Equilibrium Expression (Keq)Equilibrium Expression (Keq) Also called “Mass Action Expression”Also called “Mass Action Expression” Relates the concentration of products to reactants once Relates the concentration of products to reactants once

equilibrium has been reached.equilibrium has been reached.

For this general reaction:For this general reaction:

aA + bB aA + bB ↔ cC + dD↔ cC + dD

Keq = Keq = [C][C]cc x [D] x [D]dd

[A][A]aa x [B] x [B]bb

[ ] the brackets mean “the concentration of”

So basically concentration of products overSo basically concentration of products over

concentration reactants raised to the powerconcentration reactants raised to the power

of their coefficient in balanced equationof their coefficient in balanced equation

Keq = Keq = [C][C]cc x [D] x [D]dd

[A][A]aa x [B] x [B]bb

IMPORTANTIMPORTANT

Exclude solids and pure liquidsExclude solids and pure liquids as they do as they do

not have concentration values.not have concentration values.

Products

Reactants

Ex: Ex: Write Keq expression for: Write Keq expression for:

NN22(g) + 3H(g) + 3H22(g) ↔ 2NH(g) ↔ 2NH33(g)(g)

All gases (nothing excluded)All gases (nothing excluded)

Keq = Keq = [NH[NH33]]22

[N[N22] x [H] x [H22]]33

Ex: Write Keq expression for:Ex: Write Keq expression for:

2NO(g) + 2H2NO(g) + 2H22(g) (g) ↔ N↔ N22(g) + 2H(g) + 2H22O(l)O(l)

Keq = Keq = [[NN22]]

[[NONO]]22 x [H x [H22]]22

Take out pure liquidTake out pure liquidhttp://www.kentchemistry.com/links/Kinetics/EquilibriumExpression.htm

Ex: Write Keq expression for:Ex: Write Keq expression for:

NaCl(s) + HNaCl(s) + H22SOSO44(l) (l) ↔ HCl(g) + NaHSO↔ HCl(g) + NaHSO44(s)(s)

Keq = Keq = [[HClHCl]]

Take out solids and pure liquidTake out solids and pure liquid

Value of Equilibrium Constant (Keq)Value of Equilibrium Constant (Keq)

At equilibrium if you put the concentrationAt equilibrium if you put the concentration

values (Molarity) into the Keq expression youvalues (Molarity) into the Keq expression you

will get a specific numberwill get a specific number

((The is a unitless number and is unique to that reaction.)The is a unitless number and is unique to that reaction.)

The only thing that can change the value of KeqThe only thing that can change the value of Keq

is a is a change in temperature.change in temperature.

If Keq = 1If Keq = 1 Conc. products = reactants at equilibriumConc. products = reactants at equilibrium

If Keq > 1If Keq > 1 Favors ProductsFavors Products Large Keq = large quantities of product at equilibriumLarge Keq = large quantities of product at equilibrium

If Keq < 1If Keq < 1 Favors ReactantsFavors Reactants Small Keq = large quantities of reactant at equilibriumSmall Keq = large quantities of reactant at equilibrium

Value of Equilibrium Constant (Keq)Value of Equilibrium Constant (Keq)

Keq = Keq = [Products][Products]xx

[Reactants] [Reactants]yy

http://employees.oneonta.edu/viningwj/sims/equilibrium_constant_s.html

2A(g) + 3B(aq) 2A(g) + 3B(aq) ↔ 2AB(g)↔ 2AB(g)

At equilibrium [A] = .3M, [B] = .1M, [AB] = .8MAt equilibrium [A] = .3M, [B] = .1M, [AB] = .8M

find the Keq.find the Keq.

