chapter 17

11

CHAPTER 17CHAPTER 17

Chemical EquilibriumChemical Equilibrium

22

Chapter GoalsChapter Goals1.1. Basic ConceptsBasic Concepts

2.2. The Equilibrium ConstantThe Equilibrium Constant

3.3. Variation of KVariation of Kcc with the Form of the Balanced with the Form of the Balanced

EquationEquation

4.4. The Reaction QuotientThe Reaction Quotient

5.5. Uses of the Equilibrium Constant, KUses of the Equilibrium Constant, Kcc

6.6. Disturbing a System at Equilibrium: PredictionsDisturbing a System at Equilibrium: Predictions

7.7. The Haber Process: A Practical Application of The Haber Process: A Practical Application of EquilibriumEquilibrium

33

Chapter GoalsChapter Goals8.8. Disturbing a System at Equilibrium: CalculationsDisturbing a System at Equilibrium: Calculations

9.9. Partial Pressures and the Equilibrium ConstantPartial Pressures and the Equilibrium Constant

10.10. Relationship between KRelationship between Kpp and K and Kcc

11.11. Heterogeneous EquilibriaHeterogeneous Equilibria

12.12. Relationship between Relationship between GGoorxnrxn and the Equilibrium and the Equilibrium

ConstantConstant

13.13. Evaluation of Equilibrium Constants at Different Evaluation of Equilibrium Constants at Different TemperaturesTemperatures

44

Basic ConceptsBasic Concepts

ReversibleReversible reactionsreactions do not go to completion. do not go to completion. They can occur in either directionThey can occur in either direction Symbolically, this is represented as:Symbolically, this is represented as:

gggg D d + C cB b + A a

55


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

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

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

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

66


One example of a dynamic equilibrium can One example of a dynamic equilibrium can be shown using radioactive be shown using radioactive 131131I as a tracer I as a tracer in a saturated PbIin a saturated PbI22 solution. 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

77


This movie depicts a dynamic equilibrium.This movie depicts a dynamic equilibrium.

88

Basic ConceptsBasic Concepts Graphically, this is a representation of the Graphically, this is a representation of the

rates for the forward and reverse reactions rates for the forward and reverse reactions for this general reaction.for this general reaction.

gggg D d + C cB b + A a

99


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

1010


1111


1212

The Equilibrium ConstantThe Equilibrium Constant

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

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

rate. reverse therepresents which DCkRate

rate. forward therepresents which BAkRate

rr

ff

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

1313


When system is at equilibrium:When system is at equilibrium:

RateRateff = Rate = Raterr

BA

DC

k

k

torearrangeswhich

DCkBAk

:give toiprelationsh rate for the Substitute

r

f

rf

1414


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

kk

K and

KC DA B

f

rc

c

1515


Similarly, for the general reaction:Similarly, for the general reaction:

we can define a constantwe 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)

1616

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

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

1717


Example 17-1: Write equilibrium constant Example 17-1: Write equilibrium constant expressions for the following reactions at expressions for the following reactions at 500500ooC. All reactants and products are C. All reactants and products are gases at 500gases at 500ooC. C.

5

23c

235

PCl

ClPClK

ClPClPCl

1818


You do it!

HI 2 I + H 22

1919


22

2

c

22

IH

HIK

HI 2 I + H

2020


You do it!

OH 6+NO 4O 5 + NH 4 223

2121


52

43

62

4

c

223

ONH

OHNO=K

OH 6+NO 4O 5 + NH 4

2222


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

2323


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

Equil []’sEquil []’s 0.028 0.028 MM 0.172 0.172 MM 0.086 0.086 MM

You do it!You do it!

235 ClPClPCl

2424


53.0K

028.0

086.0172.0K

PCl

ClPClK

c

c

5

23c

2525


Example 17-3: The decomposition of PClExample 17-3: The decomposition of PCl55

was studied at another temperature. One was studied at another temperature. One mole of PClmole of PCl55 was introduced into an was introduced into an

evacuated 1.00 liter container. The evacuated 1.00 liter container. The system was allowed to reach equilibrium system was allowed to reach equilibrium at the new temperature. At equilibrium at the new temperature. At equilibrium 0.60 mole of PCl0.60 mole of PCl33 was present in the was present in the

container. Calculate the equilibrium container. Calculate the equilibrium constant at this temperature.constant at this temperature.

