Patterns of Chemical Change
Rates 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
Measuring rate of reaction
Two common ways:
1) Measure how fast the
products are formed
2) Measure how fast the
reactants are used up
Rate of reaction graph
Amount of
product
formed
Fast rate
of reaction
here
Slower rate of reaction here
due to reactants being used up
Slower reaction
Time
Biological catalysts
Our bodies contain catalysts for all the reactions that go on inside
them – they are called ENZYMES. Enzymes work best at certain
temperatures:
Optimum temperature
Enzyme
(roughly 40OC)
activity
Enzymes are denatured
beyond 40OC
Increasing activity
due to more
collisions
Temperature
Enzymes are used in the food industry. Two examples:
1) Fermentation:
Glucose
Ethanol + Carbon dioxide
Ethanol is used for alcohol, carbon dioxide for making bread rise
2) Yoghurt production – enzymes in bacteria produce yoghurt from
milk by converting sugar (lactose) into lactic acid, which tastes sour
Endothermic 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
Energy level diagrams
Energy
level
Activation
energy
Using a catalyst
might lower the
activation energy
Energy given
out by
reaction
Reaction progress
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)
Burning Methane
CH4 + 2O2
To burn methane
you have to break
all of these bonds:
And then you
have to make
these ones:
2H2O + CO2
Bond energies
C-H = 435 Kj
O=O = 497 Kj
Total for breaking bonds = 4x435 + 2x497 = 2734 Kj
C=O = 803 Kj
H-O = 464 Kj
Total for making bonds = 2x803 + 4x464 = 3462 Kj
Drawing this on an energy diagram:
2734 Kj
3462 Kj
-728 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.
Atomic mass
RELATIVE ATOMIC MASS, Ar
(“Mass number”) = number of
protons + number of neutrons
SYMBOL
PROTON NUMBER = number of
protons (obviously)
Relative formula mass, Mr
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:
Relative atomic mass of O = 16
Relative atomic mass of H = 1
Therefore Mr for water = 16 + (2x1) = 18
Work out Mr for the following compounds:
1) HCl
H=1, Cl=35 so Mr = 36
2) NaOH
Na=23, O=16, H=1 so Mr = 40
3) MgCl2
Mg=24, Cl=35 so Mr = 24+(2x35) = 94
4) H2SO4
H=1, S=32, O=16 so Mr = (2x1)+32+(4x16) = 98
5) K2CO3
K=39, C=12, O=16 so Mr = (2x39)+12+(3x16) = 138
More examples
CaCO3
40 + 12 + 3x16
HNO3
1 + 14 + 3x16
2MgO
2 x (24 + 16)
3H2O
3 x ((2x1) + 16)
4NH3
2KMnO4
3C2H5OH
4Ca(OH)2
100
80
Calculating percentage mass
If you can work out Mr then this bit is easy…
Percentage mass (%) =
Mass of element Ar
Relative formula mass Mr
x100%
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%
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
Calculating 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 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = 48
2MgO = 2 x (24+16) = 80
Step 3: LEARN and APPLY the following 2 points:
1) 48g of Mg makes 80g of MgO
2) 60g of Mg makes 60/48 x 80 = 100g of MgO
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?
1.
Work out Mr:
2H2O = 2 x ((2x1)+16) = 36
2H2 = 2x2 = 4
2. 36g of water produces 4g of hydrogen,
3. 6g of water will produce (6/36) x 4 = 0.66g of hydrogen
2) What mass of calcium oxide is produced when 10g of calcium burns?
Mr: 2Ca = 2x40 = 80
2Ca + O2
2CaO
2CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O3
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204
4Al + 3O2
4Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3
Another 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
Calculating 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
Example questions
1) What volume of hydrogen is produced when 18g of water is
electrolysed?
2H20
2H2 + O2
2) 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
3) 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?
Mg + 2HCl
MgCl2 + H2
Empirical 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 using
masses:
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 an equation 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 Fe2O3
Example questions
1) Find the empirical formula of magnesium oxide which
contains 24g of magnesium and 16g 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.
Electrolysis
Molecule of solid
copper chloride
Molecule of solid
copper chloride after
being dissolved
Chloride
ion
Copper
ion
Electrolysis
Electrolysis is used to extract a HIGHLY REACTIVE metal.
When we electrolysed
copper chloride the
negative chloride ions
moved to the positive
electrode and the positive
copper ions moved to the
negative electrode –
OPPOSITES ATTRACT!!!
= chloride ion
= copper ion
Electrolysis equations
We need to be able to write “half equations” to show what
happens during electrolysis (e.g. for copper chloride):
At the negative electrode the
positive ions GAIN electrons to
become neutral copper ATOMS. The
half equation is:
Cu2+ + 2 e-
Cu
At the positive electrode the
negative ions LOSE electrons to
become neutral chlorine
MOLECULES. The half equation is:
2 Cl- - 2 e-
Cl2
Calculating masses and volumes from electrolysis
Example question: How much chlorine is released at the positive
electrode if 2g of copper is collected at the negative electrode?
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
2) Now use this equation:
Actual mass of chlorine
Relative mass of chlorine
Actual mass of copper
Relative mass of copper
3) Write down the answer: _______g of chlorine is liberated, which
would occupy a volume of ________litres