7.1: DYNAMIC EQUILIBRIUM TERMS IN CHEMICAL SYSTEMS.
Closed system: a system separated from its surrounding by a physical boundary in which chemical
changes can be studied.
Dynamic equilibrium: a balance between the forward and reverse processes in a closed system
Forward reaction: A left-to-right movement of reaction (→) Reverse: A right-to-left movement of
reaction(←)
Solubility equilibrium: A dynamic equilibrium between a solute and a solvent in a saturated solution in
a closed system.
PHASE EQUILIBRIUM
A dynamic equilibrium between different physical states of a pure substance in a closed system. e.g.
H2O(s) ↔ H2O(l)
t = 0 oC
CHEMICAL REACTION EQUILIBRIUM
A dynamic equilibrium where the concentrations of all reactants and products remain constant with time
in a closed system.
An example:
NO2(g) + NO2(g) ↔ N2O4(g)
N2O4(g)
↔
2NO2(g)
CO2(g)
I2(s)
↔
↔
CO2(aq)
2I-(aq)
When the Rate of Regeneration of the Reactants is equal to the Rate of consumption of the Reactants,
then the system will be in a STATE OF EQUILIBRIUM
Rate of forward
=
rate of reverse
(micro activities continues)
rate of reaction does not reach Zero at Equilibrium
Note:
Once a system has reached equilibrium any factor which causes a change in the rate of forward or
reverse, disturbs the equilibrium.
The system readjusts so that the rate again becomes equal and a new equilibrium is established.
CALCULATING CONCENTRATIONS AT EQUILIBRIUM
A RICE table is a convenient way to organize the information needed to solve a stoichiometric problem
involving an equilibrium system.
R- mole ratio
I - stands for of reactants and product (before reaction),
C - stands for Change in the concentrations of reactants and products
E - stands for the concentration of reactants and products at equilibrium.
1
Example:
In the reaction for the formation of hydrogen fluoride from its elements at SATP, if the reaction begins
with 1.00 mol/L of H2(g) and F2(g) and no HF(g). Calculate the concentrations of H2(g) and HF(g) at
equilibrium if the equilibrium concentration of F2(g) is measured at 0.24 mol/L
H2(g) + F2(g)
HF(g)
Note: Concn. of gases given in moles per litre means that there is one mole of gas per litre of space.
Using ICE Table:
First list the information given in the question
[H2(g)]initial = 1.00 mol/L
[F2(g)] initial = 1.00 mol/L
[HF(g)] initial = 0.00 mol/L
[F2(g)]Equilib = 0.24 mol/L
ICE Table: For Reaction of H2(g) and F2(g)
Initial Concn. (mol/L)
H2(g)
1.00
Change Concn. (mol/L)
-x
Equilib Concn (mol/L)
1.00-x
+
F2(g)
1.00
↔
-x
2HF(g)
0.00
2x
1.00-x
2x
Calculation:
With the equilibrium concn of F2(g) = 0.24 mol/L, the value of x can be determined.
1.00 mol/L -x = 0.24 mol/L
- x = 0.24 mol/L - 1.00 mol/L
-x = - 0.76 mol/L
Calculating for other entities:
[H2] = 1.00mol/L -x
= 1.00 mol/L - 0.76 mol/L
= 0.24 mol/L
[HF] =2x
= 2(0.76 mol/L)
= 1.52 mol/L
Therefore, the equilibrium concns. Of H2(g) and HF(g) are 0.24 mol/L and 1.52 mol/L respectively
Try Q #6, 7 P. 437
7.2: EQUILIBRIUM LAW IN CHEMICAL REACTIONS.
The Equilibrium Law expression, also called the Law of Mass Action by Guldberg and Waage
states: 
That in a system at equilibrium, at a fixed temperature, the product of the equilibrium
concentration of the product divided by the concentration of the reactants, each concentration being raised
to the power equal to the coefficient of the substance in the equation, must be equal to a constant.
2
Example: For a general chemical reaction
aA
K
+ bB ↔ cC +dD
= [C]c [D]d
[A]a [B]b
A, B, C and D are chemical entities in gas or aqueous phases. a, b,c,d are the coefficient in the balanced
chemical and k is a constant called the equilibrium constant.
