Chemistry
The Science in Context
Chapter 15
Chemical Equilibrium
Climate models are a system of differential equations
derived form
basic physics and
Chemistry.
These
equations
model kinetic
processes but
often we can
approximate
parts of the
system by
assuming that
they reach
equilibrium…
For example, climate models often employ the
concept of radiative equilibrium
incoming energy from the sun = out going energy
from the earth
One “simple
model is
(1 − a)Sπr2
= 4πr2εσT4
We have observed that the oncentrations of O2, N2 and
NO eventually stop changing. So the reaction can be
written:
N2+O2 = 2NO
Equilibrium
Equilibrium is reached when the forward and
reverse rates of reaction are equal.
R
P
kf[R] = kr[P]
Equilibrium
Equilibrium is reached when the forward and
reverse rates of reaction are equal.
R
P
kf/kr= [P]/[R] = Kc
subscript “c” denotes conc.
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]
[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
.0200 .0200
[H2]
0
0
K=






2 






H CO


 2
 
 
 


2 



H O CO
The Equilibrium Constant
For the general reaction:
aA+bB
cC+dD
Kc = ([C]c[D]d)/([A]a[B]b)
The subscript “c” is used to indicate that the value
of K is calculated based upon a ratio of
concentrations.
In pressure units..
Kp = (PC)c(PD)d)/(PA)a(PB)b
The Equilibrium Constant
The equilibrium expression for a chemical
reaction is the ration of the concentration terms for
products divided by that for reactant in accordance
with the balanced chemical equation.
The equilibrium constant (K) is the numerical
value for the equilibrium expression for a chemical
reaction
The law of mass action states that the equilibrium
constant will have a characteristic value at a given
temperature.
Properties of Equilibrium Constants
•
Keq applies only at equilibrium, i.e. when the
rate forward = rate reverse
• Keq is independent of initial conditions. The
reaction can start with all products, all reactants,
and any combination in between, as long as the
temperature is constant, at equilibrium, the ratio of
products to reactants is determined by the
equilibrium constant.
• Keq is related to the stoichiometry of the balanced
(net) chemical equation. Based upon Keq the final
state of the system can be calculated.
Properties of Equilibrium Constants
•Keq varies with temperature. It will increase with
increasing temperature for an endothermic reaction.
It will decrease with increasing temperature for an
exothermic reaction.
•Keq is reported as a unitless value. Although
related to the ratio of the rate constants, K is a
thermodynamic property. The previous
thermodynamic properties; ΔH, ΔS, and ΔG, were all
described relative to Standard State conditions:1 M
for a solute and 1 atm for a gas. A more correct form
of the equilibrium expression is also written relative to
Standard State conditions.
Equilibrium Tutorial
»PC version
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.
Equilibrium in the Gas Phase Tutorial
»PC version
This unit describes the difference between Kc and Kp
and explains how to convert between them. Includes
practice exercises.
Problem 16
Kc = 8.7E6 at 700K for the reaction between NO
and O2.
The rate constant for the reverse reaction at this
temperature is 0.54M-1s-1
What is the rate constant for the forward reaction
at 700K?
Answer:
Keq = kf/kr
kf = Keq * kr = 8.7E6 *0.54M-1s-1
kf = 4.7E6 M-2s-1
Problem
Write equilibrium constant expressions for the
following reactions.
a) 2 ClO(g) = Cl2(g) + O2(g)
b) 2 O3(g) = 3 O2(g)
What are the units for an equilibrium constant?
Problem
The value of the equilibrium constant (Kp) for the
formation of ammonia
N2(g) + 3 H2(g) = 2 NH3(g)
Is 4.5E-5 at 450°C.
What is the value of Kp for the following…
NH3(g) = ½ N2(g) + 3/2 H2(g)
Answer: 149
Problem 35
Phosgene (COCl2) is used in manufacturing although it
is highly poisonous. It is formed from the reaction of
carbon monoxide and chlorine:
CO(g) + Cl2(g)
COCl2(g)
The value of Kc for the reaction is 5.0 at 325°C. What
is the value of Kp?
Answer: 0.10
Problem 28
A 100 mL reaction vessel initially contains 2.60 
10–2 moles of NO and 1.30  10–2 moles of H2. At
equilibrium, the concentration of NO in the vessel
is 0.161 M. At equilibrium the vessel also contains
N2, H2O, and H2.
What is the value of the equilibrium constant Kc for
