Chapter 16-Chemical Equilibrium

Copyright © 1999 by Harcourt Brace & Company

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CHEMICAL

EQUILIBRIUM

Chapter 16

Properties of an Equilibrium

Equilibrium systems are

• DYNAMIC (in constant motion)

• REVERSIBLE

• can be approached from either direction

2

Copyright (c) 1999 by Harcourt Brace & Company

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• REVERSIBLE


3

Pink to blue

Co(H

2

O)

6

Cl

2

---> Co(H

2

O)

4

Cl

2

+ 2 H

2

O


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• REVERSIBLE


4

Pink to blue

Co(H

2

O)

6

Cl

2

---> Co(H

2

O)

4

Cl

2

+ 2 H

2

O

Blue to pink

Co(H

2

O)

4

Cl

2

+ 2 H

2

O ---> Co(H

2

O)

6

Cl

2


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Chemical Equilibrium

Fe 3+ + SCN FeSCN 2+

+

Fe(H

2

O)

6

3+ + SCN Fe(SCN)(H

2

O)

5

3+

+ H

2

O

5


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Chemical Equilibrium

Fe 3+ + SCN FeSCN 2+

6

• After a period of time, the concentrations of reactants and products are constant.

• The forward and reverse reactions continue after equilibrium is attained.


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Examples of Chemical

Equilibria

Phase changes such as

H

2

O(s) H

2

O(liq)

7


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Examples of

Chemical

Equilibria

Formation of stalactites and stalagmites

CaCO

3

(s) + H

2

O(liq) + CO

2

(g)

Ca 2+ (aq) + 2 HCO

3

(aq)


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8

Chemical

Equilibria

CaCO

3

(s) + H

2

O(liq) + CO

2

(g)

Ca 2+ (aq) + 2 HCO

3

(aq)

At a given T and P of CO

2

, [Ca 2+ ] and

[HCO

3

] can be found from the

EQUILIBRIUM CONSTANT .


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9

THE EQUILIBRIUM CONSTANT

For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T)

10


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THE EQUILIBRIUM CONSTANT

For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T) conc. of products

[C]c [D]d

K =

[A]a [B]b conc. of reactants equilibrium constant

If K is known, then we can predict concs. of products or reactants.


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11

Determining K

2 NOCl(g) 2 NO(g) + Cl

2

(g)

Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO.

Calculate K.

Solution

Set of a table of concentrations

[NOCl] [NO]

2.00

0

[Cl

2

]

0 Before

Change

Equilibrium 0.66

12


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Determining K

2 NOCl(g) 2 NO(g) + Cl

2

(g)

Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO.

Calculate K.

Solution

Set of a table of concentrations

Before

Change

Equilibrium

[NOCl] [NO]

2.00

0

-0.66

1.34

+0.66

0.66

[Cl

0

2

+0.33

0.33

]

13


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2 NOCl(g)

Determining K

Before

Change

Equilibrium

2 NO(g) + Cl

2

(g)

[NOCl] [NO]

2.00

-0.66

1.34

0

+0.66

0.66

K



[NO]

2

[Cl

2

[NOCl]

2

]

[Cl

2

]

0

+0.33

0.33

14


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2 NOCl(g)

Determining K

Before

Change

Equilibrium

2 NO(g) + Cl

2

(g)

[NOCl] [NO]

2.00

-0.66

1.34

0

+0.66

0.66

[Cl

2

]

0

+0.33

0.33

K



[NO]

2

[Cl

2

[NOCl]

2

]

K



[NO]

2

[Cl

2

[NOCl]

2

]

=

(0.66)

2

(0.33)

= 0.080

(1.34)

2

15


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Writing and Manipulating K

Expressions

Solids and liquids

NEVER appear in equilibrium expressions.

S(s) + O

2

(g)

SO

2

(g)

16


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Expressions

Solids and liquids

NEVER appear in equilibrium expressions.

S(s) + O

2

(g)

SO

2

(g)

K



[SO

2

]

[O

2

]


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17


Expressions

Solids and liquids NEVER appear in equilibrium expressions.

NH

3

(aq) + H

2

O(liq)

NH

4

+ (aq) + OH (aq)

18


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Expressions

Solids and liquids NEVER appear in equilibrium expressions.

