chemical equilibrium chapter 16 copyright © 1999 by harcourt brace & company all rights...

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

Chapter 16Chapter 16

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

Chapter 16Chapter 16

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

2

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Equilibrium systems areEquilibrium systems are

• DYNAMIC (in constant DYNAMIC (in constant motion)motion)

• REVERSIBLE REVERSIBLE

• can be approached from can be approached from either directioneither direction

Equilibrium systems areEquilibrium systems are




3










Pink to bluePink to blueCo(HCo(H22O)O)66ClCl22 ---> Co(H ---> Co(H22O)O)44ClCl22 + 2 H + 2 H22OO

4










Pink to bluePink to blueCo(HCo(H22O)O)66ClCl22 ---> Co(H ---> Co(H22O)O)44ClCl22 + 2 H + 2 H22OO

Blue to pinkBlue to pinkCo(HCo(H22O)O)44ClCl22 + 2 H + 2 H22O ---> Co(HO ---> Co(H22O)O)66ClCl22

5


Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCNFeSCN2+2+

Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCNFeSCN2+2+

Fe(H2O)63+ Fe(SCN)(H2O)5

3++ SCN-

+

+ H2O

6


Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCN FeSCN2+2+

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

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

7


Examples of Chemical Examples of Chemical EquilibriaEquilibria

Phase changes such asPhase changes such as H H22O(s) O(s) HH22O(liq)O(liq)

8


Examples Examples of of

Chemical Chemical EquilibriaEquilibria

Formation of stalactites and stalagmitesFormation of stalactites and stalagmites

CaCOCaCO33(s) + H(s) + H22O(liq) + COO(liq) + CO22(g)(g)

CaCa2+2+(aq) + 2 HCO(aq) + 2 HCO33--(aq) (aq)

9


Chemical Chemical EquilibriaEquilibria

CaCOCaCO33(s) + H(s) + H22O(liq) + COO(liq) + CO22(g)(g)

CaCa2+2+(aq) + 2 HCO(aq) + 2 HCO33--(aq)(aq)

At a given T and P of COAt a given T and P of CO22, [Ca, [Ca2+2+] and [HCO] and [HCO33--] can be ] can be

found from the found from the EQUILIBRIUM CONSTANTEQUILIBRIUM CONSTANT. .

10


THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

For any type of chemical equilibrium of the For any type of chemical equilibrium of the typetype

a A + b B a A + b B c C + d Dc C + d D

the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)

11


THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

For any type of chemical equilibrium of the For any type of chemical equilibrium of the typetype

a A + b B a A + b B c C + d Dc C + d D

the following is a CONSTANT (at a given T)the following is a CONSTANT (at a given T)

K =[C]c [D]d

[A]a [B]b

conc. of products

conc. of reactantsequilibrium constant

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

12


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)

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

SolutionSolution

Set of a table of concentrationsSet of a table of concentrations

[NOCl][NOCl] [NO][NO] [Cl[Cl22]]

BeforeBefore 2.002.00 00 00

ChangeChange

EquilibriumEquilibrium 0.660.66

13



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

SolutionSolution

Set of a table of concentrationsSet of a table of concentrations



ChangeChange -0.66-0.66 +0.66+0.66 +0.33+0.33

EquilibriumEquilibrium 1.341.34 0.660.66 0.330.33

14





ChangeChange -0.66-0.66 +0.66+0.66 +0.33+0.33


K [NO]2[Cl2 ]

[NOCl]2 K

[NO]2[Cl2 ]

[NOCl]2

15


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)



ChangeChange -0.66-0.66 +0.66+0.66 +0.33+0.33


K [NO]2[Cl2 ]

[NOCl]2 K

[NO]2[Cl2 ]

[NOCl]2

K [NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080K

[NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080

16


Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions


Solids and liquids Solids and liquids NEVER appear in NEVER appear in equilibrium equilibrium expressions.expressions.

S(s) + OS(s) + O22(g) (g)

SO SO22(g)(g)

17




Solids and liquids Solids and liquids NEVER appear in NEVER appear in equilibrium equilibrium expressions.expressions.

S(s) + OS(s) + O22(g) (g)

SO SO22(g)(g)

K [SO2 ][O2 ]

K [SO2 ][O2 ]

18




Solids and liquids NEVER Solids and liquids NEVER appear in equilibrium appear in equilibrium expressions.expressions.

NHNH33(aq) + H(aq) + H22O(liq) NHO(liq) NH44++

(aq) + OH(aq) + OH--(aq)(aq)

19




Solids and liquids NEVER Solids and liquids NEVER appear in equilibrium appear in equilibrium expressions.expressions.

