1. Equilibrium
Keeping your balance

2. Equilibrium
Systems at equilibrium are subject to two opposite processes occurring at the same rate
Establishment of equilibrium
2NO2  N2O4
Start with 1.00 mol N2O4 and allow it to react
As NO2 appears, N2O4disappears
The reaction slows and comes to an apparent stop with 0.2 moles NO2 present and 1.6 moles N2O4 present

3. Equilibrium
Equilibrium animation

4. Equilibrium At equilibrium, G = 0.
Neither process is more spontaneous that the other.
For the amounts present at equilibrium, considerations of enthalpy and entropy exactly cancel each other
The amounts present (the “position of equilibrium”) can be changed by changing the temperature

5. Types of equilibrium systems
Phase equilibria
Vapor – liquid
H2O(l)  H2O(g)
liquid – solid
H2O(s)  H2O(l)
Solution equilibrium
Solid in liquid – saturated salt water
NaCl(s)  NaCl(aq)

6. Types of equilibrium systems
Gas in liquid – dissolved oxygen
O2(aq)  O2(g)
Gas phase reaction equilibria
H2 + I2  2HI
Acid-base equilibria
HF + H2O  H3O+ + F-
NH3 + H2O  NH4+ + OH-

7. Equilibrium constant expression
This is a characteristic of an equilibrium system that governs the position of equilibrium
The value is temperature dependent
Equilibrium constant is the concentrations of the products divided by the concentrations of the reactants at equilibrium (“Law of Mass Action”).
PCl3(g) + Cl2(g)  PCl5(g)
Keq = [PCl5]/ [PCl3][Cl2]

8. Equilibrium constant expression
In higher order reactions (those with coefficients greater than 1) the concentrations are raised to the power of the coefficients of the reactants and products
H2 + I2  2HI
Keq = [HI]2/[ H2][ I2]
Solids and pure liquids do not appear in the equilibrium constant expression because their activity is constant.

9. Equilibrium constant expression
Example: Nitrogen dioxide dimerizes to form dinitrogen tetroxide as follows:
N2O4(g)  2NO2(g)
N2O4 is placed in a flask and allowed to reach equilibrium at a temperature where Kp = At equilibrium, PN2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2.
Solution: Kp = = PNO22/PN2O4, and PN2O4 = 2.71atm.
Substitute and rearrange: = PNO22/2.71, so PNO2 = [0.133(2.71)] = 0.600atm

10. Solubility equilibrium expressions
For solution of ionic materials
NaCl(s)  Na+(aq) + Cl- (aq)
Concentration of solid does not appear, so
Ksp = [Na+][Cl-]
Solubilities can be calculated from Ksp, since at equilibrium the solution is saturated
Example: Calculate the molar solubility of lead (II) chloride at 25ºC if the solubility product constant is 1.8x10-10 at that temperature.

11. Solubility product constant
Solution: PbCl2  Pb+2 + 2Cl-
Ksp = [Pb+2][Cl-]2 = 1.8x10-10
[Pb+2] = molar solubility
[Cl-] = 2[Pb+2]
Set [Pb+2] = x
1.8x10-10 = x(2x)2 = 4x3
x = 3.6x10-4M

12. Common ion effect
The presence of a common ion decreases the solubility of a salt.
Example: Find the molar solubility of lead (II) chloride in 0.100M HCl. Assume no changes in volume.
Solution: Ksp = 1.8x10-10 = (x)( x)2
Ignore 2x since it is small compared to 0.100
Ksp = 1.8x10-10 = (x)(0.100)2
x = 1.8x10-8M
Check answer

13. LeChatelier’s Principle
Disturbing an equilibrium system results in a shift in the position of equilibrium until equilibrium is reestablished at a new position
Change in concentration – when the concentration of one product or reactant is increased, the equilibrium will shift to use up some of the excess material
H2 + I2  2HI
What is the effect of adding additional hydrogen gas?
More HI is formed – position of equilibrium is shifted “right”

14. LeChatelier’s Principle
Example: You introduce mole HI into a 1.00L flask and allow it to come to equilibrium. If Keq for the reaction is 1.75, find the concentrations of H2, I2 and HI.
Solution: Keq = [HI]2/[ H2][ I2] = 1.75
[H2] = [I2] = x
[HI] = (0.230 – 2x)
1.75 = ( x)/x2
x = [H2] = [I2] = 0.105M; [HI] = (0.230-x) = 0.125M

15. LeChatelier’s Principle
Example part 2. Find the new concentrations if mole H2 is added to the reaction vessel from the previous problem.
Solution. Make an ICE chart.
[H2]
[I2]
[HI]
Initial
0.105M
0.125M
Change x
- x
+ 2x
Equilibrium x x x

16. LeChatelier’s Principle
Plug the equilibrium values into the equilibrium constant expression and solve for x.
1.75 = [HI]2/[H2][I2] = ( x)2/(0.205-x)(0.105-x)
1.75x x = x + 4x2
0 = 2.25x x
x =
[H2] = x = 0.185M
[I2] = x = 0.085M
[HI] = x = 0.166M

17. LeChatelier’s Principle
Check your answer by re-evaluating the equilibrium constant at the new position.
0.1662/(0.185)(0.085) = 
Solubility (common-ion effect) example: George adds g solid NaC2H3O2 to 50.0mL saturated silver acetate (Ksp = 2.0x10-3). How much silver acetate precipitates? Assume no volume changes.
Solution: Find concentration of saturated silver acetate solution.
(2.0x10-3) = [Ag+] = [silver acetate] = 0.045M

18. LeChatelier’s Principle
[Ag+]
[C2H3O2-]
Initial
0.045M
Change
- x x
Equilibrium x x
Ksp = 2.0x10-3 = (0.045–x)(0.057–x)
0 = x2 – 0.102x
x = (change in molarity = amt of ppt)
Represents 2.9x10-4mol or g
Check answer

19. LeChatelier’s Principle
Change in temperature will shift the position of equilibrium so as to favor the endothermic direction
2NO2(g)  N2O4(g) kJ
The “+57.2kJ” indicates that heat has left the system, thus the forward direction is exothermic.
Raising the temperature means more energy put into the system, and shifting to the left uses the excess energy and relieves the stress.

20. LeChatelier’s Principle
Changes in pressure – increasing the pressure in gas systems favors the side with fewer particles.
If both products and reactants have the same number of particles, no change results.
Which of the following systems would produce more products as a result of increased pressure?
Cr2O7-2(aq) + H2O(l)  2CrO4-2(aq) + H3O+(aq)
H2(g) + Br2(g)  2HBr(g)
Ca(OH)2(s)  CaO(s) + H2O(g)
N2(g) + 3H2(g)  2NH3(g)