Chapter 15
Chemical Equilibrium
Equilibrium
A chemical equilibrium is a dynamic process
in which the concentrations of reactants and
products remain constant over time and the
rate of a reaction in the forward direction
matches its rate in the reverse direction.
Equilibrium Terms






Reactions that do not go to completion
Reactions that have a Rf = Rr, in this chapter
What is equal in equilibrium the rate forward is
equal to the rate reverse
For equilibrium reactions with greater
concentration of products compared to reactants,
we say shifted to the right, or products favored
For equilibrium reactions with greater
concentration of reactants compared to products,
we say shifted to the left, or reactants favored
Swimming pool example
Concentration vs Time
Equilibration Time
Note: equilibrium is when the concentrations of reactants are
not changing, i.e. zero slope
Shifted left or right?
Concentration vs. Time
Equilibration Time
Note: equilibrium is when the concentrations of reactants are
not changing, i.e. zero slope
Shifted left or right? To the rignt, products have larger
concentration.
Rate vs Time
Equilibration time
Law of Mass Action






Developed by Norwegian chemists and proposed
in 1864
This relation holds for gases and solutions, since
there concentrations are variable.
Since solids and liquids have constant
concentrations, there values are incorporated in
the equilibrium constant K
If the reaction is reversed then Kr = 1/K
If a balanced equation is multiplied by a factor n
then Kn=Kn
K is normally expressed without units, since units
are dependent on reaction stoichiometry.
Law of Mass Action
The law of mass action states that an expression
relating the concentrations of products to reactants
at chemical equilibrium has a characteristic value at
a given temperature when each concentration term
in the expression is raised to a power equal to the
coefficient of the substance in the balanced chemical
equation for the reaction.
aA + bB
[products]
K = [reactants]
cC + dD
K=
[C]c [D]d
[A]a[B]b
Experimental Data
Initial and Equilibrium Concentrations of the Reactants and Products in the
Water-Gas Shift Reaction at 500K.
H2O(g) + CO(g) <===> H2(g) + CO2(g)
Experiment
[H2O]
1
Equilibrium Concentrations
(M)
Initial Concentration (M)
[CO]
.0200 .0200
[H2]
[CO2]
0
0
[H2O]
[CO]
[H2]
[CO2]
.0034 .0034 .0166
.0166
.0200 .0200 .0034 .0034 .0166
.0166
3
.0100 .0200 .0300 .0400 .0046 .0146 .0354
.0454
4
.0200 .0100 .0200 .0100 .0118 .0018 .0282
.0182
2
K 1=
0
H2 CO 2 










0
[0.0166][0.0166]

 
 =
H O CO
[0.0034][0.0034]
 2
 

= 24
Calculation of Kc
• The mass action expression or equilibrium constant
expression describes the relation between the
concentration (or partial pressure) terms of reactants
and products when a system is at equilibrium.
• The equilibrium constant (K) is the value of the ratio
of concentration (or partial pressure) terms in the
equilibrium constant expression at a specific
temperature.
K=






2 






H CO



 
 
 


2 



H2 O CO
= 24
Kinetics and Equilibrium
• For any elementary step which reaches
equilibrium
aA + bB
cC + dD
the forward rate = kf[A]a[B]b
the reverse rate = kr[C]c[D]d
Kinetics and Equilibrium
At Equilibrium: forward rate = reverse rate
kf[A]a[B]b = kr[C]c[D]d
kf
Kc =
=
kr
C D
a
b
A  B
c
d
Equilibrium constant or mass action expression
Kinetics and Equilibrium
At Equilibrium:
rate = reverse rate
forward
kf[A]a[B]b = kr[C]c[D]d
kf
Kc =
=
kr
C D
a
b
A  B
c
d
Equilibrium constant or mass action expression
Note: This is for an elementary step, so Law of
Action is defined for the sum of all steps!
Gaseous Equilibrium Expression
aA + bB
cC + dD
If substances A, B, C, and D are all gases then
the equilibrium could be expressed with their
partial pressures (atm).
PC  PD
Kp =
a
b
PA PB
c
d
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
PA a PB b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d
[(na/V)RT]a[(nb/V)RT]b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d [Mc]c[Md]d [RT]c [RT]d
=
a
b
[(na/V)RT] [(nb/V)RT] [Ma]a[Mb]b[RT]a[RT]b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d [Mc]c[Md]d[RT]c [RT]d
=
a
b
[Ma]a[Mb]b[RT]a [RT]b
[(na/V)RT] [(nb/V)RT]
[Mc]c [Md]d [RT]c+d
Kp =
[Ma]a [Mb]b [RT]a+b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d [Mc][Md] [RT]c [RT]d
=
a
b
[Ma] [Mb] [RT]a [RT]b
[(na/V)RT] [(nb/V)RT]
[Mc][Md] [RT]c+d
[Mc]c[Md]d [RT](c+d)-(a+b)
Kp =
=
a+b
[Ma] [Mb] [RT]
[Ma]a [Mb]b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d [Mc][Md] [RT]c [RT]d
=
a
b
[Ma] [Mb] [RT]a [RT]b
[(na/V)RT] [(nb/V)RT]
[Mc][Md] [RT]c+d
[Mc]c [Md]d [RT](c+d)-(a+b)
= Kc(RT)Δn
Kp =
=
[Ma] [Mb] [RT]a+b
[Ma]a [Mb]b
Equilibrium in the Gas Phase
For a gas PV = nRT or P = (n/V)RT n/V is moles per
liter or molarity P = MRT, not volume constant for each
reactant
c
d

