Chapt. 17 – Chemical Equilibrium 17.1A State of Dynamic Balance 17.2 Factors Affecting Chemical...

Chapt. 17 – Chemical Equilibrium

17.1 A State of Dynamic Balance

17.2 Factors Affecting Chemical Equilibrium

17.3 Using Equilibrium Constants

Section 17.1 A State of Dynamic Balance

• List the characteristics of chemical equilibrium.

• Write equilibrium expressions for systems that are at

equilibrium.

• Calculate equilibrium constants from concentration data.

Chemical equilibrium is described by an equilibrium constant expression that relates the concentrations of reactants and products.

Section 17.1 A State of Dynamic Balance

Key Concepts

• A reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.

• The equilibrium constant expression is a ratio of the molar concentrations of the products to the molar concentrations of the reactants with each concentration raised to a power equal to its coefficient in the balanced chemical equation.

• The value of the equilibrium constant expression, Keq, is constant for a given temperature.

What is Equilibrium?

N2(g) + 3H2(g) 2NH3(g) DG0 = -33.1 kJ

Spontaneous, slow reaction under standard conditions

At higher temperature (723 K) and pressure reaction proceeds at a practical rate

Next slide shows reaction progress if start with 1 mol N2 and 3 mol H2

Reactant concentrations

initially decrease

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

Time

Con

cent

ratio

n

H2

NH3

N2

Product concentration initially increases

All concentrations stop changing – reactant concentrations are not zero

Reversible Reactions

Reaction going to completion: almost complete conversion of reactants to products

Most reactions do not go to completion –known as reversible reactionsForward: N2(g) + 3H2(g) 2NH3(g)

Reverse: 2NH3(g) N2(g) + 3H2(g)

Reversible: N2(g) + 3H2(g) ↔ 2NH3(g)

Double arrow

Reversible Reactions

N2(g) + 3H2(g) ↔ 2NH3(g)Time zero

• Reactants at max concentration, forward rate at maximum, reverse rate at zero

Time prior to equilibrium• Reactant concentration lower, forward

rate slower, some reverse reaction

At equilibrium• Forward and reverse rates equal – no

further concentration changes

Chemical Equilibrium

N2(g) + 3H2(g) ↔ 2NH3(g)

Forward and reverse reactions balance each other because

Rateforward = Ratereverse

Does not mean concentrations of reactants and products are equal

• Can be equal in some special cases

Chemical Equilibrium

Forward Rate

Reverse Rate

Equilibrium

Time

Forward = Reverse Rate

Rea

ctio

n R

ates

Dynamic Equilibrium

State of equilibrium not a static one• Some reactant molecules are always

changing into product molecule• Some product molecules are always

changing into reactant molecules• Only net concentrations of reactants and

products don’t change

Equilibrium Constant

Law of Chemical Equilibrium: at a given temperature, a chemical system may reach a state in which a particular ratio of reactant and product concentrations reaches a constant value

aA +bB cC +dD

Equilibrium constant

Equilibrium ConstantaA +bB ↔ cC +dD

Keq = [C]c[D]d = equilibrium constant

[A]a[B]b

[X] = molar concentration of quantity under equilibrium conditionsExponents come from coefficients in balanced chemical equationKeq is temperature dependent

Units of Keq vary


aA +bB ↔ cC +dD


[A]a[B]b

Keq >> 1 numerator >> denominator• Reaction as written above favors

production of products

Keq <<1 denominator >> numerator• Reaction as written above favors

production of reactants


aA +bB ↔ cC +dD


[A]a[B]b

Rules for writing an expression for Keq differ for homogeneous and heterogeneous equilibrium

• Homogeneous – all reactants and products in same physical state

• Heterogeneous – not homogeneous

Keq – Homogeneous Equilibrium

Homogeneous when all reactants and products are in same physical state

H2(g) + I2(g) ↔ 2HI(g)

Keq = _[HI]2

[H2] [I2]