2626


0 0 1.00 Initial

ClPClPCl g2g3g5

M

2727


MMM

M

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl g2g3g5

2828

MMM

MMM

M

0.60 0.60 0.40 mEquilibriu

0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl g2g3g5


2929

Tanother at 90.0

40.0

60.060.0K


0.60+ 0.60+ 0.60- Change

0 0 1.00 Initial

ClPClPCl

'c

g2g3g5

MMM

MMM

M


3030


Example 17-4: At a given temperature Example 17-4: At a given temperature 0.80 mole of N0.80 mole of N22 and 0.90 mole of H and 0.90 mole of H22 were were

placed in an evacuated 1.00-liter placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NHcontainer. At equilibrium 0.20 mole of NH33

was present. Calculate Kwas present. Calculate Kcc for the reaction. for the reaction.


3131


26.060.070.0

20.0

HN

NHK


0.20+ 0.30- 0.10- Change

0 0.90 0.80 Initial

NH 2 H 3 + N

3

2

322

23

c

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

MMM

MMM

MM

3232

Variation of KVariation of Kcc with the with the

Form of the Balanced EquationForm of the Balanced Equation The value of KThe value of Kcc depends upon how the balanced depends upon how the balanced

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

This reaction has a KThis reaction has a Kcc=[PCl=[PCl33][Cl][Cl22]/[PCl]/[PCl55]=0.53]=0.53

235 ClPClPCl

3333


Form of the Balanced EquationForm of the Balanced Equation Example 17-5: Calculate the equilibrium constant Example 17-5: Calculate the equilibrium constant

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

Equil. []’s 0.172 Equil. []’s 0.172 MM 0.086 0.086 M M 0.0280.028 M M

The concentrations are from Example 17-2.The concentrations are from Example 17-2.

523 PClCl PCl

3434


Form of the Balanced EquationForm of the Balanced Equation

KPCl

PCl Clc' 5

3 2

3535



KPCl

PCl Clc' 5

3 2

0 028

0172 0 08619

.. .

.

3636

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.

3737

The Reaction QuotientThe Reaction Quotient

3838


3939

The Reaction QuotientThe Reaction Quotient The mass action expression or reaction quotient The mass action expression or reaction quotient

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

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

ba

dc

BA

DCQ

dD+cC bB+aA

:reaction general For this

4040


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

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

To make this prediction we compare Q To make this prediction we compare Q with Kwith Kcc..

4141


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

4242

The Reaction QuotientThe Reaction Quotient Example 17-6: The equilibrium constant for the Example 17-6: The equilibrium constant for the

following reaction is 49 at 450following reaction is 49 at 450ooC. If 0.22 mole of C. If 0.22 mole of II22, 0.22 mole of H, 0.22 mole of H22, and 0.66 mole of HI were put , and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must system be at equilibrium? If not, what must occur to establish equilibrium?occur to establish equilibrium?

c

c

2

22

2

(g)2(g)2(g)

K<Q

49Kbut 0.9Q

0.922.022.0

66.0

IH

HI=Q

0.66 0.22 0.22

HI 2 I H

Q. calculatecan We

s.[]' mequilibriuy necessarilnot are

problem in thegiven ionsconcentrat The

MMM

4343

Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc

Example 17-7: The equilibrium constant, KExample 17-7: The equilibrium constant, Kcc, is , is 3.00 for the following reaction at a given 3.00 for the following reaction at a given temperature. If 1.00 mole of SOtemperature. If 1.00 mole of SO22 and 1.00 mole and 1.00 mole of NOof NO22 are put into an evacuated 2.00 L are put into an evacuated 2.00 L container and allowed to reach equilibrium, what container and allowed to reach equilibrium, what will be the concentration of each compound at will be the concentration of each compound at equilibrium?equilibrium?

