CHAPTER 17 Chemical Equilibria 1

7/30/2019 CHAPTER 17 Chemical Equilibria 1

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CHAPTER 17

Chemical Equilibrium


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Chapter Goals

1. Basic Concepts2. The Equilibrium Constant

3. Variation of Kc with the Form of the

Balanced Equation4. The Reaction Quotient

5. Uses of the Equilibrium Constant, Kc

6. Disturbing a System at Equilibrium:

Predictions

7. The Haber Process: A Practical

Application of Equilibrium


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Chapter Goals

8. Disturbing a System at Equilibrium:Calculations

9. Partial Pressures and the Equilibrium

Constant10. Relationship between Kp and Kc

11. Heterogeneous Equilibria

12. Relationship between Gorxn and theEquilibrium Constant

13. Evaluation of Equilibrium Constants at

Different Temperatures


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1. To explain how chemical equilibrium can be achieved when therates of two opposing processes are equal.

2. To derive an equilibrium law from a balanced chemical reaction fora reversible reaction.

3. To explain the significance of equilibrium constant, to relate Kc to Kpand its use in equilibrium calculation.

4. To differentiate between homogeneous equilibria andheterogeneous equilibria.

5. To explain the various factors that can affect chemical equilibria. 6. To explain Arrhenius, BrowstedLowry and Lewis concept of acid

and base and give examples.

7. To identify conjugate acid and base pairs in a Browsted-Lowry acid-base pair reaction.

8. To show how the strengths of Browsted acids and bases follow

periodic trends. 9. To explain the characteristic of an amphoteric substance.

10. Define pH, pOH and water dissociation constant, Kw

11.To learn how to use pH scale to measure strength of acids andbases and to differentiate between strong and weak acids or bases.

12. To explain what is a buffer and how buffers enable control of pHin solutions.

LO


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Basic Concepts

Reversible reactions do not go to completion. Chemical reaction that can occur in either direction

Symbolically, this is represented as:

gggg Dd+CcBb+Aa
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Basic Concepts

Chemical equilibrium exists when twoopposing reactions occursimultaneouslyat the same rate.

A chemical equilibrium is a reversible reactionthat the forward reaction rate is equal to thereverse reaction rate.

Chemical equilibria are dynamic equilibria.

Molecules are continually reacting, eventhough the overall composition of the reactionmixture does not change.


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Basic Concepts

One example of a dynamic equilibrium can

be shown using radioactive 131I as a tracer

in a saturated PbI2 solution.

solution.intogowilliodineeradioactivtheofSome

solution.filter thethenminutes,fewaforStir2

I2PbPbI

solution.PbIsaturatedainPbIsolidPlace1

-

(aq)

2

(aq)

OH

2(s)

2

*

22


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Basic Concepts

This movie depicts a dynamic equilibrium.
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Basic Concepts

Graphically, this is a representation of therates for the forward and reverse reactionsfor this general reaction.

gggg Dd+CcBb+Aa


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Basic Concepts

One of the fundamental ideas of chemical

equilibrium is that equilibrium can be

established from either the forward or

reverse direction.


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Basic Concepts


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Basic Concepts


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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.reversetherepresentswhichDCkRate

rate.forwardtherepresentswhichBAkRate

rr

ff

(g)(g)(g)(g) DCBA


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When system is at equilibrium:

Ratef= Rater

BA

DC

k

k

torearrangeswhich

DCkBAk:givetoiprelationshratefor theSubstitute

r

f

rf


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Because the ratio of two constants is a

constant we can define a new constant as

follows :

k

kK and

KC D

A B

f

r

c

c


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Similarly, for the general reaction:

we can define a constant

reactions.allforvalidisexpressionThis

BA

DC

K

products

reactantsba

dc

c

DdCcBbAa (g)(g)(g)(g)


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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 theproducts, each raised to a power equal to

its stoichiometric coefficient in the

balanced equation, divided by the productof the equilibrium concentrations (in M) of

the reactants, each raised to a power

equal to its stoichiometric coefficient in the

balanced e uation.


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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


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You do it!

HI2I+H 22


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22

2

c

22

IH

HI

K

HI2I+H


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You do it!

