Transcript No Slide Title

Daniel L. Reger
Scott R. Goode
David W. Ball
www.cengage.com/chemistry/reger
Chapter 14
Chemical Equilibrium
Chemical Equilibrium
• Chemical Equilibrium: A state in which
the tendency of the reactants to form
products is balanced by the tendency of
the products to form reactants.
• Could also be defined as a system in
which the rates of the forward and
reverse reactions are the same.
 No observable changes occur at equilibrium.
Equilibrium Systems
• Phase Change
H2O (g) ⇌ H2O (l)
At equilibrium, liquid water evaporates at
the same rate the water vapor condenses.
• Chemical Equilibrium
N2(g) + 3H2(g) ⇌ 2NH3(g)
The concentrations of all species become
constant before all the reactants are
consumed.
Reaching Equilibrium
• Consider the reaction
2NO2(g) ⇌ N2O4(g)
• We start with brown
NO2 in a flask and
observe as colorless
N2O4 forms.
2NO2(g) ⇌ N2O4(g)
Starting with NO2
Starting with N2O4
The Law of Mass Action
For a general chemical reaction
aA + bB ⇌ cC + dD
the equilibrium constant is given by
c
c
[C] [D]
K eq 
a
b
[A] [B]
The Equilibrium Constant
2NO2(g) ⇌ N2O4(g)
Concentrations of Nitrogen Dioxide and Dinitrogen
Tetroxide and the Equilibrium Constant at 317 K
Initial Concentrations, M
[NO2]
[N2O4]
Equilibrium Concentrations, M
[NO2]
[N2O4]
K eq 
[N2O 4 ]
[NO 2 ] 2
2.00 × 10-2
0.00
1.03 × 10-2
4.86 × 10-3
45.8
0.00
1.00 × 10-2
1.03 × 10-2
4.86 × 10-3
45.8
3.00 × 10-2
1.00 × 10-2
1.85 × 10-2
1.57 × 10-2
45.9
4.00 × 10-2
0.00
1.61 × 10-2
1.19 × 10-2
45.9
The Equilibrium Constant
• 2NO2(g) ⇌ N2O4(g)
• Equilibriu m constant : K eq 
[N2O 4 ]
[NO 2 ]
2
 45.9
Examples
• 2O3(g) ⇌ 3O2(g)
K eq 
[O 2 ]
3
[O 3 ]
2
• CO(g) + H2O(g) ⇌ H2(g) + CO2(g)
[H2 ][CO 2 ]
K eq 
[CO][H 2O]
Kinetics and Equilibrium
• For any elementary reaction that reaches
equilibrium
aA + bB ⇌ cC + dD
 the forward rate = kf[A]a[B]b
 the reverse rate = kr[C]c[D]d
• At equilibrium the forward rate is equal to the
reverse rate so
kf[A]a[B]b = kr[C]c[D]d
• The ratio of the two rate constants is a constant
that we call Keq.
c
d
kf
[C] [D]
 K eq 
kr
[A] a [B] b
Keq and the Chemical Equation
• Keq refers to a specific chemical
equation.
• Reaction 1
K1 
Reaction 2
K2 
2H2(g) + O2(g) ⇌ 2H2O(g)
[H2 O] 2
[H2 ] 2 [O 2 ]
H2(g) + 1/2O2(g) ⇌ H2O(g)
[H2O]
1/2
[H2 ][O2 ]
 K1
Keq and the Chemical Equation
• 2H2O(g) ⇌ 2H2(g) + O2(g)
• K3 
2
[H2 ] [O 2 ]
[H2 O]
2
1

K1
• The form of the equilibrium expression
depends on the balanced equation.
• For any reaction, Krev = 1/Kfor
Test Your Skill
N2O4(g) ⇌ 2NO2(g) Keq = 4.63 × 10
(a) Determine Keq for the reaction
2NO2(g) ⇌ N2O4(g)
(b) determine Keq for the reaction
NO2(g) ⇌ 1/2N2O4(g)
-3
Pressure and Concentration
• N2O4 ⇌ 2NO2
2
[NO 2 ]
• K c  [N O ]
2 4
• Subscript c indicates molar
concentration.
2
pNO
2
• Define K p 
Subscript p indicates
pN2O4
pressure.
