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
Equilibrium Constant
aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant
[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
Equilibrium Constant
aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant
[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
Chapt. 17 – Chemical Equilibrium
17.1 A State of Dynamic Balance
17.2 Factors Affecting Chemical Equilibrium
17.3 Using Equilibrium Constants
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
Le Châtelier’s P. - Concentration
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
Le Châtelier’s P. - Concentration
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
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
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
Le Châtelier’s P. - Volume
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
Chapt. 17 – Chemical Equilibrium
17.1 A State of Dynamic Balance
17.2 Factors Affecting Chemical Equilibrium
17.3 Using Equilibrium Constants
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)]
Calculating Equilibrium Concentrations
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
Calculating Equilibrium Concentrations
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
Solubility Equilibria
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
Solubility Equilibria
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
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
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
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
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
Predicting Precipitates
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
Predicting Precipitates
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
Problems 72, 73 page 627
Problems 16, 17 page 988
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
Top Related