Solution:

The value of Kp for the equilibrium reaction N2O4 ⇌ 2NO2 is given as 2. We need to find the percentage dissociation of N2O4 at a pressure of 0.5 ATM.

To determine the percentage dissociation, we first need to calculate the equilibrium concentrations of N2O4 and NO2.

Let's assume the initial concentration of N2O4 is x. At equilibrium, the concentration of N2O4 will be (x - y), where y is the extent of dissociation.

According to the balanced equation, 1 mole of N2O4 produces 2 moles of NO2. Therefore, the concentration of NO2 at equilibrium will be 2y.

Using the given value of Kp, we can write the expression for the equilibrium constant as:

Kp = (NO2)^2 / (N2O4)

Substituting the equilibrium concentrations, we have:

2 = (2y)^2 / (x - y)

Simplifying this equation, we get:

4y^2 = 2x - 2y

4y^2 + 2y - 2x = 0

This is a quadratic equation in terms of y. Solving this equation will give us the value of y, which represents the extent of dissociation.

Once we have the value of y, we can calculate the percentage dissociation using the formula:

Percentage dissociation = (y / x) * 100

Now, let's calculate the extent of dissociation and the percentage dissociation.

Calculation:

1. Solve the quadratic equation:

4y^2 + 2y - 2x = 0

2. Calculate the values of y and x using the quadratic formula.

3. Once we have the value of y, substitute it into the percentage dissociation formula.

4. Calculate the percentage dissociation.

Example:

Let's assume the initial concentration of N2O4 (x) is 1 mole and the pressure is 0.5 ATM.

1. Solve the quadratic equation:

4y^2 + 2y - 2(1) = 0

2. Calculate the values of y and x using the quadratic formula:

y = (-2 ± √(2^2 - 4(4)(-2)) / (2(4))

y = (-2 ± √(4 + 32)) / 8

y = (-2 ± √36) / 8

y = (-2 ± 6) / 8

Since the extent of dissociation cannot be negative, we take the positive value:

y = (6 - 2) / 8

y = 4 / 8

y = 0.5

2. Substitute the value of y into the percentage dissociation formula:

Percentage dissociation = (0.5 / 1) * 100

Percentage dissociation = 50%

Therefore, at a pressure of 0.5 ATM, the percentage dissociation of N2O4 is 50%.