PC   PD 
Kp =
a
b =
 PA   PB 
[(nc/V)RT]c [(nd/V)RT]d [Mc][Md] [RT]c [RT]d
=
a
b
[Ma] [Mb] [RT]a [RT]b
[(na/V)RT] [(nb/V)RT]
[Mc][Md] [RT]c+d
[Mc] [Md][RT](c+d)-(a+b)
= Kc(RT)Δn
Kp =
=
[Ma] [Mb] [RT]a+b
[Ma] [Mb]
n is the number of moles of gaseous products
minus the number of moles of gaseous reactants
in the balanced chemical equation.
Practice
Write the equilibrium constant expressions (Kc and
Kp) for the synthesis of NO2.
4NH3(g) + 7O2(g)
Kc =
[NO2]4[H2O]6
[NH3]4[O2]7
4NO2(g) + 6H2O(g)
Kp =
P4NO2P6H2O
P4NH3P7O2
Practice
Write the equilibrium constant expressions (Kc and
Kp) for the synthesis of NO2.
4NH3(g) + 7O2(g)
Kc =
[NO2]2[H2O]6
[NH3]4[O2]7
4NO2(g) + 6H2O(g)
Kp =
Give the relation between Kc and Kp
P4NO2P6H2O
P4NH3P7O2
Practice
Write the equilibrium constant expressions (Kc and
Kp) for the synthesis of NO2.
4NH3(g) + 7O2(g)
Kc =
[NO2]2[H2O]6
[NH3]4[O2]7
4NO2(g) + 6H2O(g)
Kp =
Give the relation between Kc and Kp
Δn = 10 – 11 = - 1
P4NO2P6H2O
P4NH3P7O2
Practice
Write the equilibrium constant expressions (Kc and Kp)
for the synthesis of NO2.
4NH3(g) + 7O2(g)
Kc =
[NO2]2[H2O]6
[NH3]4[O2]7
4NO2(g) + 6H2O(g)
Kp =
P4NO2P6H2O
P4NH3P7O2
Give the relation between Kc and Kp
Δn = 10 – 11 = - 1
Kp = Kc(RT)Δn
Kp = Kc(RT)-1
Practice
Write the equilibrium constant expressions (Kc and Kp)
for the synthesis of NO2.
4NH3(g) + 7O2(g)
Kc =
[NO2]2[H2O]6
[NH3]4[O2]7
4NO2(g) + 6H2O(g)
Kp =
P4NO2P6H2O
P4NH3P7O2
Give the relation between Kc and Kp
Δn = 10 – 11 = - 1
Kp = Kc(RT)Δn
Kp = Kc(RT)-1
Where R = 0.082058 L-atm/mole-K; why?
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless?
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless? Looks
like M2/M = M, right?
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless? Looks
like M2/M = M, right?
Kp = Kc(RT)Δn =
1
1
4.00 mol (0.082058 L-atm) (298.15 K)
L
mole-K
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless? Looks
like M2/M = M, right?
Kp = Kc(RT)Δn =
1
1
4.00 mol (0.082058 L-atm) (298.15 K)
L
mole-K
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless? Looks
like M2/M = M, right?
Kp = Kc(RT)Δn =
Kp = 97.9 atm
1
1
4.00 mol (0.082058 L-atm) (298.15 K)
L
mole-K
Practice
Calculate Kp for
PCl5(g)
PCl3(g) + Cl2(g)
given Kc = 4.00 at 425oC.
Is Kc really unitless? Looks
like M2/M = M, right?
Kp = Kc(RT)Δn =
1
1
4.00 mol (0.082058 L-atm) (298.15 K)
L
mole-K
Kp = 97.9 atm
Special Note: Only gases and solutions play the
equilibrium game!
Reaction Quotient
• The reaction quotient (Q) is calculated by
inserting the values for the concentrations (or
partial pressures) of the reactants and
products in the mass action expression.
• If the reaction quotient is equal to the
equilibrium constant, then the system is at
equilibrium, if not then it is proceeding to
equilibrium.
Reaction Quotient
• The reaction quotient (Q) is calculated by
inserting the values for the concentrations (or
partial pressures) of the reactants and products in
the mass action expression.
• If the reaction quotient is equal to the equilibrium
constant, then the system is at equilibrium, if not
then it is proceeding to equilibrium.
Reaction Quotient
• If Q>K then the products are too large, thus the rate of
the reverse reaction is greater than the rate of the
forward reaction, thus increasing the concentration of
the reactants and decreasing the concentration of the
products, until equilibrium is established.
• If Q<K then the reactants are too large, thus the rate
of the forward reaction is greater than the rate of the
reverse reaction, thus increasing the concentration of
the products and decreasing the concentration of the
reactants, until equilibrium is established.
Comparing Q with K
The comparison of Q with K will indicate the direction that
