Phosgene gas `("COCl"_(2))` dissociates to carbon monoxide gas and chlorine gas with a change of pressure from 450 mm Hg to 600 mm Hg. If volume and temperature remain constant through out the experiment the value of `K_(p)` is:

Options

A. 400
B. 200
C. 115
D. 75

Answer

To solve the problem, we need to find the equilibrium constant \( K_p \) for the dissociation of phosgene gas (\( COCl_2 \)) into carbon monoxide (\( CO \)) and chlorine gas (\( Cl_2 \)). The reaction can be represented as follows:

\[ COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g) \]

### Step-by-Step Solution:

1. **Write the Initial Conditions**:
- Initial pressure of \( COCl_2 \) = 450 mm Hg
- Initial pressures of \( CO \) and \( Cl_2 \) = 0 mm Hg (since they are produced from the dissociation of \( COCl_2 \))

2. **Set Up the Change in Pressure**:
- Let \( x \) be the amount of \( COCl_2 \) that dissociates at equilibrium.
- At equilibrium:
- Pressure of \( COCl_2 \) = \( 450 - x \) mm Hg
- Pressure of \( CO \) = \( x \) mm Hg
- Pressure of \( Cl_2 \) = \( x \) mm Hg

3. **Total Pressure at Equilibrium**:
- The total pressure at equilibrium is given as 600 mm Hg.
- Therefore, we can write the equation:
\[ (450 - x) + x + x = 600 \]
Simplifying this gives:
\[ 450 + x = 600 \]

4. **Solve for \( x \)**:
- Rearranging the equation:
\[ x = 600 - 450 = 150 \text{ mm Hg} \]

5. **Calculate Partial Pressures**:
- Now we can find the partial pressures at equilibrium:
- Partial pressure of \( COCl_2 \) = \( 450 - x = 450 - 150 = 300 \) mm Hg
- Partial pressure of \( CO \) = \( x = 150 \) mm Hg
- Partial pressure of \( Cl_2 \) = \( x = 150 \) mm Hg

6. **Calculate \( K_p \)**:
- The expression for \( K_p \) is given by:
\[ K_p = \frac{P_{CO} \cdot P_{Cl_2}}{P_{COCl_2}} \]
Substituting the values:
\[ K_p = \frac{150 \cdot 150}{300} \]
Simplifying this gives:
\[ K_p = \frac{22500}{300} = 75 \]

### Final Answer:
The value of \( K_p \) is **75 mm Hg**.