• P = (n/V)RT =M RT
Relating Kp and Kc
Kp 
2
pNO
2
pN2O4
n
n
For any gas p  RT . Because  [Conc]
V
V
([NO 2 ]RT ) 2
Kp 
 K c RT
[N2 O 4 ]RT
• In general Kp = Kc(RT)Dn, where
 Dn = moles gaseous product
– moles gaseous reactant.
• When Dn = 0, Kp = Kc.
Example: Converting Kp and Kc
• Kc is 5.0 × 106 at 700 K
2SO2 (g) + O2(g) ⇌ 2SO3(g)
Calculate Kp .
Test Your Skill
• Calculate Kp for
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Given Kc = 4.00 at 425° C.
Reactions Not at Equilibrium
• For the reaction conditions shown
2NO2(g) ⇌ N2O4(g)
it takes time to reach equilibrium.
The Reaction Quotient
• Reaction Quotient, Q, has the same
algebraic form as Keq, but is evaluated
with current concentrations, rather than
equilibrium concentrations.
aA + bB ⇌ cC + dD
Q
[C] c [D]d
a
[A] [B]
b
Determining Direction of Reaction
• Q < Kc:ratio of products to
reactants is too small, reaction will
proceed in forward direction to
reach equilibrium.
• Q = Kc:the system is at equilibrium.
• Q > Kc:ratio of products to
reactants is too large, reaction will
proceed in reverse direction to
reach equilibrium.
Determining Direction of Reaction
Data refers to conditions where
Keq = [N2O4]/[NO2]2 = 0.45
Test Your Skill
• CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) Kc = 5.67
• Initial concentrations:
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)
0.100 M 0.200 M 0.500 M 0.800 M
• Determine in which direction the reaction will
proceed.
The Le Chatelier’s Principle
• Any change to a chemical reaction
at equilibrium causes the reaction
to proceed in the direction that
reduces the effects of the change.
Changing Concentration
or Partial Pressure
• Adding a reactant or product causes the
reaction to proceed in the direction that
consumes the added substance.
• Removing a reactant or product causes
the reaction to proceed in the direction
that produces the missing substance.
Changing Concentration
or Partial Pressure
• 2SO3(g) ⇌ 2SO2(g) + O2(g)
An increase in SO3 partial pressure
causes the formation of more SO2.
The Effect of a Volume Change
• A decrease in volume causes the
reaction to proceed in the direction that
decreases the number of moles of gas.
• An increase in volume causes the
reaction to proceed in the direction that
increases the number of moles of gas.
The Effect of a Volume Change
• 2SO3(g) ⇌ 2SO2(g) + O2(g)
• Increasing the volume causes the reaction to
proceed to the right.
• Decreasing the volume causes the reaction to
proceed to the left.
Changing Pressure by Adding an Inert Gas
• Changing the pressure of a system by
adding an inert gas does not change the
concentration or partial pressure of the
reactants or products.
• The equilibrium does not change when
an inert gas is added.
Changes in Temperature
• Heat is a “product” in an exothermic
reaction and a “reactant” in an
endothermic reaction.
• CO(g) + 2H2(g) ⇌ CH3OH(g)
DH = -18 kJ
• The reaction is exothermic (heat is a
product).
• Increasing temperature favors reactants.
Adding a Catalyst
• A catalyst increases the reaction rate but
does not affect the equilibrium
concentrations.
• Does not affect the value of Keq.
Equilibrium Calculations
• We will use a 5-step approach:
1. Write the balanced chemical equation.
2. Fill in a table, which we will call an iCe table,
with the concentrations of the various
species.
3. Write the algebraic expression for the
equilibrium constant.
4. Substitute concentrations from the iCe table
into the algebraic expression.
5. Solve the expression for the unknown
quantity or quantities.
Equilibrium Calculations
• You will likely see two kinds of
equilibrium calculations:
1. Equilibrium concentrations are given or
computed from other data and you
determine the value for Keq.
2. You are given starting concentrations and
Keq and calculate equilibrium
concentrations.
• The same 5-step approach works for
both types of problems.
Determining K from Experiment
• Consider the chemical reaction of
hydrogen and sulfur to make hydrogen
sulfide.