the system will proceed to reach equilibrium.
Practice
Is the following reaction at equilibrium at the stated
conditions? If it is not at equilibrium, determine which
direction the reaction will proceed. (Kc = 5.6)
CH4 + H2O
0.100M 0.200M
CO + 3H2
0.500M 0.800M
Practice
Is the following reaction at equilibrium at the stated
conditions? If it is not at equilibrium, determine which
direction the reaction will proceed. (Kc = 5.6)
CH4 + H2O
0.100M 0.200M
CO + 3H2
0.500M 0.800M
[H2]3[CO]
[0.800]3[0.500]
Q=
=
[CH4][H2O]
[0.100][0.200]
= 12.8
Practice
Is the following reaction at equilibrium at the stated
conditions? If it is not at equilibrium, determine which
direction the reaction will proceed. (Kc = 5.6)
CH4 + H2O
0.100M 0.200M
CO + 3H2
0.500M 0.800M
[H2]3[CO]
[0.800]3[0.500]
Q=
=
= 12.8
[CH4][H2O]
[0.100][0.200]
Q>KC, thus not at equilibrium, but going there. The rate of
the reverse reactions is faster than the forward reaction, which
means the product concentration is getting smaller and the
reactant concentration is getting larger, thus lowering the
value of Q until it reaches Kc
Mathematical Relationships
• The equilibrium constant for the reverse of a
reaction is the reciprocal of the equilibrium
constant for the reaction in the forward direction.
• The overall equilibrium constant for a
combination of two or more reactions is the
product of the equilibrium constants of the two or
more reactions.
• If the coefficients in a chemical equation are
doubled, the value of the equilibrium constant for
the reaction is squared.
Practice
N2O4(g)
2NO2(g)
Kc = 4.63 x 10-3
a) Determine the Kc for the reaction:
2NO2(g)
N2O4(g)
b) Determine the Kc for the reaction:
NO2(g)
1/2N2O4(g)
Practice
N2O4(g)
2NO2(g)
Kc = 4.63 x 10-3
a) Determine the K’c for the reaction:
2NO2(g)
N2O4(g) K’c=1/Kc=[4.63X10-3]-1
K’c=216
b) Determine the Kc for the reaction:
NO2(g)
1/2N2O4(g)
Practice
N2O4(g)
2NO2(g)
Kc = 4.63 x 10-3
a) Determine the K’c for the reaction:
2NO2(g)
N2O4(g) K’c=1/Kc=[4.63X10-3]-1
K’c=216
b) Determine the Kc for the reaction:
NO2(g)
1/2N2O4(g)
K’’c = Kc1/2 = [216]1/2 = 14.7
Types of Equilibria
• Homogeneous equilibria involve reactants
and products in the same phase.
• Heterogeneous equilibria involve reactants
and products in more than one phase.
Example
• What are the concentrations
of CaO and CaCO3?
• Any pure solid or liquid
substance has a constant
concentration consequently,
we incorporate them into
value of the equilibrium
constant.
• Kc = [CO2] -1
Le Chatelier’s Principle
Le Chatelier’s Principle states that a system
at equilibrium responds to a stress in such a
way that it relieves that stress.
Now consider what would happen to the equilibrium system
H2 + I2
2 HI
If more hydrogen is added.
1. The concentration of hydrogen increases.
2. The rate of the forward reaction increases, while the
rate of the reverse reaction remains the same.
3. Now the concentration of HI increases, thus the rate
of the reverse reaction increases.
4. The forward reaction rate is decreasing, since hydrogen
is being consumed and the rate of the reverse reaction
is increasing.
5. Soon the rate forward = rate reverse again and the
reaction is reestablished its equilibrium, with the same
value of the equilibrium constant K
LeChatelier and Equilibrium
LeChatelier’s Principle: If an external force is applied
to an equilibrium, then the equilibrium shifts in a
direction so as to oppose that force.
External force could be adding reactants, products,
heat, or pressure.
Shift means reactants or products are bigger.
If products are bigger then the shift is to the right.
If reactants are bigger then the shift is to the left.
