Unit 10

Chemical
Reactions
10.1- Reversible reactions.
10.2- Shifting the equilibrium
10.3- Redox reactions
10.4- More about redox reactions
10.5- Oxidising agents and
reducing agents
 Chemical reaction
A chemical reaction is the process where reactants are transformed
into products.
10.1 – Reversible reactions


 dehydrating

hydrating

hydrated means that the solid crystals contain water of crystallisation.

dehydration is the removal of water of crystallisation.
anhydrous substance contains no water of crystallisation.
Hydration is the addition of water molecules.
Water of crystallization refers to water molecules that are tightly
bound within the crystal lattice of a substance.
(water that is chemically bonded into a crystal structure.)
Some chemical reactions are reversible.
 Open system and closed system

Open system Closed system

An open system can exchange both A closed system can exchange only
energy and matter with its energy with its surroundings, not
surroundings matter.
 Reversible reaction
A reversible reaction is a reaction in which the conversion of reactants
to products and the conversion of products to reactants occur
simultaneously.
Example:
the reaction of hydrogen gas and iodine vapor to form hydrogen iodide.
Forward reaction: H 2(𝑔) + I2(𝑔) → 2HI(𝑔) (combination)
Reverse reaction:2HI(𝑔) → H2(𝑔) + I2(𝑔) (decomposition)

The two reactions can be combined into one equation by the use of a
double arrow :
H2(g) + I2(g) ⇌ 2HI(g)
When a reversible reaction occurs in a close system, it reaches to
dynamic equilibrium.
 ⇌

position of equilibrium
describes the relative concentrations of reactant and product.

At equilibrium,
- If [P] > [R] , position of equilibrium is to the right. It favours the
products.
- If [R] > [p] , position of equilibrium is to the left. It favours the
reactants.
 Questions!





N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
Changing equilibrium conditions!

When a reversible reaction is in equilibrium and you make a change,

the system acts to oppose the change, and restore equilibrium.
A new equilibrium mixture forms.
 Le Châtelier's principle
- if a dynamic equilibrium is disturbed by changing the
conditions, the position of equilibrium shifts to counteract the
change to re-establish an equilibrium.

Conditions of a reaction,
- Temperature
- Pressure (for gaseous reactions)
- Concentration
- Catalyst

Pressure is caused by the gas molecules colliding with the walls

of the container.
The more molecules present, the higher the pressure.
 exo
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
1 mol 3 mol endo 2 mol
- if you heat the equilibrium mixture, it acts to oppose the change.
More NH3 breaks down in order to use up the heat you add.
Backward reaction is favour and amount of NH3 decrease.
- if you lower the temperature, the system acts to oppose the change.
Forward reaction is favour and more NH3 forms, giving out heat.
- if you increase the pressure, the equilibrium mixture acts to oppose this.
More NH3 forms, which means fewer molecules.
- if you decrease the pressure, the equilibrium mixture acts to oppose this.
More NH3 decompose, which means more molecules.
- if you remove NH3 from the reaction mixture, the position of equilibrium
will move to counteract the change.
More N2 and H2 react to produce more NH3 .
The position of equilibrium moves to the right.
- If you add a catalyst, the catalyst speeds up the forward and back reactions
equally. So the reaction reaches equilibrium faster, which saves you time.
But the amount of ammonia does not change.
Le Châtelier’s Principle:
if conditions of equilibrium are changed, the position of the equilibrium moves
to oppose change;

Temperature: Temperature decreased; equilibrium moves in exothermic direction.

Temperature increased; equilibrium moves in the endothermic direction.

Pressure: Pressure raised; equilibrium moves to the side with the fewest gas molecules.
Pressure lowered; equilibrium moves to the side with most gas molecules.

Concentration: Decreasing reactant concentration or increasing product concentration;

equilibrium moves to the reactant side.
Increasing reactant concentration or decreasing product concentration;
equilibrium moves to the product side.
Question!

1. Hydrogen and bromine react reversibly:

H2 ( g) + Br2 ( g) ⇌ 2HBr ( g)
a. Which of these will favour the formation of more hydrogen bromide?
i add more hydrogen
ii remove bromine
iii remove the hydrogen bromide as it forms
b. Explain why increasing the pressure will have no effect on the amount
of product formed.
c. However, the pressure is likely to be increased, when the reaction is carried
out in industry. Suggest a reason for this.
2. Iodine reacts with chlorine to form dark brown iodine monochloride

I2 + Cl2 → 2ICl
This reacts with more chlorine to give yellow iodine trichloride.
There is an equilibrium between these iodine chlorides.

ICl (l) + Cl2 (g) ⇌ ICl3 (s)

(a)Explain what is meant by equilibrium.
(b)When the equilibrium mixture is heated it become a darker brown colour.
Is the reverse reaction endothermic or exothermic? Give a reason for your
choice.
(c)The pressure on the equilibrium mixture is decreased.
(i) How would this affect the position of equilibrium and why?
(ii) Describe what you would observe.
3. At most temperatures, samples of nitrogen dioxide are equilibrium
mixtures.

