chemical equilibrium chapter 16 all bold numbered problems

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

Chapter 16Chapter 16

CHEMICAL CHEMICAL EQUILIBRIUMEQUILIBRIUM

Chapter 16Chapter 16

ALL BOLD NUMBERED PROBLEMS

22

Properties of an Properties of an EquilibriumEquilibrium

Properties of an Properties of an EquilibriumEquilibrium

Equilibrium systems areEquilibrium systems are

• DYNAMIC (in constant motion) & REVERSIBLEDYNAMIC (in constant motion) & REVERSIBLE

Equilibrium systems areEquilibrium systems are

• DYNAMIC (in constant motion) & REVERSIBLEDYNAMIC (in constant motion) & REVERSIBLE

33

Pink to bluePink to blueCo(HCo(H22O)O)66

2+2+ (aq) (aq) CoCl CoCl442-2- (aq) + 6 H (aq) + 6 H22O (l)O (l)

Heat

Blue to pinkBlue to pinkCoClCoCl44

2-2- (aq) + 6 H (aq) + 6 H22O (l) O (l) Co(H Co(H22O)O)662+2+ (aq) (aq)

Cool

Blue Pink

Cool

Heat

44

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

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

++

Fe(HFe(H22O)O)663+3+

Fe(SCN)(HFe(SCN)(H22O)O)552+2++ SCN+ SCN-- + H+ H22OO

colorlesscolorless red-orangered-orange

55

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

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

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

66

Examples of Chemical Examples of Chemical EquilibriaEquilibria

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

77

Examples Examples of of

Chemical Chemical EquilibriaEquilibria

Acid Base EquilibriaAcid Base Equilibria

Color Red Purple Violet Blue Blue-Green Green

pH 2 4 6 8 10 12

+Na2CO3

HH3OO++(aq) + 2(aq) + 2CO32-(aq) OH-(aq) + 2HCO3

-(aq)+CO2

88

Examples Examples of of


Formation of stalactites (ceiling) and stalagmites Formation of stalactites (ceiling) and stalagmites

(floor)(floor)

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

Ca Ca2+2+(aq) + 2 HCO(aq) + 2 HCO33--(aq) (aq)

99


At a given T and P of COAt a given T and P of CO22, [Ca, [Ca2+2+] and ] and

[HCO[HCO33--] can be found from the ] can be found from the

EQUILIBRIUM CONSTANTEQUILIBRIUM CONSTANT. .

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

Ca Ca2+2+(aq) + 2 HCO(aq) + 2 HCO33--

(aq) (aq)

1010

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 a A + b B + b B c C + d D the following is a c C + d D the following is a CONSTANT (at a given T).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 concentrations of products or reactants. concentrations of products or reactants.

1111

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

SolutionSolution

Set up a table of concentrations:Set up a table of concentrations:

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

InitialInitial 2.00 2.00 0 0 0 0

ChangeChange

EquilibriumEquilibrium 0.66 0.66

1212

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

SolutionSolution

Set of a table of concentrations:Set of a table of concentrations:


InitialInitial 2.002.00 00 00

ChangeChange

EquilibriumEquilibrium 0.660.66

Determining KDetermining KDetermining KDetermining K

+0.66+0.66 +0.33+0.33

0.330.33-0.66-0.661.341.34

1313

2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)


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[Cl 2]

[NOCl ]2

K[NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080


1414

Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions

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

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

K [SO2 ][O2 ]

1515


Solids and liquids NEVER Solids and liquids NEVER

appear in equilibrium appear in equilibrium

expressions.expressions.

NHNH33(aq) + H(aq) + H22O(O(ll) ) NH NH44++(aq) + OH(aq) + OH--(aq)(aq)

K [NH4

+ ][OH- ]

[NH3 ]

1616

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)


Knew [SO3 ]2

[O2 ]3 = (Kold )2

K [SO3 ]

[O2 ]3/2

Knew [SO3 ]2

[O2 ]3

1717

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 ]

Knew [O2 ]

[SO2 ] =

1

Kold

Knew [O2 ]

[SO2 ]

1818


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 ]

[O 2]3/2 = K1 • K2

1919Manipulating K ExpressionsManipulating K Expressions• The equilibrium expression is tied to the The equilibrium expression is tied to the

equation from which it is written. equation from which it is written.

