Q = reaction quotientQ = Keq only at equilibrium

If Q<Keq then the forward reaction is favored

If Q>Keq then the reverse reaction is favored

If Keq >1, the reaction is considered spontaneous

KKpp

Equilibrium constant when reactants Equilibrium constant when reactants and products are gases.and products are gases.

Treated same as KTreated same as Keq, eq, but using atm or but using atm or kPa rather than molarities.kPa rather than molarities.

2HI(g) 2HI(g) H H22(g) + I(g) + I22(g)(g)

Originally, a system Originally, a system contains only HI at a contains only HI at a pressure of 1.00 atm pressure of 1.00 atm at 520at 520°C. The °C. The equilibrium partial equilibrium partial pressure of Hpressure of H22 is is found to be found to be 0.10atm. Calculate 0.10atm. Calculate the equilibrium Kthe equilibrium Kpp for iodine, hydrogen, for iodine, hydrogen, and the system. and the system.

2HI2HI HH22 II22

InitialInitial 1.001.00 00 00

changechange

equilequil

-0.20-0.20 +0.10+0.10 +0.10+0.10

0.800.80 0.100.10 0.100.10

KKpp=(=(.10)(.10).10)(.10) == .0156 .0156

(.80)(.80)22

Converting KConverting Kc c to Kto Kpp

PV=nRT or P=PV=nRT or P=nnRT P = [A]RTRT P = [A]RT

V V

KKpp= K= Kcc(RT)(RT)ΔΔnn

ΔΔn = (mol gaseous product) – (mol gas n = (mol gaseous product) – (mol gas reactant)reactant)

If If ΔΔn = 0, then Kn = 0, then Kpp=K=Kcc

R is .0821 LR is .0821 L·atm/mol·K·atm/mol·K

2NO(g) + O2(g) 2NO2(g)

The value for Kc for the above reaction is

5.6 x 1012 at 290K. What is the value of Kp?

∆n=-1Kp = 5.6 x 1012 [(.0821)(290)].0821)(290)]-1-1

= = 5.6 x 105.6 x 101212 ==

2.4 x 102.4 x 1011 11

(.0821)(290).0821)(290)

KKspsp

-Solubility product constant--Solubility product constant-

Like KLike Keqeq but don’t include the but don’t include the solid solid

NaClNaCl(s)(s) Na Na++(aq)(aq) + Cl + Cl--(aq)(aq)

KKspsp = [Na = [Na++][Cl][Cl--]]

For CaClFor CaCl22(s)(s) Ca Ca2+2+(aq)(aq) + 2Cl + 2Cl--(aq)(aq)

KKspsp = [Ca = [Ca2+2+][Cl][Cl--]]22

Substance Formula Ksp

Aluminum hydroxide Al(OH)3 4.6 x 10-33

Barium carbonate BaCO3 2.6 x 10-9

Barium chromate BaCrO4 1.2 x 10-10

Barium fluoride BaF2 1.0 x 10-6

Barium sulfate BaSO4 1.1 x 10-10

Bismuth sulfide Bi2S3 1.8 x 10-99

Cadmium carbonate CdCO3 6.2 x 10-12

Cadmium hydroxide Cd(OH)2 5.3 x 10-15

Cadmium iodate Cd(IO3)2 2.4 x 10-8

Cadmium oxalate CdC2O4 1.5 x 10-8

Solubility Product ConstantSolubility Product ConstantAgBr(s) AgBr(s) Ag Ag++(aq) + Br(aq) + Br--(aq)(aq)

KKeqeq = = [Ag[Ag++][Br][Br--]]

[AgBr][AgBr]

Solids are omitted from equilibrium Solids are omitted from equilibrium expressions, so. . .expressions, so. . .

KKspsp = [Ag = [Ag++][Br][Br--]]

For AgBr (at 25°C), KFor AgBr (at 25°C), Kspsp = 5.01 x 10 = 5.01 x 10-13-13

Find [BrFind [Br--]]Hint: [Br-] = [Ag+] [Br-] = 7.08 x 10-7M

What is the solubility expression for a What is the solubility expression for a solution of Cusolution of Cu33(PO(PO44))22??

