ACFrOgAKiW4Xb2rZWfp0S8RCoRbOD-QwstF5cTvzrQCcos7GECeRNDHsojfjubRSIogbWBEPtg14l1MuizCmerxxjq-YboPrhlJbnb59eWz8kY7Qr80mML16dgpistvWzsaim_-t7olCLUjryMER

CHAPTER 6: CHEMICAL EQUILIBRIUM

CHAPTER 6: CHEMICAL EQUILIBRIUM
6.1 Dynamic Equilibrium

Learning outcome (LO) C1 C2 C3 C4

a. Explain the following terms: √

i. Reversible reaction

ii. Dynamic equilibrium

iii. Law of mass action

b. State the characteristic of a system in equilibrium √

c. Interpret the curve of concentration reactants and products against √

time for a reversible reaction.

6.1 (a) (i) Reversible reaction (C2)

Reversible Reaction Irreversible Reaction

The chemical reaction which take place in two The chemical reaction that only proceed one

directions (both forward and reverse reaction) direction that toward the formation of product

 Example:  Example:
A B A→ B

 The Reactant product
reaction indicates as reversible

 The ( → ) indicates as irreversible reaction

Forward reaction reaction proceeds from left to
Backward reaction right
Reversible reaction reaction proceeds from right
to left

reactions which take place in

both forward and reverse

reaction

 Many other reactions do not proceed to  Most chemical reactions proceed to completion.

completion.  The reaction continues until one of the reactants

 Not all reactants are changed to product because is completely used up and the reaction stops.

the product can dissociate to form the reactants

again.

 As soon as some product molecules are formed,

the reverse process begins to take place

 Reactant molecules are formed from product

molecules

 Reaction takes place in both directions.

1

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.1(a) (ii) Dynamic Equilibrium (C2)

Video 1: DYNAMIC EQUILIBRIUM

Dynamic A state of chemical equilibrium where both of forward and reverse reactions proceed at
Equilibrium : equals rates

Rate Forward = Rate Reverse

6.1 (b) 1. Concentration of reactants and products are constant over time
2. Rate of forward = Rate of Reverse
Characteristic of 3. The Reaction Quotient (Q) = The Equilibrium Constant (K)

system in ***the difference between Q & K ( refer page 16)

equilibrium :

6.1 (c) REMEMBER!!!! Equilibrium can only occur in a closed system
Graph : The concentration-time graph for reaction between A and B :

A B
Reactant product

From the graph:

• [A] decrease with time and [B] increase with time. After time, t1, [A] and [B] remains
constant.

• As reaction proceeds, the rate of forward reaction decreases, the rate of reverse
reaction increases.

• When the rate of forward and reverse reaction becomes equal, [A] and [B] remain
constant. At this point, the system is said to be in chemical equilibrium.

• After t1, the reaction continues indefinitely even after equilibrium has been attained.
This equilibrium is a dynamic equilibrium.

2

CHAPTER 6: CHEMICAL EQUILIBRIUM

Type of Equilibrium

Physical Equilibrium Chemical Equilibrium
 Involve only phase change  Involves chemical change

 Example:  Example:

H2O (I) H2O (g) N2O4(g) 2NO2 (g)
colourless brown

Number of H2O = Number of H2O

molecules molecules returning to  When we introduce some N2O4(g) into a sealed
flask kept at 100oC, a change occurs immediately.
leaving liquid phase
 The color slowly darkens, but after a few

moments, no further color change can be seen.