• Changing the equation changes the Changing the equation changes the expression and thus the numerical value of K. expression and thus the numerical value of K.

• Three general cases need to be considered. Three general cases need to be considered.

1. If the equation is multiplied by a number, “a”, 1. If the equation is multiplied by a number, “a”, then the K is raised to the “a” power. then the K is raised to the “a” power.

2. If the equation is reversed, then the new K is 2. If the equation is reversed, then the new K is the reciprocal of the old K.the reciprocal of the old K.

3. If two equations are added, the new K is the 3. If two equations are added, the new K is the product of the two old K's.product of the two old K's.

• Using these rules, new K's can be derived for Using these rules, new K's can be derived for the modified equations.the modified equations.

SAMPLE QUESTIONS

2020


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 = [ ] • RTBut with gases, P = (n/V)•RT = [ ] • RT

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

These are designated by These are designated by KKpp. . KKcc and K and Kpp may or may not be the same. may or may not be the same.

KKpp = K = Kcc(RT)(RT)nn

2121


Practice ProblemsPractice Problems• Write the equation and the KWrite the equation and the Kcc expression for expression for

the formation of two moles of gaseous the formation of two moles of gaseous ammonia from the elements in the standard ammonia from the elements in the standard state. state.

• KKc c for this reaction at 25for this reaction at 25ooC is 3.5x10C is 3.5x1088. .

• Calculate KCalculate Kpp for this reaction. for this reaction.

• Calculate KCalculate Kcc for the reaction forming the for the reaction forming the elements from one mole of ammonia gas.elements from one mole of ammonia gas.

2222

The Meaning of KThe Meaning of K

Concentration of products is Concentration 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..

Kc = [NH3 ]2

[N2 ][H2 ]3 = 3.5 x 10 8

We can tell if a reaction is product-favored We can tell if a reaction is product-favored or reactant-favored.or reactant-favored.

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

2323

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

Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

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

Concentration of products is Concentration of products is

much lessmuch less than that of than that of reactants at equilibrium. reactants at equilibrium.

The reaction is strongly The reaction is strongly

reactant-favoredreactant-favored..Ag+(aq) + Cl-(aq) AgCl(s)is product-favored.

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

2424

Comparing Q and KComparing Q and K

• The relative magnitudes of Q and The relative magnitudes of Q and K tell us which direction the K tell us which direction the reaction will proceed to reach reaction will proceed to reach equilibrium.equilibrium.

• If Q<K, not at equilibrium and If Q<K, not at equilibrium and Reactants ------> Products.Reactants ------> Products.

• If Q=K, the system is at If Q=K, the system is at equilibrium.equilibrium.

• If Q>K, not at equilibrium and If Q>K, not at equilibrium and Products ------> Reactants.Products ------> Reactants.

2525

Using QUsing Q Can tell if a reaction is at equilibrium. 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

2626

Using QUsing Q

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?

H

H

H

H

H

H

H

H

H H H

C

H

HHH H


K =

n-butane iso-butane

[iso]

[n] = 2.5

2727Using QUsing Q

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

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

Q = product concentrationsreactant concentrations

Q = conc. of isoconc. of n

= 0.350.15

= 2.3

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

In general, all reacting chemical systems are In general, all reacting chemical systems are characterized by their characterized by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..

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

SAMPLE QUESTIONS

2828

Using KUsing KUsing KUsing K

PROBLEM: Place 1.00 mol each of HPROBLEM: Place 1.00 mol each of H22 and I and I22 in in

a 1.00 L flask. Calculate the equilibrium a 1.00 L flask. Calculate the equilibrium concentrations.concentrations.

HH22(g) + (g) + II22(g) (g) 2 H 2 HII(g)(g)

Kc = [HI]2

[H2 ][I2 ] = 55.3

2929

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

[H[H22]] [[II22]] [H[HII]]

Initial Initial

ChangeChange

EquilibEquilib

where where xx is defined as amount of H is defined as amount of H22 and I and I22

consumed on approaching equilibrium.consumed on approaching equilibrium.