NH

3

(aq) + H

2

O(liq)

NH

4

+ (aq) + OH (aq)

K



[NH

4

+

][OH

-

]

[NH

3

]


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19


Expressions

Adding equations for reactions

S(s) + O

2

(g) SO

2

(g) K

1

= [SO

2

] / [O

2

]

SO

2

(g) + 1/2 O

2

(g) SO

3

(g)

K

2

= [SO

3

] / [SO

2

][O

2

] 1/2

NET EQUATION

S(s) + 3/2 O

2

(g) SO

3

(g)

20


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Expressions

Adding equations for reactions

S(s) + O

2

(g) SO

SO

2

(g) + 1/2 O

2

(g)

2

(g) K

1

SO

3

(g)

= [SO

2

] / [O

2

]

K

2

= [SO

3

] / [SO

2

][O

2

] 1/2

NET EQUATION

S(s) + 3/2 O

2

(g) SO

3

(g)

K net



[SO

3

[O

2

]

]

3/2

= K

1

• K

2

21


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Expressions

Changing coefficients

S(s) + 3/2 O

2

(g) SO

3

(g) K



[SO

3

[O

2

]

]

3/2

22

2 S(s) + 3 O

2

(g) 2 SO

3

(g)


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Expressions


S(s) + 3/2 O

2

(g) SO

3

(g) K



[SO

3

[O

2

]

]

3/2

23

2 S(s) + 3 O

2

(g) 2 SO

3

(g)

K new



[SO

3

[O

2

]

]

3

2


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Expressions


S(s) + 3/2 O

2

(g) SO

3

(g) K



[SO

3

[O

2

]

]

3/2

24

2 S(s) + 3 O

2

(g) 2 SO

3

(g)

K new



[SO

3

[O

2

]

]

3

2

K new



[SO

3

[O

2

]

]

3

2

= (K old

)

2


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Expressions

Changing direction

S(s) + O

2

(g) SO

2

(g) K



[SO

2

]

[O

2

]

SO

2

(g) S(s) + O

2

(g)

25


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Expressions


S(s) + O

2

(g) SO

2

(g) K



[SO

2

]

[O

2

]

SO

2

(g) S(s) + O

2

(g)

K new



[O

2

]

[SO

2

]

26


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Expressions


S(s) + O

2

(g) SO

2

(g) K



[SO

2

]

[O

2

]

SO

2

(g) S(s) + O

2

(g)

K new



[O

2

]

[SO

2

]

27

K new



[O

2

]

[SO

2

]

=

1

K old


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Expressions

Concentration Units

We have been writing K in terms of mol/L.

These are designated by

K c

But with gases, P = (n/V)•RT = conc • RT

P is proportional to concentration, so we can write K in terms of P. These are designated by

K p

.

K c and K p may or may not be the same.


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28

The Meaning of K

1.

Can tell if a reaction is productfavored or reactant-favored.

For N

2

(g) + 3 H

2

(g) 2 NH

3

(g)

29


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1.


For N

2

(g) + 3 H

2

(g) 2 NH

3

(g)

K c

=

[NH

3

[N

2

]

2

][H

2

]

3

= 3.5 x 10

8

30


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1.


For N

2

(g) + 3 H

2

(g) 2 NH

3

(g)

K c

=

[NH

3

[N

2

]

2

][H

2

]

3

= 3.5 x 10

8

Conc. of products is much greater than that of reactants at equilibrium.

31


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1.


For N

2

(g) + 3 H

2

(g) 2 NH

3

(g)

K c

=

[NH

3

[N

2

]

2

][H

2

]

3

= 3.5 x 10

8

Conc. of products is much greater than that of reactants at equilibrium.

The reaction is strongly productfavored

.


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32

33


For AgCl(s)

Ag + (aq) + Cl (aq)

K c

= [Ag + ] [Cl ] = 1.8 x 10 -5

Conc. of products is much less than that of reactants at equilibrium.

The reaction is strongly reactant-favored

.

Ag + (aq) + Cl (aq)

AgCl(s) is product-favored.


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2. Can tell if a reaction is at equilibrium.

If not, which way it moves to approach equilibrium.

H

H—C—C—C—C—H

H n-butane

H H

H H

H

H iso-butane

H H H

H—C—C—C—H

H

H C H

H

H

K =

[iso]

[n]

= 2.5

34


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H

H—C—C—C—C—H

H n-butane

H H

H H

H

H iso-butane

H H H

H—C—C—C—H

H

H C H

H

H

K =

[iso]

[n]

= 2.5

If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?

Which way does the reaction “shift” to approach equilibrium?

See Screen 16.9.


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35


In general, all reacting chemical systems are characterized by their REACTION

QUOTIENT, Q .

product concentrations

Q = reactant concentrations

If Q = K, then system is at equilibrium.

36


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QUOTIENT, Q .




If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?

Q = conc. of iso

= conc. of n

0.35

0.15

= 2.3

Q (2.3) < K (2.5).


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37



QUOTIENT, Q .




