CH16 Chemical Equilibrium

7/28/2019 CH16 Chemical Equilibrium

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Copyright (c) 1999 by Harcourt Brace & Company

All rights reserved

Properties of an EquilibriumEquilibrium systems are

DYNAMIC (in constantmotion)

REVERSIBLE

can be approached fromeither direction


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Properties of an EquilibriumEquilibrium systems are

DYNAMIC (in constantmotion)

REVERSIBLE

can be approached fromeither direction

Pink to blueCo(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O

Blue to pink

Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2


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Chemical Equilibrium

Fe3+

+ SCN-

FeSCN2+

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

3++ SCN-

+

+ H2O


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Chemical EquilibriumFe3+ + SCN- FeSCN2+

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

The forward and reverse reactions continueafter equilibrium is attained.


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Examples of ChemicalEquilibria

Phase changes such asH2O(s) H2O(liq)


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Examples

ofChemicalEquilibria

Formation of stalactites and

stalagmites

CaCO3(s) + H2O(liq) + CO2(g)

Ca2+(aq) + 2 HCO3-(aq)


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Chemical

Equilibria

CaCO3(s) + H2O(liq) + CO2(g)

Ca2+(aq) + 2 HCO3-(aq)

At a given T and P of CO2, [Ca2+

] and[HCO3

-] can be found from the

EQUILIBRIUM CONSTANT.


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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)


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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)

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

products or reactants.


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

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

SolutionSet of a table of concentrations

[NOCl] [NO] [Cl2]

Before 2.00 0 0

Change

Equilibrium 0.66


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

SolutionSet of a table of concentrations

[NOCl] [NO] [Cl2]

Before 2.00 0 0

Change -0.66 +0.66 +0.33

Equilibrium 1.34 0.66 0.33


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[NOCl] [NO] [Cl2]

Before 2.00 0 0

Change -0.66 +0.66 +0.33


K [NO]2[Cl2 ]

[NOCl]

2


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[NOCl] [NO] [Cl2]

Before 2.00 0 0

Change -0.66 +0.66 +0.33


K [NO]2[Cl2 ]

[NOCl]

2

K [NO]2[Cl2 ]

[NOCl]2=

(0.66)2(0.33)

(1.34)2= 0.080


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Writing and Manipulating K

ExpressionsSolids and liquids

NEVER appear in

equilibriumexpressions.

S(s) + O2(g)

SO2(g)


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ExpressionsSolids and liquids

NEVER appear in

equilibriumexpressions.

S(s) + O2(g)

SO2(g)

K [SO2 ][O2 ]


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ExpressionsSolids and liquids NEVER

appear in equilibrium

expressions.

NH3(aq) + H2O(liq)

NH4+

(aq) + OH

-

(aq)


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ExpressionsSolids and liquids NEVER

appear in equilibrium

expressions.

NH3(aq) + H2O(liq)

NH4+

(aq) + OH

-

(aq)

K [NH4+][OH- ]

[NH3 ]


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Writing and Manipulating KExpressions

Adding equations for reactions

S(s) + O2(g) SO2(g) K1 = [SO2] / [O2]

SO2(g) + 1/2 O2(g) SO3(g)

K2 = [SO3] / [SO2][O2]1/2

NET EQUATION

S(s) + 3/2 O2(g) SO3(g)

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Adding equations for reactions

S(s) + O2(g) SO2(g) K1 = [SO2] / [O2]

SO2(g) + 1/2 O2(g) SO3(g)

K2 = [SO3] / [SO2][O2]1/2

NET EQUATION

S(s) + 3/2 O2(g) SO3(g)

Knet [SO3]

[O2 ]3/2

= K1 K2

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Changing coefficients

S(s) + 3/2 O2(g) SO3(g)

2 S(s) + 3 O2(g) 2 SO3(g)

K [SO3 ][O2 ]

3/2

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S(s) + 3/2 O2(g) SO3(g)

2 S(s) + 3 O2(g) 2 SO3(g)

K [SO3 ][O2 ]

3/2

Knew

[SO3 ]2

[O2 ]3

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S(s) + 3/2 O2(g) SO3(g)

2 S(s) + 3 O2(g) 2 SO3(g)

Knew [SO3 ]

2

[O2

]3= (Kold)

2

K [SO3 ][O2 ]

3/2

Knew

[SO3 ]2

[O2 ]3

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K [SO2 ][O2 ]

Changing direction

S(s) + O2(g) SO2(g)

SO2(g) S(s) + O2(g)

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Changing direction

S(s) + O2(g) SO2(g)

SO2(g) S(s) + O2(g)

K [SO2 ][O2 ]

Knew [O2 ]

[SO2 ]

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Changing direction

S(s) + O2(g) SO2(g)

SO2(g) S(s) + O2(g)

K [SO2 ][O2 ]

Knew [O2 ][SO2 ]

= 1Kold

Knew [O2 ]

[SO2 ]

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Concentration UnitsWe have been writing K in terms of mol/L.

These are designated by

KcBut with gases, P = (n/V)RT = conc RTP is proportional to concentration, so we can

write K in terms of P. These are designatedby Kp.

Kc and Kp may or may not be the same.

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The Meaning of K

1. Can tell if a reaction is product-favored or reactant-favored.

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

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The Meaning of K


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

Kc = [NH3 ]

2

[N2 ][H2]3 = 3.5 x 108

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The Meaning of K


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

Conc. of products is much greaterthan that of reactants at equilibrium.