Examples: - Write the equilibrium law expressions for the following: a. H2(g) +
I2(g)
↔
2HI(g)
+
Heat
K = [HI(g)]2
[H2(g)][I2(g)]
b. 3H2(g)
K
+
↔
2NH3(g)
+
Heat
=
+ O2(g) ↔
c. 2NO(g)
K
N2(g)
2NO2(g)
=
d. 4NH3(g) + 7O2(g) ↔ 4NO2(g) +6H2O(g)
The Equilibrium Constant and Reaction Kinetics
This relates the rate of chemical reaction to chemical equilibrium
Example:
vf
A + B ↔ C + D
vr
vf is rate of forward reaction, and vr represents the rate of reverse reaction
vf
A + B → C + D (Forward reaction)
vf
= Kf[A][B] (Kf is rate constant of forward reaction)
For the reverse reaction,
vr
C + D → A + B
vr = Kr[[C][D] (Kr is rate constant)
At equilibrium, vf
= vr
Therefore Kf[A][B] = Kr[C][D] (Concentrations are molar)
Solving for Kf/Kr,
Kf = [C][D]
Kr
[A][B]
3
K = Kf = [C][D]
Kr [A][B]
See sample problems on pp 442-443. Try out Practice problems pp 444 #2-3
LIMITATIONS OF QUILIBRIUM CONSTANT AND PERCENT REACTION VALUES
 K varies with temperature changes. Therefore, it is necessary to specify temperature when giving
K value.

K = Kf/Kr, any factor which causes an unequal change of Kf or Kr will cause the change on K
In exothermic (-ve) reaction, heat is given off in forward reaction. Increase in temperature, then increases
Kr than Kf. K value will decrease.
Endothermic (+ve) reaction, heat is absorbed in forward reaction. Increase temperature then increase K f
than Kr. Therefore temperature will increase K value.
Changing temperature of an equilibrium system changes both the concentration of the participants and the
value of K.
E.g., N2(g) + 3H2(g) ↔ 2NH3(g)
K = 4.26 x 108 at 25oC
K = 1.02 x 10-5 at 300oC
K = 8.00 x 10-7 at 400oC
AT CONSTANT TEMPERATURE
Changing the equilibrium concentration does not affect K, because rate constants are not affected by
concentration changes.
When the concentration of one participant is changed the concentration of the other
varies in such a way as to maintain a constant value for Ke.
K
=
[Product]
[Reactant]
(Law of Mass of Action)
see table 2 pp. 440
HETEROGENEOUS EQUILIBRIA
Equilibria in which reactants and products are in more than one phases
e.g. solids, liquids, gases and aqueous solutions. Systems in this unit will be homogeneous equilibria.
Examples:
1) 2H2O(l) ↔ 2H2(g) + O2(g)
K = [H2(g)]2[O2(g)]
[H2O(l)]2
Since water concentration is a constant, it is incorporated into the K value
K[H2O(l)] = [H2(g)]2[O2(g)]
4
K = [H2(g)]2[O2(g)]
2) Q: Write the equilibrium law equation for the decomposition of solid ammonium chloride to gaseous
ammonia and hydrogen chloride gas.
NH4Cl(s) ↔ NH3(g) + HCl(g)
3) Q: Write the equilibrium law equation for the preparation of quicklime, CaO(s), by heating
calcium carbonate, CaCO3(s), from seashells.
CaCO3(s) ↔ CaO(s) + CO2(g)
K =[CO2(g)]
4) Q: Write the equilibrium law equation for the reaction of zinc in copper (II) chloride equation.
Zn(s) + Cu2+(aq) + 2Cl-(aq) ↔Zn2+(aq) + Cu(s) + 2Cl-(aq)
Zn(s) + Cu2+(aq) ↔ Zn2+(aq) + Cu(s);
K = [Zn2+(aq)][Cu(s)]; K[Zn(s)] = [Zn2+(aq)]
[Zn(s)][Cu2+(aq)]
[Cu(s)]
[Cu2+(aq)]
K = [Zn2+(aq)]
[Cu2+(aq)]
Relation of K to Reaction Equations
At 448oC: H2 + I2
↔ 2HI
K1 = [HI] 2 = 50.2
[H2][I2]
When a coefficient of a given reaction equation are multiplied by ½, the K for the new equation is square
root of the original K.