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens? Shift left
4. Add more CO2, what happens?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens? Shift left
4. Add more CO2, what happens? Shift left, more reactants.
5. Add more water?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens? Shift left
4. Add more CO2, what happens? Shift left, more reactants.
5. Add more water? Nothing, only (aq) and (g) play game.
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens? Shift left
4. Add more CO2, what happens? Shift left, more reactants.
5. Add more water? Nothing, only (aq) and (g) play game.
6. Remove heat?
We already determined that if hydrogen is added then, more
HI is produced. Since hydrogen is a reactant and on the left
of the double arrows and more product is produced, then
this equilibrium shifts right in accordance to LeChatelier’s
principle.
Consider the following equilibrium system:
H+ (aq) + HCO3- (aq)
H2O (l) + CO2 (g) + heat
1. Is this system exothermic or endothermic? Exothermic
2. If hydrogen is added, what happens? More Product, shift right.
3. If hydrogen is removed, what happens? Shift left
4. Add more CO2, what happens? Shift left, more reactants.
5. Add more water? Nothing, only (aq) and (g) play game.
6. Remove heat? Shift right
N2O4
brown
Warm water
2NO2 + heat
light brown
Cold water
Another Example, consider the following equilibrium
CO2 (g) + H2O (l)
H2CO3 (aq) + heat
1. What would happen if more CO2 were added?
Another Example, consider the following equilibrium
CO2 (g) + H2O (l)
H2CO3 (aq) + heat
1. What would happen if more CO2 were added? Shift right
Another Example, consider the following equilibrium
CO2 (g) + H2O (l)
H2CO3 (aq) + heat
1. What would happen if more CO2 is added? Shift right
2. What would happen if more water is added?
Another Example, consider the following equilibrium
CO2 (g) + H2O (l)
H2CO3 (aq) + heat
1. What would happen if more CO2 is added? Shift right
2. What would happen if more water is added? Nothing
Stress Can be a Volume Change
If volume
decreases, then
equilibrium shifts
to the side with
the lease gas
particles to
relieve the
increase in
pressure
Concentration Stress
Adding a reactant or removing a product
shifts the position of a chemical
equilibrium in the forward direction,
favoring the formation of more product.
Responses of an Exothermic Reaction [2 A(g) <==> B(g)] at Equilibrium
to Different Kinds of Stress
Kind of Stress
How Stress is Relieved
Direction of Shift
Add A
Remove A
To the right
Remove A
Add A
To the left
Remove B
Add B
To the right
Add B
Remove B
To the left
Increase T by
adding heat
Remove some heat
To the left
Decrease T by
removing heat
Add heat
To the right
Increase Pressure
Remove A to relieve
pressure
To the right
Decrease Pressure
Add A to maintain equilibrium
pressure
To the left
Catalysts and Equilibrium Systems
• Catalysts can help systems reach
equilibrium more quickly, but have no
affect on the value of K.
Practice
At equilibrium, there are 2.50 mol H2, 1.35 x
10- 5 mol S2, and 8.70 mol H2S. Calculate
Kc for the following reaction.
2H2(g) + S2(g)
2H2S(g)
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
2HI(g)
H2(g) + I2(g)
Calculate the concentration of all
species at equilibrium.
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
0.100
2HI(g)
0.00
0.00 initial
H2(g) + I2(g)
equilibrium
Calculate the concentration of all
species at equilibrium.
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
0.100
2HI(g)
0.100 - 2x
0.00
0.00 initial
H2(g) + I2(g)
x
x
Calculate the concentration of all
species at equilibrium.
[H2][I2]
K = [HI]2
equilibrium
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
0.100
2HI(g)
0.100 - 2x
0.00
0.00 initial
H2(g) + I2(g)
x
x