2NO2 (g) ⇌ N2O4 (g)

dark brown pale yellow

(i) At 25 ℃, the mixture contains 20% of nitrogen dioxide. At 100℃ this has
risen to 90%.
Is the forward reaction exothermic or endothermic? Give a reason for
your choice.

(ii) Explain why the colour of the equilibrium mixture becomes lighter when t
the pressure on the mixture is increased.
 Chapter 10.3 - Redox Reactions

Redox reactions involve both oxidation

and reduction at the same time.

Oxidation numbers give information about

the degree of oxidation or reduction of
compounds.
 Oxidation and Reduction

 Oxidation

 Reduction

 Oxidation state


 positive negative
zero
- If the oxidation number is positive, it means the atom loses electrons.
- If it is negative, it means the atom gains electrons.
- The oxidation number is zero for all elements.

Note!
Roman numerals are used after an element in the name of a compound to refer to its
oxidation state.
The oxidation state of an element is zero Ag, Cu, Cl2, H2, O2
Group I elements +1
Some elements have a fixed oxidation state in their
Group II elements +2
compounds.
Hydrogen +1
Oxygen -2
H2SO4
The oxidation state of atoms or ions in a compound add (+1x2)+(+6)+(-2x4)
up to zero. +2 +6 -8 = 0
(the total ox.no of all the elements in a compound is zero) CuO
+2 +(-2) = 0
Cu2+ = +2
The oxidation state of simple ion is the same as the charge
Cl- = -1
on the ion.
Fe3+ = +3
SO42-
+6 +(-2x4)
The sum of oxidation states of the atoms in a polyatomic +6 – 8 = -2
ion is the same as the charge on the ion. NO3-
+5 +(-2x3)
+5 -6 = -1
1.Define oxidation and reduction in terms of oxygen loss/gain
- Oxidation is gain of oxygen
- Reduction is loss of oxygen

Example: Mg(s) + FeO(s) → MgO(s) + Fe(s)

The reaction between Magnesium and Iron(II)oxide, Mg has gained oxygen

and oxidized to MgO while FeO has lost oxygen and reduced to Fe.

Mg (s) + Fe O (s) → MgO (s) + Fe (s)

oxidation no 0 +2 -2 +2 -2 0
2.Define oxidation and reduction in terms of hydrogen loss/gain
- Oxidation is loss of hydrogen
- Reduction is gain of hydrogen

Example: CH4(g) + 2Cl2(g) → C(s) + 4HCl(g)

In that reaction, carbon has lost hydrogen and oxidized to carbon while
chlorine has gained hydrogen and reduced to hydrogen chloride.

CH4 (g) + 2Cl2 (g) → C(s) + 4HCl(g)

-4 +1 0 0 +1 -1
3.Define redox in terms of electron transfer

- Oxidation Is Loss (of electrons)

- Reduction Is Gain (of electrons)

Example: 2Fe(s) + O2(g) → 2 FeO(s)

0 0 +2 -2

In the above reaction, iron has lost ēs to form Fe2+ ion and oxidized.
while oxygen has gained ēs to form O2- ion and reduced.

The half equations are;

2 Fe → 2 Fe2+ + 4ē (oxidation reaction)
O2 + 4ē → 2O2- (reduction reaction)
4.Define redox in terms of oxidation state

- Oxidation Is Increasing its oxidation state

- Reduction Is Decreasing its oxidation state

Example:
When chlorine gas is bubbled into sodium bromide solution, bromide
ions, Br-, are displaced to form bromine molecules, Br2.
How do we know which substance has been oxidized?

Cl2(g) + Br-(aq) → Cl- (aq) + Br2(g)

0 -1 -1 0
The oxidation state of bromine increases from -1 to 0.
Br- ion has been oxidized to bromine molecule (Br2).
The oxidation state of chlorine decreases from 0 to -1.
Cl2 molecule has been reduced to chlorine ion (Cl-).
Questions!
1.The element strontium forms a nitrate, Sr(NO3)2, which decomposes on heating as
shown below.

2 Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g)

Using oxidation numbers, explain why the reaction involves both oxidation and
reduction.

2. The reaction between magnesium and sulfuric acid is a redox reaction.

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Use transfer of ēs to identify which element has been oxidized. Explain your answer.
5.Define oxidising agent and reducing agent

- Oxidising agent is a substance which oxidises another substance during a redox reaction.
Oxidising agents are reduced
- Reducing agent is a substance which reduces another substance during a redox reaction.
Reducing agents are oxidized
They can be identified by recognising which elements have been oxidised/reduced in an
equation.