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

4444


0 0 0.500 0.500 Initial

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

MM

4545

MxMxMxMx

MM

+ + - - Change

0 0 0.500 0.500 Initial



4646

MxMxMxMx

MxMxMxMx

MM

500.0 500.0 mEquilibriu

+ + - - Change

0 0 0.500 0.500 Initial



4747


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

4848

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


4949


Example 17-8: The equilibrium constant is Example 17-8: The equilibrium constant is 49 for the following reaction at 45049 for the following reaction at 450ooC. If C. If 1.00 mole of HI is put into an evacuated 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium equilibrium, what will be the equilibrium concentration of each substance?concentration of each substance?

You do it!

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

5050


MMx

MMx

Mxxxxx

x

xx

x

MxMxMx

MxMxMx

M

78.0200.1HI

11.0IH

11.0 ;00.19 ;200.10.7

2-1.00=7.0=K

2-1.00 =49=

IH

HI=K

2-1.00 mEquilibriu

2- + + Change

1.00 0 0 Initial

HI 2 I + H

22

c

2

22

2

c

(g)2(g)2(g)

5151

Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions

LeChatelier’s PrincipleLeChatelier’s Principle - If a change of conditions - If a change of conditions (stress) is applied to a system in equilibrium, the (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.the stress in reaching a new state of equilibrium.

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

Some possible stresses to a system at equilibrium Some possible stresses to a system at equilibrium are:are:

1.1. Changes in concentration of reactants or products.Changes in concentration of reactants or products.

2.2. Changes in pressure or volume (for gaseous reactions)Changes in pressure or volume (for gaseous reactions)

3.3. Changes in temperature.Changes in temperature.

5252


For convenience we may express the amount of For convenience we may express the amount of a gas in terms of its partial pressure rather than a gas in terms of its partial pressure rather than its concentration.its concentration.

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

5353


1 Changes inChanges in Concentration of Reactants and/or Products Concentration of Reactants and/or Products • Also true for changes in pressure for reactions involving gases.Also true for changes in pressure for reactions involving gases. Look at the following system at equilibrium at 450Look at the following system at equilibrium at 450ooC.C.

49

IH

HIK

HI 2IH

22

2

c

g22

5454



side.product or right theshift to willEquilbrium

reaction. forward thefavors This

.K<Q added, is H some If

49IH

HIK

HI 2IH

c2

22

2

c

g22

5555

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



5656


2 Changes in VolumeChanges in Volume • (and pressure for reactions involving gases)(and pressure for reactions involving gases) Predict what will happen if the volume of this system at equilibrium Predict what will happen if the volume of this system at equilibrium

is changed by changing the pressure is changed by changing the pressure at constant temperatureat constant temperature::

22

42c

g42g2

NO

ON=K

ONNO 2

5757


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

5858

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


5959


6060


3 Changing the TemperatureChanging the Temperature

6161


3 Changing the Reaction TemperatureChanging the Reaction Temperature Consider the following reaction at equilibrium:Consider the following reaction at equilibrium:

reaction.reactant or reverse thefavors This

products. thestresses emperaturereaction t theIncreasing

reaction thisofproduct a isHeat

kJ 198+SO 2O+SO 2 g3g2g2

reaction? in thisproduct or reactant aheat Is

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

reaction. forwardor reactants thefavors This

reactants. thestresses emperaturereaction t theDecreasing

reaction. thisofproduct a isHeat

kJ 198+SO 2O+SO 2 g3g2g2

6262


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

reverse reaction equally.reverse reaction equally.

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

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

6363


Example 17-9: Given the reaction below at Example 17-9: Given the reaction below at equilibrium in a closed container at 500equilibrium in a closed container at 500ooC. How C. How would the equilibrium be influenced by the would the equilibrium be influenced by the following?following?

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)

effect no catalyst platinum a gIntroducin f.

right NH ofion concentrat theDecrease e.







