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 of finding the degree of dissociation of \( COCl_2 \) at equilibrium pressure of 2.0 atm, we can follow these steps:

### Step 1: Write the equilibrium expression
The reaction is given as:
\[ COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g) \]

The equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{P_{CO} \cdot P_{Cl_2}}{P_{COCl_2}} \]

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of \( COCl_2 \). If we start with 1 mole of \( COCl_2 \):
- At equilibrium, the moles of \( COCl_2 \) will be \( 1 - \alpha \)
- The moles of \( CO \) will be \( \alpha \)
- The moles of \( Cl_2 \) will also be \( \alpha \)

### Step 3: Calculate total moles at equilibrium
The total number of moles at equilibrium will be:
\[ n_{total} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Calculate partial pressures
The total pressure \( P_t \) is given as 2.0 atm. The partial pressures can be expressed as:
- \( P_{COCl_2} = \frac{(1 - \alpha)}{(1 + \alpha)} \cdot P_t \)
- \( P_{CO} = \frac{\alpha}{(1 + \alpha)} \cdot P_t \)
- \( P_{Cl_2} = \frac{\alpha}{(1 + \alpha)} \cdot P_t \)

### Step 5: Substitute into the equilibrium expression
Substituting the partial pressures into the \( K_p \) expression:
\[ K_p = \frac{P_{CO} \cdot P_{Cl_2}}{P_{COCl_2}} = \frac{\left(\frac{\alpha}{1 + \alpha} \cdot P_t\right) \cdot \left(\frac{\alpha}{1 + \alpha} \cdot P_t\right)}{\frac{(1 - \alpha)}{(1 + \alpha)} \cdot P_t} \]
This simplifies to:
\[ K_p = \frac{\alpha^2 \cdot P_t^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Plug in known values
Given \( K_p = 8 \times 10^{-8} \) atm and \( P_t = 2 \) atm, we can substitute these values into the equation:
\[ 8 \times 10^{-8} = \frac{\alpha^2 \cdot (2)^2}{(1 - \alpha)(1 + \alpha)} \]
This simplifies to:
\[ 8 \times 10^{-8} = \frac{4\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 7: Assume \( \alpha \) is small
Since \( K_p \) is very small, we can assume \( \alpha \) is small, which allows us to approximate \( 1 - \alpha \approx 1 \) and \( 1 + \alpha \approx 1 \):
\[ 8 \times 10^{-8} \approx 4\alpha^2 \]

### Step 8: Solve for \( \alpha \)
Rearranging gives:
\[ \alpha^2 \approx \frac{8 \times 10^{-8}}{4} = 2 \times 10^{-8} \]
Taking the square root:
\[ \alpha \approx \sqrt{2 \times 10^{-8}} = \sqrt{2} \times 10^{-4} \approx 1.414 \times 10^{-4} \]

### Final Answer
The degree of dissociation \( \alpha \) of \( COCl_2 \) at the given temperature and equilibrium pressure is approximately:
\[ \alpha \approx 1.41 \times 10^{-4} \]