1/2H2 + 1/2I2
↔ HI
K2 = [HI]
= √50.2
[H2][I2]
When doubled, K is (raised to the power)
2H2 + 2I2 ↔ 4HI
K3 = [HI]2
[H2][I2]
Reversed K is reciprocal of the original.
[H2][I2]
HI
↔
H2 + I2
K4 = [HI]
= (50.2)2
1
50.2
RELATIONSHIP OF K TO THE EXTENT OF A RACTION
The magnitude of K is a measure of the extent to which a given reaction has taken place before
equilibrium was established.
High K →
High concentration of Products and low concentration of reactants
Low K →
Low concentration of Products and high concentration of reactants.
Ke = 1 →
Means that all the coefficients are 1 at equilibrium. Half of the reactants have
been converted to products, which is 50% yield.
5
Example 1:
Calculate the equilibrium constant k, and concentrations when 0.4 mole of PCl5 is heated in 10.0 L
container and equilibrium established in which 0.25 moles of Cl2 is present.
Solution:
PCl5(g) ↔
1mole
PCl3(g) + Cl2(g)
1 mole
1 mole
a) What is the number of moles of PCl5 and PCl3 at equilibrium?
Draw ICE table:
PCl5(g) ↔ PCl3(g) + Cl2(g)
Inital
0.40
0.00
change
-x
x
x
0.4-x
x
x
Equilibrium
0.00
mole of Cl2:
mole of PCl5:
x =0.25 moles
0.40-x
0.40 -0.25 = 0.15 moles
b) What is the equilibrium concentration of all three participants
[PCl5] = 0.15mole = 0.015 mol/L
10.0 L
[PCl3] = 0.25moles ie 0.25mol =0.25 mol/L
10.0L
[Cl2] = 0.25 moles = 0.025 mol/L
10.0 L
c) Calculate the equilibrium constant for the reaction
7.3: QUALITATIVE CHANGES IN EQUILIBRIUM SYSTEMS
LE CHATELIER’S PRINCIPLE
States that: When a stress is applied to a system at equilibrium, the system readjusts so as to relieve or offset the
stress.
Stress equal:
(a) Change in concentration of one or more participants.
(b) Temperature
(c) Pressure (related to volume change)
The Effect of changing concentration
H2(g)
+
I2(g)
↔ 2HI(g)
Increase H2(g) will result in producing more HI(g). New equilibrium now shifts towards the Product side
Therefore the concentration of reactants will shift the reaction forward (to the right)
6
An Increase in the concentration of the products will shift the reaction backward (to the left)
RATE THEORY AND CONCENTRATION CHANGES
This Provides explanation of the equilibrium shift that occurs when a reactant concentration is increased.
 collisions are more frequent for a forward reaction since the reverse rate is not
immediate. Results in increased product.
as the concentration of product increases, reverse reaction is favoured until the two rates
become equal, and equilibrium is established.
the rates at the new equilibrium is faster than the original. Why?
LE CHATELIER'S PRINCIPLE AND TEMPERATURE CHANGES
The energy in a chemical equilibrium equation can be treated as part of the reactant or a product.
Endothermic reaction:
reactants + energy ↔products
Exothermic reaction:
reactants ↔ products + energy
Energy can be added or removed from a system by heating or cooling the container. In both, the
equilibrium shift to minimize the change.
If endothermic system is heated, the equilibrium shifts to the right, to yield more product and absorbing
some of the added heat. (see fig. 6 Pg. 453)
N2O4(g) + energy ↔ 2NO2(g)
If an exothermic system is heated, the equilibrium shifts in the direction of
reactants. Energy will be used to convert product to reactants. Increasing the
concentration of reactants. (see fig. 7 Pg. 453).
Removing energy (Cooling) causes the system to shift to the right to yield more
product.
As part of production process for sulphuric acid, the product is favoured if the
temperature is kept low.
2SO2(g) + O2(g) ↔ 2SO3(g) + energy
Change in Total Volume (Related to Pressure) of an Equilibrium system.
Haber process
N2(g) +
3H2(g)
↔
2NH3(g) +
92KJ
When Volume is decreased, the pressure increases, there is a greater increase in the number of collisions (
and effective collisions) between reactant molecules per unit time than the product molecules.
