1 17 Chemical Equilibrium. 2 Chapter Goals 1.Basic Concepts 2.The Equilibrium Constant 3.Variation of K c with the Form of the Balanced Equation 4.The

1

17 Chemical Equilibrium

2

Chapter Goals

1. Basic Concepts

2. The Equilibrium Constant

3. Variation of Kc with the Form of the Balanced Equation

4. The Reaction Quotient

5. Uses of the Equilibrium Constant, Kc

6. Disturbing a System at Equilibrium: Predictions

3

Chapter Goals

7. The Haber Process: A Commercial Application of Equilibrium

8. Disturbing a System at Equilibrium: Calculations

9. Partial Pressures and the Equilibrium Constant

10.Relationship between Kp and Kc

11.Heterogeneous Equilibria

4

Chapter Goals

12.Relationship between Gorxn and the

Equilibrium Constant

13.Evaluation of Equilibrium Constants at Different Temperatures

5

Basic Concepts

• Reversible reactions do not go to completion.– They can occur in either direction– Symbolically, this is represented as:

gggg D d + C cB b + A a

6

Basic Concepts

• Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.– A chemical equilibrium is a reversible reaction

that the forward reaction rate is equal to the reverse reaction rate.

• Chemical equilibria are dynamic equilibria.– Molecules are continually reacting, even

though the overall composition of the reaction mixture does not change.

7

Basic Concepts

• One example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution.

solution. into go williodine eradioactiv theof Some

solution. filter the then minutes, few afor Stir 2

I 2 Pb PbI

solution. PbI saturated ain PbI solid Place 1-(aq)

2(aq)

OH

2(s)

2*22

8

Basic Concepts

• This movie depicts a dynamic equilibrium.

9

Basic Concepts

• Graphically, this is a representation of the rates for the forward and reverse reactions for this general reaction.

gggg D d + C cB b + A a

10

Basic Concepts

• One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.

11

Basic Concepts

12

Basic Concepts

13

The Equilibrium Constant

• For a simple one-step mechanism reversible reaction such as:

• The rates of the forward and reverse reactions can be represented as:

rate. reverse therepresents which DCkRate

rate. forward therepresents which BAkRate

rr

ff

(g)(g)(g)(g) D C B A

14


• When system is at equilibrium:

Ratef = Rater

BA

DC

k

k

torearrangeswhich

DCkBAk

:give toiprelationsh rate for the Substitute

r

f

rf

15


• Because the ratio of two constants is a constant we can define a new constant as follows :

kk

K and

KC DA B

f

rc

c

16


• Similarly, for the general reaction:

we can define a constant

reactions. allfor validis expression This

BA

DCK

products

reactants ba

dc

c

D d C c B b A a (g)(g)(g)(g)

17

The Equilibrium Constant• Kc is the equilibrium constant .• Kc is defined for a reversible reaction at a given

temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

18


• Example 17-1: Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

5

23c

235

PCl

ClPClK

ClPClPCl

19


You do it!

HI 2 I + H 22

20


22

2

c

22

IH

HIK

HI 2 I + H

21


You do it!

OH 6+NO 4O 5 + NH 4 223

22


52

43

62

4

c

223

ONH

OHNO=K

OH 6+NO 4O 5 + NH 4

23


• Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity.– Activities are directly related to molarity

24


• Example 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.

• Equil []’s 0.028 M 0.172 M 0.086 M

You do it!

235 ClPClPCl

25


Kc PCl3 Cl2

PCl5 Kc

0.172 0.086 0.028

Kc 0.53

26


• Example 17-3: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.

27


0 0 1.00 Initial

ClPClPCl g2g3g5

M

28


MMM

M

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl g2g3g5

29

MMM

MMM

M

0.60 0.60 0.40 mEquilibriu

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl g2g3g5


30

Tanother at 90.0

40.0

60.060.0K

0.60 0.60 0.40 mEquilibriu

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl

'c

g2g3g5

MMM

MMM

M


31


• Example 17-4: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.

