1

Chemical EquilibriumA state of no net change in reactant & product concentrations.

BUT…

There is a lot of activity at the molecular level.

2

Kinetics EquilibriumFor an “elementary step” in the mechanism:

A + 2B AB2kfkr

ratef = kf[A][B]2

rater = kr[AB2]

Kinetics Equilibrium

3

A + 2B AB2If start with only reactants, forward rate is high. As products build up, reverse rate increases and forward rate decreases.

4

Kinetics Equilibriumratef = kf[A][B]2

rater = kr[AB2]

equal ratesRate vs.

Time

ratef = kf[A][B]2

rater = kr[AB2]

equal rates

reactants

prod.

equilibrium,no change

Conc. vs.

Time

5

At equilibrium: ratef = rater

A + 2B AB2kfkr

Kinetics Equilibrium

kf[A][B]2 = kr[AB2]

kfkr

=[AB2]

[A][B]2= Kc

EquilibriumConstant

6

Since kf and kr are constants depending only on temp.,Kc is also a constant depending on temp.

Kinetics Equilibrium

Kc describes the ratio of products to reactants at equilibrium.

Kc = kfkr

=[AB2]

[A][B]2

7

Chemical Equilibrium

reactants productsA + B C + D

Chemical equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, concentrations do not change.

8

Chemical EquilibriumN2O4(g) 2NO2(g)colorless brown

This reaction can be monitored spectroscopically.

Draw Lewis structure of NO2

9

N2O4(g) 2NO2(g)

time2 time3 time4

no change

time1

100% N2O4

10

conc

ent

ration

time time time

Exp.#1 Exp.#2 Exp.#3

N2O4(g) 2NO2(g)

N2O4N2O4

NO2NO2

N2O4

NO2

11

N2O4(g) 2NO2(g)Final concentrations depend

on what you start with.At equilibrium, forward and

reverse rates are equal.

12

Equilibrium Constant (K)aA + bB cC + dD

K is independent of concentrations; it depends only on temperature.

K = [C]c[D]d

[A]a[B]bExponents are the coefficients

13

Equilibrium Constant

K =[C]c[D]d

[A]a[B]b

K<10-3 10-3<K<103 K>103

react. prod. react. prod. react.prod.mostly similar mostly

reactants mixture products

14

Equilibrium ConstantUnits are important even though they are not written !!

Kc :concentration (M)

Kp :pressure (atm)

15

Types of EquilibriaHomogeneous: all reactants and products in same phase.

Heterogeneous: reactants and products in different phases.

16

Homogeneous EquilibriaN2O4(g) 2NO2(g)

P2NO2PN2O4

Kp =

[ ] = M = mol/LPartial P (atm)

(P a conc.)

Kc = [NO2]2

[N2O4]

17

Relationship: Kc and KpaA(g) bB(g)

Kp =PB

b

PAa

PA =(nART)/V

=[A]RT

PB =(nBRT)/V

=[B]RT

18

Kp=PB

b

PAa

([B]RT)b

([A]RT)a=

Relationship: Kc and Kp

Dngas = mol prodgas – mol reactgas

=[B]b

[A]a(RT)Dngas

=[B]b

[A]a(RT)b-aKp

19

Kp =[B]b

[A]a(RT)Dn

Relationship: Kc and Kp

If Dn = 0, then Kp = Kc

Kp = Kc(RT)Dn

20

K’cUsed when the concentration of one of the reacting chemicals doesn’t change.

21

Relationship: Kc and K’ce.g. ionization of aqueous acetic acid

(CH3COOH(aq) = HAc)

HAc + H2O Ac- + H3O+

K’c =[Ac-][H3O

+]

[HAc][H2O]

22

Relationship: Kc and K’c

K’c =[Ac-][H3O

+]

[HAc][H2O]

But [H2O], the solvent, doesn’t change significantly during the course of the reaction; it is ~constant.

23

K’c [H2O] = Kc

K’c =[Ac-][H3O

+]

[HAc][H2O]

Relationship: Kc and K’c

[Ac-][H3O+]

[HAc]=

Solvent is dropped from Kc.

