Chapter 18 Chemical Equilibrium

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Chapter 18
Chemical Equilibrium
18.1 The Nature of Chemical
Equilibrium
Reversible Reactions
• A chemical reaction in which the products
re-form the reactants
Reactants ↔ Products
• Le Chatelier’s Principle
When system at equilibrium is disturbed
by application of a stress, it attains a new
equilibrium position that minimizes the
stress.
Chemical Equilibrium
• The state where the concentrations of all
reactants and products remain constant
with time
• All reactions carried out in a closed vessel
will reach equilibrium
Chemical Equilibrium
• Sometimes, if little product is formed,
equilibrium lies far to the left (towards
reactants)
Reactants
Products
• Sometimes, if little reactant remains,
equilibrium lies far to the right (towards
products)
Reactants
Products
Dynamic Equilibrium
• Reactions continue to take place
• Reactant molecules continue to be converted
to product
• Product continues to be converted to
reactant (reverse reaction)
• Forward and reverse reactions take place at
the same rate at equilibrium
Causes of Equilibrium
• Beginning of Reaction
Only reactant molecules exist, so only reactant
molecules may collide
• Middle
As product concentration increases, collisions
may take place that lead to the reverse reaction
• At Equilibrium
Rates of forward and reverse reactions are
identical
Suppose that compound A reacts with B to
form C and D.
The symbol
is used for system at
equilibrium.
If we add more A what would happen
to the amount of C at equilibrium?
Adding A increases the amount of C and
D at equilibrium.
If we add more C what would happen
to the amount of A at equilibrium?
adding more C increases the amount of
A and B at equilibrium.
If we add A what would happen to the
amount of B at equilibrium?
A reacts with B to form C and D.
Therefore, adding A decreases B.
Example - The Haber Process
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
• Hydrogen is consumed at
3x the rate of nitrogen
• Ammonia is formed at 2x
the rate at which nitrogen
is consumed
Points to Remember
1. Rxns in a closed system always reach
equilibrium
2. When a rxn has reached equilibrium, that
means the rate of the forward rxn equals
the rate of the reverse rxn
3. A system will remain at equilibrium until
the system is disturbed in some manner
The Equilibrium Constant
• For the balanced forward equation:
jA + kB → lC + mD
(j, k, l, m) are coefficients
K = [C]l [D]m
[A]j [B]k
• K is a constant called the equilibrium constant
• K varies depending on temperature and upon the
coefficients of the balanced equation
• K is determined experimentally
• [X] represents concentration of chemical species at
equilibrium
Equilibrium Constant
4NH3(g) + 7O2(g)  4NO2(g) +6H2O(g)
K = [NO2]4[H2O]6
[NH3]4[O2]7
The value of K
will tell us the
position of
equilibrium
Practice
N2(g) + 3H2(g)  2NH3(g)
[equilibrium]
]2
K = [NH3
[N2] [H2]3
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
Calculate K
K=
[3.1x 10-2]2
[8.5 x 10-1] [3.1 x 10-3]3
K = 3.8 x 104
no units
Practice
N2(g) + 3H2(g)  2NH3(g)
[NH3] = 3.1 x 10-2 M
[N2] = 8.5 x 10-1 M
[H2] = 3.1 x 10-3 M
The value of K
will tell us the
position of
equilibrium
K = 3.8 x 104
If K >1, the rxn favors products
If K <1, the rxn favors reactants
If K = 1, equilibrium lies in the middle
Practice
Original rxn
N2(g) + 3H2(g)  2NH3(g)
New rxn
2NH3(g)  N2(g) + 3H2(g)
Same [ ]
Original K
K = 3.8 x 104
[NH3] = 3.1 x 10-2 M
Calculate new K [N ] = 8.5 x 10-1 M
2
[H2] = 3.1 x 10-3 M
The rxn was inversed so
K is inversed
K=
1
3.8 x 104
K = 2.6 x 10-5
Practice
Original rxn
N2(g) + 3H2(g)  2NH3 (g)
New rxn
½N2(g) + 3/2H2(g)  NH3 (g)
Original K
K = 3.8 x 104
Same [ ]
[NH3] = 3.1 x 10-2 M
Calculate new K [N ] = 8.5 x 10-1 M
2
[H2] = 3.1 x 10-3 M
The coefficients were
reduced by ½
K = (3.8 x 104)1/2
K = 2.0 x 102
Interpreting K
• If K = 1, then there are equal concentrations of
reactants & products
• If K is small, the forward rxn occurs slightly
before equilibrium is established, reactants are
favored
• If K is large, reactants are mostly converted
into products when equilibrium is established,
products are favored
Heterogeneous Equilibria
• The position of a heterogeneous equilibrium
does not depend on the amounts of pure
solids or liquids present
• If pure solids or liquids are involved in a
chemical reaction, their concentrations are
not included in the equilibrium expression for
the reaction
• Pure liquids are not the same as solutions,
whose concentration can change
More Practice!