You do it!

32


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

Initial 0.80 M 0.90 M 0

Change - 0.10 M - 0.30 M +0.20 M

Equilibrium 0.70 M 0.60 M 0.20 M

Kc NH3 2

N2 H2 3 0.20 2

0.70 0.60 3 0.26

33

Variation of Kc with the Form of the Balanced Equation• The value of Kc depends upon how the balanced

equation is written.• From example 17-2 we have this reaction:

• This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53

235 ClPClPCl

34

Variation of Kc with the Form of the Balanced Equation• Example 17-5: Calculate the equilibrium constant

for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.

Equil. []’s 0.172 M 0.086 M 0.028 M

The concentrations are from Example 17-2.

523 PClCl PCl

35

Variation of Kc with the Form of the Balanced Equation

Kc'

PCl5 PCl3 Cl2

36


KPCl

PCl Clc' 5

3 2

0 028

0172 0 08619

.. .

.

37

KPCl

PCl Cl

KK

or K K

c' 5

3 2

cc' c

'

c

0 0280172 0 086

19

1 1 10 53 19

.. .

.

. .


Large equilibrium constants indicate that most of Large equilibrium constants indicate that most of the reactants are converted to products.the reactants are converted to products.

Small equilibrium constants indicate that only Small equilibrium constants indicate that only small amounts of products are formed.small amounts of products are formed.

38

The Reaction Quotient

39


40

The Reaction Quotient• The mass action expression or reaction quotient

has the symbol Q. – Q has the same form as Kc

• The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.

For this general reaction :

aA +bB cC +dD

Q C c

D d

A aB b

41


• Why do we need another “equilibrium constant” that does not use equilibrium concentrations?

• Q will help us predict how the equilibrium will respond to an applied stress.

• To make this prediction we compare Q with Kc.

42


fractions. as K and Q of think thisunderstand help To

extent.greater a toright the tooccursreaction The KQ

extent.greater a toleft the tooccursreaction The KQ

m.equilibriuat is system The K=Q

:When

c

c

c

c

43


• Example 17-6: The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

The concentrations given in the problem

are not necessarily equilibrium []' s.

We can calculate Q.

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

0.22 M 0.22 M 0.66 M

Q =HI 2

H2 I2 0.66 2

0.22 0.22 9.0

Q 9.0 but Kc 49

Q < Kc

44

Uses of the Equilibrium Constant, Kc

• Example 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

(g)3(g)2(g)2(g) NO SO NO SO

45


0 0 0.500 0.500 Initial

NO SO NO SO (g)3(g)2(g)2(g)

MM

46

MxMxMxMx

MM

+ + - - Change

0 0 0.500 0.500 Initial



47

MxMxMxMx

MxMxMxMx

MM

500.0 500.0 mEquilibriu

+ + - - Change

0 0 0.500 0.500 Initial



48


equation. thesidesofboth of thecan take We

.squareperfect a isequation This

500.0500.000.3

NOSO

NOSOK


+ + - - Change

0 0 0.500 0.500 Initial

NO SO NO SO

22

3c

(g)3(g)2(g)2(g)

xx

xx

MxMxMxMx

MxMxMxMx

MM

49

22

3

22

3c

(g)3(g)2(g)2(g)

NOSO184.0500.0

NOSO316.073.2

865.0 ;73.2 0.865 ;1.73-0.865

500.0=1.73

sides.both of thecan take We

square.perfect a isequation This

500.0500.000.3

NOSO

NOSOK


+ + - - Change

0 0 0.500 0.500 Initial

NO SO NO SO

MMx

Mx

xxxx

x

x

xx

xx

MxMxMxMx

MxMxMxMx

MM


50


• Example 17-8: The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

You do it!