OH6+NO4O5+NH4 223


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22


5

2

4

3

6

2

4

c

223

ONH

OHNO=K

OH6+NO4O5+NH4


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Equilibrium constants are dimensionless

because they actually involve a

thermodynamic quantity called activity.

Activities are directly related to molarity


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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 MYou do it!

235 ClPClPCl


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53.0K

028.0

086.0172.0K

PCl

ClPClK

c

c

5

23c


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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. Thesystem was allowed to reach equilibrium

at the new temperature. At equilibrium

0.60 mole of PCl3 was present in thecontainer. Calculate the equilibrium

constant at this temperature.


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001.00Initial

ClPClPCl g2g3g5

M


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MMM

M

0.60+0.60+0.60-Change001.00Initial

ClPClPCl g2g3g5


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MMM

MMM

M

0.600.600.40mEquilibriu

0.60+0.60+0.60-Change001.00Initial

ClPClPCl g2g3g5



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Tanotherat90.0

40.0

60.060.0K


0.60+0.60+0.60-Change001.00Initial

ClPClPCl

'

c

g2g3g5

MMM

MMM

M



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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 NH3was present. Calculate Kc for the reaction.

You do it!


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26.060.070.0

20.0

HN

NHK


0.20+0.30-0.10-Change

00.900.80Initial

NH2H3+N

3

2

3

22

23c

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

MMM

MMM

MM


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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


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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 PClClPCl


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Form of the Balanced Equation

KPCl

PCl Cl

c' 5

3 2


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KPCl

PCl Cl

c' 5

3 2

0028

0172 0 086

19.

. .

.


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K PCl

PCl Cl

KK

or KK

c' 5

3 2

cc' c

'

c

0028

0172 0 08619

1 1 10 53

19

.

. ..

..



Large equilibrium constants indicate that most of

the reactants are converted to products. Small equilibrium constants indicate that only

small amounts of products are formed.


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The Reaction Quotient
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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 thatthe concentrations used in Q are notnecessarily equilibrium values.

ba

dc

BA

DCQ

dD+cCbB+aA

:reactiongeneralFor this


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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.


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fractions.asKandQofthinkthisunderstandhelpTo

extent.greateratorightthetooccursreactionTheKQ

extent.greateratoleftthetooccursreactionTheKQ

m.equilibriuatissystemTheK=Q

:When

c

c

c

c


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Example 17-6: The equilibrium constant for thefollowing reaction is 49 at 450oC. If 0.22 mole ofI2, 0.22 mole of H2, and 0.66 mole of HI were putinto an evacuated 1.00-liter container, would the

system be at equilibrium? If not, what mustoccur to establish equilibrium?

c

c

2

22

2

(g)2(g)2(g)

K


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Uses of the Equilibrium

Constant, Kc Example 17-7: The equilibrium constant, Kc, is

3.00 for the following reaction at a giventemperature. If 1.00 mole of SO2 and 1.00 moleof NO2 are put into an evacuated 2.00 L

container and allowed to reach equilibrium, whatwill be the concentration of each compound atequilibrium?

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

U f th E ilib i


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Constant, Kc

000.5000.500Initial

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

MM

U f th E ilib i


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MxMxMxMx

MM

++--Change

000.5000.500Initial



Constant, Kc

U f th E ilib i


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MxMxMxMx

MxMxMxMx

MM

500.0500.0mEquilibriu

++--Change

000.5000.500Initial



Constant, Kc

U f th E ilib i


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Constant, Kc

equation.thesidesofbothofthecan takeWe

.squareperfectaisequationThis

500.0500.000.3

NOSO

NOSOK


++--Change

000.5000.500Initial

NOSONOSO

22

3c

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

xx

xx

MxMxMxMx

MxMxMxMx

MM

U f th E ilib i


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22

3

22

3c

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

NOSO184.0500.0

NOSO316.0

73.2865.0;73.20.865;1.73-0.865

500.0=1.73

sides.bothofthecan takeWe

square.perfectaisequationThis

500.0500.000.3

NOSO

NOSOK


++--Change

000.5000.500Initial

NOSONOSO

MMx

Mx

xxxx

x

x

xx

xx

MxMxMxMx

MxMxMxMx

MM


Constant, Kc

U f th E ilib i


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Constant, Kc 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 reachequilibrium, what will be the equilibrium

concentration of each substance?

