Unit 1
PPA1
Effect of Concentration on Reaction Rate
The reaction is
H2O2
+
2H+ +
2I-

2H2O
+
I2
The mixture also contains sodium thiosulphate, with S2O32- ions.
These react with the iodine as it forms:
I2
+
2S2O32-

2I-
+
S4O62-.
When all the thiosulphate has reacted, iodine accumulates in the solution. If starch is
present, the mixture turns blue / black after a while, when all the thiosulphate has
reacted.
A set of mixtures like this is set up
Mixture
Volume of H2SO4 / cm3
Volume of Na2S2O3 / cm3
Volume of starch / cm3
Volume of KI / cm3
Volume of water / cm3
1
10
10
1
25
0
2
10
10
1
20
5
3
10
10
1
15
10
4
10
10
1
10
15
5
10
10
1
5
20
Note that the only thing that changes is the concentration of KI – we use less and less
of it, replacing it with water so that the overall volume stays the same.
This means that the concentrations of the other reactants stay the same, as does the
concentration of the hydrogen peroxide, when it is added.
To each mixture we add 5 cm3 of hydrogen peroxide from a syringe, and start timing;
we stop timing when the mixture goes blue / black in colour.
The rate is calculated as 1 / time.
A plot of rate against iodide concentration is a straight line – the rate is directly
proportional to iodide concentration.
The method works well, because there is a sharp colour change at the end, and the
length of time for the change to take place means that it can be measured accurately.
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Unit 1
PPA2
Effect of Temperature on Reaction Rate.
The reaction studied is the reaction between oxalic acid and an acidified solution of
potassium permanganate.
5(COOH)2
+
6H+
+
2MnO4-

2Mn2+
+
10CO2
+
8H2O.
In the experiments, the oxalic acid solution is added to a mixture of potassium
permanganate and sulphuric acid. The sulphuric acid supplies the H+ ions required by
the reaction. In the reaction, the purple colour of the permanganate fades until it is
colourless. When it is colourless, the reaction is taken as being complete.
Identical mixtures of sulphuric acid, potassium permanganate and water are used in
each experiment. The same amount of oxalic acid is always added. The only
difference is the temperature.
The mixture of sulphuric acid, potassium permanganate and water is heated to the
required temperature. The beaker is placed on a white tile. The oxalic acid is added,
and the mixture is stirred with a thermometer. The time taken for the mixture to go
colourless is measured. The temperature is re – measured and an average taken.
The rate is calculated as 1 / time.
The experiment is repeated at suitable temperatures e.g. 40, 50, 60, 70 Celsius.
It is found that the rate increases as the temperature increases.
The experiments are carried out in beakers which have to be dry. If they were not,
then this could affect the concentrations of the solutions.
It is not possible to get an accurate value of the rate at room temperature, because the
colour change is too gradual to allow the end of the reaction to be judged.
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Unit 1
PPA3
Enthalpy of Combustion of Ethanol
Enthalpy of Combustion is the energy released when one mole of a compound is
burned completely in oxygen.
The equation for the reaction is
CH3CH2OH
+
3O2

2CO2
+
3H2O
A spirit burner is weighed.
A known mass of water is placed in a copper can.
The temperature of the water in the can is measured.
The burner is placed below the can and the wick ignited.
Once the temperature of the water has risen by a certain amount, the burner is
extinguished and reweighed.
Data required:
Initial and final mass of burner (gives mass of alcohol burned)
Initial and final temperature of the water (gives temperature rise, T)
Mass of water.
Calculation
The heat transferred is given by
Heat = cmT
The mass of alcohol burned released this amount of heat.
This is scaled up to the amount of heat released if one mole of ethanol
was burned, that is, 46g of ethanol.
Sources of error
Incomplete combustion of the ethanol (some burns to give soot or CO)
Heat loss to surroundings (this is the main source of error)
Evaporation of ethanol from the wick is also a possibility.
If asked to draw apparatus, include
copper can, with water in it
spirit burner
thermometer in the water
draught shield
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Unit 2
PPA1
Oxidation
This PPA is about tests which will distinguish between aldehydes and ketones.
Aldehydes have the functional group -CHO and ketones have a C=O (carbonyl)
group flanked by carbon atoms.
Aldehydes are more easily oxidised (to alkanoic acids) than ketones, which are not
easy to oxidise. Therefore, oxidising agents (chemicals which cause other compounds
to be oxidised) are used to distinguish between them.
The oxidising agents used are
Benedict’s solution (blue, due to Cu2+ ions)
Tollens reagent (contains ammonia and silver ions)
Acidified potassium dichromate solution. (sulphuric acid is
used)
The aldehydes and ketones are heated in test tubes with the reagents in a water bath. A
bunsen is not used because the aldehydes and ketones are flammable.
Aldehydes cause the following changes:
Benedicts solution becomes orange / red (due to Cu2+  Cu+)
Tollens reagent gives a silver mirror on the test tube in which
the reaction is carried out.
Acidified potassium dichromate solution changes from orange
to green / blue (dichromate  Cr3+)
Ketones do not affect these reagents.
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Unit 2
PPA2
Making Esters.
Use your notes to revise the formation of esters from carboxylic acids and alcohols.
The formation of esters is a reversible reaction and is slow at room temperature. The
rate is increased by heat and by using the catalyst, concentrated sulphuric acid.
The apparatus used is like this:
Cotton wool
Wet paper towel
Mixture of acid, alcohol
and conc. sulphuric acid
Hot water
The mixture is heated with a water bath since the alcohol, acid and ester are
flammable and flames should not be close to the mixture.
The wet paper towel is to condense ester vapour back to liquid to prevent escape of
vapour.
The cotton wool is to prevent any liquid spurting out of the tube.
After a time, the contents of the tube are poured into a small beaker containing a
solution of sodium hydrogen carbonate. This neutralises the sulphuric acid catalyst.
Unreacted carboxylic acid and alcohol dissolve in this solution. The ester does not and
it floats as a separate layer on the surface of the liquid. If smelled, it will have a sweet,
fruity odour. It forms a separate layer because it is less dense than water and is
insoluble in water, being non – polar.
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Unit 2
PPA3
Factors affecting enzyme activity
The enzyme studied is catalase. It catalyses the breakdown of hydrogen peroxide to
water and oxygen.
2H2O2