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

51


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

Initial 0 0 1.00M

Change + x M + x M - 2x M

Equilibrium x M x M 1.00 - 2x M

Kc =HI 2

H2 I2 = 49 = 1.00 - 2x 2

x x

Kc = 7.0 =1.00 - 2x

x7.0x 1.00 2x; 9x 1.00; x 0.11 M

H2 I2 x M 0.11 M

HI 1.00 2x M 0.78 M

52

Disturbing a System at Equlibrium: Predictions• LeChatelier’s Principle - If a change of conditions

(stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.

– We first encountered LeChatelier’s Principle in Chapter 14.

• Some possible stresses to a system at equilibrium are:

1. Changes in concentration of reactants or products.2. Changes in pressure or volume (for gaseous reactions)3. Changes in temperature.

53

Disturbing a System at Equlibrium: Predictions• For convenience we may express the amount of

a gas in terms of its partial pressure rather than its concentration.

• To derive this relationship, we must solve the ideal gas equation.

ion.concentrat its toalproportiondirectly is

gas a of pressure partial theT,constant at Thus

[]RT=P

mol/L, units thehas V

nBecause

RTV

nP

nRTPV

54

Disturbing a System at Equlibrium: Predictions1 Changes in Concentration of Reactants and/or Products

• Also true for changes in pressure for reactions involving gases.– Look at the following system at equilibrium at 450oC.

H2 I2 2 HI g

Kc HI 2

H2 I2 49

55



H2 I2 2 HI g

Kc HI 2

H2 I2 49

If some H 2 is added, Q < Kc.

This favors the forward reaction.

Equilbrium will shift to the right or product side.

56

side.reactant or left, theshift to willEquilbrium

reaction. reverse thefavors This

K>Q,H some remove weIf

49IH

HIK

HI 2IH

c2

22

2

c

g22



57

Disturbing a System at Equlibrium: Predictions2 Changes in Volume

• (and pressure for reactions involving gases)– Predict what will happen if the volume of this system at equilibrium is

changed by changing the pressure at constant temperature:

2 NO2 g N2O4 g

Kc =N2O4 NO2 2

58

Disturbing a System at Equlibrium: Predictions

gas. of molesfewer producesreaction forward The

reaction. forward or theformation product favors This

.K<Qpressure, theincreases which decreased, is volume theIf

NO

ON=K

ONNO 2

c

22

42c

g42g2

59

produced. are gas of moles More

reaction. reverse or the reactants thefavors This

.K>Qpressure, thedecreases which increased, is volume theIf

NO

ON=K

ONNO 2

c

22

42c

g42g2


60


61

Disturbing a System at Equlibrium: Predictions3 Changing the Temperature

62

reaction. forwardor reactants thefavors This

reactants. thestresses emperaturereaction t theDecreasing

reaction. thisofproduct a isHeat

kJ 198+SO 2O+SO 2 g3g2g2

Disturbing a System at Equlibrium: Predictions3 Changing the Reaction Temperature

– Consider the following reaction at equilibrium:

reaction? in thisproduct or reactant aheat Is

kJ/mol 198H SO 2 O SO 2 orxn3(g)2(g)2(g)

63


• Introduction of a Catalyst– Catalysts decrease the activation energy of both the forward and

reverse reaction equally.

• Catalysts do not affect the position of equilibrium.– The concentrations of the products and reactants will be the

same whether a catalyst is introduced or not.– Equilibrium will be established faster with a catalyst.