• Experiment shows that at equilibrium,
there are 2.50 mol H2, 1.35 × 10-5 mol
S2, and 8.70 mol H2S in a 12.0 L flask.
• Calculate Kc.
Determining K from Experiment
• 2H2(g) + S2(g) ⇌ 2H2S(g)
1. Balanced
equation.
• At equilibrium, there are 2.50 mol H2,
-5
1.35 x 10 mol S2, and 8.70 mol H2S
in a 12.0 L flask.

[H2] = 2.50 mol/12.0 L = 0.208 M
 [S2] = 1.35 x 10-5 mol/12.0 L
= 1.12 x10-6 M
 [H2S] = 8.70 mol/12.0 L = 0.725 M
2. Calculate
equilibrium
concentrations.
An iCe table is
not needed in
this first
example.
Determining K from Experiment
• Kc 
• Kc 
[H2 S] 2
3. Expression for
equilibrium
constant.
2
[H2 ] [S 2 ]
[0.725]
2
[0.208] [1.12  10 ]
2
• K c  1.08  10
-6
4. Substitute
concentrations
into expression
for equilibrium
constant.
7
5. Solve.
Determining K from Experiment II
• A scientist places 1.0 mol of HI in a
10.0-L flask. Experiments show that the
equilibrium concentration of I2 is
0.020 M.
• Calculate Kc for
2HI(g) ⇌ H2 (g)+ I2(g)
• The initial concentration of HI is
1.0mol/10.0 L or [HI] = 0.10 M.
Determining K from Experiment II
• Introducing the iCe table
1. Chemical
equation.
2HI(g) ⇌ H2 (g) + I2(g)
initial conc., M
Change, M
equil conc., M
0.10
?
0
?
0
0.020
2. iCe table with
starting information.
Determining K from Experiment II
2HI(g) ⇌ H2 (g)+ I2(g)
Initial conc., M
Change, M
Equil. conc., M
2. iCe table,
complete.
0.10
-0.04
0.06
Kc =
Kc =
0
+0.02
0.02
[H2 ][I2 ]
[HI]2
[0.02][0.02]
[0.06]
Kc = 0.11
2
0
+0.02
0.02
3. Algebraic
expression for
K.
4. Substitute
concentrations
from table.
5. Solve.
Equilibrium Calculations Given K and Initial
Concentrations
• If the initial concentration of CO is
0.028 M and H2 is 0.14 M, and Kc = 0.50
for
CO(g) + H2(g) ⇌ CH2O(g)
calculate the equilibrium concentrations
of all species.
• We will use our 5-step approach.
1. Write the balanced equation
CO(g) + H2(g) ⇌ CH2O(g)
Step 2. Set up iCe table
i, M
C, M
e, M
CO(g) + H2(g) ⇌ CH2O(g)
0.028 0.14
0.
-y
-y
+y
0.028-y 0.14-y
y
We have defined y as the change in
concentration, and it is unknown.
3. Write algebraic expression for Kc
CO(g) + H2(g) ⇌ CH2O(g)
[CH2O]
Kc =
[CO][H2 ]
4. Substitute from iCe table
i, M
C, M
e, M
CO(g) + H2(g) ⇌ CH2O(g)
0.028 0.14
0.
-y
-y
+y
0.028-y 0.14-y
y
[CH 2O]
Kc 
[CO][H 2 ]
y
0.50 
(0.028  y )(0.14  y )
Step 5. Solve
• Solve by quadratic formula y = 0.0018 or
2.1665
• Only one root will give possible
concentrations.
Value of y
[CO] = 0.028 - y =
[H2] = 0.14 - y =
[CH2O] =
y
=
0.0018
2.1665
0.026 M -2.1385 M
0.14 M
-2.0265 M
0.0018 M 2.1665 M
Step 6. Check
• Substitute numerical values to calculate
the equilibrium constant.
Kc = [CH2O] / [CO][H2]
= 0.0018 / (0.026 x 0.14)
= 0.49
• 0.49 is quite close to the expected 0.50,
so we can be confident that we did the
problem correctly.
Test Your Skill
• If the initial concentration of PCl5 is
0.100 M, and Kc = 0.60, what are the
equilibrium concentrations for the
reaction? First, calculate y, the change
in concentration.