CuCu33(PO(PO44))2 2 3Cu 3Cu2+2+(aq) + 2PO(aq) + 2PO443-3-

KKspsp = [Cu = [Cu2+2+]]33[PO[PO443-3-]]22

What is the [BeWhat is the [Be2+2+] in a saturated solution ] in a saturated solution of Be(OH)of Be(OH)22? K? Ksp sp = 1.58 x 10 = 1.58 x 10-22-22

For every BeFor every Be2+ 2+ ion, there are 2OHion, there are 2OH-- ions. ions.

1.58 x 101.58 x 10-22-22 = x(2x) = x(2x)2 2 = 4x = 4x33

X = 3.41 x 10X = 3.41 x 10-8-8 M M

Calculate the solubility product constant for lead(II) Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride, if 50.0 mL of a saturated solution of lead(II)

chloride was found to contain 0.2207 g of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it. chloride dissolved in it.

PbClPbCl22(s) (s) Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)

KKspsp = [Pb = [Pb2+2+][Cl][Cl--]]2 2

Convert the amount of dissolved lead(II) Convert the amount of dissolved lead(II) chloride into moles per liter.chloride into moles per liter.

0.2207 g PbCl0.2207 g PbCl2 2 xx (1 mol PbCl(1 mol PbCl22/278.1 g /278.1 g PbClPbCl22)=)=

7.94 x 107.94 x 10-4-4 mol mol 7.94 x 10-4 mol = 0.0159 M PbCl2.0500L

create an "ICE" tablecreate an "ICE" table

PbCl2 (s)  Pb2+(aq) Cl-(aq)

Initial All solid 0 0

Change - 0.0159 M + 0.0159 M + 0.0318 M

equilibrium Less solid 0.0159 M 0.0318 M

Substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.

Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

Find [IOFind [IO33--] in a saturated ] in a saturated

solution of copper (II) iodate.solution of copper (II) iodate. KKspsp of Cu(IO of Cu(IO33))22 is 7.41 x 10 is 7.41 x 10-8-8

Cu(IOCu(IO33))22 Cu Cu2+2+ + 2IO + 2IO33--

If [CuIf [Cu2+2+]=x then [IO]=x then [IO33--]=]=

KKspsp=[Cu=[Cu2+2+][IO][IO33--]]22

= x= x(2x)(2x)22

= 4x= 4x33

7.41 x 107.41 x 10-8-8=4x=4x33

X= 2.65 x 10X= 2.65 x 10-3-3M and [IOM and [IO33--] = 5.30 x 10] = 5.30 x 10--

33MM

2x2x

Common Ion EffectCommon Ion Effect

KKspsp is unchanged by the addition of a solute. is unchanged by the addition of a solute. The solubility of a slightly soluble salt is The solubility of a slightly soluble salt is

reduced by the presence of a second solute reduced by the presence of a second solute that produces a common ion.that produces a common ion.

Addition of a common ion will decrease the Addition of a common ion will decrease the concentration of the ion it bonds with, concentration of the ion it bonds with, creating a precipitate.creating a precipitate.

If the ions’ concentrations > KIf the ions’ concentrations > Ksp, sp, ppt will formppt will form

What is the [TlWhat is the [Tl++] when .050mol ] when .050mol of NaBr is added to 500.0ml of of NaBr is added to 500.0ml of a saturated solution of TlBr?a saturated solution of TlBr?

What should happen according to What should happen according to LeChatelier’s Principle?LeChatelier’s Principle?

TlBr TlBr Tl Tl++ Br Br--

KKspsp= 3.39 x 10= 3.39 x 10-6-6

InitialInitial 00 0.100.10

ChangeChange +x+x +x+x

EquilibriuEquilibriumm

xx .10 + x.10 + x

KKspsp = [Tl = [Tl++][Br][Br--] = 3.39 x 10] = 3.39 x 10-6-6

3.39 x 103.39 x 10-6-6 = x (.10 +x) = x (.10 +x) X X ͌ ͌ 0.00184 so .10 + 0.00184 = .100.00184 so .10 + 0.00184 = .10 3.39 x 103.39 x 10-6-6 = x (.10) = x (.10)

3.39 x 103.39 x 10-5-5 = x = x