NHNH33(aq) + H(aq) + H22O(liq) NHO(liq) NH44++

(aq) + OH(aq) + OH--(aq)(aq)

K [NH4

+][OH- ][NH3 ]

K [NH4

+][OH- ][NH3 ]

20




Adding equations for reactionsAdding equations for reactions

S(s) + OS(s) + O22(g) SO(g) SO22(g) K(g) K11 = [SO = [SO22] / [O] / [O22]]

SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)

KK22 = [SO = [SO33] / [SO] / [SO22][O][O22]]1/21/2

NET EQUATIONNET EQUATION

S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)

21




Adding equations for reactionsAdding equations for reactions

S(s) + OS(s) + O22(g) (g) SO SO22(g) K(g) K11 = [SO = [SO22] / [O] / [O22]]

SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)

KK22 = [SO = [SO33] / [SO] / [SO22][O][O22]]1/21/2

NET EQUATIONNET EQUATION

S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)

Knet

[SO3]

[O2 ]3/2 = K1 • K2 Knet [SO3]

[O2 ]3/2 = K1 • K2

22




Changing coefficientsChanging coefficients

S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)

2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO2 SO33(g)(g)

K [SO3 ]

[O2 ]3/2 K [SO3 ]

[O2 ]3/2

23





S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SOSO33(g)(g)

2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO2 SO33(g)(g)

K [SO3 ]

[O2 ]3/2 K [SO3 ]

[O2 ]3/2

Knew [SO3 ]2

[O2 ]3 Knew

[SO3 ]2

[O2 ]3

24





S(s) + 3/2 OS(s) + 3/2 O22(g) (g) SO SO33(g)(g)

2 S(s) + 3 O2 S(s) + 3 O22(g) (g) 2 SO 2 SO33(g)(g)

Knew [SO3 ]2

[O2 ]3 = (Kold )2 Knew

[SO3 ]2

[O2 ]3 = (Kold )2

K [SO3 ]

[O2 ]3/2 K [SO3 ]

[O2 ]3/2

Knew [SO3 ]2

[O2 ]3 Knew

[SO3 ]2

[O2 ]3

25




K [SO2 ]

[O2 ] K

[SO2 ]

[O2 ]

Changing directionChanging direction

S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)

SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)

26





S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)

SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)

K [SO2 ]

[O2 ] K

[SO2 ]

[O2 ]

Knew [O2 ][SO2 ]

Knew [O2 ][SO2 ]

27





S(s) + OS(s) + O22(g) (g) SO SO22(g)(g)

SOSO22(g) (g) S(s) + O S(s) + O22(g)(g)

K [SO2 ]

[O2 ] K

[SO2 ]

[O2 ]

Knew [O2 ][SO2 ]

= 1

Kold Knew

[O2 ][SO2 ]

= 1

Kold

Knew [O2 ][SO2 ]

Knew [O2 ][SO2 ]

28




Concentration UnitsConcentration Units

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

These are designated by These are designated by KKcc

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

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

by by KKpp. .

KKcc and K and Kpp may or may not be the same. may or may not be the same.

29


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K1.1. Can tell if a reaction is product-Can tell if a reaction is product-

favored or reactant-favored.favored or reactant-favored.

For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)

30




For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH2 NH33(g)(g)

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

31




For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)

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

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

32




For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)

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

The reaction is strongly The reaction is strongly product-product-favoredfavored..

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108

33


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KFor AgCl(s) For AgCl(s)

AgAg++(aq) + Cl(aq) + Cl--

(aq)(aq)

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

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

The reaction is strongly The reaction is strongly

reactant-favoredreactant-favored..

AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.

AgAg++(aq) + Cl(aq) + Cl--(aq) (aq) AgCl(s)AgCl(s)is product-favored.is product-favored.

34


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K2. Can tell if a reaction is at equilibrium. 2. Can tell if a reaction is at equilibrium.

If not, which way it moves to approach If not, which way it moves to approach equilibrium.equilibrium.

H

H

H

H

H

H

H

H

H H H

CH

HHH H

H—C—C—C—C—H H—C—C—C—H

K =

n-butane iso-butane

[iso]

[n] = 2.5

H

H

H

H

H

H

H

H

H H H

CH

HHH H


K =

n-butane iso-butane

[iso]

[n] = 2.5

35


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K

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

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

See Screen 16.9.See Screen 16.9.

H

H

H

H

H

H

H

H

H H H

CH

HHH H


K =

n-butane iso-butane

[iso]

[n] = 2.5

36


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems In general, all reacting chemical systems

are characterized by their are characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..

If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.

Q = product concentrationsreactant concentrations


37


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems In general, all reacting chemical systems

are characterized by their are characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..


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

Q (2.3) < K (2.5). Q (2.3) < K (2.5).



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3

38


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems are In general, all reacting chemical systems are

characterized by their characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..


Q (2.33) < K (2.5). Q (2.33) < K (2.5).

