Chemical Equilibrium
IJSO Training (Phase 3)
Dr. Kendrew K. W. Mak
Department of Chemistry
The Chinese University of Hong Kong
Textbook: John Green, Sadru Damji, Chemistry for the International Baccalaureate Diploma Programme, 2nd Ed.
1
Reversible（可逆）and
Irreversible（不可逆）
Reactions go to completion:
products are much more energetically
favorable than the reactants.
Reactions occur easily:
low activation energy (活化能)
Reactions that do not occur:
the activation energy is too high, or the
reaction is not energetically favorable
2
Equilibrium Reaction
（平衡反應）
For some chemical systems, the energies of the reactants and products are
comparable so that the reactions are reversible – they can occur in either
direction.
When a reaction attained the state of chemical equilibrium (化學平衡), the
concentrations of reactants and products remain constant.
Example:
The Haber Process (哈柏法)
heat
pressure
catalyst
N 2 (g)
nitrogen
+
3 H2 (g)
2 NH 3
hydrogen
ammonia
3
Equilibrium Reaction
Chemical equilibrium is a state of dynamic equilibrium (動態平衡) that occurs
in a closed system when the forward and reverse reactions of a reversible
reaction (可逆反應) occur at the same rate.
A + B
C + D
Forward rate (正向速率) = kf[A][B]
Backward rate (逆向速率) = kb[C][D]
At equilibrium, kf[A][B] = kb[C][D]
4
Equilibrium Reaction
At equilibrium, all of the species involved (reactants and products)
are present at constant concentrations.
5
Dynamic Equilibrium（動態平衡）
When equilibrium is reached, it appears that a chemical reaction has stopped.
In reality, both forward and reverse reactions are still occurring, but there is a
balance between transformation of reactants into products, and
transformation of products into reactants.
The rate of the forward reaction equals to the rate of the reverse reaction.
Equilibrium reactions are represented with a double arrow between reactants
and products, showing their reversible and dynamic nature.
CH3COOH (aq)
CH3COO-(aq) + H+(aq)
acetic acid
acetate ion
hydrogen ion
(醋酸)
(醋酸根離子)
(氫離子)
6
Dynamic Equilibrium
For a specific reaction, the equilibrium state will be the same, no matter
the equilibrium is approached from which side.
N2O4 (g)
Initial concentration
Equilibrium concentration
1 mol dm-3
0.4 mol dm-3
2 NO2 (g)
0 mol dm-3
1.2 mol dm-3
7
Catalyst and Equilibrium
If a catalyst is present, the same equilibrium state will be attained, but it
will be attained more quickly.
A catalyst speeds up both the forward and the backward reactions.
The overall effect is to produce exactly the same concentrations at
equilibrium, whether or not a catalyst is in the reaction mixture.
8
The Equilibrium Constant
A quotient of equilibrium concentrations of reactant and product substances
that has a constant value for a given reaction at a given temperature is called
an equilibrium constant (平衡常數).
a A + b B + ...
c C + d D + ...
[C ]c [ D]d
Equilibriu m constant  κ 
[ A]a [ B]b
Example 1:
N2(g) + O2(g)
Equilibrium constant = K =
2 NO(g)
[NO]2
[N2][O2]
9
The Equilibrium Constant
Example 2:
1/8 S8(s) + O2(g)
K' =
[SO2(g)]
[S8(s)]1/8
[O2]
SO2(g)
K=
[SO2(g)]
[O2]
Which one is correct?
Since sulphur (硫) is a solid substances, and for any solid the number of
molecules per unit volume remains the same at any given temperature.
Therefore the concentration of sulphur is not changed as the reaction
proceeds.
10
The Equilibrium Constant
Example 3:
NH4+(aq) + OH-(aq)
NH3(aq) + H2O(l)
K=
[NH4+] [OH-]
[NH3]
Because the molar concentration (摩爾濃度) of water is effectively constant
for reactions involving dilute solutions, the concentration of water is not
included in the equilibrium constant expression.
11
Equilibrium Constant Expressions
for Related Reactions
N2(g) + 3 H2(g)
2 NH3(g)
1/2 N2(g) + 3/2 H2(g)
K1 =
NH3(g) K 2 =
[NH 3]2
[N2][H 2]
3
= 2.4 x 10 7
[NH 3]
[N2 ]1/2 [H 2]3/2
= (2.4 x 107 )1/2 = 4.9 x 10 3
2 NH3(g)
N2(g) + 3 H2(g)
K3 =
=
[N2 ][H 2]3
[NH 3]2
1
2.4 x 10
7
= 4.2 x 10-8
Whenever the stoichiometric coefficients (化學計量系數) of a balanced
equation are multiplied by some factor, the equilibrium constant (平衡常
數) for the new equation (K2 in this case) is the old equilibrium constant
(K1) raised to the power of the multiplication factor.