You do it!

(g)2(g)2(g) HI2I+H

U f th E ilib i


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Constant, Kc

MMx

MMx

Mxxxx

x

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

2-1.00mEquilibriu

2-++Change

1.0000Initial

HI2I+H

22

c

2

22

2

c

(g)2(g)2(g)

Di t bi S t t


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Disturbing a System at

Equlibrium: Predictions LeChateliers 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 LeChateliers Principle in Chapter

14.

Some possible stresses to a system atequilibrium are:

1. Changes in concentration of reactants or products.

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

3. Changes in temperature.

Di t bi S t t


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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 theideal gas equation.

ion.concentratitstoalproportiondirectlyis

gasaofpressurepartialtheT,constantatThus[]RT=P

mol/L,unitsthehasV

nBecause

RTV

nP

nRTPV

Di t bi S t t


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1 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.

49IHHI

K

HI2IH

22

2

c

g22

Di t bi S t t


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side.productorrighttheshift towillEquilbrium

reaction.forwardthefavorsThis

.K


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side.reactantorleft,theshift towillEquilbrium

reaction.reversethefavorsThis

K>Q,HsomeremoveweIf

49IH

HIK

HI2IH

c2

22

2

c

g22








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2 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 atconstant temperature:

2

2

42c

g42g2

NO

ON=K

ONNO2



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gas.ofmolesfewerproducesreactionforwardThe

reaction.forwardor theformationproductfavorsThis

.K


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produced.aregasofmolesMore

reaction.reverseor thereactantsthefavorsThis

.K>Qpressure,thedecreaseswhichincreased,isvolumetheIf

NO

ON=K

ONNO2

c

2

2

42c

g42g2





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3 Changing the Temperature

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3 Changing the Reaction Temperature

Consider the following reaction at equilibrium:

reaction.reactantorreversethefavorsThis

products.thestressesemperaturereaction ttheIncreasing

reactionthisofproductaisHeat

kJ198+SO2O+SO2 g3g2g2

reaction?in thisproductorreactantaheatIs

kJ/mol198HSO2OSO2o

rxn3(g)2(g)2(g)

reaction.forwardorreactantsthefavorsThis

reactants.thestressesemperaturereaction ttheDecreasing

reaction.thisofproductaisHeatkJ198+SO2O+SO2 g3g2g2



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Equlibrium: Predictions Introduction of a Catalyst

Catalysts decrease the activation energy of both the forward andreverse 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.



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Example 17-9: Given the reaction below atequilibrium in a closed container at 500oC. Howwould the equilibrium be influenced by thefollowing?

rightNHofionconcentrattheDecreasee.

rightHofionconcentrattheIncreased.

rightvolumethedecreasingbypressuretheIncreasingc.

rightemperaturereaction ttheDecreasingb.

leftemperaturereaction ttheIncreasinga.

procedurereactiononEffectFactor

kJ/mol92HNH2H3N

3

2

o

rxn3(g)2(g)2(g)

effectnocatalystplatinumagIntroducinf.

rightNHofionconcentrattheDecreasee.






kJ/mol92HNH2H3N

3

2

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






kJ/mol92HNH2H3N

2

o

rxn3(g)2(g)2(g)





kJ/mol92HNH2H3No

rxn3(g)2(g)2(g)




kJ/mol92HNH2H3N orxn3(g)2(g)2(g)








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Equlibrium: Predictions Example 17-10: How will an increase in pressure (caused

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

rightPClCl+PClc.

leftOH6+NO4O5+NH4b.

effectnoHI2I+Ha.

mEquilibriuonEffectReaction

g5g2g3

g2g2(g)g3

gg2g2

rightOH2OH2d.

rightPClCl+PClc.

leftOH6+NO4O5+NH4b.

effectnoHI2I+Ha.


g2g2g2

g5g2g3

g2g2(g)g3

gg2g2

leftOH6+NO4O5+NH4b.

effectnoHI2I+Ha.


