1 chapter 15 chemical equilibrium. 2 basic concepts reversible reactions do not go to completion...

1

CHAPTER 15

Chemical Equilibrium

2

Basic Concepts

• Reversible reactions do not go to completion– occur in either direction

– represented as:

a A + b B c C + d Dg g g g

3

Basic Concepts• Chemical equilibrium

– reversible reaction that the forward reaction rate is equal to the reverse reaction rate

– dynamic equilibria

1

2

Place solid PbI in sat'd PbI sol'n

PbI Pb + 2 I

Stir a few minutes, then filter.

Discover that some radioactive, I is in solution.

2*

2

2H O 2+ -

-*

2

,

4

Basic Concepts• Graphical representation of the rates for the forward and

reverse reactions for this general reaction

a A + b B c C + d Dg g g g

5

Basic Concepts

• One of the fundamental ideas of chemical equilibrium:

• Equilibrium can be established from either direction - forward or reverse.

6

The Equilibrium Constant• For a simple one-step mechanism reversible reaction

such as:

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

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

R k A B forward rate

R k C D reverse ratef f

r r

7

The Equilibrium Constant

• When system is at equilibrium:

Rf = Rr

thus k A B k C D

which rearranges to

kk

C DA B

f r

f

r

8


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

kk

K and

KC DA B

f

rc

c

9


• Similarly, for the general reaction:

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

• we can define a constant

KC D

A B

expression valid for all reactions

c

c d

a b reactants products

10

The Equilibrium Constant• Kc is the equilibrium constant .

• Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

11


• Write equilibrium constant expressions for the following reactions at 500oC.

PCl PCl Cl

KPCl Cl

PCl

5 3 2

c3 2

5

12


OH 6+NO 4O 5 + NH 4 223

13


4 NH + 5 O 4 NO + 6 H O

K =NO H O

NH O

3 2 2

c2

3 2

4 6

4 5

14

Partial Pressures and the Equilibrium Constant

• Gas phase reactions can have equilibrium constants expressed in partial pressures rather than concentrations.– For gases the pressure is proportional to concentration

– PV = nRT

– P = nRT/V = []RT and [] = P/RT

15


• Gas phase reactions can have equilibrium constants expressed in partial pressures rather than concentrations.– For gases the pressure is proportional to concentration– PV=nRT– P=nRT/V=[]RT and []=P/RT

• Consider this system in equilibrium at 5000C.

2 Cl + 2 H O 4 HCl + O

KHCl O

Cl H O and K

P P

P P

2 g 2 g g 2 g

c

42

22

22 p

HCl4

O

Cl

2

H O

22

2 2

16


KP P

P P

K K so for this reaction

K = K (RT) or K = K (RT)

c

PRT

PRT

PRT

PRT

HCl4

O

Cl

2

H O

2

1RT

5

1RT

4

c p1

RT

c p-1

p c1

HCl O2

Cl2 H2O

2

2 2

4

2 2

17

Relationship Between Kp and Kc

• Relationship between Kp and Kc is:

K K RT or K K RT

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

p cn

c pn

18

The Equilibrium Constant• Equilibrium constants are dimensionless because

they actually involve a thermodynamic quantity called activity.– Activities are directly related to molarity

19

The Equilibrium Constant• 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.

PCl5 PCl3 + Cl2

Equil []’s 0.028 M 0.172 M 0.086 M

20


KPCl Cl

PClc3 2

5

0172 0 0860 028

0 53

. ..

.

21

The Equilibrium Constant• The decomposition of PCl5 was studied at another

temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.

22


PCl PCl Cl

Initial 1.00 0 0

Change - 0.60 + 0.60 + 0.60

Equilibrium 0.40 0.60 0.60

K at another T

5 g 3 g 2 g

c'

M

M M M

M M M

0 60 0 600 40

0 90. .

..

23


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

24


N + 3 H 2 NH

Initial 0.80 0.90 0

Change - 0.10 - 0.30 + 0.20

Equilibrium 0.70 0.60 0.20

KNH

N H

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

c3

2 2

M M

M M M

M M M2

3

2

3

0 20

0 70 0 600 26

.

