CHEMICAL REACTIONS

Physical & Chemical changes

Physical change
 Physical changes (such as melting or evaporating) do not produce any new
chemical substances.

 These changes are often easy to reverse and mixtures produced are usually
relatively easy to separate.

Chemical change
 In chemical reactions, new chemical products are formed that have very
different properties to the reactants.
 Most chemical reactions are impossible to reverse.
 Energy changes also accompany chemical changes and energy can be given out
(exothermic) or taken in (endothermic).
 The majority of chemical reactions are exothermic with only a small number
being endothermic.

Collision theory
At an atomic level, a chemical reaction will occur when two conditions
are met. Two particles need to collide and they must have enough energy to react.
This means the reaction rate will depend on these two factors: Collision rate &
particle energy.
If two atoms collide but they don’t have enough energy, then the reaction will not
occur. If the particles have enough energy but they don’t collide, then again, the
reaction will not occur.

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It is very important to realize that there are certain factors that affect the collision
rate and particle energies, and therefore directly affect the chemical reaction rate.
The factors that affect the collision rate and particle energies are:
(a) Concentration (b) Particle size
(c) Temperature (d) Catalysts

Factors affecting rate of chemical reaction

(a) Concentration

Diagram showing increase in concentration of solution

Explanation:
Increase in the concentration of a solution, the rate of reaction will increase.

This is because there will be more reactant particles in a given volume, allowing
more frequent and successful collisions per second, increasing the rate of
reaction.

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Graphical representation:

Graph showing the effect of the concentration of a solution on the rate of reaction

Experimental Evidence:

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 Diagram showing the apparatus needed to investigate the effect of concentration on the rate of reaction

Procedure:

 Measure 50 cm3 of Sodium Thiosulfate solution into a flask

 Measure 5 cm3 of dilute Hydrochloric acid into a measuring cylinder

 Draw a cross on a piece of paper and put it underneath the flask

 Add the acid into the flask and immediately start the stopwatch

 Look down at the cross from above and stop the stopwatch when the cross can no
longer be seen

 Repeat using different concentrations of Sodium Thiosulfate solution (mix

different volumes of sodium thiosulfate solution with water to dilute it)

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 Result:
 With an increase in the concentration of a solution, the rate of reaction will
increase.

 This is because there will be more reactant particles in a given volume, allowing
more frequent and successful collisions, increasing the rate of reaction.

(b) Particle size

Diagram showing surface area to volume ratio of various sized cubes

Explanation:
Increase in the surface area of the solid, the rate of reaction will increase.

This is because more surface area particles will be exposed to the other reactant
so there will be more frequent and successful collisions per second, increasing
the rate of reaction.

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Graphical representation:

Graph showing the effect of the surface area of a solid on the rate of reaction

Experimental Evidence:

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 Diagram showing the process of downwards displacement to investigate the effect of the surface area
of a solid on the rate of reaction
Procedure:
 Add dilute hydrochloric acid into a conical flask.

 Use a capillary tube to connect this flask to a measuring cylinder upside down in
a bucket of water (downwards displacement).

 Add calcium carbonate chips into the conical flask and close the bung.

 Measure the volume of gas produced in a fixed time using the measuring cylinder.

 Repeat with different sizes of calcium carbonate chips (solid, crushed and
powdered).

Result:
 Smaller sizes of chips causes an increase in the surface area of the solid, so the
rate of reaction will increase.

 This is because more surface area of the particles will be exposed to the other
reactant so there will be more frequent and successful collisions, increasing the
rate of reaction.

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(c) Temperature

Diagram showing the effect of temperature on particles

Explanation:

Increase in the temperature, the rate of reaction will increase

This is because the particles will have more kinetic energy than the
required activation energy, therefore there will be more frequent and successful
collisions per second, increasing the rate of reaction.