the following reaction?
2H2(g) + 2NO(g) ↔ 2H2O(g) + N2(g)
Answer: 19.5
Problem 30
For the reaction: 2NO(g) + O2(g)
2NO2(g)
Kc = 5E12.
What is the value of Kc for each of the following:
a) NO(g) + 1/2O2(g)
NO2(g)
b) 2NO2(g)
2NO(g) + O2(g)
c) NO2(g)
NO(g) + 1/2O2(g)
The reaction quotient, Q, is defined in the same way
as K, the equilibrium constant. Q is the numerical
value for the mass action at some time point in the
reaction.
A ↔ B
Q = [B]/[A]
As reactions
proceed
toward
equilibrium,
Q increases
or decreases
until Q = K
Problem 54
The equilibrium constant (Kc) for the reaction
2C
D+E
Is 3E-3. At a particular measurement time, the
composition of the reaction mixture is:
[C] = [D] = [E] = 5E-4.
In which direction (forward or reverse) will the
reaction proceed to reach equilibrium.
Answer: Q = 1
Keq = 3 E-3
[C] = [D] = [E] = 5E-4
Q = [D][E]/[C]2 = [5E-4][5E-4]/[5E-4]2
Equilibrium and Thermodynamics Tutorial
»PC version
Manipulate values for G and Kc to explore how Gibbs
free energy is related to equilibrium constant and
reaction spontaneity. Includes practice exercises.
K, Q, and G
There is a direct relationship
between the equilibrium
constant, K, and the free
energy, G.
G>O; K<1 “reactant favored”
G<0; K>1 “product favored”
G=0; K=1
G = G° + RT ln(Q)
At equilibrium, G = 0; Q=Keq
G° = -RTln(Keq)
Problem 48
In glycolysis, the hydrolysis of ATP to ADP is used to
drive the phosphorylation of glucose:
glucose + ATP ↔
ADP + glucose 6-phophate
G° for the reaction equals, -16.7 kJ/mol
What is Kc for the reaction at 298K?
Answer: Kc=846
Problem 72
Under the appropriate conditions, NO forms N2O and
NO2:
3 NO(g) ↔ N2O(g) + NO2(g)
Use the values of G° for the following reactions to
calculate the values of Kp for the above reaction at
500°C.
2 NO(g) + O2(g) ↔
2NO2 (g) G° = -69.7 kJ
2 N2O(g) ↔ 2NO(g) + N2(g)
N2(g)+ O2(g) ↔ 2NO(g)
G° = -33.8 kJ
G° = 173.2 kJ
Answer: Kc=1.16E7; Kp=1.83E5
Disturbing an equilibrium and
Le Châtelier’s Principle
When a change is imposed on a
system at equilibrium, the system
will react in the direction (toward
products or reactants) that
reduces the effect and amount of
change, (i.e. re-establishes
equilibrium).
When the total
pressure is
increased upon a
gas-phase reaction,
the reaction shifts
to “relieve the
stress” i.e. to the
reactants or
products side with
smaller number of
total moles.
According to Le Châtelier’s principle, reactions will
shift an equilibrium in an endothermic
(exothermic) direction when the temperature is
increased (decreased).
Higher temperature
Lower temperature
Le Châtelier’s Principle Tutorial
»PC version
In this tutorial you can 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. Includes practice exercises.
Applications of Le Chatelier’s Principle
Problem 86
How will the following changes affect the
position of the equilibrium:
2NO2(g) ↔
NO(g) + NO3(g)
a) The concentration of NO is increased
b) The concentration of NO2 is increased
c) The volume of the system is allowed to
expand to 5-times its initial value.
d) temperature increase or decrease
Applications of Le Chatelier’s Principle
Problem 79
Patients suffering from carbon monoxide
poisoning are treated with pure oxygen
(hyperbaric) to remove CO for the hemoglobin
(Hb). The two relevant equilibria are:
Hb + 4CO(g) ↔
Hb(CO)4(g)
Hb + 4O2(g) ↔
Hb(O2)4(g)
Kc for CO binding to hemoglobin is about 150
times that of O2 binding.
How, then, does this treatment work?
Reversing the first reaction…and summing,
Hb(CO)4(g) ↔ Hb + 4CO(g)
Hb + 4O2(g) ↔
Hb(O2)4(g)
Net: Hb(CO)4(g) + 4O2(g) ↔ Hb(O2)4(g) + 4CO(g)
How do we calculate how Keq changes with
temperature?
G° = -RT ln(K)
G° = H° - TS°
ln(K) = - G°/RT
ln(K) = - H°/RT + TS°/RT
ln(K) = - (H°/R)(T-1) + S°/R
Changing K with Temperature
ln(K1) = - (H°/R)(T1-1) + S°/R
ln(K2) = - (H°/R)(T2-1) + S°/R
ln(K1) - ln(K2) = - (H°/R)(T1-1) + (H°/R)(T2-1)
ln (K2/K1) = - (H°/R)(T2-1 – T1-1)
This is known as the van’t Hoff equation and can
be used to calculate Kp or Kc at different
temperatures. It assumes that the enthalpy value
(H) does not change over the temperature of
interests.
How/why does H change with temperature?
Changing K with Temperature
Problem 86
The value of Kp for the reaction:
N2(g) + 3H2(g) ↔ 2 NH3(g)
Is 41 at 400K. If H°r= -92.2 kJ/mol,