. ..

25

Variation of Kc with the Form of the Balanced Equation

• Value of Kc depends upon how the balanced equation is written.

PCl5 PCl3 + Cl2

and

Kc= [PCl3][Cl2] = 0.53

[PCl5]

26


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

27


PCl3 + Cl2 PCl5

Equil. []’s 0.172 M 0.086 M 0.028 M

KPCl

PCl Clc' 5

3 2

28


KPCl

PCl Cl

KK

or K K

c' 5

3 2

cc' c

'

c

0 0280172 0 086

19

1 1 10 53 19

.. .

.

. .

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

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

29

Heterogeneous Equlibria• Heterogeneous equilibria have two or more phases

– pure solids and liquids have activities of unity

– solvents in very dilute solutions have activities that are essentially unity

CaCO3(s) CaO(s) + CO2(g) (at 500oC)

2COp2c P=K ][CO=K

30

Heterogeneous Equlibria

SO2(g) + H2O(l) H2SO3(aq) (at 25oC)

H2O(l) is the solvent

K =H SO

SO K =c

2 3

2p

1PSO2

31


CaF2(s) Ca2+(aq) + 2 F-

(aq) (at 25oC)

K = Ca F K is undefinedc2 2

p

32


3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) (at 500oC)

K =H

H O K

P

Pc

2

2p

H

H O

2

2

4

4

4

4

33

Uses of the Equilibrium Constant, Kp

g2gg Br + NO 2 NOBr 2

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

34

Solving for the Equilibrium Constant, Kp

2 NOBr 2 NO + Br

Initial [] atm 0 0

Change - 0.34 atm + 0.34 atm + 0.17 atm

Equilibrium - . atm 0.34 atm 0.17 atm

g g 2 g

X

X X X

X X X X0 34

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

35


P P P P

atm = - . atm 0.34 atm + 0.17 atm

0.25 atm = 1.17 atm = 0.21 atm

since NOBr is 34% dissociated, it is 66% undissociated

Tot NOBr NO Br2

0 25 0 34. X X X X

X X

36


P - . atm atm

P . atm atm

P . atm atm

KP P

P

NOBr

NO

Br

pNO Br

NOBr

2

2

X X X

X

X

0 34 0 66 0 66 0 21 014

0 34 0 34 0 21 0 071

017 017 0 21 0 036

0 071 0 036

0149 3 10

2

2

2

23

. . . .

. . .

. . .

. .

..

37


• The numerical value of Kc for this reaction is

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. . .

38

Problems using the Equilibrium Constant, Kp

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

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

39


H I 2 HI

Initial 0.33 0.33 0

Change - - +

Equilibrium 0.33 - 0.33 -

K =HI

H I 0.33- 0.33-

= . ; = .

[H ] [I ]

HI

mol I 3.0 L 0.07 0.21 mol

2 g 2 g g

c

2

2 2

2 2

2mol

L

M M

X M XM XM

X M X M XM

X

X

XX

X X M

X M M

XM M

2

2

492

7 02

9 2 3 0 26

0 33 0 07

2 0 52

2

2 .

( . ) .

.

?

40


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

41


P P []RT 0.07 0.0821 723 K 4 atm

P []RT 0.0821 723 K 31 atm

H Imol

LL atmmol K

HImol

LL atmmol K

2 2

0 52.

42


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

43


P = P P P atm = 39 atmTot H I HI2 2 4 4 31

44

Uses of the Equilibrium Constant, Kc

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

SO NO SO NO2(g) 2(g) 3(g) (g)

45


SO NO SO NO

Initial 0.500 0.500 0 0

Change - - + +

Equilibrium X X

KSO NO

SO NO X X

equation is a perfect square, take of both sides

1.73 = 0.865-1.73

SO NO

SO

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

c3

2 2

3

M M

X M X M X M X M

M M X M X M

X X

XX

X X X

X M

X M M

0 500 0 500

3 000 500 0 500

0 5002 73 0 865

0 316

0 500 0184

. .