12
Equilibrium Constant Expressions
for Related Reactions
Exercise 1
Consider the gas-phase (氣相) reaction:
H2(g) + I2(g)
2 HI(g)
If a flask initially containing 0.025 mol/L of H2 and 0.025 mol/L of I2 is
heated to 400oC, the concentrations of H2 and I2 decline and the
concentration of HI increases. The concentration of HI at equilibrium is: [HI]
= 0.039 mol/L. Calculate the equilibrium constants (平衡常數).
Answer K = 50
13
Equilibrium Constant Expressions
for Related Reactions
Exercise 2
0.0500 mol of acetic acid (醋酸) was dissolved in 1.00 L of distilled water,
and it was found that 3.05% of the acetic acid was ionized into CH3COO- ions
and H+ ions. Calculate the equilibrium constant for ionization (電離作用) of
acetic acid.
Ans = 4.80 x 10-5
14
The Meaning of the Equilibrium
Constant
The value of the equilibrium constant (平衡常數) tells how far a reaction has
proceeded by the time equilibrium has attained.
K >>1
Reaction is product-favored; equilibrium concentrations of products
are greater than equilibrium concentrations of reactants.
NO(g) + O3(g)
K <<1
NO2(g) + O2(g)
K=
[NO2][O2]
[NO][O3]
= 6 x 1034
Reaction is reactant-favored. Equilibrium concentrations of reactants
are greater than equilibrium concentrations of products.
3 O2(g)
2 O3(g)
K=
[O3]2
[O2]3
= 6.25 x 10-58
15
Calculating Equilibrium
Concentrations
Exercise 3
Hydrogen (氫) reacts with carbon dioxide (二氧化碳) at high temperature
and gives water and carbon monoxide (一氧化碳)
H2(g) + CO2(g)
H2O(g) + CO(g)
K = 0.10 (at 420oC)
A flask containing 0.200 mol/L of H2 and 0.200 mol/L of CO2 is heated at
420oC until equilibrium is attained. What are the concentrations of
reactants and products at equilibrium?
Answer: [H20] = [CO] = 0.048 mol/L;
[H2] = [CO2] = 0.152 mol/L
16
Calculating Equilibrium
Concentrations
Exercise 4
The dimerization (二聚作用) of nitrogen dioxide (二氧化氮) to
dinitrogen tetraoxide (四氧化二氮) has an equilibrium constant of 1.7 x
102 at 25oC.
2 NO2(g)
N2O4(g) K = 1.7 x 102 (at 25oC)
If 1.00 mol N2O4 and 0.500 mol NO2 are initially placed in a container
whose volume is 4.00L, calculate the concentrations of N2O4(g) and
NO2(g) present when equilibrium is achieved at 25oC.
Ans: [NO2] = 0.414 mol L-1, [N2O4] = 0.292 mol L-1
17
Shifting a Chemical Equilibrium:
Le Chatelier’s Principle
The Le Chatelier’s Principle (勒沙得利爾原理):
If a system is in equilibrium and the conditions are changed so that the
system is no longer at equilibrium, the system will adjust to a new
equilibrium state such that the effect of the change in conditions is partially
counteracted or compensated for.
Le Chatelier’s principle applies to changes in conditions such as:
•
•
•
the concentrations of reactants or products that appear in the
equilibrium constant expression
the pressure/volume of the equilibrium system
the temperature
18
Shifting a Chemical Equilibrium:
Le Chatelier’s Principle
Change
Effect on Equilibrium
Change in Kc?
Increase concentration
Shifts to the opposite side
No
Decrease concentration
Shifts to that side
No
Increase pressure
Shifts to side with lease moles No
of gas
Decrease pressure
Shifts to side with most moles No
of gas
Increase temperature
Shifts in endothermic (吸熱)
direction
Yes
Decrease temperature
Shifts in exothermic (放熱)
direction
Yes
Add a catalyst (催化劑)
No change
No
19
Effect of Concentration
If the concentration of a species is increased, then the equilibrium moves
towards the other side causing the concentration to fall to a value between
the original concentration and the increased value.
Fe(H2O)63+(aq) + SCN-(aq)
Yellow-brown
Colourless
[Fe(H2O)5SCN]2+(aq) + 6 H2O(l)
Blood-red
Addition of Fe(H2O)63+(aq)
Addition of SCN-(aq)
- equilibrium shifts to product side
- equilibrium shifts to product side
Addition of [Fe(H2O)5SCN]2+(aq)
- equilibrium shifts to reactant side
The values of K remain unchanged.
20
Effect of Pressure
If the total pressure of a system is increased then the equilibrium shifts to
the side with least moles (摩爾) of gas, so causing the pressure to fall to a
value between the original pressure and the increased value.
