Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.1
CHEMICAL EQUILIBRIA
DYNAMIC EQUILIBRIA
I.
REVERSIBLE REACTIONS
(A) Irreversible reaction
Reactions that can go to completion are called irreversible reactions.
e.g.The reaction between hydrochloric acid and calcium carbonate to give carbon dioxide.
(B)
Reversible reaction
Reactions that can not go to completion or take place in both directions are called reversible
reactions. The symbol is used to represent a reversible reaction.
e.g.
The reaction of hydrogen and iodine to form hydrogen iodide in a closed system.
H2(g) + I2(g)  2HI(g)
Note
<1> Hydrogen can combine with iodine to form hydrogen iodide. It can considered as the
forward reaction.
<2> Hydrogen iodide can also dissociate to form hydrogen and iodine. It can be considered as
the backward reaction.
<3> When the forward reaction rate is equal to the backward reaction rate, a dynamic equilibrium
is established.
<4> The following graph shows a set of results at a certain temperature and pressure.
(i)
curve (a) represents the amount of HI formed when
0.5 mol H2 is allowed to reaction with 0.5 mol I2.
If the reaction goes to completion, 1 mol HI should be formed.
actually, only 0. 78 mol of HI is collected.
This shows that the reaction does not go to completion.
(ii) curve (b) represents the amount of HI left when 1.0
mol of HI is allowed to decompose into H2 and I2
Also, 0.78 mol of HI
is left. This also shows that
the decomposition of
HI does not go to completion.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
<5>
Chpt.15:p.2
It appears that, with 0.78 mol of HI formed, no further reaction takes place. The
system has reached equilibrium. The same amount of HI is present when
equilibrium is reached from the other direction, by the dissociation of hydrogen
iodide.
II.
DYNAMIC NATURE OF CHEMICAL EQUILIBRIUM
In an equilibrium mixture of H2, I2 and HI, it is not obvious that chemical reactions are still
occurring. In fact, both the forward arid the reverse reactions are still taking place. Since the
rates of forward and the reverse reactions are equal, the concentration of each species remains
constant. The system is said to be in dynamic equilibrium.
Evidence
Iodine, which contains a small quantity of radioactive iodine-131 is injected into an equilibrium
mixture. Radioactive iodine appears in the HI, showing that the formation of hydrogen iodide
and the decomposition of hydrogen iodide are still occurring.
III.
CHARACTERISTICS OF CHEMICAL EQUILIBRIUM
It should be noted that the reaction system remains in dynamic equilibrium only as long as the
system remains isolated. An isolated system (closed system) is one in which there is no
exchange of matter or energy between the system and its surroundings.
In summary, there are four characteristics of a system in chemical equilibrium
<1> The rates of the forward and reverse (backward) reactions are equal.
<2> The system does not undergo any observable change.
e.g.
concentration, pressure et.c.
<3> The system is closed.
<4> The equilibrium can be reached from both directions, from the reactants side (forward
reaction) or form the product side (backward reaction)
Exercise 1
When potassium dichromate or potassium chromate dissolves in water, an equilibrium is
established as follows:
Cr2O72-(aq) + H2O(l)  2CrO42 (aq) + 21f(aq)
orange
yellow
(a)
(b)
ANSWER
Predict and explain the observation when
(I)
acid is added
(ii)
alkali is added
What can you conclude about equilibrium by observing the colour change?
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.3
THE EQUILIBRIUM LAW
RELATIVE CONCENTRATION AT EQUILIBRIUM
The concentration changes in a system of equilibrium can be illustrated by the experiment of the
reaction iron(III) chloride and ammonium thiocyanate.
Fe3+ (aq) +
pale brown
NCS- (aq)
colourless

FeNCS (aq)
blood red
A red solution is produced when equal volumes of iron(III) chloride and ammonium thiocyanate
solutions of some concentration are mixed. The system forms an equilibrium readily, containing
unreacted Fe3+,. unreacted NCS- and product FeNCS2+.