.. .

.; ; . .

.

. . 2 2NO

46


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

H + I 2 HI

You do it!

2(g) 2(g) (g)

47


H + I 2 HI

Initial 0 0 1.00

Change + + - 2

Equilibrium 1.00 - 2

K =HI

H I= 49 =

1.00 - 2

K = 7.0 =1.00 - 2

H I

HI

2(g) 2(g) (g)

c2 2

c

2 2

M

X M X M X M

X M X M X M

XX X

XX

X X X X M

X M M

X M M

2 2

7 0 100 2 9 100 011

011

100 2 0 78

. . ; . ; .

.

. .

48

The Reaction Quotient

• Q - Mass action expression or reaction quotient– same form as Kc

– concentrations are not necessarily equilibrium values

49


• Q - Mass action expression or reaction quotient– same form as Kc

– concentrations are not necessarily equilibrium values

aA + bB cC + dD

QC D

A B

c d

a b

50


• Compare Q with Kc

– predict direction reaction will occur to attain equilibrium

fractions as K and Q ofthink

extentgreater a toright the tooccursreaction KQ

extentgreater a toleft the tooccursreaction KQ

mequilibriuat is system K=QWhen

c

c

c

c

51

The Reaction Quotient• The equilibrium constant for the following reaction is 49 at

450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

52

The Reaction Quotient• The equilibrium constant for the following reaction is 49 at

450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

H I HI

We calculate Q 0.22 0.22 0.66 not necessarily equilibrium []'s

Q =HI

H I

Q but K

Q < K

2(g) 2(g) (g)

2 2

c

c

2

0 660 22 0 22

9 0

9 0 49

2 2

M M M

.. .

.

.

53

Factors That Affect Equlibria• LeChatelier’s Principle - If a change of conditions

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

• Some stresses are changes in:– concentration

– pressure

– temperature

54

Factors That Affect Equlibria1 Changes in Concentration (and Pressure for reactions

involving gases)

• Look at the following system at equilibrium at 450oC (Kc=49)

H I HI

KHI

H I

2 2 g

c2 2

2

492

55


involving gases)


H I HI

KHI

H I

Add some H

Q < K favors forward rxn. (to the right)

2 2 g

c2 2

2

c

2

492

56


involving gases)


H I HI

KHI

H I

Add some H

Q < K favors forward rxn. (to the right)

Remove some H

Q > K favors reverse rxn. (to the left)

2 2 g

c2 2

2

c

2

c

2

492

57

Factors That Affect Equlibria2 Changes in Volume (and Pressure for reactions involving

gases)• Change the volume by changing the pressure at constant

temperature on the following system at equilibrium:

2 NO N O

K =N O

NO

2 g 2 4 g

c2 4

2

2

58




2 NO N O

K =N O

NO

If the volume is decreased (pressure increased)

Q < K forward rxn. (produces fewer moles gas) favored

2 g 2 4 g

c2 4

2

c

2

59




2 NO N O

K =N O

NO

If the volume is decreased (pressure increased)

Q < K forward rxn. (produces fewer moles gas) favored

If the volume is increased (pressure decreased)

Q > K reverse rxn. (produces more moles gas) favored

2 g 2 4 g

c2 4

2

c

c

2

60

Factors That Affect Equlibria

3 Changing the Temperature• Consider the following reaction at equilibrium

2 SO + O 2 SO + 198 kJ2 g 2 g 3 g

61



2 SO + O 2 SO +198 kJ

Increasing the temperature

favors rxn. that consumes heat (reverse rxn. favored)

2 g 2 g 3 g

62



2 SO + O 2 SO +198 kJ

Increasing the temperature

favors rxn. that consumes heat (reverse rxn. favored)

Decreasing the temperature

favors rxn. that produces heat (forward rxn. favored)

2 g 2 g 3 g

63

Factors That Affect Equlibria• Introduction of a Catalyst• Catalysts decrease the activation energy of both the forward

and reverse reaction equally.• Does not affect the position of equilibrium.