Calculate the value of Kp at 700K.
Answer:Kp(700K) = 2.84E-4
Problem
The equilibrium constant, Kp, for the thermal
decomposition of NO2
2NO2(g) ↔ 2 NO(g) + O2(g)
is 6.5E-6 at 450K.
If a reaction vessel at this temperature initially
contains 0.500 atm NO2, what will be the partial
pressures of NO2, NO and O2 when equilibrium has
been established?
Problem 76
On a particularly smoggy day, the concentration of
NO2 in the air over an urban area reaches 2.2x10-7 M.
If the temperature of the air is 25°C, what is the
concentration of the NO2 dimer, N2O4, in the air?
N2O4(g) ↔ 2 NO2(g)
Kc = 6.1x10-3
Problem 93 (Heterogeneous Equilibria)
Passing steam over hot carbon produces a mixture of
carbon monoxide and hydrogen.
H2O(g) + C(s) ↔ CO(g) + H2(g)
Kc (1000°C) = 3.0E-2
Calculate the equilibrium partial pressures of the
reactants, if PH2O = 0.442 atm and PCO = PH2 = 0 at
the start of the reaction.
PH2O = 0.442 and PCO = PH2 = 0
H2O(g) + C(s) ↔ CO(g) + H2(g)
PH2O = 0.442 and PCO = PH2 = 0
H2O(g) + C(s) ↔ CO(g) + H2(g)
Solving Equilibrium Problems Tutorial
»PC version
Learn to determine equilibrium concentrations of
reactants and products from starting concentrations and
an equilibrium constant. Includes interactive exercises.
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).
CQ15-7.1b-Sublimation of Added Iodine
Crystals of iodine, I2(s), are placed in a
closed container and are allowed to
sublime until they come to equilibrium
with I2(g). If the temperature inside the
container is doubled, the partial
pressure of I2(g) will:
A) Stay the same B) Increase C) Decrease
CQ15-7.2a-Sublimation of Heated Iodine
Please consider the following arguments for each
answer and vote again:
A. The partial pressure of I2(g) depends only on the
equilibrium constant, so raising the temperature will
have no effect.
B. Because the sublimation of I2(s) is an endothermic
process, adding heat will drive the reaction toward
the production of more I2(g).
C. Raising the temperature increases the rate of
condensation. Therefore, the partial pressure of I2(g)
will decrease.
CQ15-7.2b-Sublimation of Heated 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
CQ15-7.3a-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.
CQ15-7.3b-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
CQ15-7.4a-Temperature and HI, H , and I Equilibrium
C) Q is lower
Consider the following arguments for each answer
and vote again:
A. The numbers of moles on both sides of the reaction
are the same, so temperature will not affect the value
for Q.
B. Increasing the temperature will increase the partial
pressures of all the gases, so the value for Q will be
higher.
C. The reaction of HI(g) to form H2(g) and I2(g) is an
exothermic process, so increasing the temperature
will lower the value for Q.
CQ15-7.4b-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)
B)
CQ15-7.5a-Dissociation of Chlorous Acid
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].
CQ15-7.5b-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.
B) It increases.
CQ15-7.6a-Dilution of Hydrofluoric Acid
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.
CQ15-7.6b-Dilution of Hydrofluoric Acid
Green Ni(H2O)62+ reacts with ammonia, NH3, to form
blue
Ni(NH3)62+.
Ni(NH3)62+
reacts
with
ethylenediamine (en), (CH2NH2)2, to form violet
Ni(en)32+.
If the ethylenediamine is added first, followed by
ammonia, what color will the solution be?
A) Green
B) Blue
CQ15-7.7a-Equilibrium Colors of Ni Complexes
C) Violet
Consider the following arguments for each answer
and vote again:
A. By altering the order in which the different
compounds are added, the solution will revert to its
original green color.
B. Ammonia is the last reactant added, so the final
equilibrium solution will be blue.
C. Changing the order in which the compounds are
added will not alter the final equilibrium state.
CQ15-7.7b-Equilibrium Colors of Ni Complexes
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)
CQ15-7.8a-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.
CQ15-7.8b-Decompression of MgCO in a Piston
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This concludes the Norton Media Library
slide set for chapter 15
Chemistry
The Science in Context
by
Thomas Gilbert,
Rein V. Kirss, &
Geoffrey Davies