Chemistry Ch09 Chemical Equilibria

8/9/2019 Chemistry Ch09 Chemical Equilibria

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99ChemicalChemical

equilibriaequilibria


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9.1 Chemical equilibrium9.1 Chemical equilibrium

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

Rates of the forward and reverse

reactions are equal There is no net change in the overall

composition of the reaction mixture

Dynamic equilibrium Reactants substances on the left

Products substances on the right


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9.2 The equilibrium constant,9.2 The equilibrium constant, K,K,

and the reaction quotient,and the reaction quotient, QQ

N2O4(g) 2NO2(g)


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For a given OVERALL system

composition equilibrium concentrations

are independent of direction of approach




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aA + bB cC + dD

The following holds when equilibrium

is established:

Equilibrium constant expression

Kc equilibrium constant

Kcdependent on temperature, always

specify temperature when Kcreported

? A ? A

? A ? A bea

e

de

ce

c

c

B

c

A

c

D

c

C

K

!

UU

UU


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aA + bB cC + dD

The following holds when equilibrium

is established:

Equilibrium constant expression

Kc equilibrium constant

Kcdependent on temperature, always

specify temperature when Kcreported

? A ? A? A ? Aba

dc

cBA

DCK !


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aA + bB cC + dD

Qc reaction quotient

Expression for systems not necessarily

at equilibrium

Kccan have only one positive value at

a specific temperature

Qccan have any positive value

? A ? A? A ? Aba

dc

cBA

DCQ !


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Qc= Kcat equilibrium

Qc> Kcsystem reacts to use up

products and generate more reactants Qc< Kcsystem reacts to use up

reactants and generate more products

c d

p a b

p p

p pK

p p

p p

U U

U U


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The relationship between K and Kc

ng= (number of moles ofgaseous products)

(number of moles ofgaseous reactants)

Important to use correct units in this

equation Equilibrium concentrations in mol m-3

Equilibrium pressures in Pa

gn

cRT

KK

(

U

13 Lmol1000mmol ( v cc


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Manipulating equilibrium constant

expressions

When the direction of an equation isreversed, the new equilibrium constant is

the reciprocal of the original

cc KK

1'

!

? A

? A? A235

ClCl

Cl!c

? A? A? Ac

' 3 2

5

PCl Cl

PCl!

PCl3 + Cl2 PCl5

PCl5 PCl3 + Cl2


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expressions

When the coefficients in an equation aremultiplied by a factor, the equilibrium

constant is raised to a power equal to

that factor.? A

? A? A235

ClPCl

PCl!cPCl3 + Cl2 PCl5

? A? A ? A

cK

2" 5

2 23 2

PCl

PCl Cl!2PCl3 + 2Cl2 2PCl5

2'' cc !


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expressions

When chemical equilibria are added,their equilibrium constants are multiplied.? A

? A ? A222

22

1ON

ON!cK

? A

? A ? AcK

42

2 2 32 2

NO

N O O!

? A? A ? A

c

42

3 2 42 2

NO

N O

!

2N2 + O2 2N2O

2N2O + 3O2 4NO2

2N2 + 4O2 4NO2

? A

? A ? A

? A

? A ? A

? A

? A ? A4222

42

32

22

42

22

2

22

ON

NO

OON

NO

ON

ON

!v 321 ccc !v


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The magnitude of the equilibrium

constant

Product concentrations are in thenumerator ofKc

The size ofKcgives a measure of how

far the reaction proceeds towardscompletion when equilibrium is reached


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The magnitude of the equilibrium

constant

reactant product




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Equilibrium constant expressions for

heterogeneous systems

Homogeneous reaction, all reactants andproducts are in the same phase

Heterogeneous reaction, more than one

phase exists in reaction mixture

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)Kc= [H2O(g)][CO2(g)]

Do not include the concentrations of pure

solids or pure liquids


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Equilibrium constant

expressions for

heterogeneous systems

For any pure liquid or solid at

constant temperature, the

ratio of amount of substance

to volume of substance isconstant




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9.3 Equilibrium and Gibbs9.3 Equilibrium and Gibbs

free energyfree energy

Free energy diagrams

Phase changes


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Free energy diagrams

Chemical reactions

12

9.3 Equilibrium and Gibbs9.3 Equilibrium and Gibbs

free energyfree energy

4

34


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9.3 Equilibrium and Gibbs9.3 Equilibrium and Gibbsfree energyfree energy