HH22(g) + (g) + II22(g) (g) 2 H 2 HII(g), K(g), Kcc = 55.3 = 55.3

1.001.00 1.001.00 0 0

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

1.00-x1.00-x 1.00-x1.00-x 2x 2x

3030

HH22(g) + (g) + II22(g) (g) 2 H2 HII(g), K(g), Kcc = 55.3 = 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] = 55.3

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

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

7.44 = 2x

1.00 - x

3131

HH22(g) + (g) + II22(g) (g) 2 H 2 HII(g), K(g), Kcc = 55.3 = 55.3

x = 0.788x = 0.788

Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [] = [II22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

[H[HII] = 2x = 1.58 M] = 2x = 1.58 M

7.44 = 2x

1.00 - x

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

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

3232

Sample ProblemSample Problem

At a certain temperature 8.00 atm of At a certain temperature 8.00 atm of

HH22SS(g)(g) comes to equilibrium with H comes to equilibrium with H2(g)2(g)

and Sand S2(g)2(g). The equilibrium pressure of . The equilibrium pressure of

the sulfur gas is 0.60 atm.the sulfur gas is 0.60 atm.

Write the equation for the Write the equation for the decomposition of the hydrogen sulfide, decomposition of the hydrogen sulfide, calculate the equilibrium pressure for calculate the equilibrium pressure for each gas, and calculate Keach gas, and calculate KPP..

Solution

3333

Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium

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

3434

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

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what is 0.50 M, what are the equilibrium concentrations?are the 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 +2x+2x

EquilibEquilib 0.50 - x0.50 - x 2x2x

-x-x

3535


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)

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

3636


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 = -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

x = -b b2 - 4ac

2a

3737


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] = 0.50 - x = 0.47 M] = 0.50 - 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)

More Sample Problems

EQUILIBRIUM AND EQUILIBRIUM AND EXTERNAL EFFECTSEXTERNAL EFFECTSEQUILIBRIUM 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 CHATELIER’S PRINCIPLELE CHATELIER’S PRINCIPLE• ““...if a system at equilibrium is ...if a system at equilibrium is

disturbed, the system tends to shift disturbed, the system tends to shift its equilibrium position to counter the its equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”

38

3939

EQUILIBRIUM 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.

4040

Figure 16.6

NO2 / N2O4 is temperature dependent.

50o C 0o C

4141

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(g) + H(g) + H22O(liq) O(liq) CO CO22(aq) + heat(aq) + heat

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

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

K increases as T goes down becauseK increases as T goes down because[CO[CO22] increases and P(CO] increases and P(CO22) decreases.) decreases.

• Increase T. Now what?Increase T. Now what?

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

4242

Temperature Effects Temperature Effects on Equilibriumon Equilibrium

NN22OO44 (colorless) + heat (colorless) + heat

2 NO2 NO22 (brown) (brown)

HHoo = + 57.2 kJ = + 57.2 kJ

Kc [NO2 ]2

[N2O4 ]

KKcc = 0.00077 at 273 K = 0.00077 at 273 K

KKcc = 0.0059 at 298 K = 0.0059 at 298 K

4343


• 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

4444

• NN22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g) (g) • K = 3.5 x 10K = 3.5 x 1088 at 298 K at 298 K• K = 0.16 at 723 KK = 0.16 at 723 K

NHNH33 ProductionProduction

Fritz Haber, 1909Fritz Haber, 1909

4545


Concentration changesConcentration changes

no change in Kno change in K

only the position of only the position of equilibrium changesequilibrium changes

4646

Le Chatelier’s PrincipleLe Chatelier’s Principle

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

4747


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

4848


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

4949


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

5050

Figure 16.7

(a) 7 n-butane molecules (b) 7 iso-butane are added

(c) What are the new equilibrium concentration?