Graphical representation:

Graph showing the effect of temperature on the rate of reaction

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Experimental Evidence:

Diagram showing the apparatus needed to investigate the effect of temperature on the rate of reaction

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 Procedure:
 Dilute Hydrochloric acid is heated to a set temperature using a water bath

 Add the dilute Hydrochloric acid into a conical flask

 Add a strip of Magnesium and start the stopwatch

 Stop the time when the Magnesium fully dissolves

 Repeat at different temperatures and compare results

Result:
 With an increase in the temperature, the rate of reaction will increase.

 This is because the particles will have more kinetic energy than the required
activation energy, therefore more frequent and successful collisions will occur,
increasing the rate of reaction.

(d) Catalyst

Graph showing the effect of the use of a catalyst on the rate of reaction

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Explanation:
When a catalyst is used, the rate of reaction will increase
Catalysts reduce the activation energy as they create alternative
pathways requiring lower activation energy, allowing more successful and
frequent collisions.

Graphical representation:

Graph showing the effect of the use of a catalyst on the rate of reaction
Experimental Evidence:

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 Diagram showing the apparatus needed to investigate the effect of a catalyst on the rate of reaction

Procedure

 Add Hydrogen Peroxide into a conical flask

 Use a capillary tube to connect this flask to a measuring cylinder upside down
in a bucket of water (downwards displacement)

 Add the catalyst Manganese(IV) Oxide into the conical flask and close the bung

 Measure the volume of gas produced in a fixed time using the measuring
cylinder

 Repeat experiment without the catalyst of Manganese(IV) Oxide and compare

results

Result:
 Using a catalyst will increase the rate of reaction.

 The catalyst will provide an alternative pathway requiring lower activation

energy so more colliding particles will have the necessary activation energy to
react.

 This will allow more frequent and successful collisions, increasing the rate of
reaction.

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 Photochemistry

Photochemical reactions
These reactions occur only when light is present

The greater the intensity of ultraviolet light then the greater the rate of reaction

E.g. the substitution of hydrogen atoms in methane by chlorine:

CH4 + Cl2 → CH3Cl + HCl

Silver salts in photography
Black and white photography film surfaces contain crystals of silver bromide
(AgBr).

When exposed to light they decompose to silver:

2AgBr → 2Ag + Br2
AgBr is colourless at low concentrations but the Ag appears grey-black

Parts of the film appear black, grey or white depending on the exposure:

 Stronger light = black or dark grey

 Weaker light = light grey

 Not exposed = white

Photosynthesis
This is the process in which plants produce food for reproduction and growth

The equation is:

6CO2 + 6H2O → C6H12O6 + 6O2
The process requires sunlight and chlorophyll

Chlorophyll is the green pigment in plants which absorbs sunlight and acts as
the catalyst for photosynthesis.

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 Reversible Reactions
Some reactions go to completion, where the reactants are used up to form the
product molecules and the reaction stops when all of the reactants are used up.

In reversible reactions, the product molecules can themselves react with each
other or decompose and form the reactant molecules again.

It is said that the reaction can occur in both directions: the forward reaction
(which forms the products) and the backward reaction (which forms the
reactants).

Chemical equations for reversible reactions

When writing chemical equations for reversible reactions, two arrows are used to
indicate the forward and reverse reactions (⇌)

Example: The reaction for the Haber Process which is the production of
ammonia from hydrogen and nitrogen:
N2 + 3H2 ⇌ 2NH3

Hydrated & anhydrous salts

 Hydrated salts are salts that contain water of crystallisation which affects their
molecular shape and colour.

 Water of crystallisation is the water that is stoichiometrically included in the

structure of some salts during the crystallisation process.

 A common example is copper(II) sulfate which crystallises forming the salt

copper(II) sulfate pentahydrate, CuSO4.5H20

 Water of crystallisation is indicated with a dot written in between the salt

molecule and the surrounding water molecules.