3

2

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







2

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





kJ/mol 92H NH 2 H 3 N orxn3(g)2(g)2(g)










6464


Example 17-10: How will an increase in pressure (caused by Example 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the decreasing the volume) affect the equilibrium in each of the following reactions?following reactions?

right PClCl+PCl c.

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

effect no HI 2I +H a.

mEquilibriuon Effect Reaction

g5g2g3

g2g2(g)g3

gg2g2

right OH 2 OH 2 d.

right PClCl+PCl c.




g2g2g2

g5g2g3

g2g2(g)g3

gg2g2




g2g2(g)g3

gg2g2



gg2g2 mEquilibriuon Effect Reaction

6565


Example 17-11: How will an increase in Example 17-11: How will an increase in temperature affect each of the following temperature affect each of the following reactions?reactions?

left kJ 92 +HCl 2 ClH b.

left 0H ON NO 2 a.


gg2g2

orxn4(g)22(g)

right kJ 25H HI 2I +H c.

left kJ 92 +HCl 2 ClH b.

left 0H ON NO 2 a.


gg2g2

gg2g2

orxn4(g)22(g)

left 0H ON NO 2 a.

mEquilibriuon Effect Reaction orxn4(g)22(g)

6666

The Haber Process: A Practical The Haber Process: A Practical Application of EquilibriumApplication of Equilibrium

The Haber process is used for the The Haber process is used for the commercial production of ammonia.commercial production of ammonia. This is an enormous industrial process in the This is an enormous industrial process in the

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

production.production. Look at Example 17-9. What conditions Look at Example 17-9. What conditions

did we predict would be most favorable for did we predict would be most favorable for the production of ammonia?the production of ammonia?

6767


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

6868


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

6969


This diagram illustrates the commercial This diagram illustrates the commercial system devised for the Haber process.system devised for the Haber process.

7070

Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations

To help with the calculations, we must determine To help with the calculations, we must determine the direction that the equilibrium will shift by the direction that the equilibrium will shift by comparing Q with Kcomparing Q with Kcc..

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

ggg C B A

7171


45.0

20.0

30.030.0

A

CBK

0.30 0.30 0.20 s[]' Equil.

C B A

c

ggg

MMM

7272


If the volume of the reaction vessel were If the volume of the reaction vessel were suddenly doubled while the temperature suddenly doubled while the temperature remained constant, what would be the new remained constant, what would be the new equilibrium concentrations?equilibrium concentrations?

1 Calculate Q, Calculate Q, after the volume has been doubledafter the volume has been doubled

22.0

10.0

15.015.0

A

CB=Q

0.15 0.15 0.10 s[]' Equil.

C B A ggg

MMM

7373


Since Q<KSince Q<Kcc the reaction will shift to the right to the reaction will shift to the right to re-establish the equilibrium.re-establish the equilibrium.

2 Use algebra to represent the new Use algebra to represent the new concentrations.concentrations.

MMM 0.15 0.15 0.10 s[]' initial New

C + B A ggg

7474


Since Q<KSince Q<Kcc the reaction will shift to the right to the reaction will shift to the right to

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

concentrations.concentrations.

MxMxMx

MMM

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A ggg

7575





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

7676





x

xx

MxMxMx

MxMxMx

MMM

10.0

15.015.045.0

A

CB=K

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

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A

c

ggg

7777


00225.075.0

+0.30+0.0225=0.45-0.045

equation quadratic thisSolve

2

2

xx

xxx

7878


Mx

x

x

0.03 and 78.02

81.075.0

12

0225.01475.075.0

2a

ac4bb-

2

2

7979


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

8080


Example 17-13: Refer to example 17-12. Example 17-13: Refer to example 17-12. If the initial volume of the reaction vessel If the initial volume of the reaction vessel were halved, while the temperature were halved, while the temperature remains constant, what will the new remains constant, what will the new equilibrium concentrations be? Recall that equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 the original concentrations were: [A]=0.20 MM, [B]=0.30 , [B]=0.30 MM, and [C]=0.30 , and [C]=0.30 MM..