5151

Butane-Butane-Isobutane Isobutane

EquilibriumEquilibrium

K = [isobutane]

[butane] 2.5

butanebutane

isobutaneisobutane

5252

Butane Butane IsobutaneIsobutane

butanebutane

isobutaneisobutane

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

5353

Butane Butane IsobutaneIsobutaneAssume you are at equilibrium with [iso] = 1.25 M Assume you are at equilibrium with [iso] = 1.25 M

and [butane] = 0.50 M. Now add 1.50 M butane. and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, When the system comes to equilibrium again, what are [iso] and [butane]? 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]

1.250.50 + 1.50

= 0.63

Q is LESS THAN K. Therefore, the reaction will Q is LESS THAN K. Therefore, the reaction will shift to the ____________. shift to the ____________. Right (products)

5454

Butane Butane IsobutaneIsobutane

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

SolutionSolution

Q 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 concentration tableSet up concentration table

[butane][butane] [isobutane][isobutane]

InitialInitial

ChangeChange

EquilibriumEquilibrium

0.50 + 1.500.50 + 1.50 1.251.25

- x - x + x+ x

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

5555

Butane Butane IsobutaneIsobutaneYou are at equilibrium with [iso] = 1.25 M and You are at equilibrium with [iso] = 1.25 M and

[butane] = 0.50 M. Now add 1.50 M butane. [butane] = 0.50 M. Now add 1.50 M butane.

SolutionSolution

K = 2.50 = [isobutane]

[butane]

1.25 + x2.00 - x

x = 1.07 Mx = 1.07 M

At 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 Equilibrium has shifted toward isobutane.isobutane.

5656


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

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

5757


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

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

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

Kc = [NO2 ]2

[N2O4 ] = 0.0059 at 298 K

5858

Write the equation for the formation of two mole of Write the equation for the formation of two mole of ammonia gas from the elements in the standard ammonia gas from the elements in the standard state. state. H for this reaction is -92kJ. Predict the H for this reaction is -92kJ. Predict the changechange in K and the in K and the directiondirection of shift of an of shift of an equilibrium mixture equilibrium mixture when:when:

1. The temperature is increased.1. The temperature is increased.

2. The P2. The PT T is increased by adding He.is increased by adding He.

3. The total volume is increased.3. The total volume is increased.

4.Hydrogen gas is removed.4.Hydrogen gas is removed.

5.Nitrogen gas is added.5.Nitrogen gas is added.

6.The temperature is decreased.6.The temperature is decreased.

7.The total volume is decreased.7.The total volume is decreased.

smaller, leftsmaller, left

none, rightnone, right

none, leftnone, left

none, leftnone, left


larger, rightlarger, right



3 H2 (g) + N2 (g) <--> 2 NH3 (g) + 92kJ

5959Practice ProblemsPractice Problems

1. 1. 2 NO (g) + O2 NO (g) + O22 (g) <--> 2 NO (g) <--> 2 NO22 (g) (g)

a) Write the Ka) Write the Kpp expression. expression.

b) Write the Kb) Write the Kcc expression. expression.

c) If Kc) If Kcc = 0.50 at 25 = 0.50 at 25ooC, calculate KC, calculate Kpp..

d) Determine Kd) Determine Kcc for the following reaction: for the following reaction:

4 NO (g) + 2 O4 NO (g) + 2 O22 (g) <--> 4 NO (g) <--> 4 NO22 (g) (g)

e) If 2.5 moles of NO, Oe) If 2.5 moles of NO, O22, and NO, and NO22 are placed in are placed in a 5.0 L container, is the system at equilibrium? a 5.0 L container, is the system at equilibrium? If not, will the reaction proceed to the left or to If not, will the reaction proceed to the left or to the right?the right?

f) If 2.5 moles of NO and Of) If 2.5 moles of NO and O22 are placed in a 5.0 L are placed in a 5.0 L

container at 350container at 350ooC, 1.1 moles of NOC, 1.1 moles of NO22 are are present at equilibrium. Calculate Kpresent at equilibrium. Calculate Kcc..

6060Practice ProblemsPractice Problems

2. 2.5 moles of carbon dioxide is placed in a 0.50 L 2. 2.5 moles of carbon dioxide is placed in a 0.50 L flask.flask.

COCO22 (g) <--> C (s) + O (g) <--> C (s) + O22 (g) K (g) Kcc = 1.82 at 25 = 1.82 at 25ooCC

Calculate all equilibrium concentrations.Calculate all equilibrium concentrations.

3. Calculate all equilibrium concentrations if the 3. Calculate all equilibrium concentrations if the initial initial

concentration of Clconcentration of Cl22 is .100. is .100.