24

Let’s Try Some ProblemsWrite Kc and Kp for:

HF(aq) +H2O H3O+(aq) + F-(aq)

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

CH3COOH(aq) + C2H5OH(aq)

CH3COOC2H5(aq) + H2O

25

Some More Fun ! At equilibrium at 230 oC:

2NO + O2 2NO2

0.0542 M 0.127 M 15.5 M

Calculate Kc and Kp

26

Yes, There’s More !PCl5(g) PCl3(g) + Cl2(g)

Kp = 1.05 at 250oC

If PPCl5 = 0.875 atm

and PPCl3 = 0.463 atm

What is PCl2 ?

27

Heterogeneous EquilibriaCaCO3(s) CaO(s) + CO2(g)

[CaCO3] and [CaO] = constant

The conc. of a solid is an intrinsic property:

K’c = [CaO][CO2]

[CaCO3]

CaCO3 CaO

28

Heterogeneous Equilibria

K’c = [CaO][CO2]

[CaCO3]

= KcK’c[CaCO3]

[CaO]= [CO2]

(as long as CaO and CaCO3 are present)

Solids and liquids dropped from Kc.

29

Try It !!!

Write Kc and Kp for:

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

P4(s) + 6Cl2(g) 4PCl3(l)

30

Keep Going

CaCO3(s) CaO(s) + CO2(g)

Calculate Kp and Kc if PCO2is 180 mmHg at 800oC.

31

Multiple EquilibriaA + B C + D K’cC + D E + F K’’c

overall

A + B E + F Kc

K’cK’’c =[C][D] [E][F][A][B] [C][D]

= Kcx

32

Multiple Equilibria

K’cK’’c = Kc

(useful in many acid-base problems)

For sequential equilibria:

If add reactions multiply K’s

33

Try it !!!H2CO3(aq) H+ + HCO3

-

Kc = 4.2 x 10-7

HCO3-(aq) H+ + CO3

-2

Kc = 4.8 x 10-11

What is the expression for and value of overall Kc ?

34

Summary1.For reverse reaction,

Kc(reverse) = 1/Kc(forward)

Kc = 4.63 x 10-3N2O4(g) 2NO2(g)

2NO2(g) N2O4(g)K’c = 216

And KcK’c = 1

35

Summary2. Kc depends how

equation is written.

N2O4(g) 2NO2(g) Kc

Kc = K’c

2N2O4(g) 4NO2(g) K’c

36

Summary3. Units of reactants &

products are mol/L for Kc and atm for Kp.

4. Kc and Kp are unit-less

5. Solvents, & pure solids & liquids are not in Kc

Summary

37

6. Must specify equation & T(the value of K can only change with temp.)

7. For sequence of reactions that add to a net reaction: Kc = K’cK’’c

38

Give It a Shot

How are the Kc values for the following equations related?

N2(g) + 3H2(g) 2NH3(g)

N2(g) + H2(g) NH3(g)13

23

Qc & Qp

39

What if the concentrations of reactants and products are not at equilibrium?

We define a new quantity Q, similar to K but using the actual concentrations [X]o rather than equilibrium concentrations [X].

40

Qc & Qp

Qc =[C]o

c[D]od

[A]oa[B]o

b

aA + bB cC + dDFor the reaction:

The reaction direction can be determined by comparing Q vs. K.

41

Reaction Quotient: Qc

If Qc > Kc, reaction proceeds to left.

If Qc = Kc, at equilibrium.If Qc < Kc, reaction proceeds

to right.

Qc =[C]o

c[D]od

[A]oa[B]o

b

42

Try It!N2(g) + 3H2(g) 2NH3(g)

At reaction start:N2 = 0.249 molH2 = 3.21 x 10-2 molNH3 = 6.42 x 10-4 mol

in 3.50 L flask

If Kc = 1.2, which direction will the reaction proceed?

43

Calculating ConcentrationsFor the reaction: A B

Kc = 24.0 and

[A]o = 0.850 M

[B]o = 0 M

What are the equilibrium concentrations of A and B?

44

Calculating Concentrationsusing “ICE” Table

A BInitial (M) 0.850 0

Change (M) -1x +1x

Equil. (M) (0.850 – x) x

Substitute into Kc

changecontrolled by stoich.

