vitamin c chemistry coursework
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Decomposition of Vitamin C Coursework
Contents
Aims and Hypotheses 2
Background Theory 3
Risk Assessment 14
References 15
Method 16
Calculations 21
Results 21
Analysis 31
Conclusion 34
Errors 35
Limitations 42
Evaluation 44
1
Aims and HypothesesMy Aims
The aims of my experiments are:
To determine a rate law for the decomposition of vitamin C naturally and in the
presence of a transition metal catalyst.
To calculate the activation enthalpy of the decomposition of vitamin C through the use of
Arrhenius’ equation.
To determine the effect of altering temperature, oxygen concentration and addition of
metal ion catalysts on the rate of reaction.
My Hypotheses
1) Temperature:
I predict that at increased temperature, the rate of decomposition will increase and so larger
titration values will be obtained. In general for reactions, an increase of 10°C will double the
rate. However, due to the denaturing at higher temperatures of the enzyme ascorbic acid
oxidase which is also responsible for the decomposition of vitamin C, I do not expect my
changes to be as significant as that.
2) Oxygen concentration:
I predict that at increased levels of oxygen, the rate of reaction will increase as there will be
higher levels of oxygen available to oxidise the ascorbic acid to dehydroascorbic acid.
3) Addition of metal ion catalysts:
I predict that the addition of a solution of copper (II) sulphate catalyst will have a significant
effect on the rate of my reaction as its ability to change oxidation stated allows it to provide an
alternate pathway for the reaction with a lower activation energy. I do not expect that the iron
(II) chloride and iron (III) sulphate catalysts will have any effect on my reaction as they do not
lie within the necessary electrode potential range.
2
Figure 1-The formula for ascorbic acid is C6H8O6
Background TheoryVitamin C (Ascorbic Acid)
Vitamin C (also known as ascorbic acid) is a water-soluble vitamin,
which has been shown to have many beneficial effects in the body.
Vitamin C strengthens and protects the immune system by stimulating
the activity of antibodies and immune system cells. Therefore, vitamin C
fights not only viruses but also tumour cells and may be a great help in
the treatment of cancers. Vitamin C seems to inhibit the secretion of the
prostaglandins that are responsible for the inflammatory and pain response. This vitamin is also
required for the building and maintenance of collagen, a protein that holds the body's cells in
place. Collagen is indispensable for bones, teeth and healing of wounds.
Vitamin C is a powerful antioxidant. It works along with vitamin E and the enzyme glutathione
peroxidase to stop free radical chain reactions. These are atoms or molecules with an unpaired
electron, which have negative effects when they damage proteins, lipids and nucleic acids. The
body has antioxidant defence mechanisms that keep them in balance but our environment
(infections, smoking, sunlight, stress etc.) increases the formation of free radicals in our body
and induces premature aging.
Vitamin C is one of the least stable of all vitamins in solution and is oxidised readily in light, air
and when heated. It is also water soluble. This means that heating in water causes the vitamin to
leach out of the food into the water and also to be oxidised, first to dehydroascorbic acid and
then to diketogulonic acid. [1]
Effect of Heat on Vitamin C
When ascorbic acid decomposes, it is generally undergoing an oxidation process, which in a
simplified form can be represented as:
Vitamin C, like any other molecule, is heavily influenced by heat. To break a chemical bond, the
molecule requires energy for it to vibrate. If the heat is too high, more bonds than is necessary
may vibrate and be broken, causing the vitamin C molecule to decompose. As stated above,
when heated in water it becomes oxidised, first to dehydroascorbic acid and then to
3
diketogulonic acid. The latter has no vitamin C activity, and hence the iodine doesn’t react with
the decomposed molecule. The starch solution should therefore pick up less iodine, and so the
titre should be lower as the temperature increase. This reaction is also irreversible.
On the other hand, the enzyme ascorbic acid oxidase is also denatured in the process of heating.
As this enzyme speeds up the oxidation of vitamin C, it may be argued that heat preserves
actually helps preserve the vitamin. I will investigate this further in my experiment.
Collision Theory
According to the collision theory, for two species to react they must both collide in the right
direction and they must collide with a certain amount of kinetic energy (known as the activation
energy of the reaction).
Orientation of the Collisions
The importance of the orientation of the collisions in a reaction can be clearly shown in the
simple reaction between ethane (CH2=CH2) and hydrogen chloride (HCl). These react to give
chloroethane. However, the reaction can only happen if the hydrogen end of the H-Cl bond
approaches the carbon-carbon double bond. Any other collision between the two molecules
doesn't work. The two simply bounce off each other.
Of the collisions shown in the diagram, only collision one may possibly lead on
to a reaction. Collision two is not possible as the double bond has a high concentration of
negative charge around it due to the electrons in the bonds. The approaching chlorine atom is
also slightly negative because it is more electronegative than hydrogen. The repulsion simply
causes the molecules to bounce off each other.
4
As my reaction also contains unsymmetrical species, the way in which they hit each other will
be important in deciding whether or not a reaction happens. [2]
Activation Energy
Even if the species are orientated properly, there will still not be a reaction unless the particles
collide with a certain minimum energy called the activation energy of the reaction. Activation
energy is the minimum energy required before a reaction can occur. You can show this on an
energy profile for the reaction. For a simple exothermic reaction, the energy profile looks like
this:
If the particles collide with less energy than the activation energy, the reaction does not take
place. The activation energy acts as a barrier to the reaction and only those collisions which
have energies equal to or greater than the activation energy result in a reaction.
Any chemical reaction results in the breaking of some bonds (needing energy) and the making
of new ones (releasing energy). Obviously some bonds have to be broken before new ones can
be made. Activation energy is involved in breaking some of the original bonds.
