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?

..............CaO + CO2 ==> CaCO3

initial..... 1M...0.5M......
equil..............0.15
0.15M in 1L = 0.15 moles
Use PV = nRT to solve for P, then
Kp = 1/pCO2 = ?

To find the value of Kp for the given reaction, we need to use the equilibrium concentrations of the reactants and products. In this case, we are given the equilibrium concentration of CO2 ([CO2] = 0.150 M).

The equilibrium expression for Kp can be written as follows:

Kp = (P(CaCO3)) / (P(CO2))

To find the pressure of CaCO3 (P(CaCO3)), we need to calculate the moles of CaCO3 and divide by the volume to get the molar concentration. Here, we are given that 1.00 mol of CaO is used, so the moles of CaCO3 formed will also be 1.00 mol.

Since the reaction is a gas-phase reaction, the molar concentration can be converted to pressure using the ideal gas law:

PV = nRT

P = (n/V)RT

P = (1.00 mol / 1.00 L) * (0.0821 L·atm/mol·K) * (200 + 273 K)

P = 50.5 atm

So, P(CaCO3) = 50.5 atm.

Now we can substitute the values into the equilibrium expression:

Kp = (P(CaCO3)) / (P(CO2))
= (50.5 atm) / (0.150 M)

Since 1 M = 1 atm, we can convert the concentration of CO2 into atm:

Kp = (50.5 atm) / (0.150 atm)
= 336.67

Therefore, the value of Kp for this reaction is 336.67.