8181


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

8282


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

8383


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

8484


Example 17-14: A 2.00 liter vessel in which the Example 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 following system is in equilibrium contains 1.20 moles of COClmoles of COCl22, 0.60 moles of CO and 0.20 mole , 0.60 moles of CO and 0.20 mole

of Clof Cl22. Calculate the equilibrium constant. . Calculate the equilibrium constant.


g2g2g COCl Cl CO

8585


20

10.030.0

60.0

ClCO

COClK

60.0 10.0 30.0 s[]' Equil.

COClCl+CO

2

2c

g2g2g

MMM

8686


An additional 0.80 mole of ClAn additional 0.80 mole of Cl22 is added to is added to

the vessel at the same temperature. the vessel at the same temperature. Calculate the molar concentrations of CO, Calculate the molar concentrations of CO, ClCl22, and COCl, and COCl22 when the new equilibrium when the new equilibrium

is established.is established.


8787


xx

x

MxMxMx

MxMxMx

MMM

M

MMM

50.030.0

60.020

ClCO

COClK

)+60.0( )-50.0( )-30.0( Equil. New

+ - - Change

rightshift K<Q 60.0 50.0 30.0 Initial New

40.0+ Add (Stress)

60.0 10.0 30.0 Equil. Orig.

COCl Cl + CO

2

2c

c

g2g2g

8888


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

8989

Partial Pressures and the Partial Pressures and the Equilibrium ConstantEquilibrium Constant

For For gas phase reactionsgas phase reactions the equilibrium the equilibrium constants can be expressed in partial pressures constants can be expressed in partial pressures rather than concentrations.rather than concentrations.

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

We can see this by looking at the ideal gas law.We can see this by looking at the ideal gas law. PV = nRTPV = nRT P = nRT/V P = nRT/V n/V = n/V = MM P= P= MMRT and RT and MM = P/RT = P/RT

9090


Consider this system at equilibrium at Consider this system at equilibrium at 50050000C.C.

2OH

2Cl

O4

HClp2

22

2

24

c

g2gg2g2

22

2

PP

PPK and

OHCl

OHClK

O+HCl 4OH 2+Cl 2

9191


K mol

atm L0.0821 R useMust

(RT)K=Kor (RT)K=K

reaction for this so KK

PP

PPK

1cp

1-pc

RT1

pc

4RT1

5RT1

2OH

2Cl

O4

HCl

2

RT

P2

RT

P

RT

P4

RTP

c

22

2

O2H2Cl

2OHCl

9292

Relationship Between KRelationship Between Kpp and K and Kcc

From the previous slide we can see that From the previous slide we can see that the relationship between Kthe relationship between Kpp and K and Kc c is: is:

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

RTKKor RTKK npc

ncp

9393


Example 17-15: Nitrosyl bromide, NOBr, is Example 17-15: Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction 34% dissociated by the following reaction at 25at 25ooC, in a vessel in which the C, in a vessel in which the totaltotal pressure is 0.25 atmosphere. What is the pressure is 0.25 atmosphere. What is the value of Kvalue of Kpp??

g2gg Br + NO 2 NOBr 2

9494


atm 0.17 atm0.34 atm34.0- mEquilibriu

atm 0.17+ atm 0.34+ atm 0.34- Change

0 0 atm [] Initial

Br + NO 2 NOBr 2 g2gg

xxxx

xxx

x

9595


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

9696


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

9797


The numerical value of KThe numerical value of Kcc for this reaction for this reaction

can be determined from the relationship of can be determined from the relationship of KKpp and K and Kcc..

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. . .

9898


Example 17-16: KExample 17-16: Kcc is 49 for the following reaction is 49 for the following reaction

at 450at 450ooC. If 1.0 mole of HC. If 1.0 mole of H22 and 1.0 mole of I and 1.0 mole of I22 are are

allowed to reach equilibrium in a 3.0-liter vessel,allowed to reach equilibrium in a 3.0-liter vessel,

(a) How many moles of I(a) How many moles of I22 remain unreacted at remain unreacted at

equilibrium?equilibrium?


gg2g2 HI 2I H

9999


mol 0.21 0.074L 3.0I mol ?