This cause a shift in equilibrium to the right with fewer molecules
4 molecules of reactants →
2 molecules of Product
Therefore decrease volume result in increase NH3
The shift causes a concentration change that increases the reverse rate and decrease the forward until a
new equilibrium is reached.
7
Effect of a catalyst on an Equilibrium State


The addition of a catalyst: does not affect an equilibrium
•A catalyst speeds both forward and reverse reactions (by lowering the activation energy)
•It allows us to get to equilibrium faster, but it does not alter equilibrium concentrations.
Adding Inert Gases
Adding an inert gas such as helium to a system at equilibrium increases the pressure of the system(at
constant volume), but does not cause a shift in the equilibrium position.
Summary of Le Chatelier’s principle
E.g. N2+ 3H2 ↔2 NH3+ 92 kJ
Amounts of products and reactants: equilibrium shifts to compensate
↑N2: N2+ 3H2 ↔2NH3+ 92 kJ shift right
↓H2: N2+ 3H2 ↔2NH3+ 92 kJ shift left
Pressure (due to decreased volume): increase in
pressure favors side with fewer molecules
Catalysts: does not influence reaction, but helps
to establish equilibrium state faster.
Inert gases: Inert gases do not disturb equilibrium, but increases the total number of entities (same as
substances in different state).
Example: - 1.
Indicate in which direction the equilibrium shift when these changes are made. Justify your answer in
terms of the Le Chatelier’s Principle?
2SO2(g)+
O2(g)
↔
2SO3(g)+
Heat
a)
b)
c)
d)
Concentration of SO2 is increased (Shift to right)
The temperature of the system is increased. (A shift to the left)
Volume of the container is increased.( A shift to the left)
Helium (He) gas is added at constant volume so that the total
pressure is increased (No effect, equilibrium establishes quickly)
e) He(g) is added but total pressure is constant.(shift to side with fewer molecules)
Example 11: - This equation represents a gaseous system at equilibrium.
Heat + 2H2O(g)
↔
2H2(g) +
O2(g)
8
Indicate in which direction the equilibrium shift when these changes are made. Justify your answer in
terms of the Le Chatelier’s Principle?
a) The [H2] is increased
b) The [H2O] is increased
c) The [O2] is decreased
d) Temperature is increased
e) The volume of the container is decreased
f) He (g) is added at constant volume so that the total pressure is increased.
g) He (g) is added but the total pressure is kept constant
h) A catalyst is added.
7.4: Read and make notes on the Haber Process - Synthesis of ammonia P. 461-462
Discussion time in class
7.5: Quantitative Changes in Equilibrium systems
Recall: Equilibrium expression equation
aA + bB ↔ cC + dD
K = [C]c[D]d
[A]a[B]b
This means that the larger value of K, the more the product is favoured. The smaller the value of K, the
more the reverse reaction.
Example:
H2(g) + I2(g) ↔ 2HI(g)
k = 50 at 450oC
CO2(g) + H2 ↔ CO(g) + H2O(g)
k = 1.1 at 900oC
The Reaction Quotient, Q
When the reactant and product of a given reaction are mixed, it is useful to know whether the mixture is
at equilibrium or not or the direction to shift to reach equilibrium.
If the concentration of one of the reactant or product is zero, the equilibrium shift to the direction that
produces the missing
component.
If they are nonzero, it is more difficult to determine the direction of the move toward equilibrium.
Therefore we use a trial value called the reaction Quotient, Q. Q is similar to K
K - concentration at equilibrium
Q - concentration may or may not be at equilibrium.
By using the law of mass action, initial concentration instead of equilibrium concentration is used.
Example:
N2(g) + 3H2(g) ↔ 2NH3(g)
Q = [NH3]02
[N2]0[H2]03
Zeros indicate initial concentrations.
9
To determine in which direction a system will shift to reach equilibrium, we compare the values of Q
and K.
There are three possible cases
1. Q is equal to K, The system is at equilibrium; no shift will occur.
2. Q is greater than K, that is, ratio of the initial concentration of products to initial
concentration of reactants is too large. To reach equilibrium, a net change of product to reactants must
occur. The system shifts to the left, consuming products and forming reactants, until equilibrium is
reached.
