At 46° C Kp for the reaction `N_(2)O_(4) (g)hArr 2NO_(2)(g)` is 0.667 atm. Compute the percent dissociation of `N_(2)O_(4)` at 46°C at a total pressure of 380 Torr.

Answer

To solve the problem of calculating the percent dissociation of \( N_2O_4 \) at 46°C and a total pressure of 380 Torr, we will follow these steps:

### Step 1: Convert Pressure from Torr to atm
We know that:
\[ 1 \text{ atm} = 760 \text{ Torr} \]
Thus, to convert 380 Torr to atm:
\[ P = \frac{380 \text{ Torr}}{760 \text{ Torr/atm}} = 0.5 \text{ atm} \]

### Step 2: Write the Reaction and Expression for \( K_p \)
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Where \( P_{NO_2} \) is the partial pressure of \( NO_2 \) and \( P_{N_2O_4} \) is the partial pressure of \( N_2O_4 \).

### Step 3: Define Initial Conditions and Changes
Let the initial amount of \( N_2O_4 \) be \( A \) atm. At equilibrium, let \( \alpha \) be the degree of dissociation. Thus, we have:
- Initial: \( N_2O_4 = A \), \( NO_2 = 0 \)
- Change: \( N_2O_4 = A - A\alpha \), \( NO_2 = 2A\alpha \)
- At equilibrium:
\[ P_{N_2O_4} = A(1 - \alpha) \quad \text{and} \quad P_{NO_2} = 2A\alpha \]

### Step 4: Total Pressure at Equilibrium
The total pressure \( P_{total} \) at equilibrium is given by:
\[ P_{total} = P_{N_2O_4} + P_{NO_2} = A(1 - \alpha) + 2A\alpha = A(1 + \alpha) \]
Given that \( P_{total} = 0.5 \) atm, we have:
\[ A(1 + \alpha) = 0.5 \]

### Step 5: Substitute into the \( K_p \) Expression
Substituting the expressions for \( P_{NO_2} \) and \( P_{N_2O_4} \) into the \( K_p \) expression:
\[ K_p = \frac{(2A\alpha)^2}{A(1 - \alpha)} = \frac{4A^2\alpha^2}{A(1 - \alpha)} = \frac{4A\alpha^2}{1 - \alpha} \]
Given that \( K_p = 0.667 \) atm, we can set up the equation:
\[ 0.667 = \frac{4A\alpha^2}{1 - \alpha} \]

### Step 6: Solve for \( A \)
From the total pressure equation \( A(1 + \alpha) = 0.5 \), we can express \( A \) as:
\[ A = \frac{0.5}{1 + \alpha} \]
Substituting this into the \( K_p \) equation:
\[ 0.667 = \frac{4 \cdot \frac{0.5}{1 + \alpha} \cdot \alpha^2}{1 - \alpha} \]
This simplifies to:
\[ 0.667(1 - \alpha)(1 + \alpha) = 2\alpha^2 \]

### Step 7: Solve the Equation for \( \alpha \)
Expanding and rearranging:
\[ 0.667(1 - \alpha^2) = 2\alpha^2 \]
\[ 0.667 - 0.667\alpha^2 = 2\alpha^2 \]
\[ 0.667 = 2\alpha^2 + 0.667\alpha^2 \]
\[ 0.667 = 2.667\alpha^2 \]
\[ \alpha^2 = \frac{0.667}{2.667} \approx 0.25 \]
\[ \alpha = \sqrt{0.25} = 0.5 \]

### Step 8: Calculate Percent Dissociation
The percent dissociation is given by:
\[ \text{Percent Dissociation} = \alpha \times 100 = 0.5 \times 100 = 50\% \]

### Final Answer
The percent dissociation of \( N_2O_4 \) at 46°C and a total pressure of 380 Torr is **50%**.