The relationship between G and K

is the gas constant, 8.3141 J K1 mol1 is the temperature in kelvin

lnQ is the natural logarithm of the

reaction quotient

Gaseous reactions: Q is calculated using

partial pressures expressed in Pa

Reactions in solution: Q is calculated from

molar concentrations

QRTGG ln(!( U


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QRTGG ln(!( U



The reaction 2NO2(g) N2O4(g) has G

= 5.40kJ mol-1 at 298 K. In a reaction mixture, the partial

pressures of NO2 and N2O4 are 0.25 105 Pa and

0.60 x 105 Pa, respectively. In which direction must

this reaction proceed to reach equilibrium?

2

2

42

!

U

U

p

p

p

p

Q

NO

ON

p

NO

NO

p

pG G RT

p

p

2 4

2

2l

UU

U

( ! (


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G5

53 -1 1 -125

5

0.60 10

1 105.40 10 J mol 8.314J K mol 298K ln

0.25 10

1 10

( ! v v

v 12

molJ100.2 v!(G

K298!T

11molKJ314.8 !R

131molJ1040.5molkJ40.5 U v!!(G

Pa1060.0 5ON42

v!p Pa1025.0 5NO2

v!p

Pa101 5v!Up

Gis positive, forward reaction is nonspontaneous

so the reverse reaction occurs


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If the system is at equilibrium

G = 0

Q = K

G RT K0 lU! (

QRTGG l(!( U

G RTlU( !


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2SO2(g) + O2(g) 2SO3(g)G= -1.40 x 102 kJ mol-1 for this reaction at 25 C.

What is the value ofKp at this temperature?

G RT lnKU( !

RT

GKp

U(!l

p

p

K

K

5 1

-1 1

1.40 10 J molln

8.314J K mol 298K

ln 56.5

v!

!

24

5.56

103 v!

!

p

p

K

eK


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Thermodynamic data collected at 25 C

may be used to calculate the values ofequilibrium constants at temperatures

other than 25 C

Values ofHand S do not change

much with temperature


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9.4 How systems at equilibrium9.4 How systems at equilibrium

respond to changerespond to change

Le Chteliers principle

If an outside influence upsets an

equilibrium, the system undergoes achange in a direction that counteracts

the disturbing influence and, if possible,

returns the system to equilibrium

Better to compare equilibrium constant, K,

and reaction quotient, Q, when examining

the effect of perturbation to a chemical

process at equilibrium


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Adding or removing a product or

reactant

When not a pure solid or liquid removalor addition of a reactant or product

instantaneously alters the concentration

of that species in the reaction mixture

The value ofQ changes so that Q K,

the system is no longer at equilibrium


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Adding or removing a product or

reactant

Q = K equilibrium

Q < K shift towards products

Q > K shift towards reactants

[Cu(H2O)4]2+(aq) + 4Cl-(aq) [CuCl4]

2-(aq) + 4H2O(l)

Q

- !

- -

24

42

2 4

CuCl

Cu OH Cl


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Changing the pressure in gaseous

reactions

Two ways of changing the total pressure

Changing the volume of the system

Adding an inert gas

Consider the equilibrium

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

3HN

2NH

22

3

!

UU

U

p

p

p

p

p

p

Qp


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reactions

Changing the volume of the system

3

HN

2

NH

22

3

!

UU

U

p

p

p

p

p

p

Qp

? A? A? A

cQ !

23

32 2

NH

N H

V

nc!

c

n

VQ

nn

V V

!

v

3

22

2

NH

2

3

HN

3

c

nQ V

n n

! vv

3

2 2

2NH 2

3N H


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reactions

Adding an inert gas

Increases total pressure of system

Does not alter the position of equilibrium

Add helium to N2/H2/NH3 equilibrium mixture

Does not react with products or reactants Qp is not changed

3

HN

2

NH

22

3

!