Keq = 49.7 at 731 K• Note: no units for this particular reaction

Keq – Homogeneous Equilibrium

N2(g) + 3H2(g) ↔ 2NH3(g)

Keq = [NH3]2

[N2] [H2]3

Units = L2/mol2

Practice

Homogeneous Equilibrium Constants

Problems 1(a-e), 2 page 601

Problem 44 (a-b) page 626

Problems 1- 5 page 988

Occurs when reactants & products present in more than one physical state

Ethanol in closed flask

C2H5OH(l) ↔ C2H5OH(g)

Heterogeneous Equilibrium

C2H5OH(l)

C2H5OH(g)

Keq – Heterogeneous Equilibrium

Rule: Pure liquids and solids don’t appear in the equilibrium expression because at constant temperature their concentrations don’t change when the amount present changes

C2H5OH(l) C2H5OH(g)

Keq = [C2H5OH(g)]I2 (s) I2(g)

Keq = [I2 (g)]

Keq – Heterogeneous Equilibrium

2NaHCO3(s) ↔

Na2CO3(s) + CO2(g) + H2O(g)

Keq = [CO2(g)] [H2O(g)]

Practice

Heterogeneous Equilibrium Constants

Problems 3 (a-e), 4 page 603

Problems 45 – 47, 49 page 626

Problems 6 – 8 page 988

Numerical Value of Keq

For a given reaction, final values of reactant and product concentrations will satisfy Keq expression regardless of initial concentrations usedEquilibrium position: set of final concentrations of reactants and products

One value of Keq, lots of possible equilibrium positions

Notation for Concentrations

[X]0 = molar concentration of X at time zero = initial concentration

[X]eq = molar concentration of X at equilibrium = final concentration

Numerical Value of Keq

H2(g) + I2(g) ↔ 2HI(g)

Keq = _[HI]2

[H2] [I2]

[H2]0 [I2]0 [HI]0 [H2]eq [I2]eq [HI]eq Keq

1.00 2.00 0.00 0.066 1.07 1.86 49.70

0.0 0.0 5.00 0.55 0.55 3.90 49.70

1.00 1.00 1.00 0.332 0.332 2.34 49.70

Three different equilibrium positions yield the same value of Keq

Position 1Position 2Position 3

Value of Keq – Problem 17.3

N2(g) + 3H2(g) ↔ 2NH3(g)

Keq = [NH3]2

[N2] [H2]3

[NH3] = 0.933 mol/L

[N2] = 0.533 mol/L [H2] = 1.600 mol/L

Value of Keq?

Keq = . [0.933]2 .= 0.399 (L2/mol2) [0.533] [1.600]3

Keq of Some Common Reactions

Reaction Keq2H2(g) + O2(g) ↔ 2H2O(l) 1.4x1083 298 K

CaCO3(s) ↔ CaO(s) + CO2(g)1.9x10-23 298 K1.0 1200 K

2SO2(g) + O2(g) ↔ 2SO3(g) 3.4 1000 K

C(s) + H2O(g) ↔ CO(g) + H2(g)1.6x10-21 298 K10.0 1100 K

Which ones favor production of products?

Practice

Calculating value of Keq

Problems 5 – 7, 11 page 605

Problem 48 page 626

Problems 9 – 10 page 988

Section 17.2 Factors Affecting Chemical Equilibrium

• Describe how various factors affect chemical equilibrium.

• Explain how Le Châtelier’s principle applies to equilibrium systems.

When changes are made to a system at equilibrium, the system shifts to a new equilibrium position.

Section 17.2 Factors Affecting Chemical Equilibrium

Key Concepts

• Le Châtelier’s principle describes how an equilibrium system shifts in response to a stress or a disturbance.

• When an equilibrium shifts in response to a change in concentration or volume, the equilibrium position changes but Keq remains constant. A change in temperature, however, alters both the equilibrium position and the value of Keq.