At Equilbrium :

 Rate of forward reaction = Rate of reverse reaction

N2O4(g)  2NO2(g) 2NO2(g)  N2O4(g)

 [N2O4(g)] and [NO2(g)] remain constant, but
[N2O4(g)] ≠ [NO2(g)]

6.1 (a) (iii) Law of Mass Action (C2)

LAW OF MASS ACTION For reversible reaction at equilibrium and a constant temperature, a certain
ratio of reactant and product concentration has a constant value, Keq

Explain the relationship between [reactant] and [product] at equilibrium

Example:

aA + bB cC + dD Where
a, b, c, d = Stoichiometric
At equilibrium: Keq = [ C ]c [ D ]d coefficient in the balanced eq.
Keq = equilibrium constant
[ A ]a [ B ]b [ ] = concentration

3

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.2 Equilibrium Constant

Learning outcome (LO) C1 C2 C3 C4

a. Define homogeneous and heterogeneous equilibria. √ √

b. Write expressions for equilibrium constants in terms of concentration, Kc √
√
and partial pressure, Kp for homogeneous and heterogeneous systems.
c. Use the equations, Kp = Kc(RT)∆n to solve equilibrium problems. √
√
d. Calculate Kc, Kp and the quantities of species present at equilibrium.

e. Determine the degree of dissociation, α.

f. Predict the direction of net reaction by comparing the values of reaction

quotient, Q and K.

6.2 a) Define homogeneous and heterogeneous equilibria (C1)

Type of Equilibria

Homogeneous Equilibria Heterogeneous Equilibria

Reaction in which all substances (products and reactants) Reaction in which the all substances are in two
are in the same phase or more phases

Example: Example:

2SO2 (g) + O2 (g) 2SO3 (g) NH4HS (s) NH3 (g) + H2S (g)

Exercise 6.2.1

State the type of equilibria for the reactions below:

a) 3CO2(g) + 4H2O(g) d) C
3H8(g) + 5 O2(g) H3(g) + 5/2 O2(g) 2NO(g) + 3H2O(g)

b) NH3(g) + HCl(g) e) N
H4Cl(s) NO3(s) KNO2(s) + 1/2 O2(g)

c) H
2O(l) + SO3(g)
H2SO4(aq)

4

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.2 (b) Write expressions for Equilibrium Constants, Keq (C2)

 Homogeneous Equilibria
 The concentrations in the equilibrium constant expression are usually given in moles per liter (M),
and so the symbol K is often written with a subscript c for “concentration” as in Kc.

Example : 2SO2 (g) + O2 (g) 2SO3 (g)

In term of Concentration (Kc)

Kc = [SO3]2
[SO2]2 [O2]

 When the reactants and products in a chemical equation are gaseous, we can formulate the

equilibrium expression in terms of partial pressures

instead of molar concInentetrramtioonfsP.ressure (Kp) where:
P = partial pressure of the particular gas;
Kp = (PSO3)2 Kp = equilibrium constant for the reaction in
(PSO2)2 PO2 term of pressure of reactants.

 When the equilibrium involved aqueous state, only Kc are exists.

Example : HF (aq) H+ (aq) + F- (aq)

In term of Concentration (Kc)

Kc = [H+][F-]
[HF]

 Heterogeneous Equilibria
 For heterogeneous reactions, the concentration terms of pure solids and pure liquids are not
included in the expression of equilibrium constant.
 The density of any given solid and liquid is constant and changes very little with temperature, we
can say it is constant.
 Thus, for heterogeneous reactions, the concentration of pure solids and liquids can be neglected.
For example:

Example 1 : CaCO3 (s) CaO (s) + CO2 (g)

In term of Concentration (Kc) In term of Pressure (Kp)
Kc = [CO2] Kp = PCO2

Example 2 : H2O (l) + CO32- (aq) OH- (aq) + HCO3 (aq)

In term of Concentration (Kc)

Kc = 5[HCO3][ OH-]
[CO32-]

CHAPTER 6: CHEMICAL EQUILIBRIUM

EXERCISE 6.2.2

Write expressions for KC and KP if applicable, for the following reversible reactions at equilibrium:

C3H8(g) + 5 O2(g) 3CO2(g) + 4H2O(g)

NH4Cl(s) NH3(g) + HCl(g)

6.2 (c) Use the equation Kp = Kc (RT) ∆n to solve equilibrium problems. (C3)

For reactions involving gases, Kp and Kc are not necessarily equal.

