`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 step-by-step, we will follow the process outlined in the video transcript, while providing detailed explanations for each step.

### Given Data:
- The equilibrium constant \( K_p \) for the reaction \( N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \) is \( 0.66 \) at \( 46^\circ C \).
- Total pressure \( P_t = 0.5 \, \text{atm} \).

### Step 1: Set Up the Initial Conditions
Let’s assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \) at the beginning (initial state).

- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

### Step 2: Define the Change in Moles
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium, the moles will be:

- Moles of \( N_2O_4 = 1 - \alpha \)
- Moles of \( NO_2 = 2\alpha \) (since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \))

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium will be:

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

### Step 4: Calculate Mole Fractions
The mole fractions for each component are:

- Mole fraction of \( N_2O_4 \):

\[ \chi_{N_2O_4} = \frac{1 - \alpha}{1 + \alpha} \]

- Mole fraction of \( NO_2 \):

\[ \chi_{NO_2} = \frac{2\alpha}{1 + \alpha} \]

### Step 5: Calculate Partial Pressures
The partial pressures can be calculated using the total pressure:

- Partial pressure of \( N_2O_4 \):

\[ P_{N_2O_4} = \chi_{N_2O_4} \times P_t = \frac{1 - \alpha}{1 + \alpha} \times 0.5 \]

- Partial pressure of \( NO_2 \):

\[ P_{NO_2} = \chi_{NO_2} \times P_t = \frac{2\alpha}{1 + \alpha} \times 0.5 \]

### Step 6: Write the Expression for \( K_p \)
The expression for \( K_p \) is given by:

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left(\frac{2\alpha}{1 + \alpha} \times 0.5\right)^2}{\frac{1 - \alpha}{1 + \alpha} \times 0.5} \]

### Step 7: Simplify the Expression
Substituting the expressions for partial pressures into the \( K_p \) equation:

\[ K_p = \frac{\left(\frac{2\alpha \times 0.5}{1 + \alpha}\right)^2}{\frac{1 - \alpha}{1 + \alpha} \times 0.5} \]

This simplifies to:

\[ K_p = \frac{(2\alpha)^2 \times 0.25}{(1 - \alpha)(1 + \alpha) \times 0.5} \]

\[ K_p = \frac{4\alpha^2 \times 0.25}{(1 - \alpha)(1 + \alpha) \times 0.5} = \frac{2\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 8: Set Up the Equation
Now, substituting \( K_p = 0.66 \):

\[ 0.66 = \frac{2\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 9: Solve for \( \alpha \)
Cross-multiplying gives:

\[ 0.66(1 - \alpha)(1 + \alpha) = 2\alpha^2 \]

Expanding and rearranging leads to a quadratic equation in terms of \( \alpha \):

\[ 0.66 - 0.66\alpha^2 = 2\alpha^2 \]
\[ (2 + 0.66)\alpha^2 = 0.66 \]
\[ 2.66\alpha^2 = 0.66 \]
\[ \alpha^2 = \frac{0.66}{2.66} \approx 0.248 \]
\[ \alpha \approx 0.498 \text{ (approximately 0.5)} \]

### Step 10: Calculate Percent Dissociation
The percent dissociation of \( N_2O_4 \):

\[ \text{Percent dissociation} = \alpha \times 100 \approx 50\% \]

### Step 11: Calculate Partial Pressures at Equilibrium
Using \( \alpha \approx 0.5 \):

- Partial pressure of \( N_2O_4 \):

\[ P_{N_2O_4} = \frac{1 - 0.5}{1 + 0.5} \times 0.5 = \frac{0.5}{1.5} \times 0.5 = \frac{0.25}{1} = 0.167 \, \text{atm} \]

- Partial pressure of \( NO_2 \):

\[ P_{NO_2} = \frac{2 \times 0.5}{1 + 0.5} \times 0.5 = \frac{1}{1.5} \times 0.5 = \frac{0.5}{1.5} = 0.333 \, \text{atm} \]

### Final Results:
- Percent dissociation of \( N_2O_4 \) is approximately **50%**.
- Partial pressure of \( N_2O_4 \) at equilibrium is **0.167 atm**.
- Partial pressure of \( NO_2 \) at equilibrium is **0.333 atm**.