64


65effect no catalyst platinum a gIntroducin f.

right NH ofion concentrat theDecrease e.

right H ofion concentrat theIncrease d.

right volume thedecreasingby pressure theIncreasing c.

right emperaturereaction t theDecreasing b.

left emperaturereaction t theIncreasing a.

procedurereaction on Effect Factor

kJ/mol 92H NH 2 H 3 N

3

2

orxn3(g)2(g)2(g)

Disturbing a System at Equlibrium: Predictions• Example 17-9: Given the reaction below at

equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

66

right OH 2 OH 2 d.

right PClCl+PCl c.

left OH 6+NO 4 O 5+NH 4 b.

effect no HI 2I +H a.

mEquilibriuon Effect Reaction

g2g2g2

g5g2g3

g2g2(g)g3

gg2g2

Disturbing a System at Equlibrium: Predictions• Example 17-10: How will an increase in pressure (caused

by decreasing the volume) affect the equilibrium in each of the following reactions?

67

right kJ 25H HI 2I +H c.

left kJ 92 +HCl 2 ClH b.

left 0H ON NO 2 a.

mEquilibriuon Effect Reaction

gg2g2

gg2g2

orxn4(g)22(g)

Disturbing a System at Equlibrium: Predictions• Example 17-11: How will an increase in

temperature affect each of the following reactions?

68

The Haber Process: A Practical Application of Equilibrium• The Haber process is used for the

commercial production of ammonia.– This is an enormous industrial process in the

US and many other countries.– Ammonia is the starting material for fertilizer

production.

• Look at Example 17-9. What conditions did we predict would be most favorable for the production of ammonia?

69

The Haber Process: A Practical Application of Equilibrium

dilemma. thisosolution t sHaber'

res. temperatulowat slow very are kineticsreaction eHowever th

e.unfavorabl is which 0<S favorable. also 0<H favorable. is which 0<G

atm. 1000 to200= N of P and C450 = T aat run isreaction This

gas. coal from obtained H air. liquid from obtained is N

kJ 22.92H NH 2 H 3N

2o

g2g2

og3

oxides metal & Feg2g2

70

The Haber Process: A Practical Application of Equilibrium

kinetics! with thehelps and yieldreaction theincreases This

removed. is NH because mequilibriu reachesnever systemreaction The

right. ly toperiodical NH Remove 4

right. toN excess Use3

right. topressurereaction Increase 2

.decreased is yieldbut rate, increase toT Increase 1

dilemma. thisosolution t sHaber'

3

3

2

71

The Haber Process: A Practical Application of Equilibrium• This diagram illustrates the commercial

system devised for the Haber process.

72

Disturbing a System at Equilibrium: Calculations• To help with the calculations, we must determine

the direction that the equilibrium will shift by comparing Q with Kc.

• Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C. What is the value of Kc for this reaction?

ggg C B A

73

Disturbing a System at Equilibrium: Calculations

A g B g C g

Equil. []'s 0.20M 0.30M 0.30M

Kc B C

A 0.30 0.30

0.20 0.45

74

Disturbing a System at Equilibrium: Calculations• If the volume of the reaction vessel were

suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?

1 Calculate Q, after the volume has been doubled

A g B g C g

Equil. []'s 0.10 M 0.15 M 0.15 M

Q =B C

A 0.15 0.15

0.10 0.22

75

Disturbing a System at Equilibrium: Calculations• Since Q<Kc the reaction will shift to the right to

re-establish the equilibrium.2 Use algebra to represent the new

concentrations.

MMM 0.15 0.15 0.10 s[]' initial New

C + B A ggg

76



concentrations.

MxMxMx

MMM

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A ggg

77



concentrations.

MxMxMx

MxMxMx

MMM

+0.15 +0.15 -0.10 s[]' Equil. New

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A ggg

78



concentrations.

A g B g + C g

New initial []'s 0.10M 0.15 M 0.15 M

Change - x M + x M + x M

New Equil. []' s 0.10 - x M 0.15 + x M 0.15 + x M

Kc =B C

A 0.45 0.15 x 0.15 x

0.10 x

79


00225.075.0

+0.30+0.0225=0.45-0.045

equation quadratic thisSolve

2

2

xx

xxx

80


Mx

x

x

0.03 and 78.02

81.075.0

12

0225.01475.075.0

2a

ac4bb-

2

2

81


disturbed.been has mequilibriu the

after ionsconcentrat new theare These

18.0 15.0CB

07.0 )10.0(A

M. 0.03 is valueposibleonly The

answer.an as 0.78- discardcan we0.10,<<0 Since

MMx

MMx

x

x

82

Disturbing a System at Equilibrium: Calculations• Example 17-13: Refer to example 17-12.

