`K_p` for the reaction `N_2O_4 (g) hArr 2NO_2(g)` is 0.66 at `46^@C`. Calculate the part per cent dissociation of `N_2O_4` at `46^@C` and a total pressure of 0.5 atm. Also calculate the partial pressure of `N_2O_4 and NO_2` at equilibrium .

Answer

To solve the problem, we need to follow these steps:

### Step 1: Define the reaction and initial conditions
The given reaction is:
\[ N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \]

Let the initial amount of \( N_2O_4 \) be 1 mole. At equilibrium, let \( \alpha \) be the degree of dissociation of \( N_2O_4 \).

### Step 2: Write the equilibrium expression
At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 3: Calculate partial pressures
Using the total pressure \( P = 0.5 \, \text{atm} \):
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} \cdot P \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{(2\alpha)}{(1 + \alpha)} \cdot P \]

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{2\alpha}{1 + \alpha} \cdot P\right)^2}{\frac{(1 - \alpha)}{(1 + \alpha)} \cdot P} \]

### Step 5: Substitute \( K_p \) and simplify
Given \( K_p = 0.66 \) and \( P = 0.5 \):
\[ 0.66 = \frac{\left(\frac{2\alpha}{1 + \alpha} \cdot 0.5\right)^2}{\frac{(1 - \alpha)}{(1 + \alpha)} \cdot 0.5} \]
This simplifies to:
\[ 0.66 = \frac{(2\alpha)^2 \cdot 0.5}{(1 - \alpha)(1 + \alpha)} \]

### Step 6: Solve for \( \alpha \)
Rearranging gives:
\[ 0.66(1 - \alpha)(1 + \alpha) = 2\alpha^2 \cdot 0.5 \]
\[ 0.66(1 - \alpha^2) = \alpha^2 \]
\[ 0.66 - 0.66\alpha^2 = \alpha^2 \]
\[ 0.66 = \alpha^2 + 0.66\alpha^2 \]
\[ 0.66 = 1.66\alpha^2 \]
\[ \alpha^2 = \frac{0.66}{1.66} \approx 0.397 \]
\[ \alpha \approx \sqrt{0.397} \approx 0.63 \]

### Step 7: Calculate percentage dissociation
Percentage dissociation of \( N_2O_4 \):
\[ \text{Percentage dissociation} = \alpha \times 100 \approx 63\% \]

### Step 8: Calculate partial pressures at equilibrium
Now substituting \( \alpha \) back to find the partial pressures:
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{2\alpha}{1 + \alpha} \cdot 0.5 = \frac{2 \times 0.63}{1 + 0.63} \cdot 0.5 \approx 0.385 \, \text{atm} \]
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} \cdot 0.5 = \frac{1 - 0.63}{1 + 0.63} \cdot 0.5 \approx 0.115 \, \text{atm} \]

### Final Results
- Percentage dissociation of \( N_2O_4 \) at \( 46^\circ C \): **63%**
- Partial pressure of \( N_2O_4 \) at equilibrium: **0.115 atm**
- Partial pressure of \( NO_2 \) at equilibrium: **0.385 atm**