Le Châtelier’s Principle

Principle: If a stress is applied to a system at equilibrium, the system shifts in the direction that partially relieves the stress.

Stresses include• Change in concentration• Change in volume (pressure)• Change in temperature

Le Châtelier’s P. - Concentration

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)_________________________________________________________________________________________________

[CO]eq [H2]eq [CH4]eq [H2O]eq Keq

0.30000 0.10000 0.05900 0.02000 3.933

Instantaneously (by injection) raise CO concentration to 1.0000 M

Equilibrium becomes unbalanced

Rate of forward rxn increases, get

0.99254 0.07762 0.06648 0.02746 3.933New equilibrium position has reduced value of CO from instantaneous value of 1.0000

Stress of additional reactant relieved (partially) by producing more product

1.0000 0.10000 0.05900 0.02000 1.178


CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)_______________________________________________________________________________________________________________________________________

What will happen to equilibrium if:

A desiccant is used to remove H2O? Equilibrium shifts to the rightCO is removed from reaction vessel? Equilibrium shifts to the left

H2O is injected into the reaction vessel? Equilibrium shifts to the left


CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

Reactant addition

Product Removal

Reactant Removal

Product Addition

Le Châtelier’s P. - Volume

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

4 moles of reactant, 2 moles of product

Decrease V (increase P) at constant T

Le Châtelier’s P. - VolumeCO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

Compress

Final

Start More product forms

Compress

Initial

Temporarily have non-equilibrium

concentrations


CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

4 moles of reactant, 2 moles of product

Decrease V (increase P) at constant T

Stress (increased pressure) relieved by formation of more product

• P = n R at constant V, T n = # moles• If n decreases, P decreases

P is lowered but not to its original value


When volume of equilibrium mixture of gases reduced, net change occurs in direction that produces fewer moles of gas

When volume increased, net change occurs in direction that produces more moles of gas

H2(g) + I2(g) ↔ 2HI(g) - no effect of V

Le Châtelier’s P. - Temperature

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

DH0 = - 206.5 kJ

Have exothermic reaction – can think of heat as a “product”

Effect of raising T is like adding more product – equilibrium shifts to the left

This approach explains why Keq has a temperature dependence

Le Châtelier’s P. - TemperatureExothermic Reaction

Endothermic Reaction

+

++

++

+

heat + Co(H2O)62+ + 4Cl- CoCl42- + 6H2O

heat + Co(H2O)62+ + 4Cl- CoCl42- + 6H2O

heat

Raise T: Product addition

Lower T: Product removal

Raise T: Reactant addition

Lower T: Reactant removal

Summary: Le Châtelier’s Principle and Temperature

Raising temperature of equilibrium mixture shifts equilibrium condition in direction of endothermic reaction

Lowering temperature causes a shift in direction of exothermic reaction

Practice

Le Châtelier’s Principle

Problems 14 - 15 page 611

Problems 54 - 63, pages 626-27

Section 17.3 Using Equilibrium Constants

• Determine equilibrium concentrations of reactants and products.

• Calculate the solubility of a compound from its solubility product constant.

• Explain the common ion effect.

Equilibrium constant expressions can be used to calculate concentrations and solubilities.

Section 17.3 Using Equilibrium Constants

Key Concepts

• Equilibrium concentrations and solubilities can be calculated using equilibrium constant expressions.

• Ksp describes the equilibrium between a sparingly soluble ionic compound and its ions in solution.

• If the ion product, Qsp, exceeds the Ksp when two solutions are mixed, a precipitate will form.

• The presence of a common ion in a solution lowers the solubility of a dissolved substance.