g2g2(g)g3

gg2g2

effectnoHI2I+Ha.


gg2g2




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Example 17-11: How will an increase in

temperature affect each of the following

reactions?

leftkJ92+HCl2ClHb.

left0HONNO2a.


gg2g2

o

rxn4(g)22(g)

rightkJ25HHI2I+Hc.

leftkJ92+HCl2ClHb.

left0HONNO2a.


gg2g2

gg2g2

o

rxn4(g)22(g)

left0HONNO2a.mEquilibriuonEffectReaction

o

rxn4(g)22(g)

The Haber Process: A Practical


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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 conditionsdid we predict would be most favorable for

the production of ammonia?



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dilemma.thisosolution tsHaber'

res.temperatulowatslowveryarekineticsreactioneHowever the.unfavorabliswhich0


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kinetics!with thehelpsandyieldreactiontheincreasesThis

removed.isNHbecausemequilibriureachesneversystemreactionThe

right.ly toperiodicalNHRemove4right.toNexcessUse3

right.topressurereactionIncrease2

.decreasedisyieldbutrate,increasetoTIncrease1

dilemma.thisosolution tsHaber'

3

3

2



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This diagram illustrates the commercial

system devised for the Haber process.



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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 thefollowing 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 forthis reaction?

ggg CBA



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Equilibrium: Calculations

45.020.0

30.030.0

A

CBK

0.300.300.20s[]'Equil.

CBA

c

ggg

MMM



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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

22.0

10.0

15.015.0

A

CB=Q

0.150.150.10s[]'Equil.

CBA ggg

MMM



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Equilibrium: Calculations Since Q


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00225.075.0

+0.30+0.0225=0.45-0.045

equationquadraticthisSolve

2

2

xx

xxx



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Mx

x

x

0.03and78.02

81.075.0

12

0225.01475.075.0

2a

ac4bb-

2

2



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disturbed.beenhasmequilibriutheafterionsconcentratnewtheareThese

18.015.0CB

07.0)10.0(A

M.0.03isvalueposibleonlyThe

answer.anas0.78-discardcanwe0.10,


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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 newequilibrium concentrations be? Recall that

the original concentrations were: [A]=0.20

M, [B]=0.30 M, and [C]=0.30 M.You do it!



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ions.concentratmequilibriuthe

determinetosexpressionalgebraictheupSet(2)

side.reactantorleftthetoshiftsmequilibriuthethusKQ

90.040.0

60.060.0

A

CB=Q

halvedisvolumeafter theQ,Calculate(1)

0.600.600.40s[]'ousInstantane

CBA

c

ggg

MMM



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018.065.1

toreducesequationthiscompleted,isalgebraAfter the)+40.0(

)-60.0)(-60.0(45.0

A

CBK

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

--+Change

0.600.600.40s[]'initialNew

C+BA

2

c

ggg

xx

x

xx

MxMxMx

MxMxMx

MMM



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MMxCB

MMx

x

x

x

48.0)60.0(

52.0)40.0(Aanswer.possibleonlytheis0.12Thus

answer.anas1.5discardcanwe0.60


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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 COClClCO



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20

10.030.0

60.0

ClCO

COClK

60.010.030.0s[]'Equil.

COClCl+CO

2

2c

g2g2g

MMM



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g y

Equilibrium: CalculationsAn 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 equilibriumis established.

You do it!



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g y


xx

x

MxMxMx

MxMxMx

MMM

M

MMM

50.030.0

60.020

ClCO

COClK

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

+--Change

rightshiftK


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g y


MMx

MMx

MMx

x

X

xx

78.0)60.0(COCl

32.0)50.0(Cl

12.0)30.0(CO

0.67discardcanwethus0.30


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Equilibrium Constant Forgas phase reactions the equilibrium

constants can be expressed in partial pressuresrather 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

Partial Pressures and the


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Equilibrium ConstantConsider this system at equilibrium at

5000C.