For general equation:

+aA(g) bB(g) +cC(g) dD(g)

Step 1: Write equilibrium-constant expression, Kc and Kp:

Kc = [D]d[C]c Kp = PDdPCc and

[A]a[B]b PAaPBb

Step 2: Write ideal gas equation and assuming that gas behave ideally.
PV = nRT

Step 3: Relate pressure with the concentration.

To obtain the concentration of gas X, in a mixture:
[X] = nx
V

Where Px is its partial pressure. From this it follows that;
Px =[X]RT

Step 4: Substitute P from ideal gas equation into the expression for Kp and substitute Kc

Kp = PDdPCc = [D]d(RT)d[C]c(RT)c or Kp = Kc(RT)n
PAaPBb [A]a(RT)a[B]b(RT)b

This can be rearranged to give:
Kp = [D]d[C]c (RT)(d+c) – ( a+b)
[A]a[B]b

6

CHAPTER 6: CHEMICAL EQUILIBRIUM

Video 2: USE THE EQUATION KP = KC (RT) ∆n

Extra Exercises
1. Determine type of equilibria and write equilibrium constant, Kc and Kp for the reaction
below:

a) 2SO2(g) + O2(g) ⇌ 2SO3(g)
b) 2C(s) + 2O2(g) ⇌ 2CO2(g)
c) NH4HS(s) ⇌ NH3(g) + H2S(g)
d) HF(aq) ⇌ H+(aq) + F-(aq)
e) H2O(l) + SO3(g) ⇌ H2SO4(aq)

2. Consider the following equation,
N2(g) + 3H2(g) ⇌ 2NH3(g)

Applying the law of mass action derive an equation to show relationship between Kc and Kp
for the above reaction. If Kc for the following reaction is 15.35 at 300°, calculate Kp at the
same temperature.

ANS : 6.939 x 10-3
7

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.2 (d) Calculate Kc, Kp and the quantities of species present at equilibrium. (C4)
2 types of Equilibrium Problem

Equilibrium quantities (concentration, partial Initial quantities (initial concentrations, initial
pressure) are given, determine KC or KP partial pressure) and KP or KC are given,
determine equilibrium quantities such as
concentration or partial pressure.

Use ICE Table

Notes:

The approach in solving equilibrium constant problems could be summarized as follows:

i) Express the equilibrium concentration of all species in terms of the initial concentrations and
a single unknown x, which represents the change in concentration.

ii) Write the equilibrium constant expression in terms of the equilibrium concentrations.
Knowing the value of the equilibrium constant, solve x

iii) Having solved for x, calculate the equilibrium concentrations all species.

Example 1:

The equal amount of hydrogen and iodine are injected into a 1.50 L reaction flask at a fixed
temperature.

H2(g) + I2(g) 2HI(g)

At equilibrium, analysis shows that the flask contains 1.80 mole of H2, 1.80 mole of I2, and
0.520 mole of HI. Calculate Kc.

Solution:

Convert the amounts(mole) to concentration (mole/L), using the flask volume of 1.50 L

[ H2 ] = 1.80 /1.50 = 1.20 M

[ I2 ] = 1.80 /1.50 = 1.20 M

[ HI ] = 0.347 /1.50 = 0.347 M

Kc = [ HI ]2
[ H2 ][ I2 ]

= ( 0.347 )2
(1.20)(1.20)
= 8.36 x 10-2

8

CHAPTER 6: CHEMICAL EQUILIBRIUM

Video 3: ICE TABLE FOR EQUILIBRIUM
Video 4: ICE TABLE FOR EQUILIBRIUM

EXERCISE 6.2.3

1. The following equilibrium process has been studied at 230oC:

2NO(g) + O2(g) 2NO2(g)

In an experiment the concentration of the reacting species at equilibrium were found to be as follows:

[NO] = 0.0542 M; [O2] = 0.127 M; [NO2] = 15.5 M. Determine the equilibrium constant (KC) of the reaction at

this temperature.

