For the system CaO(s) + CO2(g) = CaCO3(s), I added 1.00 mol of CaO(s) to 1.00L of 0.500M CO2(g) at 200 oC. At equilibrium the [CO2] = 0.150M. What is the value of Kp for this reaction?

Kp would = CaCO3/(CaO)(CO2)

I do not know what to do with the 0.500M and 1.00mol.

First you need to work on the Kp expression. SOLIDS aren't used (pure liquids aren't either) so the expression is 1/pCO2 = Kp.

The problem tells you that AT EQUILIBRIUM [CO2] = 0.150M.
I would calculate Kc from Kc = 1/([CO2]
then convert to Kp by Kp = Kc(RT)Dn

So I did 1/0.150 and got 6.66 so now I use this with RT. I am not sure what R is?

R is 0.08206

To determine the value of Kp for the given reaction, you need to know the equilibrium concentrations of the reactants and products. In this case, you are given the equilibrium concentration of CO2, which is 0.150M.

However, to calculate the equilibrium concentrations of CaO and CaCO3, you need to consider the initial amount of CaO added and the stoichiometry of the reaction.

Given that you added 1.00 mol of CaO and the reaction equation is:

CaO(s) + CO2(g) = CaCO3(s)

You can see that the stoichiometric coefficient of CaO is 1. Therefore, at equilibrium, the concentration of CaO will be equal to the initial concentration you added, which is 1.00 mol / 1.00 L = 1.00 M.

Since the stoichiometric coefficient of CaCO3 is also 1, the concentration of CaCO3 at equilibrium will be equal to the concentration of CaO at equilibrium, which is 1.00 M.

Now that you have the equilibrium concentrations of CaO and CO2, you can calculate the value of Kp using the equation:

Kp = (CaCO3) / (CaO)(CO2)

Plugging in the values, you get:

Kp = (1.00 M) / (1.00 M)(0.150 M)
= 6.67

Therefore, the value of Kp for this reaction at 200 oC is 6.67.