2

OH

2

Cl

O

4

HCl

p2

2

2

2

2

4

c

g2gg2g2

22

2

PP

PPKand

OHCl

OHClK

O+HCl4OH2+Cl2

Partial Pressures and the


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Equilibrium Constant

Kmol

atmL0.0821RuseMust

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

reactionfor thissoKK

PP

PPK

1

cp

1-

pc

RT1

pc

4

RT1

5

RT1

2

OH

2

Cl

O

4

HCl

2

RT

P2

RT

P

RT

P4

RT

P

c

22

2

O2H2Cl

2OHCl

Relationship Between K and K


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93

Relationship Between Kp and Kc

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

reactants)gaseousofmolesof(#-products)gaseousofmolesof(#=n

RTKKorRTKK

n

pc

n

cp



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94


Example 17-15: Nitrosyl bromide, NOBr, is34% dissociated by the following reaction

at 25oC, in a vessel in which the total

pressure is 0.25 atmosphere. What is thevalue of Kp?

g2ggBr+NO2NOBr2



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95


atm0.17atm0.34atm34.0-mEquilibriu

atm0.17+atm0.34+atm0.34-Change

00atm[]Initial

Br+NO2NOBr2 g2gg

xxxx

xxx

x



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96


ted.undissocia66%isit

d,dissociate34%isNOBrBecause

atm.0.21=thusatm,1.17=atm0.25

atm0.17+atm0.34atm34.0-=atm25.0

PPPP2BrNONOBrTot

xx

xxxx



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3

2

2

2

NOBr

Br

2

NO

p

Br

NO

NOBr

NOBr

103.914.0

036.0071.0

P

PPK

atm036.0atm21.017.017.0P

atm071.0atm21.034.034.0P

atm14.0atm21.066.0P

66.034.0-P

2

2

x

x

xxx



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98


The numerical value of Kc for this reactioncan be determined from the relationship of

Kp and Kc.

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

K

p c

n

c p

n

c

9 3 10 0 0821 298 38 1031 4

. . .



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99


Example 17-16: Kc is 49 for the followingreaction 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 HI2IH



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100


mol0.210.074L3.0Imol?

51.02HI

074.0)33.0(][I][H256.0=;3.2=9

-0.33

20.7

-0.33

249

IH

HI=K

2-0.33-0.33mEquilibriu

2+--Change

00.330.33Initial

HI2IH

Lmol2

22

2

2

22

2

c

gg2g2

MMx

MMx

Mxx

x

x

x

x

MxMxMx

MxMxMx

MM



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101


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

You do it!



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102


atm30K7230.082151.0RTP

atm4.4K7230.08210.074RTPP

KmolatmL

LmolHI

Kmol

atmLL

molIH 22

M

M



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103


(c) What is the total pressure in the reactionvessel?

You do it!



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104


atm39=atm304.44.4PPP=PHIIHTot

22

Heterogeneous Equlibria


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105

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 heterogeneousequilibria? Pure solids and liquids have activities of unity.

Solvents in very dilute solutions have activities that are essentiallyunity.

The Kc and Kp for the reaction shown above are:

2COp2cP=K][CO=K

C500atCOCaOCaCOo

g2ss3



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106


undefinedisKSO

SOH=K p

2

32c

You do it!

?KandKofformstheareWhat

solvent.theisOH

C)25at(SOHOHSO:reactionFor this

pc

2

o

aq322aq2



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107


What are Kc and Kp for this reaction?

K = Ca F K is undefinedc2 2

p

C)25at(F2CaCaFo-1

aq

2

aqs2

You do it!



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108


What are Kc and Kp for this reaction?

K =H

H O

KP

P

c2

2

p

H

H O

2

2

4

4

4

4

C)500at(H4OFeOH4Fe3o

g2s43g2s

Relationship BetweenGorxn


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109

and the Equilibrium Constant Gorxn is the standard free energy change.

Gorxn is defined for the completeconversion of all reactants toall products.

G is the free energy change at nonstandard conditions

For example, concentrations other than 1 Mor pressures otherthan 1 atm.

G is related to Go by the following relationship.

quotientreaction=Q

retemperatuabsolute=T

constantgasuniversal=RQlogRT303.2G=G

orlnQRTG=G

o

o



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110

and the Equilibrium Constant

At equilibrium, G=0 and Q=Kc.

