For the reactionN2O4⇌2NO2(g), the degree of dissociation at equilibrium is 0.2 at 1 atm. ThenKpwill be1/21/41/6

To find the value of Kp for the reaction N2O4 ⇌ 2NO2(g), we need to use the given degree of dissociation at equilibrium, which is 0.2.

Step 1: Write the balanced equation for the reaction: N2O4 ⇌ 2NO2(g)

Step 2: Define the equilibrium expression using the concentrations of the reactants and products: Kp = [NO2]^2 / [N2O4]

Step 3: Since the degree of dissociation at equilibrium is 0.2, we can assume that 0.2 moles of N2O4 dissociate to form 0.2 moles of NO2. Therefore, the concentrations at equilibrium are: [N2O4] = (initial concentration of N2O4) - (0.2 moles) [NO2] = (initial concentration of NO2) + (0.2 moles)

Step 4: Substitute the equilibrium concentrations into the equilibrium expression: Kp = ([NO2] + 0.2)^2 / ([N2O4] - 0.2)

Step 5: Since the reaction is at 1 atm, we can assume that the initial pressure of N2O4 is equal to the equilibrium pressure. Therefore, the initial concentration of N2O4 is equal to the initial pressure divided by the ideal gas constant (R) and the temperature (T): [N2O4] = (1 atm) / (R * T)

Step 6: Substitute the initial concentration of N2O4 into the equilibrium expression: Kp = ([NO2] + 0.2)^2 / ((1 atm) / (R * T) - 0.2)

Step 7: Simplify the expression if necessary.

Unfortunately, without specific values for the temperature and the ideal gas constant, we cannot calculate the exact value of Kp.