64

Factors That Affect Equlibria• Given the reaction below at equilibrium in a closed

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

system thefrom NH some Removing f.

system theinto H more Forcing e.

catalyst platinum some gIntroducin d.

volume thedecreasingby pressure theIncreasing c.

re temperatu theDecreasing b.

re temperatu theIncreasing a.

kJ 92NH 2H 3+N

3

2

g3g2g2

65

Factors That Affect Equlibria• Given the reaction below at equilibrium in a closed

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

N + 3 H 2 NH kJ

a. Increasing the temperature left

b. Decreasing the temperature right

c. Increasing the pressure by decreasing the volume right

d. Introducing some platinum catalyst no effect

e. Forcing more H into the system right

f. Removing some NH from the system right

2 g 2 g 3 g

2

3

92

66

Factors That Affect Equlibria• How will an increase in pressure (caused by decreasing the

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

OH 2 OH 2 d.

PClCl+PCl c.

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

HI 2I +H a.

g2g2g2

g5g2g3

g2g2(g)g3

gg2g2

67

Factors That Affect Equlibria• How will an increase in pressure (caused by decreasing the

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

a. H + I 2 HI no effect

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

c. PCl + Cl PCl right

d. 2 H O 2 H O right

2 g 2 g g

3 g 2(g) g 2 g

3 g 2 g 5 g

2 g 2 g 2 g

68


• How will an increase in temperature affect each of the following reactions?

kJ 92 +HCl 2 ClH b.

heat +ON NO 2 a.

gg2g2

g42g2

69


• How will an increase in temperature affect each of the following reactions?

a. 2 NO N O + heat left

b. H Cl 2 HCl + 92 kJ left

c. H + I 2 HI right

2 g 2 4 g

2 g 2 g g

2 g 2 g g

70

The Haber Process• Commercial production of ammonia

N H 2 NH H kJ

N from liquid air H from coal gas

run rxn. @ T = 450 C & P = 200 to 1000 atm

G < 0 favorable H < 0 favorable S < 0 unfavorable

But: Rxn. is very slow at low T's.

1 Increase T to increase rate, but decreases yield.

2 Increase P to right.

3 Use excess N to right.

4 Remove NH periodically to right.

System never reaches equilibrium.

2 g 2 gFe & metal oxides

3 go

2 g 2 g

o

2

3

3 92 22

.

72

Application of a Stress to a System at Equilibrium

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

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

A B C

Equil. []'s 0.20 0.30 0.30

KB C

A

g g g

c

M M M

0 30 0 300 20

0 45. .

..

73


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

• Calculate Q, after the volume has been doubled

A B Cg g g

74




A B C

Equil. []'s 0.10 0.15 0.15

g g g

M M M

75




A B C

Equil. []'s 0.10 0.15 0.15

Q =B C

A

g g g

M M M

015 015010

0 22. .

..

76


• Since Q<Kc the reaction will shift to the right to re-establish the equilibrium.

2 Use algebra to represent the new concentrations.

A B + C

New initial []'s 0.10 0.15 0.15

g g g

M M M

77




A B + C

New initial []'s 0.10 0.15 0.15

Change - + +

g g g

M M M

X M X M X M

78




A B + C

New initial []'s 0.10 0.15 0.15

Change - + +

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

g g g

M M M

X M X M X M

X M X M X M

79




A B + C

New initial []'s 0.10 0.15 0.15

Change - + +

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

K =B C

A

g g g

c

M M M

X M X M X M

X M X M X M

X XX

0 45015 015

010.

. ..

80


00225.075.0

+0.30+0.0225=0.45-0.045

equation quadratic thisSolve

2

2

XX

XXX

81


MX

X

X

0.03 & 78.02

81.075.0

12

0225.01475.075.0

2a

ac4bb-

2

2

82


MMX

MMX

X

18.015.0CB

07.0)10.0(A

0.78- discard 0.10,<<0 Since

1 chapter 15 chemical equilibrium. 2 basic concepts reversible reactions do not go to completion...

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