1. At 25 °C, an equilibrium mixture of gases
contains 6.4 x 10-3 M PCl3, 2.5 x 10-2 M Cl2, &
4.0 x 10-3 M PCl5. What is the equilibrium
constant for the following reaction? K = 4.0 x 10-2
PCl5 (g) ↔ PCl3 (g) + Cl2 (g)
2. At equilibrium, a 2.0 L vessel contains 0.36 mol
of H2, 0.11 mol of Br2, & 37 mol of HBr. What
is the equilibrium constant for the reaction at
this temperature?
K = 3.5 x 104
H2 (g) + Br2 (g) ↔ HBr (g)
Chapter 18
Chemical Equilibrium
18.2 Shifting Equilibrium
The Effect of Change in Pressure:
Ways to Change Pressure
1. Add or remove a gaseous reactant or product
2. Add an inert gas (one not involved in the
reaction)
An inert gas increases the total pressure but
has no effect on the concentrations or partial
pressures of the reactants or products
The Effect of Change in Pressure
3. Change the volume of the container
When the volume of the container holding a
gaseous system is reduced, the system responds
by reducing its own volume.
This is done by decreasing the total number of
gaseous molecules in the system
N2 (g) + 3H2 (g) → 2NH3 (g)
shifts to the right to decrease the total molecules of
gas present
The Effect of Change in Pressure
• When the container volume is increased,
the system will shift so as to increase its
volume
N2 (g) + 3H2 (g) ← 2NH3 (g)
shifts to the left to increase the total number
of molecules of gas present
The Effect of Change in
Concentration
• If a reactant or product is added to a system
at equilibrium, the system will shift away from
the added component (it will attempt to "use
up" the added component)
• If a reactant or product is removed from a
system at equilibrium, the system will shift
toward the removed component (it will
attempt to "replace" the removed component)
The Effect of a Change in
Temperature
• An increase in temperature increases the energy
of the system. Le Chatelier's principle predicts
that the system will shift in the direction that
consumes the energy
• For an exothermic rxn, energy is a product. The
rxn will shift to the left to use up the excess
energy
• For an endothermic rxn, energy is a reactant.
The rxn will shift to the right to use up the energy
• A decrease in temperature will cause a system
shift in the direction that "replaces" the lost
energy
Le Chatelier’s Principle
• When a stress is applied to a system at
equilibrium, the system will shift in a
direction that will relieve the stress
• [changes] do not affect the value of K
• pressure changes do not affect the value of K
• temperature changes will affect the value of K
Practice ([change])
As4O6(s) + 6C(s)  As4(g) + 6CO(g)
Which direction will the rxn shift to re-establish equilibrium if:
Add CO?
left
right
no shift
Add or
remove C or
As4O6?
left
right
no shift
Remove
As4?
left
right
no shift
Practice (change in pressure)
Which direction will the rxn shift to reestablish equilibrium if
the volume is reduced?
P4(s) + 6Cl2(g)  4PCl3(l)
left
right
no shift
PCl3(g) + Cl2(g)  PCl5(g)
left
right
no shift
O2(g) + N2(g)  2NO(g)
left
right
no shift
Practice (change in temperature)
Which direction will the rxn shift to reestablish equilibrium if
the temperature is increased?
N2(g) + O2(g)  2NO(g) ΔHo = 181kJ
left
right
no shift
2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ
left
right
no shift
How will the change affect K?
N2(g) + O2(g)  2NO(g) ΔHo = 181kJ
increase
decrease
2SO2(g) + O2(g)  2SO3(g) ΔHo = -198kJ
increase
decrease
Completed Reactions
• You do not need to consider equilibrium if
you have a reaction that goes to
completion.
Examples:
• Formation of a gas in an open container
• Formation of a precipitate
• Formation of a slightly ionized product
(polar) – neutralization
The Common-Ion Effect
• When the addition of an ion common to
two solutes brings about precipitation or
reduced ionization.