If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M.

You do it!

83


ions.concentrat mequilibriu the

determine tosexpression algebraic theupSet (2)

side.reactant or left the toshifts mequilibriu the thusKQ

90.040.0

60.060.0

A

CB=Q

halved is volumeafter the Q, Calculate (1)

0.60 0.60 0.40 s[]' ousInstantane

C B A

c

ggg

MMM

84


018.0 65.1

toreducesequation thiscompleted, is algebra After the

)+40.0(

)-60.0)(-60.0(45.0

A

CBK

)-60.0( )-60.0( )+40.0( Equil. New

- - + Change

0.60 0.60 0.40 s[]' initial New

C + B A

2

c

ggg

xx

x

xx

MxMxMx

MxMxMx

MMM

85


MMxCB

MMx

x

x

x

48.0 )60.0(

52.0 )40.0(A

answer. possibleonly theis 0.12 Thus

answer.an as 1.5 discardcan we0.60<<0 are limits theBecause

0.12 and 1.52

42.165.1

)1(2

)18.0)(1(4)65.1(65.1

.expression for thisequation quadratic theSolve

2

86

Disturbing a System at Equilibrium: Calculations• Example 17-14: A 2.00 liter vessel in which the

following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant.

You do it!

g2g2g COCl Cl CO

87


CO g +Cl2 g COCl2 g

Equil. []'s 0.30M 0.10M 0.60M

Kc COCl2

CO Cl2 0.60

0.30 0.10 20

88

Disturbing a System at Equilibrium: Calculations• An additional 0.80 mole of Cl2 is added to

the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established.

You do it!

89


CO g + Cl2 g COCl2 g

Orig. Equil. 0.30M 0.10 M 0.60M

(Stress) Add +0.40M

New Initial 0.30M 0.50M 0.60M Q < Kc shift right

Change - x M - x M + x M

New Equil. (0.30 - x)M (0.50 - x)M (0.60 + x)M

Kc COCl2

CO Cl2 20 0.60 x

0.30 x 0.50 x

90


MMx

MMx

MMx

x

X

xx

78.0)60.0(COCl

32.0)50.0(Cl

12.0)30.0(CO

0.67 discardcan we thus0.30<<0 are limits

0.18 & 67.0)20(2

)4.2)(20(4)17(17

04.21720 toreducesequation

2

2

2

2

91

Partial Pressures and the Equilibrium Constant• For gas phase reactions the equilibrium

constants can be expressed in partial pressures rather than concentrations.

• For gases, the pressure is proportional to the concentration.

• We can see this by looking at the ideal gas law.– PV = nRT– P = nRT/V – n/V = M– P= MRT and M = P/RT

92

Partial Pressures and the Equilibrium Constant• Consider this system at equilibrium at

5000C.

2OH

2Cl

O4

HClp2

22

2

24

c

g2gg2g2

22

2

PP

PPK and

OHCl

OHClK

O+HCl 4OH 2+Cl 2

93

Partial Pressures and the Equilibrium Constant

Kc PHCl

RT 4 PO2

RT PCl2

RT 2 PH2O

RT 2 PHCl 4

PO2

PCl2 2

PH2O 2 1

RT 5

1RT 4

Kc Kp1

RT so for this reaction

Kc = Kp(RT) -1 or Kp = Kc(RT)1

Must use R 0.0821L atm

mol K

94

Relationship Between Kp and Kc

• From the previous slide we can see that the relationship between Kp and Kc is:

reactants) gaseous of moles of (#-products) gaseous of moles of (#=n

RTKKor RTKK npc

ncp

95


• Example 17-15: Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?

