Chemistry- patterns of chemical change-Junior Cycle Resource

1. 29/05/13
Junior Cycle ChemistryJunior Cycle Chemistry
Patterns of Chemical ChangePatterns of Chemical Change
Edited and Reproduced by
Theresa Lowry-Lehnen
RGN, BSc (Hon’s) Specialist Clinical Practitioner (Nursing), Dip Counselling, Dip Adv Psychotherapy, BSc
(Hon’s) Clinical Science, PGCE (QTS) , H. Dip. Ed, MEd, Emotional Intelligence (Level 9) MHS Accredited

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Rates of ReactionRates of Reaction
Chemical reactions occur when
different atoms or molecules collide:
For the reaction to happen the particles must have a
certain amount of energy – this is called the
ACTIVATION ENERGY.
The rate at which the reaction happens depends on four things:
1) The temperature of the reactants,
2) Their concentration
3) Their surface area
4) Whether or not a catalyst is used

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Measuring rate of reactionMeasuring rate of reaction
Two common ways:
1) Measure how fast the
products are formed
2) Measure how fast the
reactants are used up

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Rate of reaction graphRate of reaction graph
Amount of
product
formed
Time
Slower reaction
Fast rate
of reaction
here
Slower rate of reaction here
due to reactants being used up

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Enzymes are biological catalysts. They help the reactions
that occur in our bodies by controlling the rate of reaction.
Enzymes are
denatured
beyond 40O
C
EnzymesEnzymes
Yeast is an example of an enzyme. It is used to help a process called
fermentation:
Sugar Alcohol + carbon dioxide
The alcohol from this process is used in
making drinks and the carbon dioxide can
be used to make bread rise.
Enzymes work best in certain conditions:
Enzyme activity
Temp pH pH400
C
Could be
protease (found
in the stomach)
Could be amylase
(found in the
intestine)

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Uses of enzymesUses of enzymes
1) Enzymes are used in washing
powders to help digest food stains.
Biological washing powders will only
work on 400
C or lower.
2) Enzymes are used in baby foods to
“pre-digest” the proteins.
3) Enzymes are used to convert starch
into sugar which can then be used in
food.
4) Conversion of glucose into fructose
– glucose and fructose are “isomers”
(they have the same chemical formula),
but fructose is sweeter.

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Endothermic and exothermic reactionsEndothermic and exothermic reactions
Step 1: Energy must be
SUPPLIED to break bonds:
Step 2: Energy is RELEASED
when new bonds are made:
A reaction is EXOTHERMIC if more energy is RELEASED
then SUPPLIED. If more energy is SUPPLIED then is
RELEASED then the reaction is ENDOTHERMIC

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Energy level diagramsEnergy level diagrams
Energy
level
Reaction progress
Activation
energy
Energy given
out by
reactionUsing a catalyst
might lower the
activation energy

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Exothermic vs endothermic:Exothermic vs endothermic:
EXOTHERMIC – more
energy is given out than is
taken in (e.g. burning,
respiration)
ENDOTHERMIC –
energy is taken in but
not necessarily given out
(e.g. photosynthesis)

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Reversible ReactionsReversible Reactions
Some chemical reactions are reversible. In other words, they
can go in either direction:
A + B C + D
NH4Cl NH3 + HCl
e.g. Ammonium chloride Ammonia + hydrogen chloride
If a reaction is EXOTHERMIC in one direction
what must it be in the opposite direction?
For example, consider copper sulphate:
Hydrated copper
sulphate (blue)
Anhydrous copper
sulphate (white)
+ Heat + Water
CuSO4 + H2OCuSO4.5H2O

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Reversible ReactionsReversible Reactions
When a reversible reaction occurs in a CLOSED SYSTEM (i.e. no reactants
are added or taken away) an EQUILIBRIUM is achieved – in other words,
the reaction goes at the same rate in both directions:
A + B C + D
Endothermic reactions
Increased temperature:
Decreased temperature:
A + B C + D
A + B C + D
More products
Less products
Exothermic reactions
Increased temperature:
Decreased temperature:
A + B C + D
Less products
More products
A + B C + D

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Making AmmoniaMaking Ammonia
Nitrogen + hydrogen Ammonia
N2 + 3H2 2NH3
•High pressure
•450O
C
•Iron catalyst
Recycled H2 and N2
Nitrogen
Hydrogen
Mixture of NH3, H2 and
N2. This is cooled
causing NH3 to liquefy.
Fritz Haber,
1868-1934
Guten Tag. My name is Fritz Haber and I won the Nobel
Prize for chemistry. I am going to tell you how to use a
reversible reaction to produce ammonia, a very
important chemical. This is called the Haber Process.
To produce ammonia from nitrogen and hydrogen
you have to use three conditions:

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Uses of AmmoniaUses of Ammonia
Nitrogen
monoxide
Hot
platinum
catalyst
Ammonia gas
Oxygen
Cooled
Water and
oxygen
Nitrogen
monoxide
Nitric
acid
Ammonia + nitric acid Ammonium nitrate
NH3 + HNO3 NH4NO3
Ammonia is a very important chemical as it can be
used to make plant fertilisers and nitric acid:
More ammonia can then be used to neutralise the nitric acid to
produce AMMONIUM NITRATE (a fertiliser rich in nitrogen).
The trouble with nitrogen based fertilisers is that they can also
create problems – they could contaminate our drinking water.