3. Q is less than K, the system must shift to the right (toward product) to reach equilibrium, because the
product -to-reactant ratio is too low
Example:
In the synthesis of ammonia at 500oC the equilibrium constant is 6.0 x 10-2. Predict the direction in which
the system will shift to reach equilibrium in each of the following cases:
a) [NH3]0 = 1.0 x 10-3 M; [N2]0 = 1.0 x 10-5 M; [H2]0 = 2.0 x 10-3 M
b) [NH3]0 = 2.0 x 10-4 M; [N2]0 = 1.5 x 10-5 M; [H2]0 = 3.54 x 10-1 M
c) [NH3]0 = 1.0 x 10-4 M; [N2]0 = 5.0 M; [H2]0 = 2.0 x 10-2 M
Solution:
a) First calculate the value of Q:
Q = [NH3]02
=
(1.0 x10-3)2
[N2]0[H2]03
(1.0 x 10-5)(2.0 x 10-3)3
= 1.3 x 107
Since Q is much greater than K. to attain equilibrium, the concentration of the product must be decreased
and the concentration of reactant increased. The system will shift to the left.
Complete b & c questions above.
Try Q 1& 2 in textbook P. 464
Calculations Involving Equilibrium Systems.
We can use stoichiometry to calculate concentrations for systems at equilibrium.
1. Calculating Equilibrium concentrations from Known concentrations
Example:
HI(aq) is produced by reacting hydrogen gas and iodine vapour according to the following equation:
H2(g) + I2
↔ HI(g)
The equilibrium constant, K, for this reaction is 49.70 at 458oC.
Calculate [HI(g)] at equilibrium if the concentration of the other two entities is 1.07 mol/L
Solution:
First list the known values, then set the equilibrium constant equal to the equilibrium law expression for
the reaction:
10
[H2(g)]equil. = 1.07 mol/L
[I2(g)]equil. = 1.07 mol/L
K = 49.70 at 458oC
Equil. Expression:
[HI(g)]
[H2(g)] [I2(g)]
[HI(g)]
(1.07)(1.07)
= K
= 49.70
[HI(g)]
= 49 .70 x 1.14
= 58.9 mol/L
Practice: P466 #3,4
2. Calculating Equilibrium Concentrations from initial concentrations
This is a little more complex than the earlier method.
Example:
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700oC, K is 5.1.
Calculate the equilibrium concentration of all entities if 1.0 mol of each component is mixed in a 1.00-L
flask.
Solution:
i. write the balanced Equation:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
K = 5.10
ii. Then write the equilibrium Law Equation:
K = [CO2(g)][H2(g)]
[CO(g)][H2O(g)]
iii. Calculate the initial concentrations of all entities
[CO(g)] = [H2O(g) ] = [CO2(g)] = [H2(g)] = 1.00 mol
vol =1.00 L
[CO(g)] = 1.00 mol/L
[H2O(g)] =1.00 mol/L
[CO2(g)] =1.00 mol/L
[H2(g)] =1.00 mol/L
v. Calculate the value of Q using initial conc. And compare with K to find out if the reaction is
at equilibrium.
Q=
[CO2(g)][H2(g)]
[CO(g)][H2O(g)]
11
= [1.00][1.00] = 1.00 Since Q is less than K, therefore the system is not at equilibrium it
must shift to the right to achieve equilibrium
v. Set up an ICE table to find the conc. at equilibrium.
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Initial
Change
Equilibrium
1.00
-x
1- x
1.00
-x
1- x
1.00
x
1+x
1.00
x
1+x
vi. substitute the equilibrium conc. values from the ice table into the equilibrium Law equation
K = 5.10 = [CO2(g)][H2(g)]
[CO(g)][H2O(g)]
(1+x)(1+x) = (1.00 + x)2 =
(1-x)(1-x)
(1.00 - x)2
5.10
(1.00 + x) = √ 5.10
(1.00 - x)
(1.00 + x)
= 2.26
(1.00 - x)
X = 0.387 mol/L. Thus the system shifts to the right by consuming 0.387 mol/l CO and H2O;
and forming 0.387 mol/L CO2 and H2
vii. Calculate the conc. at equilibrium
[CO] = [H2O] = 1.00 - x = 1.00 - 0.387 = 0.613 M
[CO2] = [ H2] = 1.00 + x = 1.00 + 0.387 = 1.387 M
Practice: P. 472 # 5,6
Calculating equilibrium concentration with imperfect squares (More Complex Equilibrium
Problems)
This often occurs when K value is very small.
