1 © 2009 Brooks/Cole - Cengage CHEMICAL EQUILIBRIUM Chapter 16 Pb 2+ (aq) + 2 Cl – (aq) PbCl 2...

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© 2009 Brooks/Cole - Cengage

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUMChapter 16Chapter 16

PbPb2+2+(aq) + 2 Cl(aq) + 2 Cl––(aq) (aq) PbCl PbCl22(s)(s)

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Properties of an Properties of an Equilibrium 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

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

• REVERSIBLE REVERSIBLE

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

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 O Co(H Co(H22O)O)66ClCl22

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

+

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

2++ SCN-

+ H2O

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

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Examples of Examples of Chemical EquilibriaChemical Equilibria

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

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Examples Examples of of

Chemical Chemical EquilibriEquilibri

aa

Formation of Formation of

stalactites and stalagmitesstalactites and stalagmites

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

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

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Chemical Chemical EquilibriaEquilibria

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

Ca Ca2+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 found from the can be found from the EQUILIBRIUM EQUILIBRIUM

CONSTANTCONSTANT. .

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Reaction Quotient & Reaction Quotient & Equilibrium ConstantEquilibrium Constant

Product conc. increases and then becomes constant at equilibrium

Reactant conc. declines and then becomes constant at equilibrium

Equilibrium achieved

See Active Figure 16.2See Active Figure 16.2

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At any point in the reaction

H2 + I2 2 HI

Q = reaction quotient = [HI]2

[H2 ][I2 ]

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[HI]2

[H2 ][I2 ] = 55.3 = K

K = equilibrium constant

Equilibrium achieved

In the equilibrium region

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The Reaction Quotient, The Reaction Quotient, QQ

In general, ALL reacting chemical systems are In general, ALL reacting chemical systems are

characterized by their characterized by their REACTION REACTION

QUOTIENT, QQUOTIENT, Q..

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

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

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

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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 an Set of an “ICE”“ICE” table of concentrations table of concentrations

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

IInitialnitial 2.002.00 00 00

CChangehange

EEquilibriumquilibrium 0.660.66

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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 an Set of an “ICE”“ICE” table of concentrations table of concentrations


IInitialnitial 2.002.00 00 00

CChangehange -0.66-0.66 +0.66+0.66 +0.33+0.33

EEquilibriumquilibrium 1.341.34 0.660.66 0.330.33

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InitialInitial 2.002.00 00 00

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

EquilibriumEquilibrium 1.341.34 0.660.66 0.330.33

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 = 0.080

K = [NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080

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Writing and Writing and Manipulating K Manipulating K ExpressionsExpressions


Solids and liquids Solids and liquids NEVERNEVER appear in appear in equilibrium equilibrium expressions.expressions.

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

SO SO22(g)(g)

K = [SO2 ]

[O2 ]

K = [SO2 ]

[O2 ]

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Solids and liquids Solids and liquids NEVERNEVER appear appear in equilibrium expressions.in equilibrium expressions.

NHNH33(aq) + H(aq) + H22O(liq) O(liq)

NHNH44++(aq) + OH(aq) + OH--(aq)(aq)

K = [NH4

+ ][OH- ]

[NH3 ]

K = [NH4

+ ][OH- ]

[NH3 ]

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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 NH 2 NH33(g)(g)

Kc = [NH3 ]2

[N2 ][H2 ]3 = 3.5 x 108

Kc = [NH3 ]2

[N2 ][H2 ]3 = 3.5 x 108

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-favoredproduct-favored..

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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 reactants than that of reactants at equilibrium. at equilibrium.

The reaction with small K is The reaction with small K is strongly strongly

reactant-reactant-favoredfavored..

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

ee AgCl(s) AgCl(s)is product-favored.is product-favored.

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

ee AgCl(s) AgCl(s)is product-favored.is product-favored.

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Product- or Reactant Product- or Reactant FavoredFavored

Product-favoredProduct-favoredK > 1K > 1

Reactant-favoredReactant-favoredK < 1K < 1

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The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KK comes from thermodynamics. (K comes from thermodynamics. (See See

Chapter 19)Chapter 19)

∆∆G˚ < 0: reaction is G˚ < 0: reaction is product favoredproduct favored

∆∆G˚ > 0: reaction is G˚ > 0: reaction is reactant-favoredreactant-favored

If K > 1, then ∆G˚ is negative

If K < 1, then ∆G˚ is positive

If K > 1, then ∆G˚ is negative

If K < 1, then ∆G˚ is positive

ĘGo

= -RT ln K ĘGo

= -RT ln K

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The 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

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The Meaning of KThe Meaning of K

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

If not, which way does the reaction If not, which way does the reaction “shift” to approach equilibrium?“shift” to approach equilibrium?

