1. The Magnitude of Equilibrium Constants
The equilibrium constant, K, is the ratio of products to reactants.
Therefore, the larger K the more products are present at equilibrium.
Conversely, the smaller K the more reactants are present at equilibrium.
If K >> 1, then products dominate at equilibrium and equilibrium lies to the right.

2. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

3. Types of Equilibrium Constant
Keq =
Kc =
Kp =
Equilibrium
concentrations
pressures

4. Equilibrium and Pressure
2SO2(g) + O2(g) SO3(g)
Kp = (PSO3) (PSO2)2 (PO2)
Kc = [SO3] [SO2]2 [O2]

5. General Equation jA + kB lC + mD Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m
(PA)j (PB)k (CAxRT)j(CBxRT)k
Kp= (CC)l (CD)mx(RT)l+m (CA)j(CB)kx(RT)j+k
Kp = K (RT)(l+m)-(j+k) = K (RT)Dn
Dn=(l+m)-(j+k)=Change in moles of gas

6. The Direction of the Chemical Equation and Keq
An equilibrium can be approached from any direction.
Example:
has

7. In the reverse direction:

8. Other Ways to Manipulate Chemical Equations and Keq Values
The reaction
has
which is the square of the equilibrium constant for

9. Other Ways to Manipulate Chemical Equations and Keq Values
Equilibrium constant for the reverse direction is the inverse of that for the forward direction.
When a reaction is multiplied by a number, the equilibrium constant is raised to that power.
The equilibrium constant for a reaction which is the sum of other reactions is the product of the equilibrium constants for the individual reactions.

10. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Write a K expression and determine the value for Keq for the reaction:
2 CO2 (g) + 4 H2O (g) CH4 (g) + 4 O2 (g)
Calculate K for both reactions if the equilibrium concentrations are: [CH4] = M, [O2] = M, [CO2] = M, and [H2O] = M. (include units)

11.
x 10-3

12. Heterogeneous Equilibria
When all reactants and products are in one phase, the equilibrium is homogeneous.
If one or more reactants or products are in a different phase, the equilibrium is heterogeneous.
Consider:
experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

14. The concentration of a solid or pure liquid is its density divided by molar mass.
Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant.
For the decomposition of CaCO3:
We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.
The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.

15. Example – Write the Keq expression for the following equations:
NH4Cl (s) NH3 (g) + HCl (g)
NH3 (g) + HCl (g) NH4Cl (s)

16. Example – Write the Keq expression for the following equations:
NH4Cl (s) ↔ NH3 (g) + HCl (g)
Keq = [NH3][HCl]
NH3 (g) + HCl (g) ↔ NH4Cl (s)
Keq = ([NH3][HCl])-1