45

Calculating Concentrations

Kc = 24.0 = [B][A]

= x(0.850 – x)

x = .816 M = [B]

0.850 – x = 0.034 M = [A]

and

Substitute into Kc to check.

46

Give It A Shot!

Kc = 54.3 at 430oC

H2(g) + I2(g) 2HI(g)

If start with 0.500 mol each of H2 and I2 in a 1.00 L flask, what are equilibrium concentrations of all species?

Try Another

47

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

Starting with pure NO2, it is found that partial pressure of O2 at equilibrium is 0.25 atm.

Calculate PNO and PNO2.

At 1000 K, Kp = 158

48

Quadratics are Possible!

Set up ICE table and substitute into Kc: 0.062 M = [H2]o

0.041 M = [I2]o0.22 M = [HI]o

(Don’t solve!)

H2(g) + I2(g) 2HI(g)Kc = 54.3

49

Factors Affecting Equilibrium

Changes in external variables (T, P, concentrations, etc.) can shift equilibrium to right or left.

rightproducts

leftreactants

50

Le Chatelier’s Principle

If a stress (change) is applied to a system at equilibrium, the system will adjust to partially offset the stress.Stresses:• concentration & dilution• pressure/volume (gases)• temperature

Le Chatelier’s Principle

51

N2(g) + 3H2(g) 2NH3(g)

Haber synthesis of ammonia:

If additional N2 is added at equilibrium, what happens to:• rate of forward reaction?• rate of reverse reaction?• amounts of each substance?

N2(g) + 3H2(g) 2NH3(g)

52

conc

ent

ration

time

initial

equilibrium

new

equilibrium

H2

NH3

N2 N2added

obeyK

53

Concentration StressConsider the equilibrium:

FeSCN+2 Fe+3 + SCN-

red yellow clear

Add NaSCN left redAdd Fe(NO3)3 left red

“stress”:

54

Concentration Stress

“stress”

Add H2C2O4

Fe+3 + 3C2O4-2 Fe(C2O4)3

-3

Thus right

FeSCN+2 Fe+3 + SCN-

red yellow clear

55

Concentration StressIf concentrations of reactants or products change,the system will respond in accordance with Kc.

Kc = productsreactants

56

Concentration StressN2(g) + 3H2(g) 2NH3(g)

If [NH3] is increased to 3.65 M, then Qc> Kc, so reverse reaction predominates.

At 720 oC, Kc = 2.37 x 10-3

0.683M 8.80M 1.05M (at eq.)

Concentration: Dilution

57

HF(aq) + H2O H3O+(aq) + F-(aq)

What happens to the ionization of HF if the system is diluted?

This is just a special case of Le Chatelier’s principle.

Adding H2O shifts reaction right.

58

Pressure & Volume Stress

P = (n/V) RT

P & V have no effect on solids & liquids, but for gases:

As P increases or V decreases the conc. of gases increase.

59

P & V StressN2O4(g) 2NO2(g)

At equilib, if volume is decreased, product conc. Is affected more than reactant side:

> Kc (shift left)Qc = [NO2]o2

[N2O4]o

60

P & V StressIn general, increased P or decreased V shifts reaction to side with lesser moles of gas.

Exception: if P is increased by adding an inert gas, it does not affect equilibrium.

61

Y’all Be Stressin’Predict the direction of the reaction resulting from increased P (or decreased V):

PCl5(g) PCl3(g) + Cl2(g)

H2(g) + CO2(g) H2O(g) + CO(g)

62

Temperature Stress

Unlike concentration, P, or V, temperature changes the value of Kc.

Increasing temperature will favor the reaction that is endothermic (“uses heat”).

63

Temperature Stress

DH = 58 kJ

N2O4(g) 2NO2(g)colorless brown

T=-78oC T=-10oC T=+20oC

Temperature Stress

64

58kJ + N2O4(g) 2NO2(g)

Raising temperature (adding heat) shifts equilibrium to use up the heat (Le Chatelier).

Reaction shifts right, producing more NO2. Value of Kc increases.

65

Temperature Stress

CoCl4-2 + 6H2O

Co(H2O)6+2 + 4Cl-

Based on the following clip, predict whether this reaction is endo- or exothermic.

66

Effect of CatalystCatalysts affect the rate of reaction, but NOT the equilibrium.

