1.0 mol of nitrogen oxide NO and 1.0 mol of oxygen were mixed in a container and heated to 450 oC. At equilibrium the number of moles of oxygen was found to be 0.70 mol. The total pressure in the vessel was 4.0 atm. Calculate the value of Kp for the reaction:

2NO (g) + O2 (g) 2 NO2 (g)

I NEED ANSWER PLEAS

I would approach the problem this way.

You know equilibrium moles O2 = .7 which means 0.3 mole O2 was used. Then mole NO2 formed must be 2 x 0.3 = 0.6 and moles NO must be 1-0.6 = 0.4.
Total moles = moles NO + moles O2 + moles NO2.
mole fraction O2, for example = moles O2/total moles.
Then partial pressure O2 = mole fraction O2 x total pressure.
The NO and NO2 are done the same way.

To calculate the value of Kp for the reaction, we need to use the equation of Kp which relates the partial pressures of the reactants and products at equilibrium. The general expression for Kp is:

Kp = (P_NO2)^2 / (P_NO)^2 * (P_O2)

In this reaction, we are given the initial and equilibrium number of moles of oxygen, but we don't have the initial or equilibrium number of moles of NO or NO2. So, we first need to determine the number of moles of NO and NO2 at equilibrium.

According to the balanced reaction: 2 NO (g) + O2 (g) -> 2 NO2 (g)

The stoichiometry of the reaction tells us that the number of moles of NO2 formed is twice the number of moles of NO consumed at equilibrium. Thus, if x moles of NO react, the number of moles of NO2 formed will be 2x.

Given: Initial moles of NO = 1.0 mol
At equilibrium, moles of NO = 1.0 mol - x
At equilibrium, moles of NO2 = 2x

Now, we can calculate the pressure of NO, NO2, and O2 at equilibrium using the ideal gas law:

PV = nRT

At equilibrium, total pressure (P_total) = 4.0 atm
Moles of O2 at equilibrium (n_O2) = 0.70 mol

Using the ideal gas law, we can find the partial pressure of oxygen:

P_O2 = (n_O2 * R * T) / V

Substituting the given values, assuming the volume V remains constant, and using R = 0.0821 L.atm/(mol.K), and T = 450°C converted to Kelvin (723 K), we can calculate the partial pressure of oxygen:

P_O2 = (0.70 * 0.0821 * 723) / V

Now, let's substitute the equilibrium number of moles of NO and NO2, and the partial pressure of oxygen in the Kp expression:

Kp = ((2x)^2 / (1.0 - x)^2 * P_O2

Now, you can solve this equation to find the value of x, and then substitute it back into the Kp expression to calculate the value of Kp.