Keeping your balance

EquilibriumSystems at equilibrium are subject to two

opposite processes occurring at the same rateEstablishment of equilibrium

2NO2 N2O4

Start with 1.00 mol N2O4 and allow it to react

As NO2 appears, N2O4disappearsThe reaction slows and comes to an apparent

stop with 1.6 moles NO2 present and 0.2 moles N2O4 present

Equilibrium

Equilibrium animation

EquilibriumAt 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 systemsPhase equilibria

Vapor – liquidH2O(l) H2O(g)

liquid – solid

H2O(s) H2O(l)

Solution equilibriumSolid in liquid – saturated salt water

NaCl(s) NaCl(aq)

Types of equilibrium systemsGas in liquid – dissolved oxygen

O2(aq) O2(g)

Gas phase reaction equilibriaH2 + I2 2HI

Acid-base equilibriaHF + H2O H3O+ + F-

NH3 + H2O NH4+ + OH-

Equilibrium constant expressionThis is a characteristic of an equilibrium

system that governs the position of equilibrium

The value is temperature dependentEquilibrium 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 expressionIn 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 expressionExample: 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 = PNO2

2/2.71, so PNO2 = [0.133(2.71)] = 0.600atm

Solubility equilibrium expressionsFor 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 constantSolution: 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 effectThe 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 0.100Ksp = 1.8x10-10 = (x)(0.100)2

x = 1.8x10-8MCheck answer

LeChatelier’s PrincipleDisturbing 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 2HIWhat is the effect of adding additional hydrogen

gas?More HI is formed – position of equilibrium is

shifted “right”

LeChatelier’s PrincipleExample: 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 PrincipleExample 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 PrinciplePlug 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 - 0.0121x = 0.0148

[H2] = 0.1692-x = 0.154M

[I2] = 0.0692-x = 0.0544M

[HI] = 0.0916+2x = 0.121M

LeChatelier’s PrincipleCheck 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

Ksp = 2.0x10-3 = (0.045–x)(0.057–x)

0 = x2 – 0.102x + 0.000565x = 0.005878 (change in molarity = amt of ppt) Represents 2.9x10-4mol or 0.049gCheck answer

[Ag+] [C2H3O2-]

Initial 0.045M 0.045M

Change - x + 0.0122 - x

Equilibrium 0.045 - x 0.057 - x

LeChatelier’s PrincipleChange in temperature will shift the position

of equilibrium so as to favor the endothermic direction

2NO2(g) N2O4(g) + 57.2 kJThe “+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 PrincipleChanges 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)