At 25ºC and 1 atmosphere the following equilibrium is eatablished : 

2NO (g) <=> N2O4 (g) 

The partial equilibrium pressure of N2O4 is 0.7 atm. 
A) calculate Kp of the above system at 25ºC. 
B) calculate the partial pressure of the above two gases at equilibrium when the total pressure at equilibrium becomes 9atm and the temperature is still 25ºC

To answer this question, we need to use the concept of equilibrium constant (Kp) and the given information about the partial equilibrium pressure of N2O4. Here's how you can calculate Kp and the partial pressure of the two gases at equilibrium:

A) Calculating Kp:
The equilibrium constant (Kp) is the ratio of the products' partial pressures raised to their stoichiometric coefficients divided by the reactants' partial pressures raised to their stoichiometric coefficients. In this case, the equilibrium expression can be written as:

Kp = (P(N2O4))^2 / (P(NO))^2

Given that the partial pressure of N2O4 is 0.7 atm, we can substitute it into the equation:

Kp = (0.7)^2 / (P(NO))^2

B) Calculating the partial pressure of the gases at equilibrium:
We are given that the total pressure at equilibrium is 9 atm. We can use this information to calculate the partial pressure of each gas.

Let's assume the partial pressure of NO is x atm. Then, the partial pressure of N2O4 can be expressed as (0.7 - x) atm since the total pressure is the sum of the partial pressures of the two gases.

According to the stoichiometry of the balanced equation, the mole ratio of NO to N2O4 is 2:1. Therefore, the partial pressure of NO will be twice the partial pressure of N2O4:

x = 2(0.7 - x)

Simplifying the equation:

x = 1.4 - 2x
3x = 1.4
x = 1.4 / 3

So the partial pressure of NO is approximately 0.467 atm.

The partial pressure of N2O4 can be found by substituting the value of x into the equation (0.7 - x):

(0.7 - x) = 0.7 - 0.467
(0.7 - x) = 0.233 atm

Therefore, the partial pressure of N2O4 at equilibrium is approximately 0.233 atm.

To summarize:
A) Kp = (0.7)^2 / (P(NO))^2
B) The partial pressure of NO at equilibrium is approximately 0.467 atm, and the partial pressure of N2O4 is approximately 0.233 atm when the total pressure at equilibrium is 9 atm.

A) To calculate Kp, we use the equation:

Kp = (P(N2O4))^2 / (P(NO))^2

Given that P(N2O4) = 0.7 atm, we need to find the partial pressure of NO. Let's assume it is x atm.

Thus, (x)^2 / (0.7)^2 = Kp

B) If the total pressure at equilibrium is 9 atm, and the partial pressure of N2O4 is 0.7 atm, we can use the ideal gas law to find the partial pressure of NO.

Assuming the partial pressure of NO is x atm:

P(NO) + P(N2O4) = total pressure
x + 0.7 = 9

Now, we have two equations:
(x)^2 / (0.7)^2 = Kp
x + 0.7 = 9

To solve both equations simultaneously, substitute the value of x from the second equation into the first equation:

((9 - 0.7)^2) / (0.7)^2 = Kp

Solve this equation to find the value of Kp.

Let's solve these equations using a numerical example:

A) Let's assume Kp = 1 atm:
(0.7)^2 / (x)^2 = 1

Cross-multiplying the equation:
(x)^2 = (0.7)^2
x^2 = 0.49
x ≈ 0.7 atm

So, when Kp = 1 atm, the partial pressure of NO is approximately 0.7 atm.

B) Let's solve for x when the total pressure is 9 atm:
x + 0.7 = 9
x = 9 - 0.7
x = 8.3 atm

So, when the total pressure is 9 atm, the partial pressure of NO is approximately 8.3 atm, and the partial pressure of N2O4 is 0.7 atm.