DH

Ea with catalyst is lower so rate increases. There is no effect on DH, so equilibrium doesn’t change.

67

Le Chatelier: Try It !N2F4(g) 2NF2(g)

DH = 38.5 kJ

What happens if:

NF2 is addedV is increasedHe gas if addedT is decreased

Physical Equilibria

68

Systems can also be in physical equilibria, where rate of forward and reverse processes are equal.

H2O(s) H2O(l) (at 0oC)

CO2(g) CO2(aq) (in close container)

KCl(s) KCl(aq) (in saturated soln)(demo)

Le Chatelier: Physical Change

69

H2O(s) H2O(l)

What happens at equilibrium if:• heat is added?• pressure is increased?

70

Solubility Equilibria

Even insoluble ionic compounds are very slightly soluble!

Application of equilibria to “insoluble” ionic compounds.

71

Soluble CompoundsSoluble Compounds ExceptionsAlkali metals & NH4

+ KNOW!!!

NO3-,HCO3

-, ClO3- KNOW!!!

Halides Ag+ & Hg22+ & Pb2+

Sulfates (SO42-) Ag+ & Ca2+ & Sr2+

& Ba2+ & Pb2+

72

Insoluble Compounds

Insolubles ExceptionsCO3

2- & PO43-

& CrO42- & S2-

Alkali metals & NH4+

KNOW!!!OH-

Alkali metals & Ba2+

73

Solubility EquilibriaNow let’s get quantitative.

Ksp = 1.6E-10 = [Ag+][Cl-]

Why isn’t AgCl(s) included in Ksp?

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

Ksp= 1.2E-26 = [Ca+2]3[PO4

-3]2

Ca3(PO4)2 3Ca+2 + 2PO4-3

74

Solubility and KspFor a saturated solution:Molar solubility: Sol. in mol/LSolubility: Solubility in g/L

The solubility of CaSO4 is 0.67g/L. What is Ksp?

0.67 gL

x 1 mol136.2 g

= 4.9 x 10-3 M

75

75

Ksp of CaSO4: ICE Table

Ksp = [Ca+2][SO4-2]

= (4.9 x 10-3)2

= 2.4 x 10-5

CaSO4 Ca+2 + SO4-2

I 0 0

C -4.9E-3 4.9E-3 4.9E-3E 4.9E-3 4.9E-3

76

Solubility and KspGiven Ksp of Cu(OH)2 is

2.2E-20, what is its solubility?

(s = molar solubility of Cu(OH)2)

I 0 0C +s +2sE s 2s

Cu(OH)2 Cu+2 + 2OH-

77

Solubility and KspKsp = 2.2x10-20 = [Cu+2][OH-]2

= (s)(2s)2

s = 1.8x10-7 M

Thus solubility in g/L is:

1.8E-7M x 97.57 g/mol = 1.8E-5 g/L

= 4s3

78

Solubility and KspTable 16.3 shows the relationship between ‘s’ and Kspfor several different cases.

Ksp = s2 Ksp = 27s4

AgCl Al(OH)3

Ksp = 4s3 Ksp = 108s5

PbF2 Ca3(PO4)2

79

79

Solubility and KspYour turn !

Calculate the solubility of Al(OH)3 given Ksp = 1.8E-33

Calculate the Ksp of Bi2S3 given its solubility is 1.0E-15 M.

80

Solubility Equilibria & QAs with Keq problems, calculate reaction quotient, Q, for unsaturated or supersaturated solutions to determine the direction of reaction toward equilibrium.

81

Will a Precipitate Form?Mix 200.mL of 1.0E-4 M Sr(NO3)2& 100.mL of 1.0E-4 M NaF.

Final solution volume is 0.300 L.

The possible precipitate is SrF2 (Ksp = 2.0E-10)

Do the ion concentrations of the mixture exceed Ksp?

82

Will a Precipitate Form?

mol Sr+2 = 0.200 L x 1.0E-4 M= 2.0E-5 mol

mol F- = 0.100 L x 1.0E-4 M= 1.0E-5 mol

SrF2 Sr2+ + 2F-

Q = [Sr+2][F-]2

83

Will a Precipitate Form?Q = [Sr+2][F-]

= 7.3 x 10-4

Since Q > Ksp of 2.0E-10, a precipitate will form.