Calculating Equilibrium Concentrations

If know Keq and concentrations of all but one of the reactants and products, can solve for unknown concentration

CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

0.850M 1.333M ?M 0.286M

Keq = [CH4(g)] [H2O(g)] [CO(g)] [H2(g)]3

[CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)]


CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)

0.850M 1.333M ?M 0.286M

[CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)]

Keq = 3.933

[CH4(g)] = 3.933 (0.850)(1.333)3/ (0.286)

[CH4(g)] = 27.7 mol/L = [CH4(g)]eq


Example Problem 17.4

2H2S(g) ↔ 2H2(g) + S2(g)

1.84x10-1M ?M 5.40x10-2M

Keq = [H2(g)]2 [S2(g)] = 2.27x10-3

[H2S(g)]2

[H2(g)]2 = Keq [H2S(g)]2 / [S2(g)] = 1.42x10-3

[H2] = 3.77x10-2 mol/L

Practice

Calculating equilibrium concentrations

Problems 16 (a-c), 17 page 613

Problem 71 page 627

Problems 11(a-b) page 988

Solubility Equilibria

Solubility equilibria applies to all compounds with finite solubility, but is most commonly applied to dissolution of those with low solubility

BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)

Heterogeneous equilibrium

Keq = [Ba2+(aq)] [SO42-(aq)]

Has special symbol Ksp


BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)

Ksp = [Ba2+] [SO42-] = 1.1x10-10 @298 K

Ksp = solubility product constant

Ksp = equilibrium constant for the dissolution of a sparingly soluble ionic compound in water

Only ions appear in Ksp, but some amount of solid (however small) must be present to achieve equilibrium


Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)

Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K

Table 17.3, page 615 lists solubility product constants sorted by type of anion (carbonate, phosphate, etc.)

Ksp can be used to calculate solubility of salt or concentration of one of ions involved in equilibrium

Calculating Solubility from Ksp

AgI(s) ↔ Ag+(aq) + I-(aq)

Ksp = [Ag+(aq)] [I-(aq)] = 8.5x10-17 @298 K

s = solubility of AgI(s) in mol/L

1 mol Ag+(aq) forms per mol AgI dissolved

1 mol I-(aq) forms per mol AgI dissolved

Ksp = [Ag+(aq)] [I-(aq)] = s2 = 8.5x10-17

s = 9.2x10-9 mol/L @298 K

Note: actual units of Ksp are “ignored”

Practice

Calculating solubility of 1:1 salts

Problems 20 (a-c), 21 page 616

Problem 22(a) page 617

Problem 31, page 622

Problem 70 page 627

Problem 12 page 988

Calculating Ion Concentration from Ksp


Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K[OH-] in a saturated solution?This is a non 1:1 electrolyte – trickierHave to pay attention to stoichiometry

s = solubility of Mg(OH)2 in mol/L

1 mol Mg2+ per mol Mg(OH)2 dissolved

2 mol OH- per mol Mg(OH)2 dissolved

Calculating Ion Concentration from Ksp


Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K

[OH-] in a saturated solution?

s = solubility of Mg(OH)2 in mol/L

1 mol Mg2+ per mol Mg(OH)2 dissolved

2 mol OH- per mol Mg(OH)2 dissolvedKsp = (s) (2 s)2 = 4s3 = 5.6x10-12

s = 1.1 x10-4 mol/L [OH-] = 2 s

Practice

Calculating solubility & ion concentrations for arbitrary salts

Problems 22 (b-c), 23, 24 page 617

Problems 78, 83 page 628

Problems 13 -15 page 988

Predicting Precipitates

Mix two soluble ionic compounds – sometimes a precipitate forms

Can use Ksp to predict: Use quantity called the ion product

• Symbol Qsp

• Same functional form as Ksp • Concentrations used may or may not be

equilibrium concentration

AB(s) ↔ A+(aq) + B-(aq) Qsp=[A+][B-]

Current but not necessarily

equilibrium values


AB(s) ↔ A+(aq) + B-(aq)

Qsp=[A+][B-] current concentrations

Ksp= [A+]eq[B-]eq equilibrium concentrations

Compare Ksp to Qsp

1. Qsp < Ksp unsaturated no precipitate

2. Qsp = Ksp saturated no precipitate

3. Qsp > Ksp supersaturated precipitate

For #3, solid will form until Qsp = Ksp

Predicting PrecipitatesMix equal volumes of (soluble) 0.1M iron(III) chloride and potassium hexacyanoferrate(II)

4FeCl3(aq)+ 3K4Fe(CN)6(aq) 12KCl(aq) + Fe4(Fe(CN)6)3(s)

Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-

(aq)

Ksp = [Fe3+]4[Fe(CN)64-]3 = 3.3x10-41

Question: does a precipitate form?