See Chemistry NowSee Chemistry Now

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

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The Meaning of KThe Meaning of KAll reacting chemical systems are All reacting chemical systems are

characterized by their characterized by their REACTION QUOTIENT, REACTION QUOTIENT, QQ..

Q =

product concentrations

reactant concentrations Q =

product concentrations

reactant concentrations

Q =

conc. of iso

conc. of n =

0.35

0.15 = 2.3

Q =

conc. of iso

conc. of n =

0.35

0.15 = 2.3

Q (2.33) < K (2.5)Q (2.33) < K (2.5) Reaction is Reaction is NOTNOT at equilibrium, so [iso] must at equilibrium, so [iso] must become ________ and [n] must become ________ and [n] must ________________________..

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

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ÆÆ

Typical 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.

Kc = [HI]2

[H2 ][I2 ] = 55.3

Kc = [HI]2

[H2 ][I2 ] = 55.3

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

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HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3


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

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

Initial Initial 1.001.00 1.001.00 00

ChangeChange

EquilibEquilib

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Step 1. Set up Step 1. Set up ICEICE table to define 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.



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

Kc =

[2x]2

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

Kc =

[2x]2

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

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[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

[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

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

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

x = 0.79x = 0.79Therefore, at equilibriumTherefore, at equilibrium


Kc =

[2x]2

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

Kc =

[2x]2

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

7.44 =

2x

1.00-x 7.44 =

2x

1.00-x

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Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)


NN22OO44(g) (g) ee 2 NO 2 NO22(g)(g)

eeee

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Nitrogen Dioxide EquilibriumNitrogen Dioxide Equilibrium

NN22OO44(g) (g) 2 NO 2 NO22(g)(g)


NN22OO44(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 ICE tableStep 1. Set up an ICE table

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

InitialInitial 0.500.50 00

ChangeChange

EquilibEquilib

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

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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 ICE tableStep 1. Set up an ICE table

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

InitialInitial 0.500.50 00

ChangeChange -x-x +2x+2x

EquilibEquilib 0.50 - x0.50 - x 2x2x


NN22OO44(g) (g) 2 NO 2 NO22(g)(g)


NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

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NN22OO44(g) (g) 2 NO 2 NO22(g)(g)


NN22OO44(g) (g) 2 NO 2 NO22(g)(g)Step 2. Substitute into KStep 2. Substitute into Kcc expression and solve. expression and solve.

Kc = 0.0059 = [NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x)

Kc = 0.0059 = [NO2 ]2

[N2O4 ]=

(2x)2

(0.50 - x)

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 = 0a = 4a = 4 b = 0.0059b = 0.0059 c = -0.0029c = -0.0029

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NN22OO44(g) (g) 2 NO 2 NO22(g)(g)


NN22OO44(g) (g) 2 NO 2 NO22(g)(g)Solve the quadratic equation for x.Solve the quadratic equation for x.

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

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

x =

-b ± b2 - 4ac

2a x =

-b ± b2 - 4ac

2a

x =

-0.0059 ± (0.0059)2 - 4(4)(-0.0029)

2(4) x =

-0.0059 ± (0.0059)2 - 4(4)(-0.0029)

2(4)

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

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NN22OO44(g) (g) 2 NO 2 NO22(g)(g)


NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

x = x = 0.0260.026 or or -0.028-0.028

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

Conclusion: Conclusion: x = 0.026 Mx = 0.026 M [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)

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

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Solving Quadratic Solving Quadratic EquationsEquations

• Recommend you solve the equation exactly on a calculator or use the “method of successive approximations”

• See Appendix A.

