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
 Making a mixture from 2 or more substances or dissolving a solute in a solvent
are examples of physical changes as no new substances are produced and are
usually relatively easy to separate

Chemical change

 During chemical changes (usually referred to as chemical

reactions), new chemical substances are formed that have very
different properties to the reactants
 There may be signs that a new substance has formed, such as:
o A colour change
o A precipitate being formed
o Bubbles of gas being produced
 Most chemical reactions are very difficult to reverse
 Energy changes also accompany chemical changes and energy can be given out
(exothermic) or taken in (endothermic)
o The majority of chemical reactions are exothermic with only a small
number being endothermic

Rates of Reaction Factors

There are several factors that can affect the rate of a reaction. These are:


o Concentration of the reactants in solution or the pressure of reacting gases
o Temperature at which the reaction is carried out
o Surface area of solid reactants
o The use of a catalyst
 Changes in these factors directly influence the rate of a reaction
 It is of economic interest to have a higher rate of reaction as this implies a higher rate
of production and hence a more efficient and sustainable process
The effect of increased concentration or pressure

Explanation:

 Compared to a reaction with a reactant at a low concentration, the graph line for the
same reaction but at a higher concentration/pressure has a steeper gradient at the start
and becomes horizontal sooner
 This shows that with increased concentration of a solution or increased pressure
of a gas, the rate of reaction will increase
The effect of surface area

Graph showing the effect of increased surface area on the rate of reaction

Explanation:

 Compared to a reaction with lumps of reactant, the graph line for the same reaction but
with powdered reactant has a steeper gradient at the start and becomes horizontal
sooner
 This shows that with increased surface area of the solid, the rate of reaction will
increase
The effect of temperature

Graph showing the effect of temperature on the rate of reaction

Explanation:

 Compared to a reaction at a low temperature, the graph line for the same reaction but at
a higher temperature has a steeper gradient at the start and becomes horizontal sooner
 This shows that with increased temperature, the rate of reaction will increase

The effect of using a catalyst

 Catalysts are substances which speed up the rate of a reaction without themselves
being altered or consumed in the reaction
 The mass of a catalyst at the beginning and end of a reaction is the same and they do
not form part of the equation
 Graph showing the effect of using a catalyst on the rate of reaction

Explanation:

 Compared to a reaction without a catalyst, the graph line for the same reaction but with a
catalyst has a steeper gradient at the start and becomes horizontal sooner
 This shows that with a catalyst, the rate of reaction will increase

Collision Theory
EXTENDED

 Collision theory states that in order for a reaction to occur:

o The particles must collide with each other
o The collision must have sufficient energy to cause a reaction i.e. enough
energy to break bonds
 The minimum energy that colliding particles must have to react is known as
the activation energy
 Collisions which result in a reaction are known as successful collisions
o If they have sufficient energy (i.e. energy greater than the activation energy), they
will react, and the collision will be successful
 Not all collisions result in a chemical reaction:
o Most collisions just result in the colliding particles bouncing off each other
o Collisions which do not result in a reaction are known as unsuccessful collisions
 Unsuccessful collisions happen when the colliding species do not have enough energy
to break the necessary bonds (i.e. they collide with energy less than the activation
energy)

Diagram showing a successful and an unsuccessful collision

 Increasing the number of successful collisions means that a greater proportion of

reactant particles collide to form product molecules
 The number of successful collisions depends on:
o The number of particles per unit volume - more particles in a given volume will
produce more frequent successful collisions
o The frequency of collisions - a greater number of collisions per second will give
a greater number of successful collisions per second
o The kinetic energy of the particles - greater kinetic energy means a greater
proportion of collisions will have an energy that exceeds the activation energy
and the more frequent the collisions will be as the particles are moving quicker,
therefore, more collisions will be successful
o The activation energy - fewer collisions will have an energy that exceeds higher
activation energy and fewer collisions will be successful
 These all have an impact on the rate of reaction which is dependent on the number of
successful collisions per unit of time
 Explaining Rates Using Collision Theory
EXTENDED
Temperature