2 SO2(g) + O2(g)
C(s) + H2O(g)
H2(g) + I2(g)
2 SO3(g)
3 moles gas go to 2 moles gas
Increased P ; decrease P 
CO(g) + H2(g)
1 mole gas go to 2 moles gas
Increased P ; decrease P 
2 HI(g)
2 moles gas go to 2 moles gas
Changing P has no effect
The values of K remain unchanged.
21
Effect of Temperature
If the temperature of a system is increased then the equilibrium shifts in
the direction of the endothermic (吸熱) change, so absorbing heat and
causing the temperature to fall to a value between the original temperature
and the increased value.
N2(g) + O2(g)
2 SO2(g) + O2(g)
2 NO(g)
2 SO3(g)
DH = +180 kJ mol-1
(forward reaction - endothermic)
Increased T, K increases, equilibrium 
Decreased T, K decreases, equilibrium 
DH = -197 kJ mol-1
(forward reaction - exothermic)
Increased T, K decreases, equilibrium 
Decreased T, K increases, equilibrium 
endothermic reaction – 吸熱反應; exothermic reaction – 放熱反應
22
A Classical Example of Equilibrium
The Haber Process (哈柏法)
The Haber process involves the direct combination of nitrogen (氮) and
hydrogen (氫) to produce ammonia (氨).
N2(g) + 3 H2(g)
2 NH3(g)
23
What is the Effect of Adding More
N2 gas to the System?
N2 (g) + 3 H2 (g)
2 NH3 (g)
According to the Le Châtelier’s Principle
(勒沙得利爾原理), the system would
counteract the adding of N2 by
producing more NH3 – shifting the
equilibrium position (平衡位置) to the
product side.
Adding starting materials favours
formation of products.
http://www.cdli.ca/courses/chem3202/unit01/section02/lesson03/3-lesson-a.htm
24
What is the Effect of Reducing
the Volume of the System?
N2 (g) +
3 H2 (g)
4 moles of gases
2 NH3 (g)
2 moles of gas
According to the Le Châtelier’s Principle (勒沙得利爾原理), the system would
counteract the increased pressure by forming more ammonia (氨) (reducing
the number of molecules) – shifting the equilibrium position to the product
side.
Increasing the pressure of the reaction also favours formation of ammonia.
http://www.cdli.ca/courses/chem3202/unit01/section02/lesson03/3-lesson-a.htm
25
What is the Effect of Increasing
the Temperature of the System?
N2 (g) +
3 H2 (g)
2 NH3 (g)
DH = - 92 kJ
exothermic reaction
(heat is given out)
The reverse reaction is endothermic
(heat absorbing). According to the Le
Châtelier’s Principle (勒沙得利爾原理),
the system would counteract the
increased temperature by shifting the
equilibrium position to the reactant
side.
Increasing the reaction temperature
reduces the reaction yield (反應產率).
http://www.cdli.ca/courses/chem3202/unit01/section02/lesson03/3-lesson-a.htm
26
What is the Effect of Increasing
the Temperature of the System?
N2 (g) +
3 H2 (g)
2 NH3 (g)
DH = - 92 kJ
exothermic reaction
(heat is given out)
A Balance Between Reaction Rate (反應速率) and Yield (產率)
Although carrying out the reaction at a lower temperature can increase the
product yield, the reaction rate is decreased as well. It would take a very
long time to attain the equilibrium.
To attain a compromise between reaction rate and yield, the Haber
process (哈柏法) is usually carried out at 450 – 500 °C.
Rule of Thumb:
Reaction rate becomes double as the temperature is increased by 10°C.
27
The Role of Catalyst (催化劑)
The Haber Process (哈柏法) can be speeded up by adding a catalyst.
A catalyst provides an alternative route of
reaction where the activation energy (活化
能) is lower than the original chemical
reaction.
As the activation energy is lowered, more
molecules possess sufficient energy to
overcome the barrier, hence the reaction is
accelerated.
Usually, the catalyst is not consumed by
the overall reaction.
28
What are the Best Conditions for
the Haber Process
N2(g) + 3 H2(g)
Temp (°C)
Keq
25
6.4 x 102
200
4.4 x 10-1
300
4.3 x 10-3
400
1.6 x 10-4
500
1.5 x 10-5
http://www.ausetute.com.au/haberpro.html
2 NH3(g)
Keq=
[NH3]2
[N2][H2]3
29
Chemistry of the Haber Process
Chemical Equilibrium
Typical Industrial Conditions for Manufacturing Ammonia (氨)
Temperature:
Pressure:
Catalyst:
Yield:
450 – 500 °C
about 250 atmospheres (大氣壓)
iron (鐵)
about 10 – 20%
These conditions achieve a balance between the yield and production rate,
as well as the costs involved in the building and operation of the
manufacturing facility, and safety concerns.
30