Solubility Product and the Common
Ion Effect
AgCl(s)  Ag+(aq) + Cl-(aq)
What happens if a
solution of NaCl is
added to this
system?
Ksp = [Cl-][Ag+] = 1.8 x 10-10
solublility
product
The solubility
of AgCl has
actually
decreased
NaCl(s)  Na+(aq) +
Cl-(aq)
rxn shifts 
AgCl(s) 
Ag+(aq)
+ Cl-(aq)
Chapter 18
Chemical Equilibrium
18.3 Equilibria of Acids, Bases
and Salts
Ionization Constant of a Weak Acid
• Since weak acids only partially ionize in
solution, an equilibrium exists between the
weak acid + water, reactants and the
hydronium + conjugate base ions,
products.
• Ka= acid ionization constant
• Ka, like Keq, is temperature-dependant
Ka
• For a weak acid:
HF (g) + H2O (l) ↔ H3O+ (aq) + F- (aq)
Ka = [H3O+] [F-]
[HF]
Assume [H2O] is constant
Ionization Constant of Water
Flashback to chapter 15…
The self-ionization of water:
H2O (l) + H2O (l) ↔ H3O+ (aq) + OH- (aq)
KW = [H3O+][OH-] = 1.0 x 10-14
Assume [H2O] is constant
Hydrolysis of Salts
• Recall that the conjugates of strong acids &
bases are weak, and the conjugates of weak
acids & bases are strong:
• The salt produced from a strong acid-strong
base would be neutral.
• The salt produced from a strong acid-weak
base would be acidic.
• The salt produced from a weak acid-strong
base would be basic.
Hydrolysis
• A reaction between water molecules and
ions of a dissolved salt.
• If the anions react with water, the process
is called anion hydrolysis, and produces a
basic pH
• If the cations react with water, the process
is called cation hydrolysis, and produces
an acidic pH
Anion Hydrolysis
• For a weak acid:
HA (g) + H2O (l) ↔ H3O+ (aq) + A- (aq)
Ka = [H3O+] [A-]
[HA]
• The A- can further react with water:
A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)
(HOH)
Anion Hydrolysis
A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)
• The forward reaction is dependant on the
strength of the A-, so if the Ka of HA is low,
the strength of A- will be high.
Cation Hydrolysis
• For a weak base:
B (aq) + H2O (l) ↔ BH+ (aq) + OH- (aq)
Kb = [BH+ ] [OH-]
[B]
• The BH+ can further react with water:
BH+ (aq) + H2O (l) ↔ H3O + (aq) + B (aq)
Cation Hydrolysis
BH+ (aq) + H2O (l) ↔ H3O + (aq) + B (aq)
• The forward reaction is dependant on the
strength of the BH+, so if the Kb of B is low,
the strength of BH+ will be high.
Hydrolysis in Acid-Base Reactions
• Since the product of a neutralization is a
salt & water, depending on the type of salt
formed, the pH of the solution may be
acidic, neutral or basic.
Strong Acid-Strong Base
• The dissociated ions from salts produced
by these reactions do not undergo
hydrolysis, so the pH of these resulting
solutions will be neutral.
Weak Acid-Strong Base
• The dissociated anions produced by these
reactions will undergo hydrolysis to
produce hydroxide ions that will raise the
pH, creating a basic solution
Strong Acid-Weak Base
• The dissociated cations produced by these
reactions will undergo hydrolysis to
produce hydronium ions that will lower the
pH, creating an acidic solution
Chapter 18
Chemical Equilibrium
18.4 Solubility Equilibrium
Solubility
• Substances are considered soluble if at least
1 g will dissolve per 100 g water.
• Substances are considered insoluble if less
than 0.1 g will dissolve per 100 g water.
• Substances that fall in-between 0.1-1 g are
considered partially soluble.
Ksp
• The solubility product constant indicates
the respective molar concentrations of
dissolved ions at equilibrium in a saturated
solution.
• The smaller the Ksp value, the more
insoluble the compound.
• See Appendix Table A-13, page 861
Practice
If the solubility of magnesium sulfate at 20°C is 33.7g
per 100.g of water, Calculate Ksp.