Where collisions are relatively gentle, there isn't enough energy available to start the bond-
breaking process, and so the particles don't react.
The Maxwell-Boltzmann Distribution
Due to the key role of activation energy in deciding whether a collision will result in a reaction,
it is useful to know what proportion of the particles present have high enough energies to react
when they collide. In any system, the particles present will have a very wide range of energies.
For gases, this can be shown on a graph called the Maxwell-Boltzmann Distribution which is a
plot of the number of particles having each particular energy. Although the graph only applies to
gases, the conclusions that can be drawn from it can also be applied to reactions involving
liquids.
5
The area under the curve is a measure of the total number of particles present.
As stated previously, for a reaction to happen particles must collide with energies equal to or
greater than the activation energy for the reaction. The activation energy can be marked on the
Maxwell-Boltzmann distribution:
This diagram illustrates that the large majority of the particles don't have enough energy to
react when they collide. To enable them to react we either have to change the shape of the
curve, or move the activation energy further to the left. The shape of the curve can be changed
by changing the temperature of the reaction. The position of the activation energy can be
adjusted by adding a catalyst to the reaction.
Effect of Temperature on Collision Frequency
Particles can only react when they collide. If you heat a substance, the particles gain more
kinetic energy and so collide more frequently. This speeds up the rate of reaction and shifts the
Maxwell-Boltzmann distribution to the right. This is illustrated in the diagram below. The graph
labelled T is at the original temperature. The graph labelled T+t is at a higher temperature.
6
By adding the position of the activation energy it is clear to see that although the curve hasn't
moved very much overall, there is approximately double the amount of particles with enough
energy to react i.e. the rate of the reaction has doubled. [3]
Effect of Catalytic Presence on Activation Energy
A catalyst is a substance which speeds up a reaction, but is chemically unchanged at the end of
the reaction. Therefore, when the reaction has finished the mass of the catalyst is exactly the
same as it was in the beginning. As previously stated, to increase the rate of a reaction you need
to increase the number of successful collisions. One possible way of doing this is to provide an
alternative way for the reaction to happen which has a lower activation energy. [4]
In other words, the activation energy on the Maxwell-Boltzmann distribution moves to the left:
7
Adding a catalyst has exactly this
effect on activation energy. A catalyst provides an alternative route for the reaction. That
alternative route has a lower activation energy. This can be shown on an energy profile as
follows:
Transition Metal Ion Catalysts
Catalysts can be divided into two main types - heterogeneous and homogeneous. In a
heterogeneous reaction, the catalyst is in a different phase from the reactants. In a
homogeneous reaction, the catalyst is in the same phase as the reactants. [5]
Variable Oxidation States
My reactions will use homogenous catalysis as all the reagents will be contained in a single
liquid phase. Transition metals act as homogeneous catalysts as they are able to change from
one oxidation state to another during a reaction, before returning to their original oxidation
state. This is able to happen because the differences between the successive ionisation
enthalpies in the 3d and 4s sub-shells are relatively small, so multiple electron loss is possible. [6]
8
Predicting the Feasibility of Transition Metal Ion Catalysts
The decomposition of ascorbic acid usually refers to its oxidation to dehydroascorbic acid, and
this redox process is associated with an electrode potential of +0.06 V. The relevant half
equation for the oxidation reaction involving oxygen has a redox potential of +0.40 V. Both of
these half equations are shown in bold in the standard electrode potential table below:
[7]
In
order for a transition metal to be a feasible catalyst, the metal must have a standard electrode
potential between these two values. A good example of this is Cu2+/Cu+ which has a potential of
+0.16 volt. First the ascorbic acid might reduce the Cu2+ to Cu+ whilst the Cu+ ions formed can
then reduce the O2 to OH-. If the activation energies of these two steps are lower than that of the
direct oxidation of ascorbic acid to dehydroascorbic acid, then Cu2+ can catalyse the process. In
my experiment I will use a solution of a copper (II) sulphate as a catalyst to try to achieve this
effect. If possible I will also use catalysts above and below the desired range to assess whether
or not such catalysts can have any effect. For this experiment I will be using iron (II) chloride
and iron (III) sulphate catalysts.
Reaction Equations
Orders of Reaction and Rate Equations
For a general reaction: A + B products
By doing experiments involving a reaction between A and B, it can be found that the rate of the
reaction is related to the concentrations of A and B in the following way:
9
Standard Electrode Potentials
Half-reaction Electrode Potential/ V
Fe2+ + 2e- Fe(s) -0.44
2H+(aq) + 2e- H2(g) 0.00
Dehydroascorbic acid + 2H+ + 2e- Ascorbic acid +0.06
Cu2+(aq) + e- Cu+
(aq) +0.16
O2(g) + 2H2O(l) + 4e- 4OH- +0.40
Fe3+ + e- Fe2+(aq) +0.77
This is called the rate equation for the reaction.
The concentrations of A and B have to be raised to some power to show how they affect the rate
of the reaction. These powers are called the orders of reaction with respect to A and B. [8]
If the order of reaction with respect to A is zero, this means that the concentration of A doesn't
affect the rate of reaction. Mathematically, any number raised to the power of zero (x0) is equal
to 1. This means that that particular term does not need to be written in the rate equation. The
concentration decreases equal amounts in equal time periods.
If the rate is directly proportional to the concentration of a particular reactant this means that
the order is first with respect to that particular reactant. Also, this order will give a
concentration-time graph with a constant concentration half-life.