2H2O
+ O2
The catalase is supplied in discs of potato.
The following apparatus is used.
pH buffer solution
+ potato discs
Water
The potato discs are left in the buffer solution for 3 minutes.
Then 1 cm3 of hydrogen peroxide solution is added from a syringe.
The rate at which bubbles of oxygen are produced is measured by counting the
bubbles and using a timer.
The experiment is repeated using other pH buffer solutions, and also 0.1 mol l-1 HCl
(pH = 1) and 0.1 mol l-1 NaOH (pH = 13).
It is found that there is an optimum (best) pH for the function of this enzyme.
This experiment can also be done, varying temperature instead of pH.
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Unit 3
PPA1
Hess’s Law.
Hess’s Law states that the overall energy change for a reaction is independent of the
route taken.
The reaction studied is
KOH(s)
+
HCl(aq)

KCl(aq)
+
H2O(l)
This can be carried out directly – i.e. in one step – add solid KOH to HCl – or it can
be done in two steps
(1)
(2)
Dissolve KOH in water to form a solution of KOH
Add the solution of KOH to the HCl
The energy changes of these two steps should amount to the same as that in the first
equation.
A
Direct route
A known mass of solid KOH is added to 25 cm3 of HCl solution.
Measure temperature at the start and at the end when all the solid has
dissolved.
Heat given out = cmT where m is the mass of HCl (taken as being the same
as the same volume of water, 25g) and T is the rise in temperature.
The heat has then to be scaled up to the figure for one mole of KOH, i.e. 56g
of KOH.
B
Indirect route
A known mass of solid KOH is added to 25 cm3 of water.
Measure temperature at the start and at the end when all the solid has
dissolved.
Heat given out = cmT where m is the mass of water (25g) and T is the rise
in temperature.
The heat has then to be scaled up to the figure for one mole of KOH, i.e. 56g
of KOH.
This solution is then added to 25 cm3 of HCl, and the temperature rise is
measured. The starting temperature is the average of the temperatures of the
two solutions.
Heat given out = cmT where m is the total mass of solutions (50g) and T is
the rise in temperature.
The heat has then to be scaled up to the figure for one mole of KOH, i.e. 56g
of KOH.
The sum of the energy changes in the indirect route should equal that for the
direct route.
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Unit 3
PPA2
Quantitative Electrolysis
If we electrolyse dilute sulphuric acid, hydrogen gas is produced at the
negative electrode.
The equation for the reaction is
2H+
+
2e-

H2
Two moles of electrons should be required to form one mole of hydrogen.
The following apparatus is used:
Dilute
sulphuric acid
A
The circuit includes an ammeter, a variable resistor and a power supply.
The measuring cylinder, at the start, should not be over the carbon rod.
The power should be switched on, and the current adjusted to 0.5A.
The current should be left to run for a few minutes to allow the porous carbon rod to
become saturated with hydrogen gas.
Then the current should be switched off. The measuring cylinder should be positioned
over the carbon rod.
The current is then switched on, and the time taken to collect around 50 cm3 of gas is
noted.
Measurements – current, time, volume of hydrogen
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Calculation
Coulombs = amps x seconds.
We know the number of coulombs required to form 50 cm3 of gas.
We can then calculate the number of coulombs required to form one mole of
gas. The molar volume is taken as 24 litres mol-1.
In theory, 2 x 96500 coulombs (= 193 000 coulombs) are required to form
one mole of hydrogen. (96500 coulombs is one mole of electrons).
Possible sources of error
Current not constant
Error in measuring gas volume
Hydrogen slightly soluble in water.
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Unit 3
PPA3
Redox Titration
Vitamin C is a reducing agent. The PPA involves finding out how much VitC is in a
tablet, by carrying out a redox titration, in which VitC reduces iodine in solution to
iodide ions.
The ion electron equations are:
C6H8O6 (VitC) 
I2
+
2e-

C6H6O6
+ 2H+
+
2e-
2I-
Step 1
Making a solution of VitC.
Place tablet in a beaker. Add 50 cm3 of deionised (purified) water to it.
Let the tablet dissolve.
Add the solution to a graduated (standard) flask.
Rinse the beaker out several times with water. Transfer washings to the flask.
Add water carefully to the standard flask to bring the volume of the solution up to the
graduation mark on the neck of the flask.
Stopper the flask and turn it upside down a few times to mix thoroughly.
Step 2.
The titration
Rinse pipette out with VitC solution.
Rinse burette out with iodine solution.
Pipette 25 cm3 VitC into a conical flask.
Add starch solution.
Add iodine solution from the burette. You see a blue colour because of the starch but
this disappears at once as the iodine reacts with the VitC.
At the end point, the mixture turns blue (excess of iodine) and stays blue.
The titration is repeated to give at least two additional concordant results (agreeing to
0.1 cm3).
The results can be used to work out the mass of VitC in the tablet.
In this experiment, the moles of iodine used are the same as the moles of VitC.
Knowing the moles of VitC in 25 cm3 we can calculate the moles in whole of the
graduated flask.
If we then multiply by 176 we get the mass of the VitC.
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