Then we can derive this relationship:

KlogRT2 303-=G

orKlnRT-=G:torearrangeswhich

KlogRT303.2G0

orKlnRTG0

0

0

0

0



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111


For the following generalized reaction, thethermodynamic equilibrium constantisdefined as follows:

DofactivitytheisCofactivitytheis

BofactivitytheisAofactivitytheis

where=K

dD+cCbB+aA

DC

BA

b

B

a

A

d

D

c

C

aa

aa

aa

aa



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112


The relationships among Gorxn, K, and the

spontaneity of a reaction are:

Gorxn K Spontaneity at unitconcentration

< 0 > 1 Forward reaction spontaneous

= 0 = 1 System at equilibrium

> 0 < 1 Reverse reaction spontaneous

Relationship BetweenGorxn and


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the Equilibrium Constant



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114


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 NO2ON



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115


eousnonspontanisreactionThis

1078.4G

78.4G

kJ82.97kJ30.512G

GG2G

GCalculate.1

NO2ON

rxnmolJ3o

rxn

rxnmolkJo

rxn

o

rxn

o

ONf

o

NOf

o

rxn

o

rxn

g2g42

g42g2


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C


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117


Kp for the reverse reaction at 25oC can becalculated easily, it is the reciprocal of the

above reaction.

2NO

ON

p

'p

2

42

PP90.6

145.01

K1K

kJ/mol78.4G

ONNO2o

rxn

4(g)22(g)

Relationship Between Gorxn

d h E ilib i C


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118

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 Gorxn at 25oC.

You do it.

g2gg2 ONO2NO2


d th E ilib i C t t


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rxnmolkJ

rxnmolJ4o

rxn

molJo

rxn

13-Kmol

Jorxn

p

o

rxn

6.701006.7G

)47.28)(2480(G

104.3ln)K298)(314.8(G

KlnRTG


d th E ilib i C t t


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The relationship for K at conditions otherthan thermodynamic standard state

conditionsis derived from this equation.

QlogRT2.303GG

or

lnQRTGG

o

o

Evaluation of Equilibrium Constants

t Diff t T t


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at Different Temperatures

From the value ofHo and K at onetemperature, T1, we can use the vant Hoffequation to estimate the value of K atanother temperature, T

2

.

21

o

T

T

12

12

o

T

T

T

1

T

1

R

H

K

Kln

or

TTR

)T(TH

K

Kln

1

2

1

2

Evaluation of Equilibrium Constants

t Diff t T t


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at Different Temperatures

Example 17-19: For the reaction inexample 17-18, Ho = 114 kJ/mol and Kp =

4.3 x 10-13 at 25oC. Estimate Kp at 250oC.

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

va ua on o qu r umConstants at Different


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Constants at Different

Temperatures

equationHofftvan'apply the

K523TandK298TLet 21

795.19KKln

K298K523314.8298523)1014.1(

KKln

equationHofftvan'apply the

K523TandK298TLet

1

2

1

2

T

T

KmolJ

molJ5

T

T

21


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Synthesis Question


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125

Sy t es s Quest o

Mars is a reddish colored planet because it hasnumerous iron oxides in its soil. Mars also has

a very thin atmosphere, although it is believed

that quite some time ago its atmosphere wasconsiderably thicker. The thin atmosphere

does not retain heat well, thus at night on Mars

the surface temperatures are 145 K and in thedaytime the temperature rises to 300 K. Does

Mars get redder in the daytime or at night?

Synthesis Question


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y Q

The formation of iron oxides from iron andoxygen is an exothermic process. Thus the

equilibrium that is established on Mars shifts to

the iron oxide (red) side when the planet iscooler - at night. Mars gets redder at night by a

small amount.

Group Question


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p Q

If you are having trouble getting a fire

started in the barbecue grill, a common

response is to blow on the coals until thefire begins to burn better. However, this

has the side effect of dizziness. This is

because you have disturbed an

equilibrium in your body. Whatequilibrium have you affected?

End of Chapter 17


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p

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

Make sure you understand this chapters

concepts!

CHAPTER 17 Chemical Equilibria 1

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Transcript of CHAPTER 17 Chemical Equilibria 1