If the rate quadruples as the concentration doubles then the order is second with respect to that
reactant. If concentration2 is plotted instead a straight line will be given. In a concentration-
time graph for a second order reaction the half-life increases as the concentration decreases
rapidly. [9]
10
The overall order of the reaction is found by adding up the individual orders. For example, if the
reaction is first order with respect to both A and B, then overall it is a second order reaction.
The Arrhenius Equation
The Arrhenius equation describes the relationship between the rate at which a reaction
proceeds, and the temperature in degrees K. The equation is as follows:
T= temperature in Kelvin.
R= the gas constant. This is a constant which comes from the ideal gas law equation:
Pressure x Volume = no. of moles x R x Temperature. It has the value 8.314 x 10-3 kJmol-1K-1.
EA= Activation energy. To fit this into the equation, it has to be expressed in joules per mole - not
in kJmol-1.
e= the exponential. This is a mathematical number with a value of 2.71828…
A= the frequency factor. This is a term which includes factors like the frequency of collisions and
their orientation. It varies slightly with temperature, but not by much. It is usually taken as
constant across small temperature ranges. [10]
The Arrhenius equation can be rearranged to become:
A linear graph can be plotted oh ln(k) against (1/T). The gradient of this graph will be equal to (-
Ea/R) and the y-intercept will be equal to ln(A).
11
Titrations
Redox titrations
Iodine-thiosulphate titrations are often used to determine the concentration of oxidising agents.
The oxidising agent is reacted with an excess of potassium iodide, producing iodine. The iodine
that is produced is then titrated against sodium thiosulphate, to find how much iodine was
produced by the reaction of the oxidising agent with potassium iodide. Once the amount of
iodine has been found, the amount of the original oxidising agent can be calculated.
The two half equations of the reaction are:
I2(aq) + 2 e- 2 I-(aq)
2 S2O32-(aq) S4O6
2-(aq) + 2 e-
Combined, these give the overall reaction of:
I2(aq) + 2 S2O32-(aq) 2 I-(aq) + S4O6
2-(aq) [11]
Production of Iodine
This step of my experiment is necessary to produce the iodine that can then go on to react with
the ascorbic acid in my vitamin C sample. It is therefore important that an excess of iodine is
produced, so that an accurate concentration of vitamin C is measured. To achieve this I will be
reacting potassium iodate (V) solution, sulphuric acid potassium iodide.
The overall reaction between the three reagents is:
5KI(aq) + 3H2SO4(aq) + KIO3(aq) 3I2(aq) + 3H2O(l) + 3K2SO4(aq)
Removing the spectator potassium and sulphate ions gives the ionic equation:
5I-(aq) + 6H+
(aq) + IO3-(aq) 3I2(aq) + 3H2O(l)
This can also be broken down into the two half equations for the reaction. For the potassium
iodide and potassium iodate respectively the equations are:
½ I2 (aq) + e- I-(aq)
IO3-(aq) + 6H+
(aq) + 5e- ½ I2(aq) + 3H2O(l)
From the first of these equations it is clear to see that the iodide gets oxidised in the reaction
and so is acting as the reducing agent. As such it will have the more negative electrode potential
(EØ). Its EØ value is actually +0.54V.
The second half-cell shows that the oxidising agent iodate is reduced as it gains electrons. This
suggests that it is this cell that has the less negative EØ, which is the case as its value is +1.19V.
12
In the first equation, the iodide ions (oxidation state of -1) are oxidised to iodine molecules
(oxidation state 0) by electron loss to the iodate (V) ion. The iodate(V) ions (oxidation state of
+5) are reduced to iodine molecules (oxidation state of 0) by electron gain from the iodide ions.
Hydrogen (+1) and oxygen (-2) do not change oxidation state. [12]
Iodine and ascorbic acid
The iodine produced in the above reaction can then go on to react with ascorbic acid to produce
iodide ions and dehydroascorbic acid:
Ascorbic acid + I2 2 I- + dehydroascorbic acid
In this reaction the iodine is immediately reduced to iodide as long as there is any ascorbic acid
present. The two iodide ions produced are then recycled back into the previous reactions as
long as there is enough iodate and hydrogen ions remaining. Once all the ascorbic acid has been
oxidised, the excess iodine is able to react with the sodium thiosulphate as the titration begins.
The equation for this reaction is:
2S2O32−
(aq) + I2(aq) → S4O62−
(aq) + 2I−(aq)
Once all the iodine has reacted with the sodium thiosulphate titrated into the solution, the
starch indicator will become colourless to show that the end point of the titration has been
reached. [13]
Starch-Iodine Complex
Starch is a carbohydrate and exists in two types of molecules: amylose (linear) and amylopectin
(branched). Amylose molecules consist of single, mostly un-branched chains of glucose
molecules, shaped like a spring. The reaction between amylose and iodine is said to account for
the colour change of iodine from brown to blue/black. [14] This is thought to be due to the iodine
molecules (in the form of I5- ions) getting stuck in the coils of the amylose molecules. The starch
forces the iodine atoms into a linear arrangement within the centre of the amylose coil. There is
some transfer of charge between the starch and the iodine. This changes the way electrons are
arranged, and so, changes the spacing of the energy levels. The energy level spacing within the
starch-iodine complex is such that it can absorb certain frequencies of visible light, giving the
deep blue/black colour of the starch-iodine complex. [15]
13
Risk AssessmentName of Reagent
Concentration Hazards Safety Precautions
Emergency Procedure
Disposal Ref.