2.0E-5mol0.300 L

x 1.0E-5mol0.300 L

=2

2

84

Will a Precipitate Form?For this problem, what are [Sr+2] and [F-] at equilibrium?

SrF2 Sr2+ + 2F-

I 2.0E-5/.300 1.0E-5/.300= 6.67E-5 = 3.33E-5

limiting

detn.by Ksp

C(stoich) -1.67E-5 -3.33E-5

E 5.00E-5 ~0

85

Will a Precipitate Form?Next plug into Ksp:

2.0E-10 = (5.00E-5)2

[F-]

Ksp = 2.0E-10 = [Sr+2][F-]2

detn.by Ksp

[F-] = 2.0E-3Mand [Sr+2] = 5.0E-5M

86

Will a Precipitate Form?Try It !!

200.mL of 0.200 M NaOH is mixed with 1.00 L of 0.100 M CaCl2.

• Will a precipitate form?• What are [Ca+2] and [OH-]

at equilibrium?

For Ca(OH)2 Ksp = 8.0E-6

87

Common Ion EffectE.g., AgCl is less soluble in aqueous AgNO3 than in water.

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

As Ag+ increases, Cl- must decrease in accordance with Ksp. Thus less AgCl will dissolve.

Le Chatelier

88

Common Ion EffectWhat is the solubility of AgCl in 0.0065 M AgNO3 ?

AgCl Ag+ + Cl-

I 0.0065 0C +s +sE 0.0065 +s +s~E 0.0065 +s

Let s = molar solubility of AgCl

89

Common Ion EffectKsp = [Ag+][Cl-]

1.6 x 10-10 = (0.0065)(s)

s = 2.5 x 10-8 MThus the amount of dissolved AgCl is 2.5 x 10-8 M

or 3.6 x 10-6 g/L Note ‘s’ is small compared to 0.0065M so approximation was valid.

90

Your turn !!

What is the molar solubilityof silver carbonate: •in water•in 0.0030 M sodium carbonate?

Ksp (Ag2CO3) = 8.1 x 10-12

Common Ion Effect

One More

91

What is the Ksp of bismuth sulfide (Bi2S3) given its solubility is 8.8E-13 g/L?

(M = 514.2 g/mol)

What is the molar solubility of bismuth sulfide in a 0.020 M Na2S solution?

92

92

The End

See Ksp summary in problem set.

93

Warm-up

94

For the reaction:2CO(g) + O2(g) 2CO2(g)Write the algebraic expressions for Kc and Kp

Warm-up

95

For the reaction at 170oC,2H2(g) + O2(g) 2H2O(g)

Calculate the ratio of Kp/Kc. Are these constants large (>>1) or small (<<1)?

Write Kc for PbCl2(s) Pb+2(aq) + 2Cl-(aq)

Warm-up

96

2NOCl(g) = 2NO(g) + Cl2(g)

1.0 mol NOCl was placed in a 2.0 L flask at 200oC. At equilibrium it was found that 40.% of NOCl had decomposed.

What is Kp?

Warm-up

97

N2(g) + O2(g) 2NO(g)

If 1.0 mol each of N2, O2, and NO are placed in a 2.0 L vessel, what is the equilibrium concentration of NO?

(Kc = 3.4E-2)

Warm-up

98

The solubility of SrF2(s) in water is 5.8E-4M. Calculate Kc.

Warm-up

99

Are they soluble in water?

•silver nitrate•calcium carbonate•copper(II) sulfide•barium hydroxide•calcium hydrogen carbonate

100

Warm-up

Write the expression for Ksp of PbF2.

What is the Ksp of PbF2, given solubility is 0.54 g/L?

Warm-upWhat is molar solubility of PbF2 in water and in 0.020 M Pb(NO3)2?

(given Ksp = 4.1E-8)

101

Warm-up

102

Silver sulfide has Ksp = 6.0E-51.What is its molar solubility in water?

What is its molar solubility in 1.0E-7 sodium sulfide solution?

If 100.mL each of 4.0E-8M silver nitrate and 1.0E-7M sodium sulfide solutions are mixed, what is [Ag+] and [S-2]?