1. Chemical Equilibrium - General Concepts (Ch. 14)
Dynamic Equilibrium
a A + b B ⇌ c C + d D
This is called a reversible reaction.
At Equilibrium:
Reaction is proceeding in both directions at the same rate.
There is no net change in concentrations of reactants and products.

2. Equilibrium Constant Expression
(a) Mass action expression: Q = “reaction quotient”
where [A], [B], etc are “non-equilibrium” concentrations
(b) At Equilibrium:
where Kc = “equilibrium constant”
[A], [B], etc. are the “equilibrium” concentrations!
If Q > K, then reaction will go backward to attain equilibrium.
If Q < K, then reaction will go forward to attain equilibrium.
Q =
[C]c[D]d
[A]a[B]b
Kc =
[C]c[D]d
[A]a[B]b
“Law of Mass Action”

3. Equilibrium Constant Expressions
(c) Magnitude of K
K ≈ [products]/[reactants]
if K is very large: mostly products at equilibrium
if K is very small: mostly reactants at equilibrium
if K ≈ 1: roughly equal conc’s of reactants and products
(d) Manipulating equilibrium constant expressions
Reverse equation --> invert Kc
Add two equations --> multiply their Kc’s
Multiply equation by coefficient “n” --> (Kc)n

4. Equilibrium Constants for Gas Phase Reactions
(a) since P ∝ molar conc., the equilibrium constant can be expressed in terms of partial pressures (in atm):
(b) relationship of Kc and Kp
based on ideal gas law…
Kp = Kc(RT)Dn
Dn = moles gaseous products - moles gaseous reactants
R = L atm / mol K
Kp =
(Pc)c(Pd)d
(Pa)a(Pb)b

5. Heterogeneous Equilibria
Concentrations of pure liquids and solids are effectively constants, ∴ they do not appear in K expressions
For example:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Kc = [NH4+][OH–]/[NH3]
CaO(s) + SO2(g) ⇌ CaSO3(s)
Kp = 1/PSO2

6. Equilibrium Calculations
Two basic types of problems:
(a) Given the equilibrium concentrations (or partial pressures), determine Kc (or Kp)
General Method: Put conc values into K expression!
Or, if necessary, set up a “concentration table” based on reaction stoichiometry, e.g.
PCl3(g) + Cl2(g) ⇌ PCl5(g)
In an experiment, 0.20 mol of PCl3 and 0.10 mol Cl2 were placed in a 1.00 L container. After the reaction had come to equilibrium the concentration of PCl3 was found to be 0.12 mole/L. Determine the Kc value.

7. Equilibrium Calculations, cont.
(b) Given K and one or more initial concentrations, determine the equilibrium concentrations
General Method:
Set up a “concentration table” based on an unknown “x” -- usually one of the equilibrium concentrations or a change in a conc.
Express all equilibrium concentrations in terms of “x” and initial values
Insert these equilibrium values (in terms of “x”) into the proper equation for K, and
If possible, solve directly for “x” using simple math
!No need for quadratic formula on Exams!
Or, make a chemically reasonable approximation!

8. Sample Problem
Initially, 0.20 mole PCl3 and 0.10 mol Cl2 are placed in a 1.00-L container. The following reaction occurs.
PCl3(g) + Cl2(g) ⇌ PCl5(g)
Consider the two different types of equilibrium problems:
(a) if the equilibrium concentration of PCl3 is found to be 0.12 M, calculate Kc.
(b) Alternatively, if Kc is known to be 33.3, calculate the concentration of PCl5 at equilibrium.