 Increasing the temperature will increase the rate of reaction

 This is because the particles will have more kinetic energy than the required activation
energy
 Therefore there will be more frequent collisions and a higher proportion of particles have
energy greater than the activation energy
 This causes more successful collisions per second, increasing the rate of reaction
 The effect of temperature on collisions is not so straightforward as concentration or
surface area; a small increase in temperature causes a large increase in rate
 For aqueous and gaseous systems, a rough rule of thumb is that for every 10 oC
increase in temperature, the rate of reaction approximately doubles

Diagram showing the increased kinetic energy that particles have at higher
temperatures

Concentration

o Increasing the concentration of a solution will increase the rate of reaction
o 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
o For a gaseous reaction, increasing the pressure has the same effect as the same
number of particles will occupy a smaller space, increasing the concentration
o If you double the number of particles you will double the number of collisions per
second
 o The number of collisions is proportional to the number of particles present

Diagram showing the decrease in space between particles at higher

concentrations

Surface Area

 Increasing the surface area of a solid will increase the rate of reaction
 This is because more surface area of the particles will be exposed to the other
reactant, producing a higher number of collisions per second
 If you double the surface area you will double the number of collisions per second

An increase in surface area means more collisions per second

Catalysts

 Catalysts are substances which speed up the rate of a reaction without themselves
being altered or consumed in the reaction
 The mass of a catalyst at the beginning and end of a reaction is the same and they do
not form part of the equation
 Different processes require different types of catalysts but they all work on the same
principle of providing a different pathway for the reaction to occur that has a lower
activation energy
 This means a higher proportion of the reactant particles have energy greater than the
activation energy and will result in more successful collisions per second
 An important industrial example is iron, which is used to catalyse the Haber Process for
the production of ammonia
 Iron beads are used to increase the surface area available for catalysis
 Enzymes are biological catalysts, they work best at specific temperature and pH ranges
 Normally only small amounts of catalysts are needed to have an effect on a reaction
 A catalyst lowers the activation energy of a reaction by providing an alternative
reaction pathway

Exam Tip
When answering questions on the effect of concentration on the rate of reaction, you
should mention that there are more particles per unit volume (usually cm3) and this
causes an increase in the rate of collisions.
 Investigating The Rate of a Reaction
 To measure the rate of a reaction, we need to be able to measure either how quickly the
reactants are used up or how quickly the products are formed
 The method used for measuring depends on the substances involved
 There are a number of ways to measure a reaction rate in the lab; they all depend on some
property that changes during the course of the reaction
 That property is taken to be proportional to the concentration of the reactant or product, e.g.,
colour, mass, volume

o faster reactions can be easier to measure when the reaction is over, by averaging a
collected measurement over the course of the reaction some reaction rates can be
measured as the reaction proceeds (this generates more data)
 Three commonly used techniques are:
o measuring mass loss on a balance
o measuring the volume of a gas produced
o measuring a reaction where there is a colour change at the end of the reaction
Investigating the effect of surface area on the rate of reaction
 Diagram showing the process of downwards displacement to investigate the effect of the
surface area of a solid on the rate of reaction

Method:

 Add dilute hydrochloric acid into a conical flask

 Use a delivery 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 quickly put the bung back into the flask
 Measure the volume of gas produced in a fixed time using the measuring cylinder
 Repeat with different sizes of calcium carbonate chips (lumps, crushed and powdered)

Result:

 Smaller sizes of chips cause 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
Effect of concentration of a solution on the rate of reaction:
 Diagram showing the apparatus needed to investigate the effect of concentration on the rate
of reaction

Method:

 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)

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

Method:

 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 reacts and disappears
 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
Effect of a catalyst on the rate of reaction:
 Diagram showing the apparatus needed to investigate the effect of a catalyst on the rate of
reaction

Method:

 Add hydrogen peroxide into a conical flask

 Use a delivery tube to connect this flask to a measuring cylinder upside down in a tub of water
(downwards displacement)
 Add the catalyst manganese(IV) oxide into the conical flask and quickly place the bung into the
flask
 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

Monitoring changes in mass

 Many reactions involve the production of a gas which will be released during the reaction
 The gas can be collected and the volume of gas monitored as per some methods above
 Alternatively, the reaction can be performed in an open flask on a balance to measure the loss in
mass of reactant
 Cotton wool is usually placed in the mouth of the flask which allows gas out but prevents any
materials from being ejected from the flask (if the reaction is vigorous)
 Diagram showing the set-up for measuring the rate of reaction by loss in mass

 This method is not suitable for hydrogen and other gases with a small relative formula
mass, Mr as the loss in mass may be too small to measure

Exam Tip
There are many different methods of investigating the rate of reaction. Another method of gas
collection you may see uses a gas syringe.

Those students taking the extended course may be required to devise and evaluate methods of
investigating rates of reaction.
Evaluating Investigations of Rates of Reactions
EXTENDED

 When investigating rates of reaction, there are a number of different methods that can be used
to carry out the same investigation
 Evaluating what is the best method to use is part of good experimental planning and design
 This means appreciating some of the advantages and disadvantages of the methods available

Table showing some Examples of Advantages and Disadvantages of Methods of

Investigating Rates of Reaction
 Interpreting Data
 Data recorded in rate studies is used to plot graphs to calculate the rate of a reaction
 Plotting a graph until the completion of the reaction shows how the rate changes with
time
 Over time the rate of reaction slows as the reactants are being used up so the line
becomes less steep and eventually becomes horizontal, indicating the reaction has
finished
 You can plot more than one run of a variable on the same graph making it easier to see
how the variable influences the rate
o For example, plotting the effect of concentration on a reaction between the acid
and marble chips

 The steeper the curve, the faster the rate of the reaction
  The curve is steepest initially so the rate is quickest at the beginning of the reaction
 As the reaction progresses, the concentration of the reactants decreases and the rate
decreases shown by the curve becoming less steep
 When one of the reactants is used up, the reaction stops, the rate becomes zero and the
curve levels off to a horizontal line
 The amount of product formed in a reaction is determined by the limiting reactant:
o If the amount of limiting reactant increases, the amount of product formed
increases
o If the amount of the reactant in excess increases, the amount of product remains
the same
 Drawing a tangent to the slope allows you to show the gradient at any point on the
curve
 The steeper the slope, the quicker the rate of reaction
 The volume of a gaseous product would increase to a maximum over time, so the line
levels out indicating the reaction is over
 Since the volume and mass would be proportional, this could also be a graph of the
mass of product versus time

Worked example
0.2 g of manganese(IV) oxide was added to 25 cm3 of 0.1 mol/dm3 hydrogen peroxide
solution. The volume of oxygen produced every minute was recorded and the results
are shown on the graph.
The experiment was repeated using the same mass of manganese(IV) oxide and at the
same temperature but using 25 cm3 of 0.2 mol/dm3 of hydrogen peroxide solution.

Sketch the curve for the results of this experiment on the same grid.
Answer

Step 1 - Deduce how the initial gradient will be different from the original graph

The hydrogen peroxide solution is twice as concentrated so the rate of reaction will be
greater and the initial gradient will be steeper

Step 2 - Deduce how much product will be formed compared to the original experiment

The amount of hydrogen peroxide determines the amount of oxygen produced. In the
2nd experiment, there are twice as many hydrogen peroxide molecules in the same
volume so the amount of oxygen gas produced will be doubled
Step 3 - Sketch the graph

Calculating the Rate of Reaction at a Particular Point

 To do this you need to find the gradient of the curve at that point
 To do this a tangent is drawn to the curve and then the gradient of the tangent
calculated

Worked example
Iodine and methanoic acid react in aqueous solution.