 Anhydrous salts are those that have lost their water of crystallisation, usually
by heating, in which the salt becomes dehydrated.
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Dehydration of Hydrated Copper (II) Sulfate:

Hydrated Copper (II) Sulfate ⇌ Anhydrous Copper (II) Sulfate + Water

CuSO4.5H2O (s) ⇌ CuSO4 (s) + 5H2O (l)

Diagram showing the dehydration of Hydrated Copper (II) Sulfate

Explanation:
 When anhydrous copper (II) sulfate crystals are added to water they turn blue and
heat is given off (exothermic); this reaction is reversible.
 When Copper (II) Sulfate crystals are heated in a test tube, the blue crystals turn
into a white powder and a clear, colourless liquid (water) collects at the top of the
test tube.
The form of Copper (II) Sulfate in the crystals is known as Hydrated Copper
(II) Sulfate because it contains water of crystallisation.
When Hydrated Copper (II) Sulfate is heated, it loses its water of crystallisation
and turns into anhydrous Copper (II) Sulfate:
CuSO4.5H2O (s) ⇌ CuSO4 (s) + 5H2O (l)

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 Dehydration of Hydrated Cobalt (II) Chloride:

Hydrated Cobalt (II) Chloride ⇌ Anhydrous Cobalt (II) Chloride + Water

CoCl2.6H2O (s) ⇌ CoCl2 (s) + 6H2O (l)

Diagram showing the dehydration of Hydrated Cobalt (II) Chloride

Explanation:
 When anhydrous blue cobalt(II) chloride crystals are added to water they turn
pink and the reaction is reversible.

 When the cobalt(II) chloride crystals are heated in a test tube, the pink crystals
turn back to the blue colour again as the water of crystallisation is lost

The form of cobalt(II) chloride in the crystals that are pink is known as hydrated
cobalt (II) chloride because it contains water of crystallisation

When hydrated cobalt(II) chloride is heated, it loses its water of crystallisation

and turns into anhydrous cobalt(II) chloride:

CoCl2.6H2O (s) ⇌ CoCl2 (s) + 6H2O (l)

 CuSO4 + 5H2O ⇌ CuSO4.5H2O White to blue

 CoCl2 + 6H2O ⇌ CoCl2.6H2O Blue to pink
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 The Concept of Equilibrium
Reversible reactions and equilibrium
When during the course of reaction, the rate of the forward reaction equals the
rate of the reverse reaction, then the overall reaction is said to be in a state of
equilibrium.

Characteristics of a reaction at equilibrium

 It is dynamic i.e: the molecules on the left and right of the equation are changing
into each other by chemical reactions constantly and at the same rate.

 The concentration of reactants and products remains constant (given there is no

other change to the system such as temperature and pressure).

 It only occurs in a closed system so that none of the participating chemical

species are able to leave the reaction vessel.

Equilibrium can only be reached in a closed vessel which prevents reactants or products from escaping
system

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The reaction between H2 & N2 in the Haber process
 When only nitrogen and hydrogen are present at the beginning of the
reaction, the rate of the forward reaction is at its highest, since the concentrations
of hydrogen and nitrogen are at their highest.

 As the reaction proceeds, the concentrations of hydrogen and nitrogen

gradually decrease, so the rate of the forward reaction will decrease.

 However, the concentration of ammonia is gradually increasing and so the

rate of the backward reaction will increase (ammonia will decompose to reform
hydrogen and nitrogen).

 Since the two reactions are interlinked and none of the gas can escape, the
rate of the forward reaction and the rate of the backward reaction will eventually
become equal and equilibrium is reached:

Diagram showing when the rates of forward and backward reactions become equal

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 The position of equilibrium
Equilibrium position refers to the relationship between the concentration of
reactants and products at the equilibrium state

 When the position of equilibrium shifts to the left, it means the concentration of
reactant increases

 When the position of equilibrium shifts to right, this means the concentration of
product increases
Effect of catalyst on equilibrium position
The presence of a catalyst does not affect the position of equilibrium but it does
increase the rate at which equilibrium is reached.

This is because the catalyst increases the rate of both the forward and backward
reactions by the same amount (by providing an alternative pathway requiring
lower activation energy).

As a result, the concentration of reactants and products is nevertheless the

same at equilibrium as it would be without the catalyst.

Diagram showing the effect of catalyst on equilibrium position

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 Le Chatelier’s Principle

 Le Chatelier’s principle states that when a change is made to the conditions of a

system at equilibrium, the system automatically moves to oppose the change.
 The principle is used to predict changes to the position of equilibrium when there
are changes in temperature, pressure or concentration.