ClCl22 (g) <--> 2 Cl (g) (g) <--> 2 Cl (g) KKcc = 0.036 = 0.036

4. Calculate the equilibrium pressure of NO4. Calculate the equilibrium pressure of NO22 if the if the initial initial

pressure of Npressure of N22OO44 is 1.0 atm at 25 is 1.0 atm at 25oo C. C.

NN22OO44 (g) <--> 2 NO (g) <--> 2 NO22 (g) (g) K Kcc = 0.11 = 0.11

6161

Practice ProblemsPractice Problems

5. The initial concentrations of CO5. The initial concentrations of CO22 and H and H22 are 0.100 are 0.100 M. M.

Calculate all equilibrium concentrations.Calculate all equilibrium concentrations.

COCO22 (g) + H (g) + H22 (g) <--> CO (g) + H (g) <--> CO (g) + H22O (g) KO (g) Kcc = 0.64 = 0.64

6. BaCO6. BaCO33(s) + 2H(s) + 2H++(aq) <--> H(aq) <--> H22O(l) + COO(l) + CO22(g) +Ba(g) +Ba2+2+(aq)(aq)

Explain the effects of the following on the given Explain the effects of the following on the given equation:equation:

a) adding HCla) adding HCl b) adding BaClb) adding BaCl22

c) adding Hc) adding H22OO d) adding BaCOd) adding BaCO33

e) adding NaOHe) adding NaOH

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Practice Problems AnswersPractice Problems Answers

1. a)1. a) b) b)

c) 0.02c) 0.02 d) 0.25d) 0.25

e) lefte) left f) 1.6f) 1.6

2. 1.8, 3.22. 1.8, 3.2 3. 0.074, 0.0523. 0.074, 0.052

4. 1.1 atm4. 1.1 atm 5. 0.044, 0.044, 5. 0.044, 0.044, 0.056, 0.056 0.056, 0.056

6. a) right6. a) right b) left b) left c) nonec) none d) d) nonenone

e) right e) right

Kc = [NO2]2

[NO]2[O2]Kp =

P2 NO2

P2 NO P O2

The End!!!The End!!!

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Sample QuestionsSample QuestionsWrite the expression.Write the expression.

1. 1. 2 H2 H22 (g) + O (g) + O22 (g) <--> 2 H (g) <--> 2 H22O O (g)(g)

Kc = [H2O]2

[H2]2[O2]

6464

Sample QuestionsSample QuestionsWrite the expression. Write the expression.

2. 2. 4 H4 H22 (g) + 2 O (g) + 2 O22 (g) <--> 4 H (g) <--> 4 H22O O (g)(g)

Kc = [H2O]4

[H2]4[O2]2

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Sample QuestionsSample QuestionsWrite the expression. Write the expression.

3. 3. 2 H2 H22O (g) <---> 2 HO (g) <---> 2 H22 (g) + O (g) + O22 (g) (g)

Kc = [H2]

2 [O2]

[H2O]2

6666

Sample QuestionsSample Questions

3. 3. 2 H2 H22O (g) <---> 2 HO (g) <---> 2 H22 (g) + O (g) + O22 (g) (g)

Kc = [H2]2 [O2]

[H2O]2

2. 2. 4 H4 H22 (g) + 2 O (g) + 2 O22 (g) <--> 4 H (g) <--> 4 H22O O (g)(g)

Kc = [H2O]4

[H2]4 [O2]2

1. 1. 2 H2 H22 (g) + O (g) + O22 (g) <--> 2 H (g) <--> 2 H22O O (g)(g)

Kc = [H2O]2

[H2]2 [O2]= 5.0

= 25

= 0.20

If

6767


Write the equation and the KWrite the equation and the Kcc expression for the expression for the formation of two moles of gaseous ammonia from formation of two moles of gaseous ammonia from the elements in the standard state. the elements in the standard state.

3 H3 H22 (g) + N (g) + N22 (g) <--> 2 NH (g) <--> 2 NH33 (g) (g)

Kc = [NH3] 2

[H2] 3 [N2]

6868


KKc c for this reaction at 25for this reaction at 25ooC is 3.5x10C is 3.5x1088. .

Calculate KCalculate Kpp for this reaction. for this reaction.