ANS : 6.44 x 103
9

CHAPTER 6: CHEMICAL EQUILIBRIUM

2. The equilibrium constant, Kp for the reaction is 158 at 1000K. What is the equilibrium pressure for O2 if

PNO2 = 0.400 atm and PNO = 0.270 atm?

2NO2(g) 2NO(g) + O2(g)

ANS : 346.8 atm
10

CHAPTER 6: CHEMICAL EQUILIBRIUM

Example 2

At initial, 0.200 mole of hydrogen halide is injected into a 2.00L reaction flask at fixed temperature:

2HI(g) H2(g) + I2(g)

If [ HI ]=0.078 M at equilibrium, calculate Kc.

Solution:

[ HI ] = 0.200 mol/2.00L = 0.100M

2HI(g) H2(g) I2(g)

Initial 0.100 0 0

Change -2x +x +x

Equilibrium 0.100-2x x x

concentration(M)

Solving for x, using the known [ HI ] at equilibrium:

[ HI ] = 0.100 M – 2x = 0.078M

x = 0.011 M

Therefore, the equilibrium concentrations are:

[ H2 ] = [ I2] = 0.011 M and [ HI ] = 0.078 M

Kc = [ H2 ][ I2 ] = 0.020
[ HI ]2

= ( 0.011 )( 0.011 )
0.0782

11

CHAPTER 6: CHEMICAL EQUILIBRIUM

Example 3:

At 440C, the equilibrium constant Kc for reaction,

H2(g) + I2(g) 2HI(g)

has a value of 49.5. If 0.200 mole of H2 and 0.200 mole of I2 are placed into a 10.0 L vessel and
permitted to react at this temperature, what will be the concentration of each substance at
equilibrium?

Solution:

Kc = [ HI ]2
[ H2 ][ I2 ]
= 49.5

[ H2 ] = 0.200 /10.0 = 0.0200 M

[ I2 ] = 0.200 /10.0 = 0.0200 M

[ HI ] = 0.0 M

H2(g) I2(g) 2HI(g)

Initial 0.0200 0.0200 0

Change -x -x +2x

Equilibrium 0.0200-x 0.0200-x 0 + 2x

concentration(M)

= 49.5

= 7.04

x = 0.0156

The equilibrium concentrations are :

[ H2 ] = 0.0200 – 0.0156 = 0.0044 M

[ I2 ] = 0.0200 – 0.0156 = 0.0044 M

[ HI ] = 2(0.0156) = 0.0312 M

12

CHAPTER 6: CHEMICAL EQUILIBRIUM

EXERCISE 6.2.4

1. At 25C, Kp=7.13 atm-1 for the following reaction.

2NO2(g) N2O4(g)

If the partial pressure of NO2 in a container is 0.15 atm at equilibrium, what is the partial

pressure of N2O4 in the mixture?

ANS : 0.1604 atm

2. At a temperature of 500C, the equilibrium constant, Kc, for the nitrogen fixation reaction for

the production of ammonia ,

3H2(g) + N2(g) 2NH3(g)

has a value of 6.0 x 10-2. If in a particular reaction vessel at this temperature, there are

0.250 mol/L of H2 and 0.0500 mol/L of NH3 present at equilibrium, what is the concentration

of N2?

ANS : 2.67M

3. At 25C, 0.0560 mole of O2 and 0.020 mol N2O were placed in a 1.00 L vessel and allowed to

react according to the following equation.

2N2O(g) + 3O2(g) 4NO2(g).

When the system reached equilibrium, the concentration of the NO2 was found to be 0.020

mol/L.

i) What are the equilibrium molar concentration of N2O and O2?

ii) What is the value of Kc for this reaction?

ANS : [N2O] = 0.01M and [O2] = 0.0412M , Kc = 23.21

4. A mixture of H2 and I2 was prepared by placing 0.200 mol of H2 and 0.200 mol of I2 in a 2.00L

flask at a certain temperature. After a period of time, equilibrium was established.

H2(g) + I2(g) 2HI(g)

At equilibrium the concentration of I2 dropped to 0.020M. Determine the value of Kc for this

reaction at this temperature.