Predicting PrecipitatesFe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)6

4-(aq)

Ksp = [Fe3+]eq4[Fe(CN)6

4-]eq3

Equal volumes of 0.10M FeCl3 & K4Fe(CN)6

solutions used

[Fe3+] = 0.050 M [Fe(CN)64-] = 0.050 M

Qsp = [Fe3+]4[Fe(CN)64-]3 = [0.050]4[0.050]3

Qsp = 7.8x10-10 Ksp = 3.3x10-41

Qsp > Ksp so precipitate will formPossible to compute moles of solid formed


Example Problem 17.7

Mix 100 mL 0.0100M NaCl with 100 mL 0.0200M Pb(NO3)2

Does PbCl2 precipitate?

PbCl2(s) ↔ [Pb2+][Cl-]2 Ksp = 1.7x10-5

Qsp = [0.0100][0.0050]2 = 2.5x10-7

Qsp < Ksp

No precipitate

Practice

Determining if precipitate will form

Problems 25(a-b), 26 page 619



Common Ion Effect

Common ion is an ion common to two or more ionic compounds

• KI, AgI I- is the common ion• CaCl2, Ca(OH)2 Ca2+ is the common ion

Common ion effect is the lowering of the solubility of an ionic substance by the presence of a common ion

Common Ion EffectLead chromate solubility at 298 K

PbCrO4(s) Pb2+(aq) + CrO42-(aq)

Pure water, s = 4.8x10-7 mol/L

Much lower in 0.01M K2CrO4

Ksp = [Pb2+][CrO42-] = 2.3x10-13

Pure water: s = [CrO42-] = [Pb2+] = 4.8x10-7

K2CrO4 solution: [Pb2+][CrO42- + 0.01] = Ksp

Common CrO42- has reduced Pb2+ solubility

Total CrO42-

Ksp= 2.3x10-11x1.0x10-2 = 2.3x10-13

[Pb2+]= 2.3x10-11

Quantities produced just from dissolution of solid

Common Ion Effect & L-C’s Principle

PbCrO4(s) ↔ Pb2+(aq) + CrO42-(aq)

Pure water: [Pb2+] = [CrO42-] = s = 4.8x10-7

K2CrO4 solution: [Pb2+][CrO42- + 0.01] = Ksp

[Pb2+]= 2.3x10-11

CrO42- is common ion

Higher common ion concentration shifts equilibrium to leftSame effect if add Pb(NO3)2 to solution – common ion now Pb2+

Using Ksp and Common IonCuCO3(s) ↔ Cu2+(aq) + CO3

2-(aq)

Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10

s = solubility of CuCO3 (s) in solution of 0.10 M K2CO3 (Prob. 17.5 & strategy on p 621)

(s)(0.10 + s) = 2.5x10-10 (use quad. eqn.)

(s)(0.10) = 2.5x10-10 (approximation)

Using Ksp and Common IonCuCO3(s) ↔ Cu2+(aq) + CO3

2-(aq)

Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10

(s)(0.10 + s) = 2.5x10-10 (using quad. eqn.)

(s)(0.10) = 2.5x10-10 (using approximation)

For this problem, exact method (using quadratic equation) and approximate method yield same answer; s = 2.5x10-9

Commonly used “trick” to quickly solve variety of equilibrium problems

Chapt. 17 – Chemical Equilibrium 17.1A State of Dynamic Balance 17.2 Factors Affecting Chemical...

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