Effects of temperature

CHANGE HOW THE EQUILIBRIUM SHIFTS

EQUILIBRIUM MOVES IN THE ENDOTHERMIC

INCREASE IN TEMPERATURE
DIRECTION TO REVERSE THE CHANGE

EQUILIBRIUM MOVES IN THE EXOTHERMIC

DECREASE IN TEMPERATURE
DIRECTION TO REVERSE THE CHANGE

Example: Iodine Monochloride reacts reversibly with Chlorine to form Iodine

Trichloride
ICl + Cl2 ⇌ ICl3
Dark Brown Yellow

When the equilibrium mixture is heated, it becomes dark brown in colour.

Explain whether the backward reaction is exothermic or endothermic:

 Equilibrium has shifted to the left as the colour dark brown means that more of
ICl is produced

 Increasing temperature moves the equilibrium in the endothermic direction

So equilibrium shifts to the left

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 Effects of pressure:

CHANGE HOW THE EQUILIBRIUM SHIFTS

EQUILIBRIUM SHIFTS IN THE DIRECTION THAT
PRODUCES THE SMALLER NUMBER OF
INCREASE IN PRESSURE
MOLECULES OF GAS TO DECREASE THE PRESSURE
AGAIN

EQUILIBRIUM SHIFTS IN THE DIRECTION THAT

PRODUCES THE LARGER NUMBER OF
DECREASE IN PRESSURE
MOLECULES OF GAS TO INCREASE THE PRESSURE
AGAIN

Example: Nitrogen Dioxide can form Dinitrogen Tetroxide, a colourless gas

2NO2 ⇌ N2O4
Brown Gas Colourless Gas

Predict the effect of an increase in pressure on the position of equilibrium:

 Number of molecules of gas on the left = 2

 Number of molecules of gas on the right = 1

 An increase in pressure will cause equilibrium to shift in the direction that

produces the smaller number of molecules of gas

So equilibrium shifts to the right

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 Effects of concentration
CHANGE HOW THE EQUILIBRIUM SHIFTS
EQUILIBRIUM SHIFTS TO THE RIGHT TO REDUCE
INCREASE IN
THE EFFECT OF INCREASE IN THE
CONCENTRATION
CONCENTRATION OF A REACTANT

EQUILIBRIUM SHIFTS TO THE LEFT TO REDUCE

DECREASE IN THE EFFECT OF A DECREASE IN REACTANT (OR
CONCENTRATION AN INCREASE IN THE CONCENTRATION OF
PRODUCT

Example: Iodine Monochloride reacts reversibly with Chlorine to form Iodine

Trichloride.

ICl + Cl2 ⇌ ICl3

Dark Brown Yellow

Predict the effect of an increase in concentration on the position of equilibrium:

 An increase in the concentration of ICl or Cl2 causes the equilibrium to shift to

the right so more of the yellow product is formed
 A decrease in the concentration of ICl or Cl2 causes the equilibrium to shift to the
left so more of the dark brown reactant is formed

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 Oxidation and Reduction
There are three definitions of oxidation. It is a reaction in which:

 oxygen is added to an element or a compound

 an element, ion or compound loses electrons
 the oxidation state of an element is increased

There are three definitions of reduction. It is a reaction in which:

 oxygen is removed from an element or a compound

 an element, ion or compound gains electrons

 the oxidation state of an element is decreased

If oxidation and reduction take place together at the same time in the same
reaction. Those are called redox reactions.

Oxidation state
The oxidation state (also called oxidation number) is a number assigned to
an atom or ion in a compound which indicates the degree of oxidation (or
reduction)

 The oxidation state helps you to keep track of the movement of electrons in a
redox process

 It is written as a +/- sign followed by a number.

 E.g: O2- means that it is an atom of oxygen that has an oxidation state of -2.