UU

U

p

p

p

p

p

p

Qp


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Changing the temperature of a

reaction mixture

Value of the equilibrium constant, K, can

only be changed by altering the

temperature

The vant Hoff equation states:

The slope of the plot of lnKversus Thas

the same sign as H

2dlnd

RTH

TK U(!


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9.4 How systems at equilibrium9.4 How systems at equilibriumrespond to changerespond to change

Changing the temperature of a reaction

mixture

H is positive

Endothermic

Q < K

Products are favoured

H is negative

Exothermic

Q > K

Reactants are favoured


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Changing the temperature of a

reaction mixture

Can use the vant Hoff equation to

calculate the Kat a specifiedtemperature if the value ofKat

another temperature is known

T THK KR T T2 1 1 2

1 1ln ln U ( !


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Addition of a catalyst

Catalysts affect the rates of chemical

reactions without being used up

Addition of a catalyst may help bring a

system to chemical equilibrium more

rapidly

Addition of a catalyst does not affect the

position of equilibrium


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9.5 Equilibrium calculations9.5 Equilibrium calculations

Calculating Kc from equilibrium

concentrations

Measure the concentrations of reactantsand products after equilibrium reached

Substitute these equilibrium values into

the equilibrium constant expression to

calculate Kc


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At a certain temperature, a mixture of H2 and

I2 was prepared by placing 0.020 mol of H2and 0.020 mol I2 into a 2.00 litre flask. After a

period of time, the equilibrium:H2(g) + I2(g) 2HI(g)

was established. The purple colour of the

I2 vapour was used to monitor the reaction,

and it was determined that, at equilibrium, theI2 concentration had dropped to 0.020 mol L

1.

What is the value of Kc for this reaction at this

temperature?


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? A? A? A22

2

IH

HI!cK

= +equilibrium

concentration

initial

concentration

change in

concentration

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

Initial concentration (mol L-1) 0.100 0.100 0.000

Change in concentration (mol L-1) -0.080 -0.080 +2(0.080)

Equilibrium concentration (mol L-1) 0.020 0.020 0.160

c

c

K

K

20.160

0.020 0.020

64

!

!


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The concentration table a summary

1. Only equilibrium concentrations can be

substituted into the equilibrium constant

expression

2. Initial concentrations should be in mol L1.

The initial concentrations are those present

in the reaction mixture when it is prepared3. Changes in concentrations always occur in

the same ratio as the coefficients in the

balanced equation


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The concentration table a summary

4. In constructing the Change in

concentration row reactant concentrations

should all change in the same direction

and have the same sign.

Row product concentrations should all

change in the opposite direction and havethe opposite sign.


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Calculating equilibrium concentrations

from initial concentrations

Involves the use of initial concentrationsand Kcto calculate equilibrium

concentrations

Require a little applied algebra

Concentration tables can be very helpful


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The water-gas shift reaction:

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

has Kc= 4.06 at 500 C.

If 0.100 mol of CO and 0.100 mol of H2O(g)

are placed in a 1.00-litre reaction vessel at

this temperature, what are the concentrationsof the reactants and products when the

system reaches equilibrium?


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? A? A? A? A

06.4OHCO

HCO

2

22 !!cK

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

Initial concentration (mol L-1) 0.100 0.100 0.0 0.0

Change in concentration (mol L-1) -x -x +x +x

Equilibrium concentration (mol L-1) 0.100-x 0.100-x x x

06.4100.0100.0

!

!xx

xxKc

06.4

100.0 2

2

! x

x

01.206.4

100.0!!

xx

xx

xx

01.2201.0

100.001.2

!

!

201.001.2 ! xx201.001.3 !x0668.0!x


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? A? A? A? A

c

c

Q

Q

2 2

2

2

2

CO H

CO H O

0.0668

4.10.033

!

! !

? A M033.00668.0100.0100.0CO !!! x? A x2H O 0.100 0.100 0.0668 0.033M! ! !? A M0668.0CO2 !! x

? A x2H 0.0668M! !

Rounding Kc to 2 significant figures gives 4.1, so the

calculated concentrations are correct.


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Chemistry Ch09 Chemical Equilibria