3 H3 H22 (g) + N (g) + N22 (g) <--> 2 NH (g) <--> 2 NH33 (g) (g)

Kp = Kc (RT)n

Kp = 3.5 x 108 (0.0821x 298)

Kp = 5.8 x 105

6969


Calculate KCalculate Kcc for the reaction forming the for the reaction forming the elements from one mole of ammonia gas.elements from one mole of ammonia gas.

NHNH33 (g) <--> 3/2 H (g) <--> 3/2 H22 (g) + 1/2 N (g) + 1/2 N22 (g) (g)

Kc = [H2] 3/2 [N2]1/2

[NH3]

= 5.3 x 10-5

= 1

(3.5 x 108)1/2

7070

The Meaning of QThe Meaning of Q

• Q, Reaction QuotientQ, Reaction Quotient

• Same expression as the Same expression as the equilibrium expression.equilibrium expression.

• Non-equilibrium concentration Non-equilibrium concentration numbers may be used.numbers may be used.

• Used to determine if a system is at Used to determine if a system is at equilibrium and if not, which equilibrium and if not, which direction it will move to reach direction it will move to reach equilibrium.equilibrium.

7171


1. If [NO1. If [NO22] = 0.050 and [N] = 0.050 and [N22OO44] = 0.050, is the ] = 0.050, is the

system at equilibrium? If not, in which system at equilibrium? If not, in which direction does the reaction move to direction does the reaction move to come to equilibrium? come to equilibrium?

NN22OO44 (g) <--> 2 NO (g) <--> 2 NO22 (g) (g) KKcc = 0.36 = 0.36

Q = [NO2]

2

[N2O4]= = 0.050

(0.050) 2

(0.050)

Q < Kc forward direction (to right, products)

7272



system at equilibrium? If not, in which system at equilibrium? If not, in which

direction does the reaction move to direction does the reaction move to come to equilibrium?come to equilibrium?


Q = [NO2]

2

[N2O4]= = 0.62

(0.50) 2

(0.40)

Q > Kc reverse direction (to left, reactants)

7373



system at equilibrium? If not, in which system at equilibrium? If not, in which

direction does the reaction move to direction does the reaction move to come to equilibrium?come to equilibrium?


Q = [NO2]

2

[N2O4]= = 0 .36

(0.50) 2

(0.69)

Q = Kc equilibrium

7474

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. (Review of slides 9-11.)(Review of slides 9-11.)

SolutionSolution

Set up a table of concentrations:Set up a table of concentrations:


InitialInitial 2.00 2.00 0 0 0 0

ChangeChange

EquilibriumEquilibrium 0.660.66

+ 0.33+ 0.33- 0.66- 0.66

1.341.34

+0.66+0.66

0.330.33

7575

2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)


BeforeBefore 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[Cl 2]

[NOCl ]2

K[NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080


7676

Sample QuestionsSample Questions1. PCl1. PCl55 dissociates to produce PCl dissociates to produce PCl33 and Cl and Cl22. If . If

2.00 moles of PCl2.00 moles of PCl55 is placed in a 1.00 L is placed in a 1.00 L flask it is found that 10.0% of the original flask it is found that 10.0% of the original PClPCl55 had dissociated, at equilibrium. had dissociated, at equilibrium. Calculate KCalculate Kcc..

PClPCl55 <--> PCl <--> PCl33 + Cl + Cl22

2.002.00 0 0 0 0- 0.200 + 0.200 +0 .200

0.2000.2001.80

Kc = [PCl3][Cl2]

[PCl5]= = 0.0222

(.200)(.200)

(1.80)

7777


2. If 1.50 moles of H2. If 1.50 moles of H22O are placed in a O are placed in a 3.00 L flask 1.20 moles of H3.00 L flask 1.20 moles of H22O remain O remain at equilibrium. Calculate Kat equilibrium. Calculate Kcc. .

2 H2 H22O <--> 2 HO <--> 2 H22 + O + O22

0.5000.500 0 0 0 0- 0.100 + 0.100 + 0.050

0.050 0.100 0.400

Kc = [H2]2 [O2]

[H2O]2= = 0.0031

(.100)2(.050)

(0.400)2

Can we use only Moles?...

No….reason is all K’s are relative, that is since all reactions are done in different containers, moles and size of the container determine the K.

The End!!!The End!!!