Potassium iodate (V) solution
0.01 moldm-3 Irritant, harmful if swallowed
Wear safety glasses and
lab coat
In case of contact with bare skin
wash thoroughly with running
water
Sink 16
Potassium iodide
solution
1.00 moldm-3 Slight irritant Wear safety glasses and
lab coat
In case of contact with bare skin
wash thoroughly with running
water
Sink 17
Sodium thiosulphate
solution
0.06 moldm-3 Irritant, harmful if swallowed
Wear safety glasses and
lab coat
In case of contact with bare skin
wash thoroughly with running
water
Sink fine for small
amount, larger
amounts should be stored for
later disposal
18
Sulphuric acid
0.50 moldm-3 Irritant, harmful if
swallowed, slightly
corrosive
Wear safety glasses and
lab coat
In case of contact with bare skin
wash thoroughly with running water. Seek
medical help if ingested
Sink 19
General safety precautions:
In case of spillage of any of these chemicals, clean up with a towel immediately.
Do not run in the lab or while holding glassware or any chemicals.
Do not sit down while working with chemicals on a desk.
Do not handle electrical equipment with wet hands.
In the case that any of these chemicals make contact with the skin, wash immediately
with cold water. If irritation occurs, seek medical help.
14
ReferencesReference Date Information Covered
1 http://www.dietobio.com/vegetarisme/en/vit_c.html 10/2/12 Background information on vitamin C
2 http://www.chemguide.co.uk/physical/basicrates/introduction.html#top
4/2/12 Collision theory
3 http://www.chemguide.co.uk/physical/basicrates/temperature.html#top
6/2/12 Maxwell-Boltzmann distributions
4 http://www.chemguide.co.uk/physical/basicrates/catalyst.html#top
6/2/12 Effect of catalysts on activation energy
5 http://www.chemguide.co.uk/physical/catalysis/introduction.html#top
10/2/12 Different types of catalysts
6 Salters Advanced Chemistry: Revise A2, Pearson Education Ltd, 2009 p30
N/A Transition metal catalysts
7 Salters Advanced Chemistry: Support Pack, Pearson Education Ltd, 2009 p222
N/A Standard electrode potentials
8 http://www.chemhume.co.uk/A2CHEM/Unit%202a/12%20How%20fast/Ch12Howfastc.htm
7/2/12 Calculating a rate equation
9 Salters Advanced Chemistry: Chemical Ideas (3rd), Pearson Education Ltd, 2008 pp 218-221
N/A Order of reactions
10 http://www.chemguide.co.uk/physical/basicrates/arrhenius.html#top
9/2/12 The Arrhenius equation
11 www.chemsheets.co.uk/ASPrac208.doc 8/2/12 Equation for iodine-thiosulphate titration
12 http://www.docbrown.info/page07/redox2.htm#6.6 8/2/12 Production of iodine
13 http://www.outreach.canterbury.ac.nz/chemistry/vitaminCiodate.shtml
2/2/12 Reaction between iodine and ascorbic acid
14 http://www.webexhibits.org/causesofcolor/6AC.html 10/2/12 Cause of colour in starch indicator
15 http://antoine.frostburg.edu/chem/senese/101/redox/faq/starch-as-redox-indicator
10/2/12 Structure of starch indicator
16 http://hazard.com/msds/mf/baker/baker/files/p5898 .htm 28/2/12 Potassium iodate (V) hazards
17 http://hazard.com/msds/mf/baker/baker/files/p5906.htm 28/2/12 Potassium iodide hazards
18 http://hazard.com/msds/mf/baker/baker/files/s5236.htm 28/2/12 Sodium thiosulphate hazards
19 http://hazard.com/msds/mf/baker/baker/files/s8236.htm 28/2/12 Sulphuric acid hazards
20 http://www.sciencedirect.com/science/article/pii/S0260877403000062
20/4/12 Externally calculated activation energy
Knowledge from OCR Chemistry B Salters Course
AS Topics A2 TopicsRates of reaction C.I. 10.1 Rate equations C.I. 10.3Effect of temperature on rate C.I. 10.2 Transition metals C.I. 11.5Catalysts C.I. 10.6 Redox titrations C.I. 9.2
Electrode potentials C.I. 9.2 and 9.3
15
MethodEquipment
4x 200cm3 beaker 3x 250cm3 conical flask 10x boiling tube 2x 25cm3 volumetric pipette (±0.06cm3) 3x 5cm3 volumetric pipette (±0.05cm3) 1x 2cm3 volumetric pipette (±0.01cm3) 1x dropping pipette 1x 50cm3 burette (±0.05cm3) 1x burette stand 3x 100cm3 volumetric flask (±0.08cm3) 1x 250cm3 measuring cylinder (±0.5cm3) 1x stopwatch (±0.005s) 1x vacuum pump 1x oxygen probe 1x thermometer (±0.05°C) 1x funnel 1x white tile 1x spatula Range of water baths Top-pan balance (±0.00005g)
Reagents
Potassium iodate (V) solution, 0.01 moldm-3 Potassium iodide solution, 1.00 moldm-3 Sulphuric acid, 0.50 moldm-3 Sodium thiosulphate solution, 0.06 moldm-3 Starch indicator solution, 1% 4 litres of orange juice Distilled water Ascorbic acid powder Copper (II) sulphate Iron (II) chloride Iron (III) sulphate
16
Method 1—Modifying the Temperature
1. Use a 25cm3 bulb pipette to accurately measure out 25cm3 of the orange juice and pour
into a boiling tube.
2. Put the conical flask into the water bath set at temperatures of 0°C, 20°C, 40°C, 60°C and
80°C.
3. Using a stopwatch, monitor the time and remove three boiling tubes from the water
bath at time intervals of 10, 20, 30 and 40 minute intervals.
4. Transfer the sample to a conical flask, using distilled water to rinse the tube and ensure
no ascorbic acid is left in the boiling tube.