g2gg Br + NO 2 NOBr 2

96


2 NOBr g 2 NO g + Br2 g

Initial [] x atm 0 0

Change - 0.34x atm +0.34 x atm +0.17x atm

Equilibrium x - 0.34 x atm 0.34 xatm 0.17x atm

97


ted.undissocia 66% isit

d,dissociate 34% isNOBr Because

atm. 0.21= thusatm, 1.17=atm 0.25

atm 0.17+atm 0.34atm 34.0-=atm 25.0

PPPP2BrNONOBrTot

xx

xxxx

98


32

2

2NOBr

Br2

NOp

Br

NO

NOBr

NOBr

103.914.0

036.0071.0

P

PPK

atm 036.0atm 21.017.017.0P

atm 071.0atm 21.034.034.0P

atm 14.0atm 21.066.0P

66.034.0-P

2

2

x

x

xxx

99


• The numerical value of Kc for this reaction can be determined from the relationship of Kp and Kc.

K = K RT or K = K RT n = 1

K

p cn

c pn

c

9 3 10 0 0821 298 38 103 1 4. . .

100


• Example 17-16: Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,

(a) How many moles of I2 remain unreacted at equilibrium?

You do it!

gg2g2 HI 2I H

101


H 2 g I2 g 2 HI g

Initial 0.33M 0.33M 0

Change - x M - x M +2x M

Equilibrium 0.33- x M 0.33 - x M 2x M

Kc =HI 2

H2 I2 49 2x 2

0.33- x 2 7.0 2x

0.33- x

9x = 2.3; x = 0.256M

[H2] [I2] (0.33 x)M 0.074M

HI 2x M 0.51M

? mol I2 3.0 L 0.074 molL 0.21 mol

102


(b) What are the equilibrium partial pressures of H2, I2 and HI?

You do it!

103


atm 30K 723 0.0821 51.0RTP

atm 4.4K 723 0.0821 0.074RTPP

K molatm L

Lmol

HI

K molatm L

Lmol

IH 22

M

M

104


(c) What is the total pressure in the reaction vessel?

You do it!

105


atm 39=atm 304.44.4PPP=P HIIHTot 22

106

Heterogeneous Equlibria• Heterogeneous equilibria have more than one phase

present.– For example, a gas and a solid or a liquid and a gas.

• How does the equilibrium constant differ for heterogeneous equilibria?– Pure solids and liquids have activities of unity.– Solvents in very dilute solutions have activities that are essentially unity.– The Kc and Kp for the reaction shown above are:

2COp2c P=K ][CO=K

CaCO3 s CaO s CO2 g at 500oC

107

Heterogeneous Equlibria

2SO

p

2

32

c P

1 K

SO

SOH=K

You do it!

?K and K of forms theareWhat

solvent. theis OH

C)25at (SOHOHSO

:reaction For this

pc

2

o

aq322g2

108


• What are Kc and Kp for this reaction?

Kc = Ca2 F 2 Kp is undefined

C)25at (F 2CaCaF o-1aq

2aqs2

You do it!You do it!

109


• What are Kc and Kp for this reaction?

Kc =H2 4

H2O 4 Kp PH2 4

PH2O 4

C)500at (H 4OFe OH 4Fe 3 og2s43g2s

110

Relationship Between Gorxn

and the Equilibrium Constant Go

rxn is the standard free energy change. Go

rxn is defined for the complete conversion of all reactants to all products.

G is the free energy change at nonstandard conditions• For example, concentrations other than 1 M or pressures other

than 1 atm.