Kc = [NH3 ]

2

[N2 ][H2]3 = 3.5 x 108

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The Meaning of K


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

Conc. of products is much greaterthan that of reactants at equilibrium.

The reaction is strongly product-favored.

Kc = [NH3 ]

2

[N2 ][H2]3 = 3.5 x 108

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

The Meaning of K

For AgCl(s)Ag+(aq) + Cl-(aq)

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

Conc. of products is muchless than that ofreactants at equilibrium.

The reaction is stronglyreactant-favored.

Ag+

(aq) + Cl-

(aq)AgCl(s)

is product-favored.

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

If not, which way it moves to approachequilibrium.

H

H

H

H

H

H

H

H

H H H

C

H

HHH H

HCCCCH HCCCH

K =

n-butane iso-butane

[iso][n]

= 2.5

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The Meaning of K

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

Which way does the reaction shift toapproach equilibrium?

See Screen 16.9.

H

H

H

H

H

H

H

H

H H H

C

H

HHH H

HCCCCH HCCCH

K =

n-butane iso-butane

[iso]

[n]= 2.5

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Th M i f K


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The Meaning of KIn general, all reacting chemical systems

are characterized by theirREACTIONQUOTIENT, Q.

If Q = K, then system is at equilibrium.

Q =product concentrations

reactant concentration

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Th M i f K


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

Q (2.3) < K (2.5).



Q = conc. of isoconc. of n

= 0.350.15

= 2.3

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Th M i f K


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Q (2.33) < K (2.5).

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

____________.



Q =conc. of iso

conc. of n=

0.35

0.15= 2.3

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Typical Calculations

PROBLEM: Place 1.00 mol each of H2 and I2 ina 1.00 L flask. Calc. equilibriumconcentrations.

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

Kc =[HI]2

[H2 ][I2 ]= 55.3

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

Step 1. Set up table to defineEQUILIBRIUM concentrations.

[H2] [I2] [HI]

Initial 1.00 1.00 0

Change

Equilib

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

Step 1. Set up table to defineEQUILIBRIUM concentrations.

[H2] [I2] [HI]

Initial 1.00 1.00 0

Change -x -x +2x

Equilib 1.00-x 1.00-x 2x

where xis defined as amt of H2 and I2

consumed on approaching equilibrium.

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

Step 2. Put equilibrium concentrationsinto Kc expression.

Kc =[2x]2

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

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

Step 3. Solve Kc expression - takesquare root of both sides.

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


7.44 =2x

1.00 - x

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


x = 0.79

7.44 =2x

1.00 - x

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


x = 0.79

Therefore, at equilibrium

[H2] = [I2] = 1.00 - x = 0.21 M

7.44 =2x

1.00 - x

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


x = 0.79

Therefore, at equilibrium

[H2] = [I2] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M

7.44 =2x

1.00 - x

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Nitrogen DioxideEquilibrium

N2O4(g) 2 NO2(g)

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Nitrogen Dioxide EquilibriumN

2O

4(g) 2 NO

2(g)

If initial concentration of N2O4 is 0.50 M, what arethe equilibrium concentrations?

Step 1. Set up an equilibrium table

[N2O4] [NO2]

Initial 0.50 0

Change

Equilib

Kc =[NO2 ]

2

[N2O4 ]= 0.0059 at 298 K

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Nitrogen Dioxide EquilibriumN

2O

4(g) 2 NO

2(g)

If initial concentration of N2O4 is 0.50 M, what arethe equilibrium concentrations?

Step 1. Set up an equilibrium table

[N2

O4

] [NO2

]

Initial 0.50 0

Change -x +2x

Equilib 0.50 - x 2x

Kc =[NO2 ]

2

[N2O4 ]= 0.0059 at 298 K

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Nitrogen Dioxide EquilibriumN2O4(g) 2 NO2(g)

Step 2. Substitute into Kc expression and solve.

Rearrange: 0.0059 (0.50 - x) = 4x2

0.0029 - 0.0059x = 4x2

4x2 + 0.0059x - 0.0029 = 0

This is a QUADRATIC EQUATIONax2 + bx + c = 0

a = 4 b = 0.0059 c = -0.0029

Kc = 0.0059 =[NO2 ]

2

[N2O4 ]=

(2x)2

(0.50 - x)

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Solve the quadratic equation for x.

ax2 + bx + c = 0

a = 4 b = 0.0059 c = -0.0029

x =-b b2 - 4ac

2a

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Solve the quadratic equation for x.ax2 + bx + c = 0

a = 4 b = 0.0059 c = -0.0029

x =-b b2 - 4ac

2a

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

2(4)

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Ni Di id E ilib i


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Solve the quadratic equation for x.ax2 + bx + c = 0

a = 4 b = 0.0059 c = -0.0029

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

x =-b b2 - 4ac

2a

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

2(4)

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Nit Di id E ilib i


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x = -0.00074 1/8(0.046)1/2 = -0.00074 0.027x = 0.026 or -0.028

But a negative value is not reasonable.

Conclusion[N2O4] = 0.050 - x = 0.47 M

[NO2] = 2x = 0.052 M

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

2(4)

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CH16 Chemical Equilibrium

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