5. Using another volumetric pipette, add 25cm3 of potassium iodate (V) solution to the
conical flask. With two 5cm3 bulb pipettes, add 5cm3 of sulphuric acid and potassium
iodide to the conical flask.
6. Run sodium thiosulphate through a 50cm3 burette, ensuring the nozzle is also filled.
Read the bottom of the meniscus and record the starting volume.
7. Place a white tile under the burette and place the conical flask on it.
8. Run the burette until the solution in the flask becomes a pale yellow colour. Then add
starch indicator and continue titrating slowly whilst continually swirling the flask.
9. Once the solution of iodine gradually fades from black to grey, start to drop the sodium
thiosulphate into the flask. A sudden change to colourless should occur when the
reaction is complete.
10. Record the titre to two decimal places and repeat three times.
Example Table of Results:
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0
5
10
15
20
25
30
17
Method 2—Aeration
1. Measure out 250cm3 of orange juice using a 250cm3 measuring cylinder and pour into a
conical flask.
2. Set up the vacuum filter so that oxygen is being bubble through the orange juice.
3. At ten minute intervals, use a 25cm3 volumetric pipette to draw up 25cm3 of the orange
juice.
4. Transfer this to another 250cm3 conical flask.
11. Using a volumetric pipette, add 25cm3 of potassium iodate (V) solution to the conical
flask. With two 5cm3 volumetric pipettes, add 5cm3 of sulphuric acid and potassium
iodide to the conical flask.
12. Run sodium thiosulphate through a 50cm3 burette, ensuring the nozzle is also filled.
Read the bottom of the meniscus and record the starting volume.
13. Place a white tile under the burette and place the conical flask on it.
14. Run the burette until the solution in the flask becomes a pale yellow colour. Then add
starch indicator and continue titrating slowly whilst continually swirling the flask.
15. Once the solution of iodine gradually fades from black to grey, start to drop the sodium
thiosulphate into the flask. A sudden change to colourless should occur when the
reaction is complete.
16. Record the titre to two decimal places and repeat three times for each length of oxygen
exposure. Carry out experiments for 10, 20, 30, 40, 50 and 60 minute exposures.
Example Table of Results:
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0
10
20
30
40
50
60
18
Method 3—Addition of Metal Ion Catalysts
1. Using the following calculations to work out the correct mass of copper (II) sulphate
solid that is necessary to catalyse the vitamin C in the orange juice.
a. Calculate the number of moles of ascorbic acid in 25cm3:
Moles of C6H8O6= mass/ RFM = 0.05/ 176.12 = 2.84 x 10-4
b. Find 10% of this to work out the moles of catalyst required:
Moles of catalyst need = 2.84 x 10-5
Therefore moles of catalyst needed in 100cm3 = 2.84 x 10-3
c. Calculate the mass of catalyst needed:
Mass of catalyst = RFM x 2.84 x 10-3
2. Using a balance, accurately weigh out this mass and put in a 100cm3 beaker.
3. Add distilled water to the beaker until the powder has dissolved.
4. Pour the solution into a 100cm3 volumetric flask and add distilled water until the
meniscus of the reaches the line.
5. Using a 2cm3 volumetric pipette, measure out 1cm3 of copper (II) sulphate solution and
put it in a conical flask containing 25cm3 of orange juice.
6. Using a volumetric pipette, add 25cm3 of potassium iodate (V) solution to the conical
flask. With two 5cm3 volumetric pipettes, add 5cm3 of sulphuric acid and potassium
iodide to the conical flask.
7. Run sodium thiosulphate through a 50cm3 burette, ensuring the nozzle is also filled.
Read the bottom of the meniscus and record the starting volume.
8. Place a white tile under the burette and place the conical flask on it.
9. Run the burette until the solution in the flask becomes a pale yellow colour. Then add
starch indicator and continue titrating slowly whilst continually swirling the flask.
10. Once the solution of iodine gradually fades from black to grey, start to drop the sodium
thiosulphate into the flask. A sudden change to colourless should occur when the
reaction is complete.
11. Repeat the titrations three times and record values to two decimal places. Repeat the
experiment for iron (II) chloride and iron (III) sulphate also.
Example Table of Results:
Transition Metal Catalyst
Titrations (cm3)1 2 3 Average
Cu2+
Fe2+
Fe3+
19
Method 4— Change in Concentration of Cu2+ Catalyst
1. Using calculation described in method 3 to work out the correct mass of copper (II)
sulphate solid that is necessary to catalyse the vitamin C in the orange juice.
2. Using a balance, accurately weigh out this mass and put in a 100cm3 beaker.
3. Add distilled water to the beaker until the powder has dissolved.
4. Pour the solution into a 100cm3 volumetric flask and add distilled water until the
meniscus of the reaches the line.
5. Using a 2cm3 volumetric pipette, measure out 2cm3 of copper (II) sulphate solution and
put it in a conical flask containing 25cm3 of orange juice.
6. Using a volumetric pipette, add 25cm3 of potassium iodate (V) solution to the conical
flask. With two 5cm3 volumetric pipettes, add 5cm3 of sulphuric acid and potassium
iodide to the conical flask.
7. Run sodium thiosulphate through a 50cm3 burette, ensuring the nozzle is also filled.
Read the bottom of the meniscus and record the starting volume.
8. Place a white tile under the burette and place the conical flask on it.
9. Run the burette until the solution in the flask becomes a pale yellow colour. Then add
starch indicator and continue titrating slowly whilst continually swirling the flask.
10. Once the solution of iodine gradually fades from black to grey, start to drop the sodium
thiosulphate into the flask. A sudden change to colourless should occur when the
reaction is complete.