Example:
H2(g) + Cl2(g) → 2HCl(g)
Oxidation no 0 0 +1 -1

The oxidation no of Hydrogen increases 0 to +1 so that it oxidized while those no to Chlorine

reduces 0 to -1 and it reduced.
Hydrogen oxidises by itself and reduces to other so Hydrogen is called reducing agent.
Deduce the oxidising agent and reduceing agent for the following reacations.

1. The reaction between copper(II) oxide and hydrogen.

2. The reaction between magnesium and solution containing silver ions, Ag+.
3. The reaction between iron(III)oxide and carbon monoxide.

Oxidizing agents Reducing agents

O2 oxygen H2 hydrogen
Br2 bromine C carbon
Cl2 chlorine CO carbon monoxide
H2SO4 sulfuric acid SO2 sulfur dioxide
HNO3 nitric acid H2S hydrogen sulfide
KMnO4 KI potassium iodide
potassium manganate(VII)
K2Cr2O7
metals
potassium dichromate(VI)
 Colour changes in redox reactions

The colour changes of potassium manganate(VII), KMnO4 and potassium iodide, KI,
can be used to test for oxidizing and reducing agents.

To determine if a substance X is a reducing agent, we react the substance with an

oxidizing agent, eg. KMnO4
- If the purple KMnO4 turns colourless, the X is reducing agent.
- If the purple KMnO4 remains purple, then X is not a reducing agent.

To determine if a substance X is an oxidizing agent, we react the substance with an

reducing agent, eg KI.
- If colourless KI turns brown, the X is oxidising agent.
- If colourless KI remains unchanged, then X is not a reducing agent.
Test for reducing agent
When aqueous solution of acidic KMnO4 is added to iron(II) solution,
solution change colour from light green to yellow.

https://www.youtube.com/watch?v=Xb3zb9E7xlw

MnO- ₄ + 5Fe ²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

purple light green colourless yellow
Ox.no +7 +2 +2 +3

Test for oxidising agent

When aqueous solution of acidic KI is added to hydrogen peroxide solution,
solution change from colourless to brown.

2 I⁻ + H₂O₂ + 2H⁺ → I₂ + 2H₂O

colourless colourless brown
Ox.no -1 -1 0 -2
• →
•

• →
Answers for practice questions
1 A. Hydrated salt

2 B. increasing the temperature shift the equilibrium to the left

3 (a) (i) Add water to the white copper sulfate

(ii) CuSO4 + 5H2 O → CuSO4 .5H2 O

(b) Blue / hydrated;

crystallisation;
anhydrous

(c) (i) Cu
(ii) Removal of oxygen / addition of hydrogen / gain of electrons /
decrease in oxidation number

(d) Fe2O3 gets reduced as it loses the oxygen and change to Iron.
4 (a) Concentrations of hydrogen and iodine are decreasing with time,
fewer reactant molecules collide so that forward reaction rate
decreases.

(b) Concentration of hydrogen iodide molecules is increasing so

that more decomposition occur to maintain equilibrium.

(c) Rate of forward reaction = rate of reverse reaction.

(d) Shifts the position of equilibrium to the right until the relative
concentrations of the reactants and products are at the correct
equilibrium concentrations.

(e) There are equal number of moles of gases on each side of the
equation.

(f) Position of equilibrium moves to the left as increasing the

temperature favour the endothermic reaction.
5 (a) (i) 0
(ii) -1
(iii) Chlorine as Cl atoms have gained electrons and itself reduced.

(b) Cl2 + 2I− → 2Cl− + I2

(c) Cl2 + 2e− → 2Cl− ;
2I− → I2 + 2e−

(d) (i) Equilibrium

(ii) Brown liquid reforms / yellow crystals disappear as position of
equilibrium shifts to the left to produce more chlorine forms.
(iii) Position of equilibrium shifts to the left; decreasing pressure
shifts reaction in direction of more gas molecules
6 (a) Reduction as well as oxidation has taken place at the same time
as N2 has been oxidized while oxygen reduced.
N2 is reducing agent

(b) When temperature increases, it favours the endothermic reaction

Position of equilibrium shifts to the right.
more NO formed

(c)(i) S oxidation number increase from 0 to +6, it oxidized when it

reacts with HNO3. So, HNO3 is oxidising agent.
(ii) ox. no. of N in HNO3 = +1 + (N) + (3 x -2) =0
So, N = +5
(iii) S loses the electrons and it oxidized.
N gains the electrons and it reduced.
7. (a) An equilibrium reaction can take place only in a closed system.
- At equilibrium, the concentrations of reactants and products are
no longer changing.
- At equilibrium, the rate of the forward reaction equals the rate
of the backward reaction.

(b) (i) when concentration of Cl2 decreases, HClO and HCl react more
and position of equilibrium will shift to left.

(ii) position of equilibrium shift to right. Increasing the pressure

more HClO and HCl will formed.
Thank you