37


Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions


Adding equations for reactionsAdding equations for reactions

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

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

K1 = [SO2 ]

[O2 ]

K1 = [SO2 ]

[O2 ]

K2 = [SO3 ]

[SO2 ][O2 ]1/2

K2 = [SO3 ]

[SO2 ][O2 ]1/2

Net equationNet equation

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

Knet = [SO3 ]

[O2 ]3/2

Knet = [SO3 ]

[O2 ]3/2 Knet = K1 x K2 Knet = K1 x K2

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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 SO 2 SO33(g)(g)

K = [SO3 ]

[O2 ]3/2

K = [SO3 ]

[O2 ]3/2

Knew= [SO3 ]2

[O2 ]3

Knew= [SO3 ]2

[O2 ]3

Knew= [SO3 ]2

[O2 ]3 = (Kold)2

Knew= [SO3 ]2

[O2 ]3 = (Kold)2

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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)

K = [SO2 ]

[O2 ]

K = [SO2 ]

[O2 ]

Knew= [O2 ]

[SO2 ] =

1

Kold

Knew= [O2 ]

[SO2 ] =

1

Kold

Knew= [O2 ]

[SO2 ]

Knew= [O2 ]

[SO2 ]

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

P is proportional to concentration, so we can write K P is proportional to concentration, so we can write K

in terms of P. These are designated by in terms of P. These are designated by KKpp. .

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

41




K using concentration and pressure unitsK using concentration and pressure units

KKpp = K = Kc c (RT)(RT)∆n∆n

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

∆∆n = 0 and Kn = 0 and Kpp = K = Kcc

For SOFor SO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)

∆∆n = –1/2 and Kn = –1/2 and Kpp = K = Kcc(RT)(RT)–1/2–1/2

42


EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS

• Temperature, catalysts, and changes in Temperature, catalysts, and changes in concentration affect equilibria.concentration affect equilibria.

• The outcome is governed by The outcome is governed by LE LE CHATELIER’S PRINCIPLECHATELIER’S PRINCIPLE

• ““...if a system at equilibrium is ...if a system at equilibrium is disturbed, the system tends to shift its disturbed, the system tends to shift its equilibrium position to counter the equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”

43


EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTSEQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS

Henri Le ChatelierHenri Le Chatelier

1850-19361850-1936

Studied mining Studied mining engineering.engineering.

Interested in glass Interested in glass and ceramics.and ceramics.

44


EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS

• Temperature change Temperature change change in K change in K• Consider the fizz in a soft drinkConsider the fizz in a soft drink

COCO22(aq) + (aq) + HEATHEAT COCO22(g) + H(g) + H22O(liq) O(liq)

• K = P (COK = P (CO22) / [CO) / [CO22] ]

• Increase T. What happens to equilibrium Increase T. What happens to equilibrium position? To value of K?position? To value of K?

• K increases as T goes up because P(COK increases as T goes up because P(CO22) )

increases and [COincreases and [CO22] decreases.] decreases.

• Decrease T. Now what?Decrease T. Now what?

• Equilibrium shifts left and K decreases.Equilibrium shifts left and K decreases.

45


Temperature Temperature Effects on Effects on EquilibriumEquilibrium

NN22OO44 (colorless) + (colorless) + heatheat

2 NO 2 NO22 (brown) (brown)

∆H∆Hoo = + 57.2 kJ = + 57.2 kJ

Kc = [NO2 ]2

[N2O4 ]

Kc = [NO2 ]2

[N2O4 ]

KKc (273 K) = 0.00077 (273 K) = 0.00077

KKc (298 K) = 0.0059 (298 K) = 0.0059PLAY MOVIE

46


Temperature Effects on Temperature Effects on EquilibriumEquilibrium

See Figure 16.8See Figure 16.8

47


EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS

• Add catalyst Add catalyst no change in K no change in K• A catalyst only affects the RATE of A catalyst only affects the RATE of

approach to equilibrium.approach to equilibrium.

Catalytic exhaust systemCatalytic exhaust system

48


• NN22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g) + (g) + heatheat

• K = 3.5 x 10K = 3.5 x 1088 at 298 K at 298 K

Haber-Bosch Process for NHHaber-Bosch Process for NH33

49


Haber-Bosch Haber-Bosch Ammonia SynthesisAmmonia Synthesis

Fritz HaberFritz Haber1868-19341868-1934Nobel Prize, 1918Nobel Prize, 1918

Carl BoschCarl Bosch1874-19401874-1940Nobel Prize, 1931Nobel Prize, 1931

50


EQUILIBRIUM AND EXTERNAL EFFECTSEQUILIBRIUM AND EXTERNAL EFFECTS

• Concentration changes Concentration changes

–no change in K no change in K

–only the equilibrium only the equilibrium composition changes.composition changes.

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Le Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s PrincipleLe Chatelier’s Principle

Adding a “reactant” to a chemical Adding a “reactant” to a chemical system.system.