1) Write the rxn
MgSO4 (s) + H2O (l)  Mg+2 (aq) + SO4-2 (aq)
2) Calculate
solubility in mol/L
33.7 g x 1 g H2O x 1000 mL x 1 mol MgSO4
100. g 1 mL H2O
1L
120.4 gMgSO4
= 2.80 mol/L
3) [Mg+2] = [SO4-2]
[Mg+2] = [SO4-2] = 2.80 mol/L
4) Solve Ksp
expression
Ksp = [Mg+2] [SO4-2] = [2.80] [2.80] = 7.84
Practice
Calculate the solubility of CaCO3, in mol/L at 25°C
given Ksp = 2.8 x 10-9
1) Write the rxn &
Ksp expression
CaCO3 (s) + H2O (l)  Ca+2 (aq) + CO3-2 (aq)
Ksp = [Ca+2] [CO3-2]
2) [Ca+2] = [CO3-2]
So, [Ca+2] = x and [CO3-2] = x
Then, Ksp = x2
3) Solve for x
Ksp = 2.8 x 10-9 = x2
X = √ 2.8 x 10-9
Solubility of calcium
carbonate = 5.3 x 10-5
More Practice!
1. What is the value of Ksp for tin (II) sulfide,
given that its solubility is 5.2 x 10-12 g / 100. g
water? 1) 1.2 x 10-25
2. What is the solubility in mol/L of manganese
(II) sulfide given that its Ksp value is 2.5 x 10-13?
2) 5.0 x 10-7
mol/L
3. Calculate the concentration of Zn+2 in a
saturated solution of zinc sulfide given that Ksp
of zinc sulfide = 1.6 x 10-24.
3) 1.3 x 10-12
mol/L
Practice
[initial]
500 mL of 0.010 M Ba(NO3)2 and 300 mL of 0.020
M NaF solutions are mixed. Will a precipitate of
BaF2 (Ksp = 1.7 x 10-6) form?
BaF2(s)  Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2 = 1.7 x 10-6
Qsp = (.00625)(.0075)2 = 3.52 x 10-7
Q < K, so
the rxn will
shift in
favor of the
products =
no
precipitate
More Practice!
1) Will a precipitate form if 20. mL of 0.034 M
sodium chloride and 15 mL of 0.083 M copper
(I) nitrate are mixed?
Yes, CuCl precipitates
2) Does a precipitate form if 100. mL of 0.0014 M
calcium nitrate and 200. mL of 0.00020 M
sodium sulfate are mixed?
No, CaSO4 does not precipitate
What if you are asked to find an
[equilbrium]?
• Use the R.I.C.E method!
Practice
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g) [NH3]o = .0150 M
When 0.0150 mol of NH3 and 0.0150 mol of O2
are placed in a 1.00 L container at a certain
temperature, the N2 concentration at equilibrium
is 1.96 x 10-3 M. Calculate Kc for the reaction at
this temperature.
4NH3(g) +
.0150
- 4x
3O2(g)

.0150 
- 3x

[O2]o = .0150 M
[N2]eq = 1.96 x 10-3 M
2N2g)
+
6H2O(g)
0
0
+ 2x
+ 6x
.0150 - 4x .0150 - 3x  1.96 x 10-3
6x
Practice
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g) [NH3]o = .0150 M
When 0.0150 mol of NH3 and 0.0150 mol of O2
are placed in a 1.00 L container at a certain
temperature, the N2 concentration at equilibrium
is 1.96 x 10-3 M. Calculate Kc for the reaction at
this temperature.
[O2]o = .0150 M
[N2]eq = 1.96 x 10-3M
Substitute x
for this value
.0150 - 4x .0150 - 3x  1.96 x 10-3
K = [H2O]6[N2]2
[NH3]4 [O2]3
K=
6x
2x =1.96 x 10-3 x =9.80 x 10-4
(6x)6(1.96 x 10-3 )2
(.0150 – 4x)4 (.0150 – 3x)3
K =5.95 x 10-6
More Practice
Mustard gas, used in chemical warfare in WWI, has been
found to be an effective agent in the chemotherapy of
Hodgkins disease. It is produced by the following reaction:
SCl2(g) + 2C2H4(g)  S(CH2CH2Cl)2(g)
An empty five-liter tank is filled with 0.258 mol SCl2 and
0.592 mol C2H4. After equilibrium is established, 0.0349 mol
of mustard gas is present.
a) Calculate [equilibrium] for each reactant gas.
b) Determine the value of Keq for the reaction.
a)[SCl2] = 0.0446 M
[C4H4] = 0.104 M
b)Keq = 14.3
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