Keeping your balance
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
1.6 moles NO2 present and 0.2 moles N2O4 present
Equilibrium
 Equilibrium animation
Equilibrium
 At equilibrium, G = 0.
 Neither process is more spontaneous than 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
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)
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+ + FNH3 + H2O  NH4+ + OH-
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]
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.
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 = 0.133. At
equilibrium, PN2O4 was found to be 2.71 atm. Calculate
the equilibrium pressure of NO2.
 Solution: Kp = 0.133 = PNO22/PN2O4, and PN2O4 = 2.71atm.
 Substitute and rearrange: 0.133 = PNO22/2.71, so PNO2 =
[0.133(2.71)] = 0.600atm
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.
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
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)(0.100 + 2x)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
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”
LeChatelier’s Principle
 Example: You introduce 0.230 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 = (0.230-2x)2/x2
x = [H2] = [I2] = 0.0692M; [HI] = (0.230-2x) = 0.0916M
LeChatelier’s Principle
 Example part 2. Find the new concentrations if 0.100
mole H2 is added to the reaction vessel from the
previous problem.
 Solution. Make an ICE chart.
[H2]
[I2]
[HI]
Initial
0.0692M
0.0692M
0.0916M
Change
+ 0.100 - x
-x
+ 2x
Equilibrium
0.0692 + 0.100 - x 0.0692 - x
0.0916 + 2x
LeChatelier’s Principle
 Plug the equilibrium values into the equilibrium
constant expression and solve for x.
1.75 = [HI]2/[H2][I2] = (0.0916+2x)2/(0.1692-x)(0.0692-x)
1.75x2 - 0.417x + 0.0205 = 0.00839 + 0.366x + 4x2
0 = 2.25x2 + 0.783x - 0.0121
x = 0.0148
[H2] = 0.1692-x = 0.154M
[I2] = 0.0692-x = 0.0544M
[HI] = 0.0916+2x = 0.121M
LeChatelier’s Principle
 Check your answer by re-evaluating the equilibrium
constant at the new position.
0.1212/(0.154)(0.054) = 1.75 
 Solubility (common-ion effect) example: George adds
0.0500g 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
LeChatelier’s Principle
[Ag+]
[C2H3O2-]
Initial
0.045M
0.045M
Change
-x
+ 0.0122 - x
Equilibrium
0.045 - x
0.057 - x
 Ksp = 2.0x10-3 = (0.045–x)(0.057–x)
0 = x2 – 0.102x + 0.000565
 x = 0.005878 (change in molarity = amt of ppt)
 Represents 2.9x10-4mol or 0.049g
 Check answer
LeChatelier’s Principle
 Change in temperature will shift the position of
equilibrium so as to favor the endothermic direction
2NO2(g)  N2O4(g) + 57.2 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.
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)