PLAY MOVIE

52



Removing a “reactant” from a chemical Removing a “reactant” from a chemical system.system.

PLAY MOVIE

53



Adding a “product” to a chemical Adding a “product” to a chemical system.system.

PLAY MOVIE

54



Removing a “product” from a chemical Removing a “product” from a chemical system.system.

PLAY MOVIE

55


EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTS

• Temperature, catalysts, and changes in Temperature, catalysts, and changes in concentration affect equilibria.concentration affect equilibria.

• The outcome is governed by The outcome is governed by LE LE CHATELIER’S PRINCIPLECHATELIER’S PRINCIPLE

• ““...if a system at equilibrium is ...if a system at equilibrium is disturbed, the system tends to shift its disturbed, the system tends to shift its equilibrium position to counter the equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”

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Butane-Butane-Isobutane Isobutane EquilibriEquilibri

umum

Butane-Butane-Isobutane Isobutane EquilibriEquilibri

umum

K =

[isobutane]

[butane]= 2.5

K =

[isobutane]

[butane]= 2.5

butanebutane

isobutaneisobutane

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ButaneButane ee IsobutaneIsobutane

• At equilibrium with [iso] = 1.25 M and At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.[butane] = 0.50 M. K = 2.5.

• Add 1.50 M butane. Add 1.50 M butane. • When the system comes to equilibrium When the system comes to equilibrium

again, what are [iso] and [butane]? again, what are [iso] and [butane]?

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Assume you are at equilibrium with [iso] = 1.25 M and [butane] = Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? equilibrium again, what are [iso] and [butane]? K = 2.5K = 2.5

SolutionSolution

Calculate Q immediately after adding more Calculate Q immediately after adding more butane and compare with K.butane and compare with K.

Q =

[isobutane]

[butane] =

1.25

0.50 + 1.50 = 0.625

Q =

[isobutane]

[butane] =

1.25

0.50 + 1.50 = 0.625

Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________.reaction will shift to the ____________.Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________.reaction will shift to the ____________.


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You are at equilibrium with [iso] = 1.25 M and [butane] You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. = 0.50 M. Now add 1.50 M butane.

SolutionSolutionQ is less than K, so equilibrium shifts right — Q is less than K, so equilibrium shifts right —

away from butane and toward isobutane.away from butane and toward isobutane.Set up ICE tableSet up ICE table [butane][butane] [isobutane][isobutane]InitialInitial ChangeChangeEquilibriumEquilibrium

0.50 + 1.500.50 + 1.50 1.251.25

- x- x + x+ x

2.00 - x2.00 - x 1.25 + x1.25 + x


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You are at equilibrium with [iso] = 1.25 M and [butane] You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. = 0.50 M. Now add 1.50 M butane.

SolutionSolution

K = 2.50 =

[isobutane]

[butane] =

1.25 + x

2.00 - x K = 2.50 =

[isobutane]

[butane] =

1.25 + x

2.00 - x

x = 1.07 Mx = 1.07 MAt the new equilibrium position, At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. [butane] = 0.93 M and [isobutane] = 2.32 M.

Equilibrium has shifted toward isobutane.Equilibrium has shifted toward isobutane.


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Le Chatelier’s PrincipleLe Chatelier’s Principle

• Change T – change in K – therefore change in P or concentrations at

equilibrium

• Use a catalyst: reaction comes more quickly to equilibrium. K not changed.

• Add or take away reactant or product: – K does not change– Reaction adjusts to new equilibrium “position”

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NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

Increase P in the system Increase P in the system by reducing the by reducing the volume (at constant T). volume (at constant T).

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

ee

PLAY MOVIE

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NN22OO44(g) (g) 2 NO 2 NO22(g)(g)

Increase P in the system by reducing the volume. Increase P in the system by reducing the volume.

In gaseous system the equilibrium will shift to the side In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). with fewer molecules (in order to reduce the P).

Therefore, reaction shifts Therefore, reaction shifts LEFTLEFT and P of NOand P of NO22 decreases decreases and P of Nand P of N22OO44 increases. increases.

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

1 © 2009 Brooks/Cole - Cengage CHEMICAL EQUILIBRIUM Chapter 16 Pb 2+ (aq) + 2 Cl – (aq) PbCl 2...

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Transcript of 1 © 2009 Brooks/Cole - Cengage CHEMICAL EQUILIBRIUM Chapter 16 Pb 2+ (aq) + 2 Cl – (aq) PbCl 2...