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 Assigning the oxidation number
Oxidation number refers to a single atom or ion

 The oxidation number of a compound is 0 and of an element (for example Br in

Br2) is also 0

 The oxidation number of oxygen in a compound is always -2 (except in peroxide

R-O-O-R, where it is -1)

 For example in FeO, oxygen is -2 then Fe must have an oxidation number of +2

as the overall oxidation number for the compound must be 0

Ionic Equations
Ionic equations are used to show only the particles that actually take part in a
reaction.

These equations show only the ions that change their status during a chemical
process, i.e: their bonding or physical state changes.

The other ions present are not involved and are called spectator ions.

Writing ionic equations

For the neutralisation reaction between hydrochloric acid and sodium hydroxide:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
If we write out all of the ions present in the equation and include the state
symbols, we get:

H+(aq) + Cl- (aq) + Na+(aq) + OH-(aq) → Na+ (aq) + Cl-(aq) + H2O(l)

The spectator ions are thus Na+ and Cl–. Removing these from the previous
equation leaves the overall net ionic equation:

H+(aq) + OH-(aq) → H2O(l)

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This ionic equation is the same for all acid-base neutralisation

Example redox equation: oxygen loss/gain

Zinc oxide + carbon → zinc + carbon monoxide
ZnO + C → Zn + CO

In this reaction the zinc oxide has been reduced since it has lost oxygen. The
carbon atom has been oxidised since it has gained oxygen

Example redox equation: electron loss/gain and oxidation state

Zinc + copper sulphate → zinc sulphate + copper
Zn + CuSO4 → ZnSO4 + Cu
Writing this as an ionic equation:
Zn(s) + Cu2+ (aq) + SO42 - (aq) → Zn2+ (aq) + SO42- (aq) + Cu(s)
By analysing the ionic equation, it becomes clear that:
Zinc has become oxidised as its oxidation state has increased and it has lost
electrons:
Zn(s) → Zn2+(aq)
Copper has been reduced as its oxidation state has decreased and it has gained
electrons:
Cu2+ (aq) → Cu(s)

OIL-RIG : Oxidation Is Loss – Reduction Is Gain.

Oxidising agent: A substance that oxidises another substance, in so doing

becoming itself reduced
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Common examples include hydrogen peroxide, fluorine and chlorine

Reducing agent: A substance that reduces another substance, in so doing

becoming itself oxidised
Common examples include carbon and hydrogen

The process of reduction is very important in the chemical industry as a means of

extracting metals from their ores
Example:

CuO + H2 → Cu + H2O
In the above reaction, hydrogen is reducing the CuO and is itself oxidised, so the
reducing agent is therefore hydrogen

The CuO is reduced to Cu and has oxidised the hydrogen, so the oxidising agent
is therefore copper oxide

Identifying redox reactions

Redox reactions can be identified by the changes in the oxidation states when a
reactant goes to a product
Example:
Chlorine + potassium iodide → potassium chloride + iodine
Cl2 + 2KI → 2KCl + I2
Chlorine has become reduced as its oxidation state has decreased from 0 to -1 on
changing from the chlorine molecule to chloride ions:

Cl2 (g) → 2Cl - (aq)

Iodine has been oxidised as its oxidation state has increased from -1 to 0 on
changing from iodide ions to the iodine molecule:

2I - (aq) → I2 (s)
Identifying redox reactions by colour changes

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  The tests for redox reactions involve the observation of a colour change in
the solution being analysed.

 Two common examples are acidified potassium manganate(VII), and

potassium iodide.

 Potassium manganate

 Potassium manganate (VII), KMnO4, is an oxidising agent which is often used

to test for the presence of reducing agents

 When acidified potassium manganate (VII) is added to a reducing agent its

colour changes from pink-purple to colourless

 Potassium iodide

 Potassium iodide, KI, is a reducing agent which is often used to test for the
presence of oxidising agents

 When added to an acidified solution of an oxidising agent such as aqueous

chlorine or hydrogen peroxide, the solution turns a brown colour due to the
formation of iodine

KMnO4 ⇒ presence of reducing agents ⇒ pink-purple to colourless

KI⇒ presence of oxidising agents ⇒ solution turns a brown colour

Prepared by: Varun Jayachandran

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