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
Basic ConceptsBasic Concepts
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
Basic ConceptsBasic Concepts
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
Basic ConceptsBasic Concepts
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
Basic ConceptsBasic Concepts
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
Basic ConceptsBasic Concepts
1111
Basic ConceptsBasic Concepts
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
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
You do it!
HI 2 I + H 22
1919
The Equilibrium ConstantThe Equilibrium Constant
22
2
c
22
IH
HIK
HI 2 I + H
2020
The Equilibrium ConstantThe Equilibrium Constant
You do it!
OH 6+NO 4O 5 + NH 4 223
2121
The Equilibrium ConstantThe Equilibrium Constant
52
43
62
4
c
223
ONH
OHNO=K
OH 6+NO 4O 5 + NH 4
2222
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
53.0K
028.0
086.0172.0K
PCl
ClPClK
c
c
5
23c
2525
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
0 0 1.00 Initial
ClPClPCl g2g3g5
M
2727
The Equilibrium ConstantThe Equilibrium Constant
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
The Equilibrium ConstantThe Equilibrium Constant
2929
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
The Equilibrium ConstantThe Equilibrium Constant
3030
The Equilibrium ConstantThe Equilibrium Constant
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.
You do it!You do it!
3131
The Equilibrium ConstantThe Equilibrium Constant
26.060.070.0
20.0
HN
NHK
0.20 0.60 0.70 mEquilibriu
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
Variation of KVariation of Kcc with the with the
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
Variation of KVariation of Kcc with the with the
Form of the Balanced EquationForm of the Balanced Equation
KPCl
PCl Clc' 5
3 2
3535
Variation of KVariation of Kcc with the with the
Form of the Balanced EquationForm of the Balanced Equation
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
.. .
.
. .
Variation of KVariation of Kcc with the with the
Form of the Balanced EquationForm of the Balanced Equation
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
The Reaction QuotientThe Reaction Quotient
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
The Reaction QuotientThe Reaction Quotient
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
The Reaction QuotientThe Reaction Quotient
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
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
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
NO SO NO SO (g)3(g)2(g)2(g)
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
4646
MxMxMxMx
MxMxMxMx
MM
500.0 500.0 mEquilibriu
+ + - - Change
0 0 0.500 0.500 Initial
NO SO NO SO (g)3(g)2(g)2(g)
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
4747
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
equation. thesidesofboth of thecan take We
.squareperfect a isequation This
500.0500.000.3
NOSO
NOSOK
500.0 500.0 mEquilibriu
+ + - - 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
500.0 500.0 mEquilibriu
+ + - - Change
0 0 0.500 0.500 Initial
NO SO NO SO
MMx
Mx
xxxx
x
x
xx
xx
MxMxMxMx
MxMxMxMx
MM
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
4949
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
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
Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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.
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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.
5656
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: 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
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
5959
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
6060
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
3 Changing the TemperatureChanging the Temperature
6161
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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.
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)
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
2
orxn3(g)2(g)2(g)
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 orxn3(g)2(g)2(g)
right emperaturereaction t theDecreasing b.
left emperaturereaction t theIncreasing a.
procedurereaction on Effect Factor
kJ/mol 92H NH 2 H 3 N orxn3(g)2(g)2(g)
left emperaturereaction t theIncreasing a.
procedurereaction on Effect Factor
kJ/mol 92H NH 2 H 3 N orxn3(g)2(g)2(g)
procedurereaction on Effect Factor
kJ/mol 92H NH 2 H 3 N orxn3(g)2(g)2(g)
6464
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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.
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
left OH 6+NO 4 O 5+NH 4 b.
effect no HI 2I +H a.
mEquilibriuon Effect Reaction
g2g2(g)g3
gg2g2
effect no HI 2I +H a.
mEquilibriuon Effect Reaction
gg2g2 mEquilibriuon Effect Reaction
6565
Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: Predictions
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.
mEquilibriuon Effect Reaction
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.
mEquilibriuon Effect Reaction
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
The Haber Process: A Practical The Haber Process: A Practical Application of EquilibriumApplication 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
6868
The Haber Process: A Practical The Haber Process: A Practical Application of EquilibriumApplication 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
6969
The Haber Process: A Practical The Haber Process: A Practical Application of EquilibriumApplication of Equilibrium
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
45.0
20.0
30.030.0
A
CBK
0.30 0.30 0.20 s[]' Equil.
C B A
c
ggg
MMM
7272
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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.
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
00225.075.0
+0.30+0.0225=0.45-0.045
equation quadratic thisSolve
2
2
xx
xxx
7878
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
Mx
x
x
0.03 and 78.02
81.075.0
12
0225.01475.075.0
2a
ac4bb-
2
2
7979
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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..
You do it!You do it!
8181
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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.
You do it!You do it!
g2g2g COCl Cl CO
8585
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
20
10.030.0
60.0
ClCO
COClK
60.0 10.0 30.0 s[]' Equil.
COClCl+CO
2
2c
g2g2g
MMM
8686
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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.
You do it!You do it!
8787
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations
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
Partial Pressures and the Partial Pressures and the Equilibrium ConstantEquilibrium Constant
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
Partial Pressures and the Partial Pressures and the Equilibrium ConstantEquilibrium Constant
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
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
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?
You do it!You do it!
gg2g2 HI 2I H
9999
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
(b) What are the equilibrium partial (b) What are the equilibrium partial pressures of Hpressures of H22, I, I22 and HI? and HI?
You do it!You do it!
101101
Relationship Between KRelationship Between Kpp and K and Kcc
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
Relationship Between KRelationship Between Kpp and K and Kcc
(c) What is the total pressure in the reaction (c) What is the total pressure in the reaction vessel?vessel?
You do it!You do it!
103103
Relationship Between KRelationship Between Kpp and K and Kcc
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
Heterogeneous EqulibriaHeterogeneous Equlibria
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
You do it!You do it!
107107
Heterogeneous EqulibriaHeterogeneous Equlibria
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
Relationship BetweenRelationship Between GGoorxnrxn
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
Relationship BetweenRelationship Between GGoorxnrxn
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
Relationship BetweenRelationship Between GGoorxnrxn
and the Equilibrium Constantand the Equilibrium Constant
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
Relationship BetweenRelationship Between GGoorxnrxn
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
Relationship BetweenRelationship Between GGoorxnrxn
and the Equilibrium Constantand the Equilibrium Constant
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
Relationship BetweenRelationship Between GGoorxnrxn
and the Equilibrium Constantand the Equilibrium Constant
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
Relationship BetweenRelationship Between GGoorxnrxn
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
Relationship Between Relationship Between GGoorxnrxn
and the Equilibrium Constantand the Equilibrium Constant
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
Relationship Between Relationship Between GGoorxnrxn
and the Equilibrium Constantand the Equilibrium Constant
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
Evaluation of Equilibrium Constants Evaluation of Equilibrium Constants at Different Temperaturesat Different Temperatures
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
Evaluation of Equilibrium Constants Evaluation of Equilibrium Constants at Different Temperaturesat Different Temperatures
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!