Calculate the concentration of all
species at equilibrium.
[H2][I2]
K = [HI]2
X2
=
= 0.11
2
[0.100 - 2X]
equilibrium
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
0.100
2HI(g)
0.100 - x
0.00
0.00 initial
H2(g) + I2(g)
x
x
equilibrium
Calculate the concentration of all
species at equilibrium.
[H2][I2]
K = [HI]2
X2
=
= 0.11
2
[0.100 - 2X]
X
= [0.11]1/2
0.100 - 2X
Practice
The initial concentrations of HI is
0.100 M and Kc is 0.11 for the reaction:
0.100
2HI(g)
0.100 - x
0.00
0.00 initial
H2(g) + I2(g)
x
x
equilibrium
Calculate the concentration of all
species at equilibrium.
[H2][I2]
K = [HI]2
X2
=
= 0.11
2
[0.100 - 2X]
0.3317(0.100-2X) = X
X
= [0.11]1/2
0.100 - 2X
X = [H2] = [I2] = 0.050 M
Practice
X = [H2] = [I2] = 0.050 M (from the previous slide)
[HI] = 0.100 – X = 0.100 – 0.050 = 0.050 M
Practice
The initial concentration of PCl5 is
0.100 M and Kc = 0.60 for the reaction:
PCl5(g)
PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations
for each species in the reaction.
Practice
In an experiment, 1.0 mol of N2O4 is
placed in a 10.0 L vessel and is allowed
to react as described below.
N2O4(g)
2NO2(g) Kc = 4.0 x 10-7
Calculate the concentration of each
species at equilibrium.
Chapter #17
Equilibrium Review
ChemTour: Equilibrium
Click to launch animation
PC | Mac
In this ChemTour, students explore the concept of
dynamic equilibrium and learn to relate the
equilibrium constant to molar concentrations and
partial pressures of products and reactants.
Includes Practice Exercises.
ChemTour: Equilibrium in the
Gas Phase
Click to launch animation
PC | Mac
This unit describes the difference between Kc and Kp and
explains how to convert between them.
ChemTour: Equilibrium and
Thermodynamics
Click to launch animation
PC | Mac
Students can manipulate values for DG and Kc to explore
how Gibbs free energy is related to equilibrium constant and
reaction spontaneity.
ChemTour: Le Châtelier’s
Principle
Click to launch animation
PC | Mac
In this ChemTour, students subject a system at equilibrium
to a stress and observe changes in the reaction quotient
and how the system reacts according to Le Châtelier’s
principle.
ChemTour: Solving Equilibrium
Problems
Click to launch animation
PC | Mac
Students learn to determine equilibrium concentrations of
reactants and products from starting concentrations and an
equilibrium constant.
Crystals of iodine, I2(s), are placed in a
closed container and allowed to sublime
until they come to equilibrium with I2(g).
If more I2(s) is added to the vessel, the
partial pressure of I2(g) will:
A) Stay the same B) Increase C) Decrease
Sublimation of Added Iodine
Please consider the following arguments for each
answer and vote again:
A. The quantity of I2(s) present never enters into the
equilibrium expression, so it cannot affect the partial
pressure of I2(g) once the system reaches equilibrium.
B. Doubling the amount of I2(s) present will double the
amount of solid that can sublime to form I2(g).
C. Adding more I2(s) will provide an increased surface
area for the I2(g) to stick to, thus decreasing the
partial pressure of I2(g).
Sublimation of Added Iodine
Crystals of iodine, I2(s), are placed in a
closed container and allowed to sublime
until they come to equilibrium with I2(g).
If more I2(s) is added to the vessel, the
partial pressure of I2(g) will:
A) Stay the same B) Increase C) Decrease
Remember solids do not participate in equilibrium
Sublimation of Added Iodine
Nitrogen dioxide, NO2(g), exists in equilibrium with
N2O4(g) according to the reaction
Suppose a mixture of NO2(g) and N2O4(g) is allowed to
come to equilibrium in a sealed vessel. If the volume of
the vessel is doubled at constant temperature, what will