51.0 2HI

074.0)33.0(][I][H

256.0= ;3.2=9

-0.33

20.7

-0.33

249

IH

HI=K

2 -0.33 -0.33 mEquilibriu

2+ - - Change

0 0.33 0.33 Initial

HI 2 I H

Lmol

2

22

2

2

22

2

c

gg2g2

MMx

MMx

Mxx

x

x

x

x

MxMxMx

MxMxMx

MM

100100


(b) What are the equilibrium partial (b) What are the equilibrium partial pressures of Hpressures of H22, I, I22 and HI? and HI?


101101


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

102102


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


103103


atm 39=atm 304.44.4PPP=P HIIHTot 22

104104

Heterogeneous EqulibriaHeterogeneous Equlibria Heterogeneous equilibria have more than one Heterogeneous equilibria have more than one

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

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

2COp2c P=K ][CO=K

C500at CO CaO CaCO og2ss3

105105

Heterogeneous EqulibriaHeterogeneous Equlibria

undefined is K SO

SOH=K p

2

32c

You do it!

?K and K of forms theareWhat

solvent. theis OH

C)25at (SOHOHSO

:reaction For this

pc

2

oaq322aq2

106106


What are KWhat are Kcc and K and Kpp for this reaction? for this reaction?

K = Ca F K is undefinedc2 2

p

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

2aqs2


107107


What are KWhat are Kcc and K and Kpp for this reaction? for this reaction?

K =H

H O K

P

Pc

2

2p

H

H O

2

2

4

4

4

4

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

108108

Relationship BetweenRelationship Between GGoorxnrxn

and the Equilibrium Constant and the Equilibrium Constant GGoo

rxnrxn is the standard free energy change. is the standard free energy change. GGoo

rxnrxn is defined for the is defined for the completecomplete conversion of all reactants to conversion of all reactants to

all products.all products.

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

than 1 atm.than 1 atm.

G is related to G is related to GGoo by the following relationship. by the following relationship.

quotientreaction =Q

re temperatuabsolute = T

constant gas universal =R

Q log RT 303.2G=G

or lnQ RTG=Go

o

109109


and the Equilibrium Constantand the Equilibrium Constant

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

110110


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

thermodynamic equilibrium constantthermodynamic equilibrium constant is is defined as follows: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

111111



The relationships among The relationships among GGoorxnrxn, K, and the , K, and the

spontaneity of a reaction are:spontaneity of a reaction are:

GGoorxnrxn KK Spontaneity at Spontaneity at unitunit concentration concentration

< 0< 0 > 1> 1 Forward reaction spontaneousForward reaction spontaneous

= 0= 0 = 1= 1 System at equilibriumSystem at equilibrium

> 0> 0 < 1< 1 Reverse reaction spontaneousReverse reaction spontaneous

112112

Relationship BetweenRelationship Between GGoorxnrxn and and

the Equilibrium Constantthe Equilibrium Constant

113113


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

constant, Kconstant, Kpp, for the following reaction at 25, for the following reaction at 25ooC C

from thermodynamic data in Appendix K.from thermodynamic data in Appendix K.

Note: this is a gas phase reaction.Note: this is a gas phase reaction.

g2g42 NO 2ON

114114



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

115115



J! tokJ from G econvert th not to

is mistakecommon A very

P

P145.0K

93.1K 2988.314-

1078.4

RT

GK ln

Kln RTG fromK Calculate 2.

orxn

ON

2NO93.1

p

K molJ

molJ3o

rxnp

porxn

42

2

e

116116


and the Equilibrium Constantand the Equilibrium Constant KKpp for the reverse reaction at 25 for the reverse reaction at 25ooC can be C can be

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

2NO

ON

p

'p

2

42

P

P90.6

145.0

1

K

1K

kJ/mol 78.4G

ONNO 2orxn

4(g)22(g)

117117

Relationship Between Relationship Between GGoorxnrxn

and the Equilibrium Constantand the Equilibrium Constant Example 17-18: At 25Example 17-18: At 25ooC and 1.00 atmosphere, C and 1.00 atmosphere,

KKpp = 4.3 x 10 = 4.3 x 10-13-13 for the decomposition of NO for the decomposition of NO22. .