Keq = Keq = [.8][.8]22 = 7111 = 7111 [.3][.3]22 x [.1] x [.1]33

Favors ProductsFavors Products

Plugging in ValuesPlugging in Values

Find concentration of ClFind concentration of Cl22 at equilibrium if, at equilibrium if, [PCl[PCl55] = .015M, [PCl] = .015M, [PCl33] = .78M and Keq = 35] = .78M and Keq = 35

PClPCl55 (g) (g) ↔ ↔ PClPCl33(g) + Cl(g) + Cl22(g)(g)

35 = 35 = [.78] x [“X”][.78] x [“X”] [.015][.015]

[Cl[Cl22] = .67M] = .67M

Plugging in ValuesPlugging in Values

Note: Note: The Keq value for the “reverse” reaction The Keq value for the “reverse” reaction

will be the inverse of the “forward” reactionwill be the inverse of the “forward” reactionProducts become reactantsProducts become reactants

Keq interactiveKeq interactivehttp://glencoe.com/sites/common_assets/http://glencoe.com/sites/common_assets/

advanced_placement/advanced_placement/chemistry_chang10e/animations/chemistry_chang10e/animations/kim2s2_5.swfkim2s2_5.swf

ICE Problems (Honors)ICE Problems (Honors) Keq problems where you are Keq problems where you are

given given INITIALINITIAL concentrations. concentrations.

Use stoich ratios to find the Use stoich ratios to find the CHANGECHANGE in concentration in concentration

Subtract this from initial Subtract this from initial concentration find the concentration find the EQUILIBRIUMEQUILIBRIUM concentration concentration that can then go into Keq that can then go into Keq expression expression

http://www.youtube.com/watch?v=rog8ou-ZepE&safe=activeCrash Course: Equilibrium Equations (mostly Honors)Crash Course: Equilibrium Equations (mostly Honors)http://www.youtube.com/watch?v=DP-vWN1yXrY&safe=activehttp://www.youtube.com/watch?v=DP-vWN1yXrY&safe=active

Solubility Equilibrium for IonicsSolubility Equilibrium for Ionics

Ionic Solids:Ionic Solids: DissociateDissociate when placed in solution. when placed in solution.

Positive and negative ions are pulled apart.Positive and negative ions are pulled apart.Polyatomic Ions stay together!!Polyatomic Ions stay together!!

If an ionic solid dissolves in a polar liquid, this If an ionic solid dissolves in a polar liquid, this process is called process is called dissolutiondissolution..

Dissolution EquationDissolution Equation: :

AgCl(s) AgCl(s) ↔↔ Ag Ag+1+1(aq) + Cl(aq) + Cl-1-1(aq)(aq)

http://programs.northlandcollege.edu/biology/Biology1111/animations/dissolve.html

Try to write a dissolution equation for CaClTry to write a dissolution equation for CaCl22(s)(s)

CaClCaCl22(s) (s) ↔↔ Ca Ca+2+2(aq) + 2Cl(aq) + 2Cl-1-1(aq)(aq)

Ksp ExpressionKsp Expression Equilibrium expressions for ionic solutions are called Equilibrium expressions for ionic solutions are called

Ksp (sp = “solubility product”).Ksp (sp = “solubility product”). Set up “K” expression as beforeSet up “K” expression as before

include (g) and (aq), cross out (s) and (l)include (g) and (aq), cross out (s) and (l)

AgCl(s) AgCl(s) ↔↔ Ag Ag+1+1(aq) + Cl(aq) + Cl-1-1(aq)(aq)

AgCl(s) AgCl(s) ↔↔ Ag Ag+1+1(aq) + Cl(aq) + Cl-1-1(aq)(aq)Cross out solidCross out solid

Ksp = [AgKsp = [Ag+1+1] x [Cl] x [Cl-1-1]]

Answer is the “product” of the concentrations of the ions at equilibrium or “ion product”

Try Writing the Ksp ExpressionTry Writing the Ksp Expression

AlPOAlPO44 CaCa33(PO(PO44))22

AlPOAlPO4 4 (s)(s) ↔ Al ↔ Al+3+3 (aq) + (aq) + POPO44-3 -3 (aq)(aq)

Ksp = [AlKsp = [Al+3+3 ] x [ ] x [POPO44-3-3 ] ]

CaCa33(PO(PO44))2 2 (s)(s) ↔ 3Ca↔ 3Ca+2 +2 (aq) + 2(aq) + 2POPO44-3 -3 (aq)(aq)

Ksp = [CaKsp = [Ca+2+2]]33 x [ x [POPO44-3-3 ] ]22

Value of KspValue of Ksp

Higher KspHigher Ksp = more soluble = more soluble Lower KspLower Ksp = less soluble = less soluble

Ex: Ex: Al(OH)Al(OH)33 Ksp = 5 x 10Ksp = 5 x 10-33-33

BaCOBaCO33 Ksp = 2 x 10Ksp = 2 x 10-9-9

Large Ksp = more solid is dissolved at equilibrium Large Ksp = more solid is dissolved at equilibrium It would also indicate a higher level of conductivity since It would also indicate a higher level of conductivity since

ionics are electrolytes!ionics are electrolytes!