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Haber Process: The economicsHaber Process: The economics
A while ago we looked at reversible reactions:
A + B C + D
Endothermic, increased temperature
A + B C + D
Exothermic, increase temperature
ExothermicEndothermic
1) If temperature was DECREASED the amount of ammonia formed would
__________...
2) However, if temperature was INCREASED the rate of reaction in both
directions would ________ causing the ammonia to form faster
3) If pressure was INCREASED the amount of ammonia formed would
INCREASE because there are less molecules on the right hand side of
the equation
Nitrogen + hydrogen Ammonia
N2 + 3H2 2NH3

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Haber Process SummaryHaber Process Summary
•200 atm pressure
•450O
C
•Iron catalyst
Recycled H2 and N2
Nitrogen
Hydrogen
Mixture of NH3, H2
and N2. This is
cooled causing NH3
to liquefy.
To compromise all of these factors, these conditions are used:
A low temperature increases the yield of ammonia but is
too slow
A high temperature improves the rate of reaction but
decreases the yield too much
A high pressure increases the yield of ammonia but costs a lot
of money

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Burning MethaneBurning Methane
CH4 + 2O2 2H2O + CO2
To burn methane
you have to break
all of these bonds:
And then you
have to make
these ones:

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Burning MethaneBurning Methane
CH4 + 2O2 2H2O + CO2
Methane
Carbon dioxide Water
Oxygen

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Bond energiesBond energies
C-H = 435 Kj
O=O = 497 Kj
Total for breaking bonds = 4x435 + 2x497 = 2734 KJ/mol
H-O = 464 KjC=O = 803 Kj
Total for making bonds = 2x803 + 4x464 = 3462 KJ/mol
Total energy change = 2734-3462 = -728 KJ/mol

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Drawing this on an energy diagram:Drawing this on an energy diagram:
2734 Kj
3462 Kj
More energy is given out (3462) than is given in (2734) –
the reaction is EXOTHERMIC. The total (“nett”) energy
change is –728 Kj. An endothermic reaction would have a
positive energy change.
-728 Kj

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Bond energy valuesBond energy values
C-H = 435 KJ/mol
O-H = 464 KJ/mol
O=O = 497 KJ/mol
C=O = 803 KJ/mol
C-O = 360 KJ/mol
C-C = 346 KJ/mol

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Burning MethanolBurning Methanol
Total for breaking bonds = 6x435
(C-H) + 2x360 (C-O) + 2x464 (O-H)
+ 3x497 (O=O) = 5749 KJ/mol
2CH3OH + 3O2 2CO2 + 4H2O
Total for making bonds =
4x803 (C=O) + 8x464 (O-H) =
6924 KJ/mol
Energy change = 5749-6924 (divide this by two as we are
dealing with two molecules of methanol) = -587.5 KJ/mol

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Atomic massAtomic mass
SYMBOL
PROTON NUMBER = number of
protons (obviously)
RELATIVE ATOMIC MASS, Ar
(“Mass number”) = number of
protons + number of neutrons

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Relative formula mass, MRelative formula mass, Mrr
The relative formula mass of a compound is blatantly the relative
atomic masses of all the elements in the compound added together.
E.g. water H2O:
Therefore Mr for water = 16 + (2x1) = 18
Work out Mr for the following compounds:
1) HCl
2) NaOH
3) MgCl2
4) H2SO4
H=1, Cl=35 so Mr = 36
Na=23, O=16, H=1 so Mr = 40
Mg=24, Cl=35 so Mr = 24+(2x35) = 94
H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98
Relative atomic mass of O = 16
Relative atomic mass of H = 1

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More examplesMore examples
CaCO3 40 + 12 + 3x16 100
HNO3 1 + 14 + 3x16
2MgO 2 x (24 + 16) 80
3H2O 3 x ((2x1) + 16)
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2

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Calculating percentage massCalculating percentage mass
If you can work out Mr then this bit is easy…
Calculate the percentage mass of magnesium in magnesium oxide, MgO:
Ar for magnesium = 24 Ar for oxygen = 16
Mr for magnesium oxide = 24 + 16 = 40
Therefore percentage mass = 24/40 x 100% = 60%
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
Calculate the percentage mass of the following:
1) Hydrogen in hydrochloric acid, HCl
2) Potassium in potassium chloride, KCl
3) Calcium in calcium chloride, CaCl2
4) Oxygen in water, H2O