I2 (aq) + HCOOH (aq) → 2I− (aq) + 2H+ (aq) + CO2 (g)

The rate of reaction can be found by measuring the volume of carbon dioxide produced
per unit time and plotting a graph as shown:
Calculate the rate of reaction at 20 seconds
Answer:


o Draw a tangent to the curve at 20 seconds:

o Complete the triangle and read off the values of x and y

o Determine the gradient of the line using change in y / change in x
o Rate of reaction = 24 ÷ 40 = 0.60 cm3/s

Exam Tip
If the amount of reactant used up is being monitored, then the graph will fall with the
steepest gradient at the start, becoming less steep until it levels off to a horizontal line.
 Reversible Reactions
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 reverse 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

bottom one points to the left: ⇌

 Each one is drawn with just half an arrowhead – the top one points to the right, and the

Example

 The reaction for the Haber process which is the production of ammonia from hydrogen
and nitrogen:

N2 + 3H2 ⇌ 2NH3

Hydrated and 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, CuSO4.5H2O
 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

Dehydration of hydrated cobalt(II) chloride:

hydrated cobalt(II) chloride ⇌ anhydrous cobalt(II) chloride + water

 Diagram showing the dehydration of hydrated cobalt(II) chloride

Hydration of cobalt(II) chloride

 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)

Exam Tip
The hydration of CoCl2 (and CuSO4) are chemical tests which are commonly used to
detect the presence of water. You should remember the equations and colour changes:

CoCl2 + 6H2O ⇌ CoCl2.6H2O Blue to pink

CuSO4 + 5H2O ⇌ CuSO4.5H2O White to blue


 The Concept of Equilibrium
EXTENDED

 We have already seen that a reversible reaction is one that occurs in both directions
 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
 Equilibrium is dynamic e.g. 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 the system

 An example of dynamic equilibrium is the reaction between H2 and 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:

3H2 (g) + N2 (g) ⇌ 2NH3 (g)

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

 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

Le Chatelier’s Principle
EXTENDED

 The relative amounts of all the reactants and products at equilibrium depend on the conditions
of the reaction
 This balance is framed in an important concept known as Le Chatelier's Principle, named after
Henri Le Chatelier who was a French military engineer in the 19th century
 This 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
 Knowing the energy changes, states and concentrations involved allows us to use the principle to
manipulate the outcome of reversible reactions
 For example, if the pressure is increased, the position of equilibrium moves in the direction
which has the smallest amount of gaseous molecules
 The position of equilibrium is said to shift to the right when the forward reaction is favoured and
there is an increase in the amount of products formed
 The position of equilibrium is said to shift to the left when the reverse reaction is favoured and
there is an increase in the amount of reactants formed

The Effect of Temperature on Equilibrium

 When the equilibrium mixture is heated, it becomes dark brown in colour. You can use this
observation to deduce whether the backward reaction is exothermic or endothermic
 Equilibrium has shifted to the left as the colour dark brown means that more ICI is produced
 Increasing temperature moves the equilibrium in the endothermic direction
 So the backward reaction is endothermic

Table showing the Effects of Temperature on Equilibrium

The Effect of Pressure on Equilibrium

Table showing the Effects of Pressure on Equilibrium

Example: Nitrogen dioxide can form dinitrogen tetroxide, a colourless gas

2NO2 ⇌ N2 O 4
brown gas colourless gas

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

o Number of molecules of gas on the left = 2
o 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
 The reaction mixture becomes paler as more colourless N2O4 is produced

The Effect of Concentration on Equilibrium

Table showing the Effects of Concentration on Equilibrium

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:

o 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
o 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

The Effect of a Catalyst on Equilibrium

 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 a catalyst on the time taken for equilibrium to be established
Exam Tip
When the conditions at equilibrium are changed, the system always responds by doing
the opposite. For example if the concentration is increased the system tries to reduce it by
changing the direction of the reaction or if the temperature is increased the system will try to
reduce the temperature by absorbing the extra heat.