9. Sample Problem
Initially, 0.20 mole PCl3 and 0.10 mol Cl2 are placed in a 1.00-L container. The following reaction occurs.
PCl3(g) + Cl2(g) ⇌ PCl5(g)
Consider the two different types of equilibrium problems:
(a) if the equilibrium concentration of PCl3 is found to be 0.12 M, calculate Kc.
(b) Alternatively, if Kc is known to be 33.3, calculate the concentration of PCl5 at equilibrium.
Answer: a. 3 x 101
b M

10. Another Sample Problem
For the following reaction, Kc = 64. Initially, 0.10 mol H2 and 0.10 mol I2 are placed in a 500-mL (3 sig fig) container. Calculate the equilibrium concentrations of all three species.
H2(g) + I2(g) ⇌ 2 HI(g)

11. Another Sample Problem
For the following reaction, Kc = 64. Initially, 0.10 mol H2 and 0.10 mol I2 are placed in a 500-mL (3 sig fig) container. Calculate the equilibrium concentrations of all three species.
H2(g) + I2(g) ⇌ 2 HI(g)
Answer:
0.32 M, 0.04 M, 0.04 M

12. Yet More Sample Problems
(1) For the following reaction, which occurs in air, Kc = 4.8 x 10–31. If the initial concentrations are [N2] = M and [O2] = M, calculate the concentration of NO at equilibrium.
N2(g) + O2(g) ⇌ 2 NO(g)
(2) For the following reaction, Kc = 4.3 x Initially, mole of H2CO2 is placed in a 1.0-L container. Calculate the equilibrium concentrations of all three species.
H2CO2(g) ⇌ CO(g) + H2O(g)

13. Yet More Sample Problems
(1) For the following reaction, which occurs in air, Kc = 4.8 x 10–31. If the initial concentrations are [N2] = M and [O2] = M, calculate the concentration of NO at equilibrium.
N2(g) + O2(g) ⇌ 2 NO(g)
Answer: 1.1 x M
(2) For the following reaction, Kc = 4.3 x Initially, mole of H2CO2 is placed in a 1.0-L container. Calculate the equilibrium concentrations of all three species.
H2CO2(g) ⇌ CO(g) + H2O(g)
Answer: 0.20 M CO, 0.20 M H2O, 9.3 x 10-7 M H2CO2

14. Le Chatelier’s Principle
“When a system at equilibrium is subjected to an external stress, the system will shift in a direction to counteract the stress and achieve a new equilibrium state.”
Common factors (“external stresses”) to consider:
Addition or removal of some reactant or product
Change in volume (important for gaseous reactions)
Change in temperature
Addition of a catalyst

15. Example
Consider a series of possible changes to the following equilibrium reaction:
PCl3(g) + Cl2(g) ⇌ PCl5(g) + 88 kJ/mol
(the reaction is exothermic : DH° = -88 kJ/mol)
What happens to [Cl2] if
(a) some PCl3 is added?
the “stress” is more PCl3
system needs to remove PCl3
∴ reaction shifts forward and [Cl2] decreases
(but no change in Kc)

16. Example, cont. PCl3(g) + Cl2(g) ⇌ PCl5(g) + 88 kJ
(the reaction is exothermic : DHd = -88 kJ)
(b) some PCl5 is added?
the “stress” is more PCl5
system needs to remove PCl5
∴ reaction shifts in reverse and [Cl2] increases
(no change in Kc)
(c) the temp is increased? (this requires a DHº value)
the “stress” is added heat
the system needs to remove heat
since the reaction is exothermic, heat is a “product”
adding a “product” causes reaction to shift in reverse
∴ [Cl2] increases (Note: Kc decreases)
Only temperature affects the actual value of Kc!

17. Example, cont. PCl3(g) + Cl2(g) ⇌ PCl5(g) + 88 kJ
(d) the volume is increased (or pressure is decreased)
the “stress” is more space for gas molecules
system tries to produce more gas to fill the space
reaction shifts in direction of more moles of gas
in this case, it shifts in reverse
(2 mol gaseous reactants vs 1 mol gaseous product)
∴ [Cl2] increases (but no change in Kc)
(e) a catalyst is added?
Addition of a catalyst does not affect the position of equilibrium. ∴ there is no change in any of the equilibrium concentrations. The catalyst merely lowers the time needed to attain equilibrium.
(no change in Kc)