7878


3. When 0.020 mole N3. When 0.020 mole N22O and 0.0560 mole O and 0.0560 mole OO22 are placed in a 1.00 L flask it is are placed in a 1.00 L flask it is found that there is 0.020 mole NOfound that there is 0.020 mole NO22 at at equilibrium. Calculate Kequilibrium. Calculate Kcc..

2 N2 N22O + 3 OO + 3 O22 <--> 4 NO <--> 4 NO22

0.0200.020 0.0560 0 0.0560 0- 0.010 - 0.015 +0 .020

0.020 0.041 0.010

Kc = [NO2]4

[N2O]2 [O2]3= = 23

(0.020)4

(.010)2(.041)3

7979


4. Some solid ammonium hydrogen sulfide is 4. Some solid ammonium hydrogen sulfide is placed in a flask at 298 K. At equilibrium, the placed in a flask at 298 K. At equilibrium, the pressure was 0.660 atm. Calculate Kpressure was 0.660 atm. Calculate Kpp and K and Kcc..

NHNH44HS <--> NHHS <--> NH33 + H + H22S S somesome 0 0 0 0

- some + 0.330 + 0.330

0.330 0.330 some - some

Kp = PNH3PH2S = (0.330)2 = 0.109 Kp = Kc(RT)2 0.109 = Kc (0.0821x298)2

Kc = 0.000182

8080


2 H2 H22S (g) <--> 2 HS (g) <--> 2 H22 (g) + S (g) + S22 (g) (g)8.008.00 00 0 0

- 1.20 + 1.20 + 0.60

0.60 1.20 6.80

Kp = PH

22 PS

2

PH2S2

= = 0.019(1.20)2 (0.60)

(6.80)2

8181

Sample ProblemsSample Problems 1. Calculate the equilibrium pressure of 1. Calculate the equilibrium pressure of

CO if 25 g COBrCO if 25 g COBr22 is placed in a 10.0 L is placed in a 10.0 L flask at 298 K.flask at 298 K.

CO (g) + BrCO (g) + Br22 (g) <--> COBr (g) <--> COBr22 (g) K (g) Kpp = 0.19 = 0.19

P = nRT/V

P = (0.13 molmol)(0.0821 L atm/molKL atm/molK)(298KK)/10.0 LL)

P = 0.32 atm

25 g mole

187.8 g= 0.13 mole

8282

Sample ProblemsSample Problems 1. Calculate the equilibrium pressure of 1. Calculate the equilibrium pressure of

CO if 25 g COBrCO if 25 g COBr22 is placed in a 10.0 L is placed in a 10.0 L flask at 298 K.flask at 298 K.

CO (g) + BrCO (g) + Br22 (g) <--> COBr (g) <--> COBr22 (g) K (g) Kpp = 0.19 = 0.19

- x

0.32 - x

0 0 0.32

+ x+ x

x x

Kp = PCOBr2

PCOPBr2

= = 0.19 (0.32 - x)

x2

x = 0.30atm = PCO

8383

Sample ProblemSample Problem 2. Calculate all equilibrium concentrations 2. Calculate all equilibrium concentrations

if 0.500 moles of HI is placed in a 2.00 L if 0.500 moles of HI is placed in a 2.00 L flask.flask.

2 H2 HII <--> H <--> H22 + + II22 KKcc = 0.018 = 0.018

- 2x + x + x

x x0.250 - 2x

0.250 0 0

Kc = [H2][I2]

[HI]2= = 0.018

x2

(0.250 - 2x)2

x = 0.026 M = [H2] = [I2]

[HI] = 0.20 M

8484

Sample ProblemSample Problem 3. Calculate all equilibrium concentrations when 3. Calculate all equilibrium concentrations when

the initial concentrations of SOthe initial concentrations of SO22 and NO and NO22 are are 0.0500.0.0500.

SOSO22 + NO + NO22 <--> NO + SO <--> NO + SO33 K Kcc = 85.0 = 85.0

- x

0.0500 - x

0.0500 0 0 0.0500 - x + x + x

0.0500 - x x x

Kc = [NO][SO3]

[SO2][NO2]= = 85.0

x2

(0.0500 - x)2

x = 0.0451 M = [NO] = [SO3]

[SO2] = [NO2] = 0.0049 M

chemical equilibrium chapter 16 all bold numbered problems

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