Example:
The equilibrium constant, K is 4.20 x 10-6 at a temperature of 1100 K for the reaction,
2H2S(g) ↔ 2H2(g) + S2(g)
K = 4.20 x 10-6
What concentration of S2(g) can be expected when 0.200 mol of H2S(g) comes to equilibrium at 1100 K
in a 1-L vessel?
n = 0.20 mol
V = 1.00L
12
[H2S(g)]0 = 0.20 mol
1.00 L
= 0.200 mol/L
K = [H2(g)]2 [S2(g) ]
[H2S(g)]2
Q = 0, since there are no initial concentration. The reaction will proceed to the right since Q is less than K
.
ICE Table:
2H2S(g) ↔
2H2(g) + S2(g)
Initial conc.
Change conc
Equil conc
0.2
-2x
0.2-2x
0
2x
+2x
0
x
x
The equilibrium concentration must satisfy the equilibrium expression
K = 4.20 x 10-6 = (2x)2 (x) since the value of K is small, we can approx. 0.2-2x
(0.2-2x)2 to 0.2 to avoid complicated equations of x3,x2 and x
So K = 4.20 x 10-6 = (2x)2 (x)
( 0.2)2
(0.04)(4.20 x10-6) = 4x3 ;
0. 168 x 10-6 =4x3
x3 = 0.042 x 10-6
x = 3√ 4.2 x 10-6
x = 3.48 x 10-3
Practice Q # 8 P. 476
Calculating Equilibrium concentrations involving a quadratic equation
In some cases, the value of K is too large to ignore. In such cases,
the calculation may involve solving by quadratic equation in the form of
ax2 + bx + c = 0
By using
x = - b ± √b2 - 4ac
2a
Example:
In a sealed container, nitrogen dioxide is in equilibrium with dinitrogen tetroxide in the equation below.
2NO2(g) ↔ N2O4(g)
K = 1.15 at 55oC
Find the equilibrium concentration of nitrogen dioxide and dinitrogen tetroxide, if the initial concentration
of nitrogen dioxide is 0.650 mol/L.
[NO2(g)]0 = 0.650 mol/L
[N2O4(g)]0 = 0.00 mol/L
K = [N2O4(g)]
[NO2(g)]2
13
Q = (0.00)
(0.650)
=0
Since Q is less than K, the reaction will proceed to the right.ICE Table for the reaction:
2NO2(g)
↔
N2O4(g)
Initial Conc.
0.65
0.00
Change
-2x
x
Equili.
0.65 - 2x
x
At equilibrium,
K = [x ]
[0.65-2x]2
1.15 = [ x ]
[0.65-2x]2
The hundred rule apply here:
If the value is less than 100, the assumption that 0.65-2x ~0.65 is rejected. Example:
[2NO2(g]0
K
1.15 = ( x )
(0.65 - x)2
= 0.65 = 0.57. This is less than 100.
1.15
1.15(0.65-2x)2 = x
1.15 (4x2 - 2.6x + 0.42) = x
4.6x2 -3x + 0.48 = x;
4.6x2 -3x-x + 0.48 = 0
4.6x2 -4x + 0.48 = 0
Solve for x using
x = - b ± √b2 - 4ac
2a
Where a = 4.6; b = -4; c 0.48
x = -4 ± √42 - (4x4.6 x 0.48)
2 x 4.6
4 + √ 16 - 8.8; = 4 + 2.68 = - 0.73
9.2
9.
In quadratic calculation, only positive values are used. OR
x =
4- √ 16 - 8.8 = 4 - 2.68 = 0.14
9.2
9.2
[NO2(g)] = x = 0.14 mol/L
x =
[N2O4 ] =
0.65 - 2x; = 0.65 - (2 x 0.14)
= 0.65 - 0.28
=0.37 mol/L
Practice: Q # 10 P. 480
14