11. Repeat the titration three times and record values to two decimal places.
12. Carry out experiments for 4 cm3, 6 cm3, 8 cm3 and 10cm3 volumes of catalysts.
Example Table of Results:
Amount of Cu2+
(cm3)Titrations (cm3)
1 2 3 Average2
4
6
8
10
20
Calculations1) In order to work out the number of moles of vitamin C in each of my experiments I must first
work out the number of moles of sodium thiosulphate by:
Moles of sodium thiosulphate = concentration of sodium thiosulphate x volume of titre (cm 3 ) 1000
The concentration of sodium thiosulphate that I used in this experiment was 0.06moldm-3.
2) The reaction between sodium thiosulphate can be shown in the ionic equation:
I2 (aq) + 2S2O32-
(aq) 2I-(aq) + S4O6
2-(aq)
As the ratio of iodine to sodium thiosulphate is 1:2, I must therefore divide the number of moles
I get in step 1 by two in order to work out the number of moles of iodine that reacted with the
sodium thiosulphate.
3) In order to work out the number of moles of iodine that reacted with the vitamin C, I must
then work out my initial number of moles of iodine added. In my experiment I formed the iodine
by reacting 25cm3 of potassium iodate (V) solution (0.01moldm-3) with 5cm3 potassium iodide
solution (1.0 moldm-3). This allows me to work out the moles of each reactant using:
Moles = concentration x volume (cm 3 ) 1000
Moles of potassium iodate = (0.01 x 25) /1000 = 2.5 x10-4
Moles of potassium iodide = (1 x 5)/ 1000 = 5 x10-3
Using the equation:
KIO3 (aq) + 5KI (aq) + 3H2SO4 (aq) 3I2 (aq) + 3K2SO4 (aq) + 3H2O (l)
I can see that the ratio of potassium iodide to potassium iodate is 5:1; therefore I must divide
my total number of moles of potassium iodide by five to give 1 x10 -3 moles. This shows clearly
that the potassium iodide was added in excess of the potassium iodate. Therefore, it is the moles
of potassium iodate that determines how much iodine was produce. As three moles of iodine are
produced for every mole of potassium iodate, I should theoretically get 2.5 x10-4 x 3 = 7.5 x10-4
moles of iodine.
4) Therefore, to determine how many moles of iodine have reacted with vitamin C in my
experiment I will take away the moles of iodine found in step 2 from the initial moles of iodine
added (7.5 x10-4 moles) as found in step 3.
5) As the reaction between vitamin C and iodine is:
C6H8O6 + I2 C6H6O6 + 2H+ + 2I-
The ratio of moles of iodine to moles of vitamin C is 1:1; therefore the number of moles of
vitamin C will be equal to the value found in step 4.
21
6) In order to calculate the rate of the reactions I will then need to plot concentration- time
graphs for each of the temperatures. The concentration of vitamin C can be found using:
Concentration = number of moles x 1000 Volume (cm3)
The volume of each of my samples was 25cm3.
7) This will allow me to plot concentration- time graphs for each of the temperatures. By
drawing tangents and calculating the gradients of these graphs at their steepest point, I will be
able to find the initial rate of decomposition of vitamin C.
8) I can then use one of these concentration-time graphs to find the rate equation for the
decomposition of vitamin C. I will be able to do this by calculating the half-life of the vitamin C
and using the theory explained previously.
9) I will then use my results and Arrhenius’ equation to find the activation energy for the
decomposition of vitamin C. To do this I will need to calculate k, so that I can find ln(k) and I will
need to convert my temperatures to Kelvin (by adding 273 to the temperature in °C) to allow
me to work out 1/T. By using the formula in its y= mx+c format of ln(k)= ln(A) + [-Ea/R][1/T], I
can then plot a graph with ln(k) along the y axis and 1/T along the x axis, with a gradient equal
to [-Ea/R]. As R is a constant at 8.314 x 10-3 kJmol-1K-1, I will then be able to find a value for Ea.
10) In order to calculate the concentration of copper (II) sulphate catalyst that I have added in
my fourth experiment I will first need to find the concentration of my catalyst on its own as
calculated earlier. This gives me a value of 2.84x10-2moldm-3. Therefore, to work out the
concentration of catalyst in my final reaction mixture I must do:
_____ Volume of catalyst } Total volume of reaction mixture
22
x 2.84 x10-2
ResultsTemperature: 19 ° C (Room temperature)
Time Intervals(min)
Titrations (cm3)1 2 3 Average
10 19.10 19.20 19.20 19.17
20 19.00 19.20 19.20 19.13
30 19.10 19.30 19.20 19.20
Initial Titration Results
Molar Calculations (Steps 1 to 5)
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
600 19.17 1.1502 5.751 1.749
1200 19.13 1.1478 5.739 1.761
1800 19.20 1.1520 5.760 1.740
Vitamin C Concentration Calculations (Step 6)
It is clear to see there is little change in the concentration of vitamin C over time at room
temperature. This experiment acts as a control for the rest of my findings as it confirms that
temperature does indeed affect concentration and therefore rate of decomposition, and it
doesn’t just change over time.