Q = conc. of iso

= conc. of n

0.35

0.15

= 2.3

Q (2.33) < K (2.5).

Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must

____________.


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38

Typical Calculations

PROBLEM: Place 1.00 mol each of H

2 a 1.00 L flask. Calc. equilibrium concentrations.

and I

2 in

H

2

(g) + I

2

(g) ¸ 2 HI(g)

[HI]

2

K c

=

[H

2

][I

2

]

= 55.3

39


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3

Step 1. Set up table to define

EQUILIBRIUM concentrations.

Initial

[H

2

]

1.00

[I

2

]

1.00

[HI]

0

Change

Equilib

40


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3

Step 1. Set up table to define

EQUILIBRIUM concentrations.

Initial

[H

2

]

1.00

[I

2

]

1.00

[HI]

0

Change -x -x +2x

Equilib 1.00-x 1.00-x 2x where x is defined as am’t of H

2 and I

2 consumed on approaching equilibrium.


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41

H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3

Step 2. Put equilibrium concentrations into K c expression.

K c

=

[2x]

2

[1.00 - x][1.00 - x]

= 55.3

42


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3

Step 3. Solve K c expression - take square root of both sides.

43


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3


2x

7.44 =

1.00 - x

44


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3


2x

7.44 =

1.00 - x x = 0.79

45


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3


2x

7.44 =

1.00 - x x = 0.79

Therefore, at equilibrium

[H

2

] = [I

2

] = 1.00 - x = 0.21 M

46


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H

2

(g) + I

2

(g) 2 HI(g), K c

= 55.3


2x

7.44 =

1.00 - x x = 0.79

Therefore, at equilibrium

[H

2

] = [I

2

] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M


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47


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Nitrogen Dioxide

Equilibrium

N

2

O

4

(g) 2 NO

2

(g)

48

49

Nitrogen Dioxide Equilibrium

N

2

O

4

(g) 2 NO

2

(g)

K c

=

[NO

2

]

2

[N

2

O

4

]

= 0.0059 at 298 K

If initial concentration of N

2

O

4 is 0.50 M, what are the equilibrium concentrations?

Step 1. Set up an equilibrium table

Initial

Change

[N

2

O

4

]

0.50

[NO

0

2

]

Equilib


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50


N

2

O

4

(g) 2 NO

2

(g)

K c

=

[NO

2

]

2

[N

2

O

4

]

= 0.0059 at 298 K

If initial concentration of N

2

O

4 is 0.50 M, what are the equilibrium concentrations?

Step 1. Set up an equilibrium table

Initial

Change

[N

2

O

4

]

0.50

-x

[NO

0

+2x

2

]

Equilib


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0.50 - x 2x


N

2

O

4

(g) 2 NO

2

(g)

Step 2. Substitute into K c expression and solve.

K c

= 0.0059 =

[NO

2

]

2

[N

2

O

4

]

(2x)

2

=

(0.50 - x)

51

Rearrange: 0.0059 (0.50 - x) = 4x 2

0.0029 - 0.0059x = 4x 2

4x 2 + 0.0059x - 0.0029 = 0

This is a

QUADRATIC EQUATION ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029


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N

2

O

4

(g) 2 NO

2

(g)

Solve the quadratic equation for x.

ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029

x =

-b



b

2

- 4ac

2a

52


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N

2

O

4

(g) 2 NO

2

(g)


ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029

x =

-b



b

2

- 4ac

2a x =

-0.0059



(0.0059)

2

- 4(4)(-0.0029)

2(4)

53


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N

2

O

4

(g) 2 NO

2

(g)


ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029

x =

-b



b

2

- 4ac

2a x =

-0.0059



(0.0059)

2

- 4(4)(-0.0029)

2(4) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

54


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N

2

O

4

(g) 2 NO

2

(g) x =

-0.0059



(0.0059)

2

- 4(4)(-0.0029)

2(4) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

x = 0.026 or -0.028

But a negative value is not reasonable.

Conclusion

[N

2

O

4

] = 0.050 - x = 0.47 M

[NO

2

] = 2x = 0.052 M


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55

Chapter 16-Chemical Equilibrium

CHEMICAL

EQUILIBRIUM

Chapter 16

Examples of Chemical

Equilibria

Chemical

Equilibria

Typical Calculations

Nitrogen Dioxide

Equilibrium

Related documents

Products

Support

Chapter 16-Chemical Equilibrium

CHEMICAL

EQUILIBRIUM

Chapter 16

Examples of Chemical

Equilibria

Chemical

Equilibria

Typical Calculations

Nitrogen Dioxide

Equilibrium

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