1. Equilibrium Pressure
If the values at equilibrium are given in partial pressure, then solving for the constant is the same, but use Kp instead of Kc.
What makes them different: Kc = equilibrium constant based on molarity and concentration.
Kp = equilibrium constant based on partial pressures in atm.

2. The mole ratio in a mixture gas is proportional to the partial pressures of each of the gases.
Dalton’s Law of Partial Pressures – the total pressure in a system is equal to the sum of the individual pressures.
Total = Pressure of + Pressure of …. Pressure Gas #1 Gas #2

3. Kp = 0.0404 Kp = [NO]2 [O2] [NO2]2 Kp = [0.270]2 [0.100] [0.425]2
2 NO2 (g) NO (g) + O2 (g)
Determine the Kp for the above reaction when at equilibrium, the pressure of nitrogen dioxide is atm, nitrogen monoxide is atm, and oxygen is atm.
Kp = [NO]2 [O2] [NO2]2
Kp = [0.270]2 [0.100] [0.425]2
Kp =

4. Relationship between Kc and Kp
We use the Ideal Gas Law, and derive the relationship between Kc and Kp.
Kp = Kc (RT)n
Kp = Equilibrium constant in partial pressure
Kc = Equilibrium constant in molar concentrations
R = Ideal Gas constant (always 0.082)
T = Temperature (in Kelvin)
n = Change in moles of GASES ONLY
(moles of gas products) – (moles of gas reactants)

5. 2 NO2 (g) NO (g) + O2 (g)
At equilibrium, the pressure of nitrogen dioxide is atm, nitrogen monoxide is atm, and oxygen is atm. Determine the Kp for the above reaction.

6. CO(g) + Cl2(g)  COCl2(g) [CO][Cl2] [0.012][0.054]
At equilibrium the concentrations at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
Start with the equilibrium expression for Kc:
Kc = [COCl2]
[CO][Cl2]
Kc = [0.14 M] = 216
[0.012][0.054]

7. CO(g) + Cl2(g)  COCl2(g) Kp = 216 x (.082 x 347)(1-2) Kp = 7.6
At equilibrium the concentrations at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
To solve for Kp we use the Ideal gas law eq. Kp = Kc (RT)n
Kp = 216 x (.082 x 347)(1-2)
Kp = 7.6