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



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3

39


Typical CalculationsTypical CalculationsTypical CalculationsTypical Calculations

PROBLEM: Place 1.00 mol each of HPROBLEM: Place 1.00 mol each of H22 and I and I22 in a in a 1.00 L flask. Calc. equilibrium concentrations.1.00 L flask. Calc. equilibrium concentrations.

HH22(g) + I(g) + I22(g) (g) ¸̧ 2 HI(g) 2 HI(g)

Kc = [HI]2

[H2 ][I2 ] = 55.3Kc =

[HI]2

[H2 ][I2 ] = 55.3

40


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 1. Set up table to define Step 1. Set up table to define EQUILIBRIUM concentrations.EQUILIBRIUM concentrations.

[H[H22]] [I[I22]] [HI][HI]

Initial Initial 1.001.00 1.001.00 00

ChangeChange

EquilibEquilib

41


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 1. Set up table to define Step 1. Set up table to define EQUILIBRIUM concentrations.EQUILIBRIUM concentrations.

[H[H22]] [I[I22]] [HI][HI]

Initial Initial 1.001.00 1.001.00 00

ChangeChange -x-x -x-x +2x+2x

EquilibEquilib1.00-x1.00-x 1.00-x1.00-x 2x2x

where where xx is defined as am’t of H is defined as am’t of H22 and I and I22

consumed on approaching equilibrium.consumed on approaching equilibrium.

42


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 2. Put equilibrium concentrations Step 2. Put equilibrium concentrations into Kinto Kcc expression. expression.

Kc = [2x]2

[1.00 - x][1.00 - x] = 55.3Kc =

[2x]2

[1.00 - x][1.00 - x] = 55.3

43


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 3. Solve KStep 3. Solve Kcc expression - take expression - take

square root of both sides.square root of both sides.

44


H2(g) + I2(g) 2 HI(g), Kc = 55.3



7.44 = 2x

1.00 - x7.44 =

2x1.00 - x

45


H2(g) + I2(g) 2 HI(g), Kc = 55.3



x = 0.79x = 0.79

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x

46


H2(g) + I2(g) 2 HI(g), Kc = 55.3



x = 0.79x = 0.79

Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x

47


H2(g) + I2(g) 2 HI(g), Kc = 55.3



x = 0.79x = 0.79

Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x

48


Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

NN22OO44(g) (g) 2 2 NONO22(g)(g)

Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

NN22OO44(g) (g) 2 2 NONO22(g)(g)

49


Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)

If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the is 0.50 M, what are the equilibrium concentrations?equilibrium concentrations?

Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table

[N[N22OO44]] [NO[NO22]]

InitialInitial 0.500.50 00

ChangeChange

EquilibEquilib

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 KKc =

[NO2 ]2

[N2O4 ] = 0.0059 at 298 K

50



If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what are the is 0.50 M, what are the equilibrium concentrations?equilibrium concentrations?

Step 1. Set up an equilibrium tableStep 1. Set up an equilibrium table

[N[N22OO44]] [NO[NO22]]

InitialInitial 0.500.50 00

ChangeChange -x-x +2x+2x

EquilibEquilib 0.50 - x0.50 - x 2x2x

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 KKc =

[NO2 ]2

[N2O4 ] = 0.0059 at 298 K

51



Step 2. Substitute into KStep 2. Substitute into Kcc expression and solve. expression and solve.

Rearrange: Rearrange: 0.0059 (0.50 - x) = 4x0.0059 (0.50 - x) = 4x22

0.0029 - 0.0059x = 4x0.0029 - 0.0059x = 4x22

4x4x22 + 0.0059x - 0.0029 = 0 + 0.0059x - 0.0029 = 0

This is a This is a QUADRATIC EQUATIONQUADRATIC EQUATION axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

Kc = 0.0059 = [NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x) Kc = 0.0059 =

[NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x)

52




Solve the quadratic equation for x.Solve the quadratic equation for x.

axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

53




axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)

54




axax22 + bx + c = 0 + bx + c = 0

a = 4a = 4 b = 0.0059 b = 0.0059 c = -0.0029c = -0.0029

x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027

x = -b b2 - 4ac

2ax =

-b b2 - 4ac2a

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)

55



x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = -0.00074 ± 0.027 = -0.00074 ± 0.027

x = 0.026 or -0.028x = 0.026 or -0.028

But a negative value is not reasonable.But a negative value is not reasonable.

ConclusionConclusion [N[N22OO44] = 0.050 - x = 0.47 M] = 0.050 - x = 0.47 M

[NO[NO22] = 2x = 0.052 M] = 2x = 0.052 M

x = -0.0059 (0.0059)2 - 4(4)(-0.0029)

2(4)x =

-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)

chemical equilibrium chapter 16 copyright © 1999 by harcourt brace & company all rights...

Documents

equilibrium constant

harcourt brace company

constant motiondynamic

o slide

copyright c

direction equilibrium

aq slide

chemical equilibrium