ANS : 64

5. At 1280°C the equilibrium constant, Kc for the reaction is 1.1 x 10-3.

Br2(g) 2Br(g)

If the initial concentration is [Br2] = 0.063M and [Br] = 0.012M, determine the concentration

of these species at equilibrium.

ANS : [Br2] = 0.06478M and [Br] = 0.008442M

6. 2SO2(g) + O2(g) 2SO3(g)
In one experiment at 750°C an equilibrium was achieve in closed vessel with partial pressure
of SO2, O2 and SO3 are 0.27 atm, 0.40 atm and 0.32 atm respectively. Calculate Kp and initial
partial pressure of SO2 and O2.
ANS : Kp = 3.512, PSO2 = 0.59 atm, PO2 = 0.56 atm

13

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.2 (e) Determine the Degree of Dissociation, α (C3)

 Dissociation reaction is a chemical reaction whereby a molecule is broken down into smaller molecules,
atoms or ions.

 Degree of dissociation, α is the fraction or the percentage of molecules that dissociate
 If dissociation occurs completely, α = 1 or 100%
 If dissociation is incomplete, then the value α of can be calculated as follows:

Fraction: α = [ ] change
[ ] initial

% dissociation : α = [ ] change
X 100

[ ] initial

Example :

If 5.00 ml of PCl5 gas is placed in a 1.00 L container at 25○C, calculate:

a) The number of moles of Cl2 and PCl3 formed at equilibrium.
b) The degree of dissociation, α, of PCl5. Given the equilibrium constant for the decomposition at

25○C is 0.40.

Solution :

Initial (mo) PCl5 (g) PCl3 (g) Cl2 (g)
Change (mol) 5.00 0 0
Final (mol) +x +x
Final [] -x x x
5.00-x x X
5.00-x

a) Kc = [PCl3][ Cl2] = x2 = 0.40

[PCl5] 5.00 – x

x2+ 0.4x – 2 = 0

x = 1.23

nPCl3 = nCl2 = 1.23 mol

b) α = x = 1.23 = 0.246

initial mol 5.00

14

CHAPTER 6: CHEMICAL EQUILIBRIUM

EXERCISE 6.2.5

1. Consider the decomposition of hydrogen bromide, HBr in the following equation:

2HBr(g) H2(g) + Br2(g)

Initially, 1.0 mol HBr is placed inti 2.0 L flask and allow to reach equilibrium. At equilibrium, the
concentration of H2 is 0.15 M. Calculate the degree of dissociation of HBr.

ANS : 0.6
15

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.2 (f) Predict the direction of net reaction by comparing the values of reaction quotient, Q and K. (C3)

To predict the direction of reaction, we compare the equilibrium constant expression Kc or Kp with
the value of reaction quotient, Qc or Qp.

The expression for Q is same as that for the equilibrium constant, K. The difference is that the Q is
determined from non-equilibrium conditions.

Reaction Quotient, Q The mass-action expression under any condition

aA + bB cC + Dd

Q = [ C ]c [ D ]d
[ A ]a [ B ]b

Comparing Q and K aA + bB cC + Dd

Q=K reactant product
System is at equilibrium!
 Concentration (or pressure) of reactants and products has attained their

equilibrium values.

 Equilibrium is reached. No net reaction occurs.

Q<K aA + bB cC + Dd

System is not at equilibrium! reactant product

 Denominator (reactants) is large relative to numerator (products)

 To reach equilibrium,, products must increase and reactants must decrease

 Thus, the equilibrium position shifts to RIGHT (to form more product) until

equilibrium is reached(Q = K)

Q>K aA + bB cC + Dd
System is not at equilibium!
reactant product

 Numerator (products) is large relative to Denominator (reactants)

 To reach equilibrium, products must decrease and reactants must increase

 Thus, the equilibrium position shifts to LEFT (to form more reactant) until

equilibrium is reached(Q = K)

Video 5: COMPARING THE VALUES OF REACTION QUOTIENT, Q AND K.

