The `K_P` for the reaction `N_2O_4 hArr 2NO_2` is 640 mm at 775 K.The percentage dissociation of `N_2O_4` at equilibrium pressure of 160 mm is :

Options

A. `80%`
B. `30%`
C. `50%`
D. `70%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation and identify the initial conditions.
The reaction given is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

At the start (t = 0):
- Initial moles of \( N_2O_4 = 1 \) mole
- Initial moles of \( NO_2 = 0 \) moles

### Step 2: Define the change in moles at equilibrium.
Let \( x \) be the amount of \( N_2O_4 \) that dissociates. Thus, at equilibrium:
- Moles of \( N_2O_4 = 1 - x \)
- Moles of \( NO_2 = 2x \)

### Step 3: Write the expression for \( K_P \).
The expression for \( K_P \) is given by:
\[ K_P = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Where:
- \( P_{NO_2} = 2x \) (since pressure is proportional to the number of moles)
- \( P_{N_2O_4} = 1 - x \)

### Step 4: Substitute the equilibrium pressures into the \( K_P \) expression.
Given that the total pressure at equilibrium is 160 mm, we can express \( K_P \):
\[ K_P = \frac{(2x)^2}{(1 - x)} \cdot \frac{P}{(1 + x)} \]
Where \( P = 160 \) mm.

### Step 5: Substitute the known values into the equation.
We know \( K_P = 640 \) mm at 775 K. Thus, we can write:
\[ 640 = \frac{(2x)^2}{(1 - x)} \cdot \frac{160}{(1 + x)} \]

### Step 6: Simplify the equation.
Rearranging gives:
\[ 640 = \frac{4x^2 \cdot 160}{(1 - x)(1 + x)} \]
This simplifies to:
\[ 640(1 - x)(1 + x) = 640 \cdot 4x^2 \]
\[ 1 - x^2 = 4x^2 \]
\[ 1 = 5x^2 \]

### Step 7: Solve for \( x \).
From the equation \( 5x^2 = 1 \):
\[ x^2 = \frac{1}{5} \]
\[ x = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \approx 0.447 \]

### Step 8: Calculate the percentage dissociation.
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = x \times 100 = 0.447 \times 100 \approx 44.7\% \]

### Final Answer:
The percentage dissociation of \( N_2O_4 \) at equilibrium pressure of 160 mm is approximately **44.7%**.

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