23
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
1.749 6.9961.761 7.0441.740 6.960
Temperature: 46 ° C
Initial Titration Results
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0 19.00 19.10 18.90 19.00
5 19.30 19.10 19.30 19.23
10 19.50 19.50 19.40 19.47
15 19.60 19.70 19.80 19.70
20 19.50 19.80 19.90 19.73
25 19.90 19.80 20.00 19.90
30 19.90 20.10 20.00 20.00
Molar Calculations
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
0 19.00 1.1400 5.700 1.800
300 19.23 1.1538 5.769 1.731
600 19.47 1.1682 5.841 1.659
900 19.70 1.1820 5.910 1.590
1200 19.73 1.1838 5.919 1.581
1500 19.90 1.1940 5.970 1.530
1800 20.00 1.2000 6.000 1.500
Vitamin C Concentration Calculations
Time(s)
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
0 1.800 7.200
300 1.731 6.924
600 1.659 6.636
900 1.590 6.360
1200 1.581 6.324
1500 1.530 6.120
1800 1.500 6.000
Temperature: 53 ° C
24
Initial Titration Results
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0 19.10 19.20 19.00 19.10
5 19.30 19.40 19.40 19.37
10 19.60 19.80 19.80 19.73
15 20.60 20.80 20.70 20.70
20 20.90 20.80 21.10 20.93
25 21.20 21.10 21.50 21.27
30 21.30 21.40 21.40 21.37
Molar Calculations
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
0 19.10 1.1460 5.730 1.770
300 19.37 1.1622 5.811 1.689
600 19.73 1.1838 5.919 1.581
900 20.70 1.2420 6.210 1.290
1200 20.93 1.2558 6.279 1.221
1500 21.27 1.2762 6.381 1.119
1800 21.37 1.2822 6.411 1.089
Vitamin C Concentration Calculations
Time(s)
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
0 1.770 7.080
300 1.689 6.756
600 1.581 6.324
900 1.290 5.160
1200 1.221 4.884
1500 1.119 4.476
1800 1.089 4.356
Temperature: 65 ° C
Initial Titration Results
25
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0 19.10 18.90 19.10 19.03
5 19.50 19.20 19.50 19.40
10 19.80 19.80 20.40 20.00
15 20.90 20.80 20.80 20.83
20 21.30 21.20 21.20 21.23
25 21.40 21.60 21.70 21.57
30 21.80 22.00 21.80 21.87
Molar Calculations
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
0 19.03 1.1420 5.710 1.790
300 19.40 1.1640 5.820 1.680
600 20.00 1.2000 6.000 1.500
900 20.83 1.2498 6.249 1.251
1200 21.23 1.2738 6.369 1.131
1500 21.57 1.2942 6.471 1.029
1800 21.87 1.3122 6.561 0.939
Vitamin C Concentration Calculations
Time(s)
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
0 1.790 7.160
300 1.680 6.720
600 1.500 6.000
900 1.251 5.004
1200 1.131 4.524
1500 1.029 4.116
1800 0.939 3.756
Temperature: 78 ° C
Initial Titration Results
Time Intervals Titrations (cm3)
26
(min) 1 2 3 Average0 19.00 19.20 19.20 19.13
5 19.70 19.60 19.80 19.70
10 20.60 20.80 20.60 20.67
15 21.60 21.50 21.50 21.57
20 22.40 22.40 22.20 22.33
25 22.90 23.10 22.80 22.93
30 23.20 23.20 23.40 23.27
Molar Calculations
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
0 19.13 1.1478 5.739 1.761
300 19.70 1.1820 5.910 1.590
600 20.67 1.2402 6.201 1.299
900 21.57 1.2942 6.471 1.029
1200 22.33 1.3398 6.699 0.801
1500 22.93 1.3758 6.879 0.621
1800 23.27 1.3962 6.981 0.519
Vitamin C Concentration Calculations
Time(s)
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
0 1.761 7.044
300 1.590 6.360
600 1.299 5.196
900 1.029 4.116
1200 0.801 3.204
1500 0.621 2.484
1800 0.519 2.076
Aeration
Time Intervals(min)
Titrations (cm3)1 2 3 Average
0 19.10 19.20 19.20 19.17
10 19.20 19.40 19.40 19.33
27
20 19.60 19.80 20.00 19.80
30 19.90 19.10 20.10 19.70
40 20.20 20.20 20.10 20.17
50 20.30 20.30 20.20 20.27
60 20.20 20.20 20.30 20.23
Molar Calculations
Time(s)
Average Titre (cm3)
Moles of sodium thiosulphate (x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
0 19.17 1.1502 5.751 1.749
600 19.33 1.1598 5.799 1.701
1200 19.80 1.1880 5.940 1.560
1800 19.70 1.1820 5.910 1.590
2400 20.17 1.2102 6.051 1.449
3000 20.27 1.2162 6.081 1.419
3600 20.23 1.2138 6.069 1.431
Vitamin C Concentration Calculations
Time(s)
Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
0 1.749 6.996
600 1.701 6.804
1200 1.560 6.240
1800 1.590 6.360
2400 1.449 5.796
3000 1.419 5.676
3600 1.431 5.724
Catalyst Reactions
Mass of catalysts
Transition Metal Catalyst Mass of catalyst (g)Cu2+ 0.7086Fe2+ 0.5645Fe3+ 1.3696
Changing catalyst
28
Transition Metal Catalyst
Titrations (cm3)1 2 3 Average
Cu2+ 20.40 20.20 20.40 20.33
Fe2+ 19.40 19.30 19.50 19.40
Fe3+ 19.90 19.50 19.40 19.60
Molar Calculations
Transition Metal Catalyst
Average Titre (cm3)
Moles of sodium thiosulphate
(x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
Cu2+ 20.33 1.2198 6.099 1.401
Fe2+ 19.40 1.1640 5.820 1.680
Fe3+ 19.60 1.1760 5.880 1.620
Vitamin C Concentration Calculations