16

CHAPTER 6: CHEMICAL EQUILIBRIUM

Example:

At 375C, the initial concentrations of N2, H2 and NH3 are 0.0711 M , 9.17 x 10-3 M and 1.83x10-4 M
respectively. If the equilibrium constant (Kc) for the reaction is 1.2 at this temperature, decide
whether the system is at equilibrium. If it is not, predict which way will the reaction proceed.

+3H2(g) N2(g) 2NH3(g)

Qc = [ NH3 ]2
[ N2 ][ H2 ]3

= ( 1.83 x 10-4)2
(0.0711)(9.17 x 10-3)3

= 0.611

Since Qc < Kc, the system is not at equilibrium.

The reaction will shift to the right until Q=K.

The net result will be an increase in the concentration of NH3 and a decrease in the concentrations
of N2 and H2.

EXERCISE 6.2.6

1. KC for the formation of nitrosyl chloride, an orange–yellow compound, from nitric oxide and
molecular chlorine is 6.5 x 104 at 35oC.

2NO(g) + Cl2(g) 2NOCl(g)

In an experiment, 2.0 x 10–2 mol of NO, 8.3 x 10–3 mol of Cl2, and 6.8 moles of NOCl are mixed in

a 2.0L flask. In which direction will the system proceed to reach the equilibrium?

2. When the numerical value of Q is less than K, in which direction does the reaction proceed to
reach equilibrium? Explain.

3. For the synthesis of ammonia,

+3H2(g) N2(g) 2NH3(g)

the equilibrium constant Kc at 375C is 1.2. Starting with [ H2 ]= 0.76 M, [ N2 ]= 0.60 M and [ NH3] =
0.48M, which gases will have decreased in concentration when the mixture comes to equilibrium ?

17

CHAPTER 6: CHEMICAL EQUILIBRIUM

6.3 LE CHATELIER’S PRINCIPLE

Learning outcome C1 C2 C3 C4
a. State Le Chatelier’s Principle. √
b. Apply the Le Chatelier’s principle to explain the effect of the following √
√
factors on a system at equilibrium:
i. concentration of reacting species √
ii. pressure by changing volume √
iii. addition of inert gas at constant volume √
iv. addition inert gas at constant pressure √
v. temperature
vi. catalyst

6.3 a) State Le Chatelier’s Principle (C1)
b) Apply the Le Chatelier’s principle to explain the effect of the following factors on a system at
equilibrium

Le Chatelier’s Principle

If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one
of the components, it reattains equilibrium by undergoing a net reaction (the system will shift its
equilibrium position) that reduces the effect of the disturbance.

Factors that affect the equilibrium position of a system.

Concentration (C2) Pressure (C2) Catalyst (C2) Temperature (C2)
Endothermic
Change Volume Exothermic
Add inert gas (C4)

Constant Volume
Constant Pressure

Video 6: LE CHATELIER’S PRINCIPLE

Video 7: LE CHATELIER’S PRINCIPLE

18

CHAPTER 6: CHEMICAL EQUILIBRIUM

i. Concentration o0f reacting species

 At equilibrium : 2SO3 (g) 2SO2 (g) + O2 (g) ∆H = -197.8 kJ

Change [reactant] Change [product]
Reactant is added, Product is added

 When reactant is added to the  When product is added to the
system, the equilibrium will shift to system, the equilibrium will shift to
the right and produced more product the left and produce more reactant

 As a result:  As a result:
 Amount of reactant decrease.  Amount of reactant increase.
 Amount of product increase.  Amount of product decrease.

Reactant is removed Product is removed
 Decreasing the concentration of a  Decreasing the concentration of a
reactant from the system, the product from the system, the
equilibrium will shift to the left and equilibrium will shift to the right and
produce more reactant produce more product
 As a result:  As a result:
 Amount of reactant increase.  Amount of reactant decrease.
 Amount of product decrease.  Amount of product increase.

