For `COCl_(2)(g) hArr CO(g) +Cl_(2)(g), K_(p)=8xx10^(-8)` atm. units at a certain temperature. Find the degree of dissociation of `COCl_(2)` at given temperature if the equilibrium pressure is 2.0 atm.

Answer

To solve the problem, we need to find the degree of dissociation (α) of COCl₂ at equilibrium given the reaction:

\[ \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \]

with the equilibrium constant \( K_p = 8 \times 10^{-8} \) atm and total equilibrium pressure \( P_{total} = 2.0 \) atm.

### Step-by-Step Solution:

**Step 1: Set up the initial conditions.**
- Let the initial number of moles of COCl₂ be 1.
- Initially, the moles of CO and Cl₂ are 0.

**Step 2: Define the degree of dissociation (α).**
- If α is the degree of dissociation, then at equilibrium:
- Moles of COCl₂ = \( 1 - \alpha \)
- Moles of CO = \( \alpha \)
- Moles of Cl₂ = \( \alpha \)

**Step 3: Calculate the total moles at equilibrium.**
- Total moles at equilibrium = \( (1 - \alpha) + \alpha + \alpha = 1 + \alpha \)

**Step 4: Write the expression for Kp.**
- The expression for \( K_p \) is given by:

\[ K_p = \frac{P_{CO} \cdot P_{Cl_2}}{P_{COCl_2}} \]

**Step 5: Calculate the partial pressures.**
- The partial pressure of each component can be calculated as follows:
- \( P_{CO} = \frac{\alpha}{1 + \alpha} \times P_{total} \)
- \( P_{Cl_2} = \frac{\alpha}{1 + \alpha} \times P_{total} \)
- \( P_{COCl_2} = \frac{1 - \alpha}{1 + \alpha} \times P_{total} \)

**Step 6: Substitute into the Kp expression.**
- Substitute the partial pressures into the \( K_p \) expression:

\[ K_p = \frac{\left(\frac{\alpha}{1 + \alpha} \times 2\right) \cdot \left(\frac{\alpha}{1 + \alpha} \times 2\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} \times 2} \]

**Step 7: Simplify the expression.**
- This simplifies to:

\[ K_p = \frac{\alpha^2 \cdot 2^2}{(1 - \alpha) \cdot 2} \cdot \frac{1}{(1 + \alpha)^2} \]

- Therefore:

\[ K_p = \frac{2\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

**Step 8: Substitute the known values.**
- Given \( K_p = 8 \times 10^{-8} \):

\[ 8 \times 10^{-8} = \frac{2\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

**Step 9: Assume α is small.**
- Since \( K_p \) is very small, we can assume \( \alpha \) is small, thus \( 1 - \alpha \approx 1 \) and \( 1 + \alpha \approx 1 \):

\[ 8 \times 10^{-8} \approx 2\alpha^2 \]

**Step 10: Solve for α.**
- Rearranging gives:

\[ \alpha^2 = \frac{8 \times 10^{-8}}{2} = 4 \times 10^{-8} \]

- Taking the square root:

\[ \alpha = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \]

### Final Answer:
The degree of dissociation of COCl₂ at the given temperature is \( \alpha = 2 \times 10^{-4} \).

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