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Calculating the mass of a productCalculating the mass of a product
E.g. what mass of magnesium oxide is produced when 60g of
magnesium is burned in air?
Step 1: READ the equation:
2Mg + O2 2MgO
IGNORE the
oxygen in step 2 –
the question
doesn’t ask for it
Step 3: LEARN and APPLY the following 3 points:
1) 48g of Mg makes 80g of MgO
2) 1g of Mg makes 80/48 = 1.66g of MgO
3) 60g of Mg makes 1.66 x 60 = 100g of MgO
Step 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80

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Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4
1. 36g of water produces 4g of hydrogen
2. So 1g of water produces 4/36 = 0.11g of hydrogen
3. 6g of water will produce (4/36) x 6 = 0.66g of hydrogen
Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al O
1) When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O 2H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O3 4Al + 3O2
2) What mass of calcium oxide is produced when 10g of calcium burns?
2Ca + O2 2CaO

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Another methodAnother method
Try using this equation:
Mass of product IN GRAMMES
Mass of reactant IN GRAMMES
Mr of product
Mr of reactant
Q. When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O 2H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
Mass of product IN GRAMMES
6g
4
36
So mass of product = (4/36) x 6g = 0.66g of hydrogen

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Calculating the volume of a productCalculating the volume of a product
At normal temperature and pressure the Relative Formula Mass (Mr)
of a gas will occupy a volume of 24 litres
e.g. 2g of H2 has a volume of 24 litres
32g of O2 has a volume of 24 litres
44g of CO2 has a volume of 24 litres etc
Q. When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O 2H2 + O2
What VOLUME of hydrogen is produced by the electrolysis of 6g of
water?
• On the previous page we said that the MASS of hydrogen produced
was 0.66g
• 2g of hydrogen (H2) will occupy 24 litres (from the red box above),
• So 0.66g will occupy 0.66/2 x 24 = 8 litres

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Example questionsExample questions
1) What volume of hydrogen is produced when 18g of water is
electrolysed?
2H20 2H2 + O2
1) Marble chips are made of calcium carbonate (CaCO3). What volume
of carbon dioxide will be released when 500g of CaCO3 is reacted
with dilute hydrochloric acid?
CaCO3 + 2HCl CaCl2 + H2O + CO2
1) In your coursework you reacted magnesium with hydrochloric acid.
What volume of hydrogen would be produced if you reacted 1g of
magnesium with excess acid?

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Empirical formulaeEmpirical formulae
Empirical formulae is simply a way of showing how many atoms are in a
molecule (like a chemical formula). For example, CaO, CaCO3, H20 and
KMnO4 are all empirical formulae. Here’s how to work them out:
A classic exam question:
Find the simplest formula of 2.24g of iron
reacting with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = 0.04 For oxygen 0.96/16 = 0.06
Step 2: Write this as a ratio and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe O

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Example questionsExample questions
1) Find the empirical formula of magnesium oxide which
contains 48g of magnesium and 32g of oxygen.
2) Find the empirical formula of a compound that contains
42g of nitrogen and 9g of hydrogen.
3) Find the empirical formula of a compound containing 20g
of calcium, 6g of carbon and 24g of oxygen.

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ElectrolysisElectrolysis
Molecule of solid
copper chloride
CuCl2 (s)
Molecule of solid
copper chloride after
being dissolved
CuCl2 (aq)
Chloride
ion
Copper
ion

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ElectrolysisElectrolysis
Electrolysis is used to separate a metal from its compund.
= chloride ion
= copper ion
When we electrolysed
copper chloride the _____
chloride ions moved to the
______ electrode and the
______ copper ions moved
to the ______ electrode –
OPPOSITES ATTRACT!!!

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Electrolysis equationsElectrolysis equations
We need to be able to write “half equations” to show what
happens during electrolysis (e.g. for copper chloride):
2 2
2
At the negative electrode the
positive ions GAIN electrons to
become neutral copper ATOMS. The
half equation is:
Cu2+
+ e-
Cu
At the positive electrode the
negative ions LOSE electrons to
become neutral chlorine
MOLECULES. The half equation is:
Cl-
- e-
Cl2

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Calculating masses and volumes from electrolysisCalculating masses and volumes from electrolysis
Consider those two half equations again:
Cu2+
+ 2e-
Cu 2Cl-
- 2e-
Cl2
1) Write down the relative atomic mass:
Copper = 63
2 molecules of chlorine (Cl2) = 2x35 = 70
Example question: How much chlorine is released at the positive
electrode if 2g of copper is collected at the negative electrode?
3) Work out the volume: 70g of chlorine would occupy 24 litres, so 2.22g
of chlorine would occupy a volume of (2.22/70) x 24 = ______ litres
2) Follow the steps:
63g of copper makes 70g of chlorine…
…so 1g of copper would make (70/63) g of chlorine… (=________g)
…so 2g of copper would make (70/63) x 2g of chlorine (=________g)