Example

Synthesis of HI

H2(g) + I2(g) 2HI(g)

Increasing the concentration of H2 to a reaction mixture will disturb the equilibrium of the system.
The system will reduce the disturbance by eliminating some of the added H2. The additional H2
reacts with some of the I2 to form more HI. As a result, the position of equilibrium shifts to the right.

19

CHAPTER 6: CHEMICAL EQUILIBRIUM

ii. Pressure by changing volume

 However, changes in pressure have a measurable effect only in systems in which gases are

involved, and then only when the chemical reaction produces a change in the total number of

gas molecules in the system (look for different numbers of moles of gas on the reactant and

product sides of the equilibrium)

 Example : At equilibrium : 2NO2 (g) N2O4 (g)

2 mol 1 mol

Change Volume

Increase volume will cause pressure to decrease Decrease volume will cause pressureto increase

(assume no change in Temperature) (assume no chane in Temperature)

2 NO2 (g) N2O4 (g) 2 NO2 (g) N2O4 (g)
2 mol 1 mol 2 mol 1 mol

 By increase the volume of container cause the  By decrease the volume of container cause the

pressure to decrease. pressure to increase.

 The equilibrium will shift to the left to increase the  The equilibrium will shift to the right to decrease

pressure by increase the number of moles of gases. the pressure by decrease the no of moles of

 As a result: gases.

 Amount of reactant increase.  As a result:
 Amount of product decrease.  Amount of reactant decrease.
 Amount of product increase.

 Now consider this reaction:

N2(g) + O2(g) ⇌ 2NO(g)

Because there is no change in the total number of molecules in the system during reaction, a
change in pressure does not favor either formation or decomposition of gaseous nitrogen
monoxide.

20

CHAPTER 6: CHEMICAL EQUILIBRIUM

iii. Addition of inert gas

Adding Inert Gas

(A gas that does not involve in the reaction (such as helium, neon and argon)

i) At constant Pressure ii) At constant volume

 When inert gas is added to the system in  When inert gas is added to the system in
equilibrium at constant pressure causes total equilibrium at constant volume causes total
number of moles of gaseous molecules in the number of moles of gaseous molecules in the
system increases. system increases.

 The partial pressure of each gas will decrease.  The total pressure of the equilibrium system
 Thus, the equilibrium will shift towards the increases.

direction in which there is increase in number of  However, the partial pressures of each of the
moles of gases to increase the partial pressure. gases in the equilibrium system remains
unchanged and no change in equilibrium.

Consider the following reaction in equilibrium: From Dalton’s law:
2NH3(g) ⇌ N2(g) + 3H2(g)
P = n (RT/V)
AA

 The addition of an inert gas at constant pressure partial pressure of any gas, A
PA =

to the above reaction will shift the equilibrium

towards the forward direction because the  Adding inert gas at constant volume will not
number of moles of products is more than the change any of the quantity on the right side of
number of moles of the reactants. the above equation. So, PA remains constant.

 Therefore, addition of inert gas at constant

volume has no effect on the equilibrium

position and the composition of the

equilibrium mixture.

21

CHAPTER 6: CHEMICAL EQUILIBRIUM

iv. Temperature

Endothermic : Reactants + heat products

Exothermic : Reactants products + heat

Endothermic system - A reaction that absorbs heat in the forward reaction
- Enthalpy change of the reaction, (∆H) is positive

At equilibrium : N2O4 (g) + HEAT 2NO2 (g) ∆H = + 58 kJ

Increase Temperature of the system Decrease Temperature of the system

N2O4 (g) 2NO2 (g) ∆H = + 58 kJ N2O4 (g) 2NO2 (g) ∆H = + 58 kJ

 The forward reaction is endothermic (+ve  The forward reaction is endothermic (+ve

 When temperature increase, the equilibrium will  When temperature decrease, the equilibrium will

shift to absorbing the heat, which is causes the shift to release more heat, which is causes the

position of equilibrium to shift to the right. position of equilibrium to shift to the left.