G is related to Go by the following relationship.

quotientreaction =Q

re temperatuabsolute = T

constant gas universal =R

Q log RT 303.2G=G

or lnQ RTG=Go

o

111


and the Equilibrium Constant

• At equilibrium, G=0 and Q=Kc. • Then we can derive this relationship:

K log RT 2.303 -=G

orK ln RT -=G

: torearrangeswhich

K log RT 303.2G0

orK ln RTG0

0

0

0

0

112


and the Equilibrium Constant• For the following generalized reaction, the

thermodynamic equilibrium constant is defined as follows:

D ofactivity theis C ofactivity theis

B ofactivity theis A ofactivity theis

where=K

dD + cC bB +aA

DC

BA

bB

aA

dD

cC

aa

aa

aa

aa

113



• The relationships among Gorxn, K, and the

spontaneity of a reaction are:

Gorxn K Spontaneity at unit concentration

< 0 > 1 Forward reaction spontaneous

= 0 = 1 System at equilibrium

> 0 < 1 Reverse reaction spontaneous

114

Relationship Between Gorxn and

the Equilibrium Constant

115


and the Equilibrium Constant• Example 17-17: Calculate the equilibrium

constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.

• Note: this is a gas phase reaction.

g2g42 NO 2ON

116



eous.nonspontan isreaction This

1078.4G

78.4G

kJ 82.97kJ 30.512G

GG2G

G Calculate .1

NO 2ON

rxn molJ3o

rxn

rxn molkJo

rxn

orxn

oON f

oNO f

orxn

orxn

g2g42

g42g2

117



2. Calculate K from G rxno RT ln K p

ln Kp G rxn

o

RT

4.78 103 Jmol

- 8.314 Jmol K 298 K 1.93

Kp e 1.93 0.145 PNO2 2

PN2O4

A very common mistake is

to not convert the G rxno from kJ to J!

118


and the Equilibrium Constant• Kp for the reverse reaction at 25oC can be

calculated easily, it is the reciprocal of the above reaction.

Kp'

1

Kp

1

0.1456.90

PN2O4 PNO2 2

2 NO2(g) N2O4(g)

G rxno 4.78 kJ/mol

119


and the Equilibrium Constant• Example 17-18: At 25oC and 1.00 atmosphere,

Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate Go

rxn at 25oC.

You do it.

g2gg2 O NO 2 NO 2

120



rxn molkJ

rxn molJ4o

rxn

molJo

rxn

13-K mol

Jorxn

porxn

6.70 1006.7G

)47.28)(2480(G

104.3ln )K 298)(314.8(G

Kln RT G

121



• The relationship for K at conditions other than thermodynamic standard state conditions is derived from this equation.

Q log RT 2.303 GG

or

lnQ RT GG

o

o

122

Evaluation of Equilibrium Constants at Different Temperatures• From the value of Ho and K at one

temperature, T1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T2.

21

o

T

T

12

12o

T

T

T

1

T

1

R

H

K

Kln

or

TT R

)T(TH

K

Kln

1

2

1

2

123

Evaluation of Equilibrium Constants at Different Temperatures

• Example 17-19: For the reaction in example 17-18, Ho = 114 kJ/mol and Kp = 4.3 x 10-13 at 25oC. Estimate Kp at 250oC.

2(g)(g)2(g)

O2NONO 2

124

795.19K

Kln

K 298K 523314.8

298523)1014.1(

K

Kln

equation Hofft van'apply the

K 523 T andK 298 TLet

1

2

1

2

T

T

K molJ

molJ5

T

T

21


125


Take the antilog of both sides of equation.

KT2

KT1

e19.80 4.0 108

Solve for KT2 & substitute the known value of KT1

KT24.0 108KT1

4.0 108 4.310-13 KT2

1.7 10 4 @ 250oC vs KT14.310-13 @ 25oC

The reaction is more product favored at the higher T.

126

17 Chemical Equilibrium

Documents

1 17 Chemical Equilibrium. 2 Chapter Goals 1.Basic Concepts 2.The Equilibrium Constant 3.Variation of K c with the Form of the Balanced Equation 4.The