Value is temperature dependant, (usually given for 25 Value is temperature dependant, (usually given for 25 °°C)C)

Much more soluble!

Just ReadJust Read

Ionic compounds have different degrees of Ionic compounds have different degrees of solubility (none are truly insoluble as it solubility (none are truly insoluble as it may indicate on your ref. tables), Ksp may indicate on your ref. tables), Ksp allows us to compare solubility.allows us to compare solubility.

This is useful when looking at how much This is useful when looking at how much relatively insoluble compounds will relatively insoluble compounds will dissolve in things like drinking water or dissolve in things like drinking water or blood plasma.blood plasma.

Ksp Problems (Honors)Ksp Problems (Honors)

Find Ksp from SolubilityFind Ksp from Solubility::A sat. solution of BaSOA sat. solution of BaSO44 has a conc. of 3.9 x 10 has a conc. of 3.9 x 10-5-5MM

of Baof Ba+2+2 ions, find Ksp. ions, find Ksp.

BaSOBaSO44 (s) (s) ↔↔ Ba Ba+2 +2 (aq)(aq)

+ SO+ SO44-2 -2

(aq)(aq)

Ksp = [BaKsp = [Ba+2+2] x [SO] x [SO44-2-2]]

Ksp = [3.9 x 10Ksp = [3.9 x 10-5-5M] x [3.9 x 10M] x [3.9 x 10-5-5M] M]

Ksp = 1.5 x 10Ksp = 1.5 x 10-9-9

Concentration of ions is the same. (1:1 ratio)

If [PbIf [Pb+2+2] = 1.9 x 10] = 1.9 x 10-3-3 in a saturated solution in a saturated solution

of PbFof PbF22 find Ksp. find Ksp.

PbFPbF22 ↔↔ PbPb+2+2 + 2F + 2F-1-1

Ksp = [PbKsp = [Pb+2+2] x [F] x [F-1-1]]22

Ksp Ksp = [X] x [2X]= [X] x [2X]22

= [1.9 x 10= [1.9 x 10-3-3] x [2(1.9 x 10] x [2(1.9 x 10-3-3)])]22

= 2.7 x 10= 2.7 x 10-8-8

Don’t know either but one is double the other

Ksp Problems (Honors)Ksp Problems (Honors)

Find Solubility from KspFind Solubility from Ksp

If the Ksp of RaSOIf the Ksp of RaSO44 = 4 x 10 = 4 x 10-11-11 calculate its calculate its solubility in pure water.solubility in pure water.

RaSORaSO44 (s) (s) ↔↔ Ra Ra+2 +2 (aq)(aq)

+ SO+ SO44-2 -2

(aq)(aq)

Ksp = [RaKsp = [Ra+2+2] x [SO] x [SO44-2-2]]

4 x 104 x 10-11-11= = [Ra[Ra+2+2] x [SO] x [SO44-2-2]]

4 x 104 x 10-11-11= [X] x [X]= [X] x [X]4 x 104 x 10-11-11= X= X22

X = the square root of 4 x 10X = the square root of 4 x 10-11 -11 == 6 x 106 x 10-6-6 M M

We don’t know either concentration but they are the same

If Ksp of PbClIf Ksp of PbCl22 = 1.6 x 10 = 1.6 x 10-5-5, calculate solubility., calculate solubility.

PbClPbCl22 ↔↔ PbPb+2+2 + 2Cl + 2Cl-1-1

Ksp = [PbKsp = [Pb+2+2] x [Cl] x [Cl-1-1]]22

1.6 x 101.6 x 10-5-5 = [X] x [2X] = [X] x [2X]22

1.6 x 101.6 x 10-5 -5 = 4X= 4X33

““X” the cube root ofX” the cube root of 1.6 x 101.6 x 10-5-5 = .016 = .016 44

[Pb[Pb+2+2] = .016M, [Cl] = .016M, [Cl-1-1] = .032M] = .032M