• PCl5 (g) ⇌ PCl3(g) + Cl2(g)
Test Your Skill
• If the initial concentration of PCl5 is 0.100 M,
and Kc = 0.60, what are the equilibrium
concentrations for the reaction? First,
calculate y, the change in concentration.
• PCl5 (g) ⇌ PCl3(g) + Cl2(g)
• y = 0.087 or -0.687
 Choose 0.087 to get a result in which all the
concentrations are positive.
• [PCl5] = 0.100 - 0.087 = 0.013 M
[PCl3] = [Cl2] = 0.087 M
Heterogeneous Equilibria
• A heterogeneous system is one in which
the reactants and products are present in
more than one phase.
• The concentration of a pure solid or
liquid is a constant and is not included in
the equilibrium expression.
Heterogeneous Equilibria
• CaCO3(s) ⇌ CaO(s) + CO2(g)
[CaO][CO 2 ]
Kc 
[CaCO 3 ]
but [CaO] and [CaCO3] are solids,
and their concentrations do not
change, so
Kc = [CO2] and/or
Kp  PCO2
Test Your Skill
2NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g)
Kc =
2Hg(l) + Cl2(g) ⇌ Hg2Cl2(s)
Kc =
NH3(g) + HCl(g) ⇌ NH4Cl(s)
Kp =
Solubility Equilibria
• Solubility equilibria: reactions that
involve dissolving or forming of a solid
from solution.
AgNO3(aq) + NaCl(aq) ⇌ AgCl(s) + AgNO3(aq)
The net ionic equation is
Ag+(aq) + Cl-(aq) ⇌ AgCl(s)
The Solubility Product Constant
• For a partly soluble or insoluble solid
such as AgCl,
+
AgCl(s) ⇌ Ag (aq) + Cl (aq)
we define Ksp, the solubility product
constant, as
Ksp = [Ag+][Cl-]
Write Expressions for Ksp
Fe(OH)3(s) ⇌ Fe (aq) + 3OH (aq)
Ksp =
2+
PbCl2(s) ⇌ Pb (aq) + 2Cl (aq)
Ksp =
2+
3Ca3(PO4)2(s) ⇌ 3Ca (aq) + 2 PO4 (aq)
Ksp =
3+
-
Calculating the Solubility Product Constant
• A scientist prepares a saturated
solution of lead(II) iodide.
Independent measurements show
that the concentration of lead is
1.3 x 10-3 M.
• Calculate Ksp.
Test Your Skill
-5
• The solubility of Pb(IO3)2 is 4.5 × 10 M.
Calculate Ksp.
Solubility Calculations
-49
• Ksp for Ag2S is 1.6 × 10 Calculate the
-49
solubility in mol/L. Ksp = 1.6 × 10
Ag2S(s) ⇌ 2 Ag (aq) + S (aq)
+
2-
Test Your Skill
• Ksp for Ca(OH)2 is 1.3 × 10-6.
• Calculate the solubility in mol/L.
Solubility
• Solubilities of different compounds cannot be
predicted by ranking them in order of Ksp.
Compound
AgIO3
Ksp
Solubility, M
-8
3.1 × 10
-9
-4
1.8 × 10
-4
Ba(IO3)2
1.5 × 10
7.2 × 10
La(IO3)3
6.2 × 10-12
6.9 × 10-4
Solubility and the Common Ion Effect
• Common ion effect: the effect of adding a
solute to a solution that contains an ion in
common. In a precipitation reaction, the
common ion effect decreases the solubility of
the solid.
• The effect is consistent with Le Chatelier’s
principle:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
adding NaCl decreases the solubility of AgCl.
The Common Ion Effect
What is the solubility of Mg(OH)2 in a solution
-12
of 0.100 M NaOH(aq)? Ksp is 8.9 × 10 for
Mg(OH)2
2+
Mg(OH)2 (s) ⇌ Mg (aq) + 2OH (aq)
The Common Ion Effect
• Calculate the solubility for the same
solute in water with no added OH-.
• Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
Example: Ca(NO3)2 + Na2CO3
• 25.00 mL of 0.050 M Na2CO3 and
10.00 mL of 0.0020 Ca(NO3)2 are
mixed. Will a precipitate form?
• Using the solubility rules, the likely
precipitate is CaCO3.
Ksp(CaCO3) = 8.7 × 10
-9