happen to the reaction quotient, Q, before the system
reacts to approach the new equilibrium state?
A) Q is unchanged
B) Q is higher
Volume and NO /N H Equilibrium
C) Q is lower
Consider the following arguments for each answer
and vote again:
A. The value for Q will not change until the system
begins to react.
B. The instantaneous effect of the volume increase will
be to decrease all of the partial pressures. This will
result in a higher value for Q.
C. Doubling the volume affects the partial pressure of
NO2(g) more than the partial pressure of N2O4(g), so
Q will be lower.
Volume and NO /N H Equilibrium
The decomposition of hydrogen iodide, HI(g), to form
H2(g) and I2(g) is an exothermic process.
Suppose a mixture of HI(g), H2(g), and I2(g) is allowed to
come to equilibrium in a sealed vessel. If the temperature
in the vessel is increased, what will happen to the reaction
quotient, Q, before the system reacts to approach the new
equilibrium?
A) Q is unchanged
B) Q is higher
C) Q is lower
Assume concentration of both to be 2, so K = 0.25. Now
divide both concentrations by 2 and K = 1
Temperature and HI, H , and I Equilibrium
Chlorous acid, HClO2, dissociates in water to form
H3O+ and ClO2-:
Which of the following plots shows [ClO2-] as a function
of [HClO2]?
A)
Dissociation of Chlorous Acid
B)
C)
Consider the following arguments for each answer
and vote again:
A. The dissociation of HClO2 is the only source of
ClO2-, so [ClO2-] must be linearly proportional to
[HClO2].
B. Two products are formed by the dissociation of
HClO2, so [ClO2-] will increase as the square of
[HClO2].
C. Because both ClO2- and H3O+ are products of the
dissociation of HClO2, [ClO2-] equals [H3O+] and
therefore, [ClO2-] is proportional to the square root
of [HClO2].
Dissociation of Chlorous Acid
When dissolved in water, HF(aq) partially
dissociates to form H3O+ and F-.
HF(aq) + H2O(l)
H3O+(aq) + F-(aq)
K = 1x10-3
Suppose enough HF is added to 1 L of water so that
[HF] = [F-] = [H3O+] = 0.001 M at equilibrium. If an
additional 1 L of water is added to the solution, what will
happen to the F- concentration after equilibrium is
regained compared to the F- concentration before the
dilution?
A) It stays the same.
Dilution of Hydrofluoric Acid
B) It increases.
C) It decreases.
Please consider the following arguments for each
answer and vote again:
A. H2O(λ) does not enter into the equilibrium
expression and so cannot affect the equilibrium
concentration of F-(aq).
B. By doubling the volume of water, the equilibrium is
shifted toward the production of more products.
C. Although the equilibrium will shift toward the
production of more F-, this shift cannot compensate
for the decrease in concentration caused by adding
the water.
Dilution of Hydrofluoric Acid
Magnesium carbonate, MgCO3(s),
decomposes to form MgO(s) and CO2(g):
MgCO3(s)
MgO(s) + CO2(g)
Suppose some MgCO3(s) is placed in a cylinder with a
movable piston and the volume is slowly increased at a
constant temperature. Which of the following plots
shows the correct relationship between the partial
pressure of CO2 and volume?
A)
B)
Decompression of MgCO in a Piston
C)
Consider the following arguments for each answer
and vote again:
A. As the volume increases, more CO2(g) will be
produced. Once the MgCO3(s) is depleted, no more
CO2(g) will be produced and the pressure will
stabilize.
B. The partial pressure of CO2(g) will not change until
the MgCO3(s) is completely decomposed, at which
point the pressure will decrease as the volume
increases.
C. As the volume increases, more CO2(g) will be
produced. However, once the MgCO3(s) is depleted,
the partial pressure of CO2(g) will drop.
Decompression of MgCO in a Piston
THE END