Download Equilibrium

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
Chemical equilibrium is described by an
equilibrium constant expression that
relates the concentrations of reactants and
products.
• List the characteristics of chemical equilibrium.
• Write equilibrium expressions for systems that are at
equilibrium.
• Calculate equilibrium constants from concentration data.
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
Concentration
N2(g) + 3H2(g)  2NH3(g)
Reactant
All concentrations
stop changing –
concentrations
reactant concentrations
are not zero
initially decrease
Product concentration
initially increases
H
2
NH3
N2
Time
Reversible Reactions
Reaction going to completion: almost
complete conversion of reactants to
Double arrow
products
Most reactions do not go to completion
–known as reversible reactions
Forward:
Reverse:
N2(g) + 3H2(g)  2NH3(g)
2NH3(g)  N2(g) + 3H2(g)
Reversible: N2(g) + 3H2(g) ↔ 2NH3(g)
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
Reaction Rates
Chemical Equilibrium
Forward
Rate
Reverse
Rate
Equilibrium
Forward = Reverse Rate
Time
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 Constant
aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant
[A]a[B]b
[X] = molar concentration of quantity
under equilibrium conditions
Exponents come from coefficients in
balanced chemical equation
Keq 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
Heterogeneous Equilibrium
Occurs when reactants & products present
in more than one physical state
C2H5OH(g)
C2H5OH(l)
Ethanol in closed flask
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 used
Equilibrium 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
Position 1
Position 2
Position
3
[HI]
K
eq
eq
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
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
[0.533] [1.600]3
(L2/mol2)
Keq of Some Common Reactions
Reaction
Keq
2H2(g) + O2(g) ↔ 2H2O(l)
1.4x1083 298 K
CaCO3(s) ↔ CaO(s) + CO2(g)
1.9x10-23 298 K
1.0 1200 K
3.4 1000 K
2SO2(g) + O2(g) ↔ 2SO3(g)
-21 298 K
1.6x10
C(s) + H2O(g) ↔ CO(g) + H2(g)
10.0 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
When changes are made to a system at
equilibrium, the system shifts to a new
equilibrium position.
• Describe how various factors affect chemical equilibrium.
• Explain how Le Châtelier’s principle applies to
equilibrium systems.
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
1.0000 0.10000 0.05900 0.02000 1.178
0.99254 0.07762 0.06648 0.02746 3.933
Instantaneously
New
equilibrium (by
position
injection)
has reduced
raise COvalue
concentration
of
CO from instantaneous
to 1.0000 M value of 1.0000
Equilibrium
Stress
of additional
becomes
reactant
unbalanced
relieved
(partially)
by producing
more product
Rate of forward
rxn increases,
get
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 right
CO 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)
Product
Removal
Reactant
addition
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. - Volume
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
Start
Compress
Initial
More
Compress
product
Final
forms
Temporarily
have nonequilibrium
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. - Temperature
Exothermic Reaction
+
+
Lower T: Product
removal
+
+
Raise T:
Product addition
+
+
Endothermic Reaction
heat + Co(H2O)62+ + 4ClCoCl42- + 6H2O
Raise T: Reactant addition
heat
heat + Co(H2O)62+ + 4ClCoCl42- + 6H2O
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
Equilibrium constant expressions can be
used to calculate concentrations and
solubilities.
• Determine equilibrium concentrations of reactants and
products.
• Calculate the solubility of a compound from its solubility
product constant.
• Explain the common ion effect.
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 – trickier
Have 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 dissolved
Ksp = (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
Current but not
called the ion product
necessarily
equilibrium values
• 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-]
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 Precipitates
Mix 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 Precipitates
Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-(aq)
Ksp = [Fe3+]eq4[Fe(CN)64-]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 form
Possible 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 Effect
Lead chromate solubility at 298 K
2+(aq) + CrO 2-(aq)
PbCrO4(s)

Pb
4
Quantities
-7 mol/L
Pure water,
s
=
4.8x10
produced just from
2Total
CrO
4
Much lower
in 0.01Mof
K2solid
CrO4
dissolution
Ksp = [Pb2+][CrO42-] = 2.3x10-13
Pure water: s = [CrO42-] = [Pb2+] = 4.8x10-7
K2CrO4 solution: [Pb2+][CrO42- + 0.01] = Ksp
[Pb2+]= 2.3x10-11
-2 = 2.3x10
-13
2-x1.0x10
2+ solubility
Common
CrO-11
has
reduced
Pb
Ksp= 2.3x10
4
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 left
Same effect if add Pb(NO3)2 to solution –
common ion now Pb2+
Using Ksp and Common Ion
CuCO3(s) ↔ Cu2+(aq) + CO32-(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 Ion
CuCO3(s) ↔ Cu2+(aq) + CO32-(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