<1> If a soluble Fe3+ salt is added to the equilibrium mixture. The colour of the solution becomes
______________indicating the increase of concentration of ____________.
A new state of equilibrium is said to be reached.
<2> Similarly, the colour of the solution will become darker when a soluble thiocyanate is added.
Increase of the concentration of NCS- can also cause the increase of FeNCS2+.
<3> On the other hand, removal of Fe3+ or NCS- from the equilibrium mixture causes the solution to
become paler, indicating the decrease in concentration of FeNSC2+
Conclusion
If the concentration of one or more of the reacting substances in an equilibrium is changed, then
the equilibrium will shift in such a way as to preserve the constancy of the relative
concentrations of the products and reactants.
This can be expressed mathematically,
[ FeNCS 2 (aq)
 constant
[ Fe 3 (aq)][ NCS  (aq)]
Note:
at a particular temperature
Any change in individual concentration terms would only change the remainder of
the concentration terms so that the above expression has the same constant value
as that before the change.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
II.
Chpt.15:p.4
THE EQUILIBRIUM LAW
Consider a general reversible reaction,
aA + bB  cC + dD
It is possible to show experimentally that an equilibrium
Kc  (
[C ]c [ D]d
) eq
[ A] a [ B]b
where Kc is the equilibrium constant and [
This is known as the equilibrium law.
Note:
(A)
] is the molar concentration
The concentration terms should not include for pure solid, pure liquid and solvent.
Significance of the Equilibrium Constant in the Equilibrium law
<1> Kc depends only on the particular reaction and on the temperature. At a fixed temperature,
Kc should be a constant.
<2> For a large value of Kc:
Products will have a large proportion in the equilibrium mixture. It implies that the
reaction has gone to a large extent to the right.
For a small value of Kc:
Reactants will have a large proportion in the equilibrium mixture.
<3> When the temperature of the system kept constant, addition of either more reactants or
products will result in the adjustment of the equilibrium system by forming more products
or more reactants respectively, so that the value of Kc will remain a constant.
(B)
Relationship of Kc to the balanced equation
It is important to state that the equation on which the equilibrium constant Kc.. is based when
stating the constant for a particular reaction.
Exercise 2
The reversible reaction involving SO2(g), O2(g) and SO3(g) at 700°C.
2SO2(g) + O2(g)  2SO3(g)
Kc1 = 2.8x102 dm3 mol-1
What will be the values of the equilibrium constant following expressions:
(a) 2SO3(g)  2SO2(g) + O2(g)
Kc2= ?
(b) SO2(g) + 1/2O2(g)  SO3(g)
Kc3 =?
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.5
(C) Simple calculations of Kc
Exercise 3
The esterification cf ethanoic acid by ethanol to make ethyl ethanoate is reversible reaction since the
ester can be hydrolysed by water to ethanoic acid and ethanol.
CH3COOH(1) + C2H5OH(l)  CH3COOC2H5(l) + H2O(l)
Calculate the amount of ethyl ethanoate formed when 1.0 mol ethanoic acid and 1.0 mol ethanol reach
equilibrium. At room temperature,
ANSWER
Exercise 4
A 10.0 cm mixture contains the initial amounts/mol ethanol 0.0515; ethanoic acid 0.0525;
water 0.0167; ester 0.0314 ; H+(aq) 1. 00x10-3.
The equilibrium amount of ethanoic acid is 0.0255 mol
Calculate the equilibrium constant Kc of the esterification.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.6
(D) Experimental determination of
Example : Esterification of ethanoic acid and ethanol
CH3COOH(1) + C2H5OH(l)  CH3COOC2H5(l) + H2O(1)
The reaction is catalysed by acid.
Procedure:
<1> A number of mixtures are made up, containing ethanol, ethanoic acid, ethyl ethanoate
and water. Each mixture is different. Every mixture contains a small amount of
hydrochloric acid to catalyse the reaction.