 Rate of forward reaction > Rate of reverse reaction  Rate of reverse reaction > Rate of forward reaction

 As a result:  As a result:
 Amount of N2O4 decrease.  Amount of N2O4 increase.
 Amount of NO2 increase.  Amount of NO2 decrease.

Example:

Co(H2O)62+ (aq) + 4Cl- (aq) CoCl42- (aq)
+ 6H2O (l) ∆H >0
Pale pink deep blue

The reaction is endothermic as ∆H > 0
In an endothermic reaction, heat is absorbed;
Increasing the temperature causes the equilibrium to shift to the right, in the direction of products

and leads to the formation of more CoCl42-; and Kc increases.
Cooling the reaction shifts the equilibrium to the left, decreasing Kc.

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CHAPTER 6: CHEMICAL EQUILIBRIUM

Exothermic system - A reaction that release heat in the forward reaction

- Enthalpy change of the reaction, (∆H) is negative

At equilibrium : 2 SO3 (g) 2SO2 (g) + O2 (g) + HEAT ∆H = -197.8 kJ

Increase Temperature of the system Decrease Temperature of the system

2 SO3 (g) 2SO2 (g) + O2 (g) ∆H = -197.8 kJ 2 SO3 (g) 2SO2 (g) + O2 (g) ∆H = -197.8 kJ

 The forward reaction is exothermic reaction (∆H = -  The forward reaction is exothermic reaction (∆H = -
ve), heat is released during the reaction. ve), heat is released during the reaction.

 When temperature increase, the equilibrium will shift  When temperature decrease, the equilibrium will
to absorbing the heat, which is causes the position of shift to release more heat, which is causes the
equilibrium to shift to the left. position of equilibrium to shift to the right.

 Rate of reverse reaction > Rate of forward reaction  Rate of reverse reaction > Rate of forward reaction
 As a result;  As a result;
 Amount of SO3 increase.  Amount of SO3 decrease.
 Amount of SO2 and O2 decrease.  Amount of SO2 and O2 increase.

v. Catalyst

 A catalyst provides an alternative pathway of lower activation energy for reaction to occur. A
catalyst will not shift an equilibrium position because both rates are equally increased. The
 equilibrium is achieved faster.
 As a result, a catalyst increases the rate at which equilibrium is achieved, but it does not change the
composition of the equilibrium mixture.
 For instance, if a catalyst is added to a mixture of PCl3 and Cl2 at 523 K, the system will attain the
 same equilibrium concentrations of PCl3, Cl2 and PCl5 faster than that without catalyst. Therefore,
catalyst often plays key roles in optimizing reaction systems.
i) Catalyst has no effect on the equilibrium position and value of KC.
ii) Function catalyst:
Catalyst speeds up rate of reaction by providing alternative mechanism with lower activation energy.
Shorten the time taken to reach equilibrium

Video 8: LE CHATELIER’S PRINCIPLE

Video 9: LE CHATELIER’S PRINCIPLE
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CHAPTER 6: CHEMICAL EQUILIBRIUM

Summary

Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in
some cases, volume and pressure; volume and pressure changes will disturb equilibrium if
the number of moles of gas is different on the reactant and product sides of the reaction.
The system's response to these disturbances is described by Le Chatelier's principle: The
system will respond in a way that counteracts the disturbance.
Not all changes to the system result in a disturbance of the equilibrium.
Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and
changing pressure or volume will not significantly disturb systems with no gases or with
equal numbers of moles of gas on the reactant and product side.

Effects of Disturbances of Equilibrium and K

Disturbance Observed Change as Direction of Shift Effect on K
Equilibrium is toward products none
Restored

reactant added added reactant is
partially consumed

product added added product is toward reactants none
partially consumed

decrease in pressure decreases toward side with fewer none
volume/increase moles of gas
in gas pressure

increase in pressure increases toward side with fewer none
volume/decrease moles of gas
in gas pressure

temperature heat is absorbed toward products for changes
increase endothermic, toward
reactants for exothermic

temperature heat is given off toward reactants for changes
decrease endothermic, toward
products for exothermic

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