Calculate Calculate GGoorxnrxn at 25 at 25ooC.C.

You do it.You do it.

g2gg2 O NO 2 NO 2

118118



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

119119



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

Q log RT 2.303 GG

or

lnQ RT GG

o

o

120120

Evaluation of Equilibrium Constants Evaluation of Equilibrium Constants at Different Temperaturesat Different Temperatures

From the value of From the value of HHoo and K at one and K at one temperature, Ttemperature, T11, we can use the van’t Hoff , we can use the van’t Hoff equation to estimate the value of K at equation to estimate the value of K at another temperature, Tanother temperature, T22..

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

121121


Example 17-19: For the reaction in Example 17-19: For the reaction in example 17-18, example 17-18, HHoo = 114 kJ/mol and K = 114 kJ/mol and Kpp = =

4.3 x 104.3 x 10-13-13 at 25 at 25ooC. Estimate KC. Estimate Kpp at 250 at 250ooC. C.

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

122122

Evaluation of Equilibrium Evaluation of Equilibrium Constants at Different Constants at Different

TemperaturesTemperatures

equation Hofft van'apply the

K 523 T andK 298 TLet 21

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

123123


T.higher at the favoredproduct more isreaction The

C25 @ 103.4K vsC250 @ 107.1K

103.4100.4K100.4K

K of eknown valu thesubstitute & Kfor Solve

100.4K

K

equation. of sidesboth of antilog theTake

o13-T

o4T

13-8T

8T

TT

880.19

T

T

12

12

12

1

2

e

124124

Synthesis QuestionSynthesis Question

Mars is a reddish colored planet because it has Mars is a reddish colored planet because it has numerous iron oxides in its soil. Mars also has a numerous iron oxides in its soil. Mars also has a very thin atmosphere, although it is believed that very thin atmosphere, although it is believed that quite some time ago its atmosphere was quite some time ago its atmosphere was considerably thicker. The thin atmosphere does considerably thicker. The thin atmosphere does not retain heat well, thus at night on Mars the not retain heat well, thus at night on Mars the surface temperatures are 145 K and in the daytime surface temperatures are 145 K and in the daytime the temperature rises to 300 K. Does Mars get the temperature rises to 300 K. Does Mars get redder in the daytime or at night?redder in the daytime or at night?

125125

Synthesis QuestionSynthesis Question

The formation of iron oxides from iron and The formation of iron oxides from iron and oxygen is an exothermic process. Thus the oxygen is an exothermic process. Thus the equilibrium that is established on Mars shifts to equilibrium that is established on Mars shifts to the iron oxide (red) side when the planet is the iron oxide (red) side when the planet is cooler - at night. Mars gets redder at night by a cooler - at night. Mars gets redder at night by a small amount.small amount.

126126

Group QuestionGroup Question

If you are having trouble getting a fire If you are having trouble getting a fire started in the barbecue grill, a common started in the barbecue grill, a common response is to blow on the coals until the response is to blow on the coals until the fire begins to burn better. However, this fire begins to burn better. However, this has the side effect of dizziness. This is has the side effect of dizziness. This is because you have disturbed an because you have disturbed an equilibrium in your body. What equilibrium in your body. What equilibrium have you affected?equilibrium have you affected?

127127

End of Chapter 17End of Chapter 17

This chapter is the key to the This chapter is the key to the understanding of Chapters 18, 19, & 20.understanding of Chapters 18, 19, & 20.

Make sure you understand this chapter’s Make sure you understand this chapter’s concepts!concepts!

chapter 17

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equilibrium constant

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