<2> The mixtures are put into stoppered bottles and left for a weak in a thermostat bath.
<3> At the end of the weak, the contents of each flask are titrated against standard sodium
hydroxide solution, using phenolphthalein as indicator. The titration gives the amount of
CH3COOH + amount of HCl. The amount of HCl is still the same as that present initially.
The amount of CH3COOH is found by subtraction,
<4> The amounts of the other substances present are calculated.
(E)
Equilibrium constant in gaseous system
Equilibrium constant for gaseous systems is usually expressed as the partial pressure of gases
rather than as the the molar concentration. Such equilibrium constant is called partial pressure
equilibrium constant.. Kp.
Example : In the reaction between hydrogen and iodine H2(g) + I2(g)  2HI(g)
2
PHI
Kp 
PH 2 xPI 2
Kp is the equilibrium constant in terms of partial pressure and PHI, PH2 and PI2
are partial pressures of HI(g), H2(g) and I2(g) respectively.
Note :
<1> Partial pressure can be expressed as atm or Nm-2
<2> Kp may have units
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.7
Exercise 5
when 1.00 mol of hydrogen and 1.00 mol of iodine are allowed to
reach equilibrium in a 1.00 dm flask at 450 0C and 1.0lxl05 Nm-2, the amount of hydrogen
iodide at equilibrium is 1.56 mol.
Calculate Kp at 450°C.
ANSWER
Exercise 6
In the Haber process for making ammonia,
N2(g) + 3H2(g)  2NH3
Nitrogen and hydrogen are mixed in a molar ratio 1:3. At equilibrium, at 600°C and 10 atm,
the mole percentage of ammonia in the mixture of gases is 15% . Find Kp at 600 C
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.8
Exercise 7
In the reaction between steam and heated iron
3Fe(s) + 4H2O(q)  Fe3O4(s) + 4H2(g)
a mixture of iron and steam was allowed to reach equilibrium at 600°C. The equilibrium
pressures of hydrogen and steam were 3.2 kPa and 2.4 kPa respectively.
(a) Express Kp for this reaction.
(b) Calculate the value of Kp.
ANSWER
(F) Thermal dissociation
Another type of reaction which reaches an equilibrium position is thermal dissociation. For
example, when phosphorus(V) chloride is heated, it dissociates partially into phosphorus(III)
chloride and chlorine:
PC15(g)  PC13(g) + Cl2(g)
As result, the pressure of the gas (at constant volume) or, alternatively, the volume of the gas
(at constant pressure) is greater than expected.
The fraction of molecules that dissociate is called degree of dissociation, and is represented
by the letter 
The amounts of substances in the equilibrium mixture formed from 1 mol of PCl5 are as follow:
Species
PCl5 (g)
PCl3(g)
Cl2(g)
Total
(i) at start
1
0
0
1
(ii) at eqm.

1-

1+
If the total pressure at equilibrium is P, then according to the Dalton’s Law of partial pressure,
PPCl5
=(
1
)P
1
, PPCl3 = (

1
)P
, PCl2 = (

)P
1
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
P  PPCl 3
Then Kp = Cl 2
=
PPCl 5
Note:


(
P)(
P)
1
1
1
(
)P
1
=
Chpt.15:p.9
2
xP
1 2
<1> The unit of Kp is a pressure unit.
<2> If  is small, 1-2  1 and 2  Kp /P , i.e. the higher the pressure, the lower is the
degree of dissociation.
Exercise 8
When 0.100 mol of phosphorus (V) chloride is heated at 150 0C in a 1.00 dm3 vessel, a pressure of
4.38 x 105 Nm-2 is measured. Calculate the degree of dissociation and the value of Kp at this
temperature.
