (342 - x) + (351.4 - x) + x = 439.5 \]
This simplifies to:
\[ 693.4 - x = 439.5 \]
From this, we can solve for \( x \):
\[ x = 693.4 - 439.5 = 253.9 \, \text{mmHg} \]

### Step 6: Calculate the equilibrium constant \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{COCl}_2}}{P_{\text{CO}} \cdot P_{\text{Cl}_2}} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{x}{(342 - x)(351.4 - x)} \]
Substituting \( x = 253.9 \):
\[ K_p = \frac{253.9}{(342 - 253.9)(351.4 - 253.9)} \]
Calculating the values:
- \( 342 - 253.9 = 88.1 \)
- \( 351.4 - 253.9 = 97.5 \)
Thus,
\[ K_p = \frac{253.9}{(88.1)(97.5)} \]

### Step 7: Effect of volume change on \( K_p \)
When the volume of the reacting vessel is decreased by three times, the pressure increases by three times (according to Boyle's Law). Therefore, the new pressures will be:
- \( P'_{\text{CO}} = 3(342 - x) \)
- \( P'_{\text{Cl}_2} = 3(351.4 - x) \)
- \( P'_{\text{COCl}_2} = 3x \)

### Step 8: Calculate new \( K_p' \)
The new equilibrium constant \( K_p' \) will be:
\[ K_p' = \frac{3x}{(3(342 - x))(3(351.4 - x))} \]
This simplifies to:
\[ K_p' = \frac{3x}{9(342 - x)(351.4 - x)} = \frac{1}{3} \cdot \frac{x}{(342 - x)(351.4 - x)} = \frac{1}{3} K_p \]

### Conclusion
Thus, when the volume of the reacting vessel is decreased by three times, the equilibrium constant \( K_p \) decreases to one-third of its original value:
\[ K_p' = \frac{1}{3} K_p \]

### Final Answer
The correct option is:
**C: K_p decreases by one-third.**