Answer
Exercise 9
At room temperature and a total pressure of 1.01 x 105 Nm-2 , the average molar mass of partially
dissociated N2O4 is 76.6 g mol-1 . Calculate
(a) the degree of dissociation ,
(b) Kp for N2O4  2NO2
(c)  at room temperature if the total pressure now is reduced to 1.01 x 104 Nm-2.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.10
THE EFFECT OF CHANGES IN CONCENTRATION. PRESSURE AND
TEMPERATURE ON EQUILIBRIUM
I. POSITION OF EQUILIBRIUM
The proportion of products to reactants in the equilibrium mixture is described as the position of
equilibrium.
For the reaction between P and Q to form R and S:
aP
+ bQ 
cR
+
dS
<1>
If the conversion of P and Q into R and S is small, the position of equilibrium lies to the left.
Kc is small.
<2>
If the equilibrium mixture is largely composed R and S. the position of equilibrium lies to the
right. Kc is large.
Note
The equilibrium constant Kc is not the same as the position of equilibrium. While Kc is constant at a
particular temperature, a change in external conditions can alter the position of equilibrium. Chemists
always want to shift equilibrium reactions in the direction of forming the products in manufacturing
process.
II.
THE EFFECT OF CONDITIONS ON THE POSITION OF EQUILIBRIUM
(A) Le Chateliers principle
Le Chatellier’s Principle states that in any equilibrium, when a change is made to some external factor
(such as temperature or pressure), the change in the position of equilibrium is such as to tend to change
the external factor in the opposite direction.
Note:
The system cannot complet3ly cancel the change in the external factor, but it moves in the
direction that will minimize the change.
(B) Changes in concentration
If the concentration of one of the reactants or producLs in a system is increased, the effect of
this change may be counterbalanced by setting up a new equilibrium so that the concentration
of the added substance is reduced.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.11
Example
An aqueous solution of bismuth(III) chloride is cloudy because of the following hydrolysis:
BiCl3(aq) + H2O(l) 
(i)
BiOC13(s) + 2HCl(aq)
If a little concentrated hydrochloric acid is added, the position of equilibrium shifts in
the direction that will absorb the acid, i.e. from right to left.
Observation:
______________________________________________________
(ii)
If more BiC13(aq) is added, the position of equilibrium shifts in the direction that will
use up BiC13(aq). i.e. from left to right.
Observation:
______________________________________________________
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.12
(C) Changes in pressure
It is mainly with gaseous reactions that changes in pressure are important. Le Chateliers
Principle can be applied to the formation of ammonia.:
N2(g) + 3H2(g)  2NH3(g)
(i)
If the pressure of an equilibrium mixture is increased, the equilibrium shifts in the direction that
tends to decrease the pressure. It does this by decreasing the total number of molecules present.
i.e. by moving from left to right of the equation. Although the position of equilibrium has
moved towards the ammonia side of the equation. the equilibrium constant has not changed.
increase in pressure, reaction tends to shift to the right
N2(g) + 3H2(g)  2NH3(g)
decrease in pressure , reaction tends to shift to the left
(ii)
In general, increasing the pressure on a system in equilibrium usually favours a decrease in
volume. While reducing the pressure favours an increase in volume.
(D) Changes in temperature
A change in temperature results in a change in the value of equilibrium constant.
<1> Exothermic reactions
Example : Synthesis of ammonia
N2(g) + 3H2(g)  2NH3(g)
Kp 
T/K
400
500
600
700
H0298K = -92 kJ mol-1
PNH 3
2
PN 2  PH 2
3
Kp/atm-2
1.0x102
1.6x10-1
3.1x10-3
6.3x10-5
From the above table, the equilibrium constant _________________ with the increase in
temperature for an exothermic reaction.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.13
<2> Endothermic reactions
Example : The dissociation of dinitrogen tetroxide
N2O4(g)  2NO2(g)
KP 
H0298K = +57 kJ mol-1
PNO2
2
PN 2O 4
T/K
Kp /atm-2
275
2.2x10-2
350
4.5
500
1.5x103
From the above table, the equilibrium constant ____________________ with the increase in
temperature for an endothermic reaction.
SUMMARY
Raiding the temperature (adding heat to a system) of an equilibrium mixture causes the
equilibrium condition to shift in a direction of the endothermic reaction; while lowering the
temperature (removal of heat from a system) causes a shift in the direction of the exothermic reaction.
That is
Higher temperature favours endothermic reaction, whilst lower temperature favours exothermic
reaction.
Note
<1>
Changes in concentrations, pressure results in altering the position of equilibrium but preserving
the equilibrium constant Kc or Kp.
<2>
Temperature change results in adjustment of the system to a new equilibrium constant Kc(or
Kp). The relation of temperature and the value of K for exothermic and endothermic reactions
can be illustrated by the following equation,
ln K  constant -
log K = constant where
H
RT
H
2.303RT
K = equilibrium constant at temperature T
H = enthalphy change, T = Kelvin temperature
R = gas constant
The above equation can be demonstrated by the experiment of the reaction between sulphur
dioxide and oxygen to form sulphur trioxide in the Contact process.
2SO3(g) + O2(g)  2SO3(g)
H0298K = -197 kJ mol-1
The Kp values of this system at various temperature is obtained together with a plot of log Kp
against
1
as follows:
T
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.14
T(K)
1 -1
(K ) x 10-4
T
Kp
800
12.5
9.1x 102
2.96
11.8
2
2.23
1
850
Log Kp
1.7 x 10
900
11.1
4.2 x 10
1.62
950
10.5
1.0 x 101
1.00
1000
10.0
3.2
0.51
1050
9.5
1.0
0.00
1100
9.1
3.9x 10-1
-0.41
1170
8.5
1.2 x 10-1
-0.92
Interpretation:
<1>
<2>
An increase in temperature for exothermic reaction (H negative) results in a decrease in the
values. For an endothermic react ion (H = positive), an increase in temperature of the reacting
system will result in an increase in the values.
For an exothermac reaction (H = negative), the term 
As temperature rises, the value of 
H
is positive.
2.303RT
H
will decrease and
2.303RT
the equilibrium constant, K ,will decrease,
resulting in less product being formed.
For an exothermic reaction (H=positive) , the term 
As temperature rises, the value of 
H
is negative.
2.303RT
H
will increase and
2.303RT
the equilibrium constant, K, will increase,
which results in an increase in product formation.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.15
Exercise 1
A transparent glass syringe with its tip sealed is filled with NO2 is in equilibrium with N204. Explain the
following phenomenon.
(a) When the plunger is pushed in rapidly from position 1 to 2
(i) the colour momentarily deepens, then
(ii) goes paler after a shbrt while.
When the plunger is drawn back rapidly from position 2 to 1
(i)
(ii)
(b)
the colour momentarily become paler. then
deepens to its original intensity after a short while.
Two thin-walled glass bulbs of equal volume are filled with equal amounts of the same brown
gas as in (a) and are sealed.
One bulb is immersed in a 80°C water bath and the other is kept at 25°C. After immersing for
1 minute. withdrawal of the bulb from the water bath shows that the colour of the gas in the hot
bulb has deepened. Explain why this is so.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 16: Chemical Equilibrium
Chpt.15:p.16
Exercise 2
The graphs shown in the a gaseous reversible system at different conditions of temperature and
pressure respectively.
(a) Indicate two effects on the system that are due to a fall in temperature. Is the formation
of the product exothermic or endothermic? Explain briefly.
(b) Indicate two effects on the system that are due to a fall in pressure. Can the above
system H2(g) + I2(g)  2HI(g) ?
(c) If fig(b) shows the results that require catalytic action to obtain sketch how would the
curve at 250 atmospheres change if the catalyst were poisoned.
(d) In practice. the conditions chosen for the synthesis of the product in an industrial plant
are 450°C and 250 atmosphere.