Transition Metal Catalyst Moles of vitamin C (x10-4) Concentration of vitamin C (x10-3 moldm-3)
Cu2+ 1.401 5.604
Fe2+ 1.680 6.720
Fe3+ 1.620 6.480
29
Concentration of copper (II) sulphate catalyst
Amount of Cu2+
(cm3)Titrations (cm3)
1 2 3 Average2 20.30 20.30 20.40 20.33
4 20.70 20.50 20.60 20.60
6 21.00 21.10 21.10 21.07
8 21.20 21.30 21.20 21.23
10 21.40 21.30 21.30 21.33
Molar Calculations
Amount of Cu2+
(cm3)
Average Titre (cm3)
Moles of sodium thiosulphate
(x10-3)
Moles of iodine (x10-4)
Moles of vitamin C (x10-4)
2 20.33 1.2198 6.099 1.401
4 20.60 1.2360 6.180 1.320
6 21.07 1.2642 6.321 1.179
8 21.23 1.2738 6.369 1.131
10 21.33 1.2798 6.399 1.101
Vitamin C Concentration Calculations
Amount of Cu2+
(cm3)Moles of vitamin C (x10-4) Concentration of vitamin C
(x10-3 moldm-3)
2 1.401 5.604
4 1.320 5.280
6 1.179 4.716
8 1.131 4.524
10 1.101 4.404
Catalyst Concentration Calculations
Amount of Cu2+
(cm3)Concentration of
Cu2+ Catalyst (x10-2
moldm-3)
Concentration of Cu2+
Catalyst in Mixture (x10-3 moldm-3)
Concentration of vitamin C (x10-3 moldm-3)
2 2.84 0.92 5.604
4 2.84 1.78 5.280
6 2.84 2.58 4.716
8 2.84 3.34 4.524
10 2.84 4.06 4.404
Analysis
30
In my hypothesis I predicted that at increased temperature, the rate of decomposition would
increase and so larger titration values would be obtained. My results seem to support this
hypothesis, however in order to get a clearer understanding I must first measure quantitatively
the rate of increase by calculating rate of each equation. I will do this by drawing tangents and
calculating the gradients of these graphs at their steepest point using the formula:
M = Y2 - Y1
X2 - X1
Where M is the gradient of the line, Y2 and Y1 are y-coordinates on the tangent and X2 and X1 are
the corresponding x-coordinates.
Changing temperature
At 46°C:
M= (6.08 x 10 -3 ) – (7.2 x 10 -3 ) = - 1.24 x 10-6 moldm-3s-1
900 – 0
At 53°C:
M= (4.80 x 10 -3 ) – (7.08 x 10 -3 ) = - 2.53 x 10-6 moldm-3s-1
900 – 0
At 65°C:
M= (4.45 x 10 -3 ) – (7.2 x 10 -3 ) = - 3.06 x 10-6 moldm-3s-1
900 – 0
At 78°C:
M= (3.00 x 10 -3 ) – (7.3 x 10 -3 ) = - 4.78 x 10-6 moldm-3s-1
900 – 0
In my hypothesis I stated that in general, increases of 10°C will double the rate. Although this
may not be the case with my experiment due to the denaturing at higher temperatures of the
enzyme ascorbic acid oxidase, by plotting a graph of temperature against rate I can analyse how
close to this model I have come.
From my graph showing how the rate of reaction varies with temperature I can see that the rate
of decrease of vitamin C increases significantly, although not quite at double the rate. For
example at 50˚C the rate would be -1.24moldm-3s-1 and 10˚C later this has increased to
2.95moldm˗ -3s-1. Therefore, the rate has increased by a factor of 1.64 which is not quite double,
but supports my original hypothesis.
31
Activation Energy
I am also going to work out the Activation Energy for the decomposition of vitamin C by using
Arrhenius’ equation of: k = Ae‾ᴱᵃ/RT
In order to find the activation energy (Ea), the equation needs to be rearranged into a suitable
form so that a linear graph can be plotted and a definite gradient obtained.
ln(k) = ln(Ae‾ᴱᵃ/RT)ln(k) = ln(A) + ln(e‾ᴱᵃ/RT)ln(k) = ln(A) + ln(e)(‾ᴱᵃ/RT)
In the form y=mx + c this is:
ln(k) = (‾ᴱᵃ/RT)(1/T) + ln(A)
Therefore, I will plot a graph with ln(k) along the y-axis, (1/T) on the x-axis. The gradient of the
graph will be equal to –Ea/R. As R is a constant at 8.314 x 10-3 kJmol-1K-1, I will then be able to
find a value for Ea.
In order to plot this graph I will first need to calculate k, so that I can find ln(k). To do this I must
use the rate equation and the rates found earlier when changing the temperature. The rate
equation is determined by:
Rate = k [A] [B]ᵃ ᵇ
As there was only one reactant, ascorbic acid, involved in the rate equation for my experiment I
will instead just use: rate = k [A] where A is Cᵃ 6H8O6.
By using my concentration-time graph for 78°C I am able to calculate the half-life of my reaction
to be constant at 1040 seconds. This shows that the reaction is first order with respect to
ascorbic acid at the rate equation is therefore:
Rate= k [C6H8O6]1
To work out activation energy I will need to calculate: K= rate [C6H8O6]
I will also need to convert my temperatures to Kelvin (by adding 273 to the temperature in °C)
to allow me to work out 1/T.
Temp. (°C)
Temp.(K)
1/T(x10-3K-1)
Rate(x10-6
moldm-3s-1)
Concentration of vitamin C
(x10-3 moldm-3)
Rate coefficient (x10-4s-1)
ln(k)(no
units)
46 319 3.14 1.24 7.20 1.72 -8.6753 326 3.07 2.53 7.08 3.57 -7.9465 338 2.96 3.06 7.16 4.27 -7.7676 349 2.87 4.78 7.04 6.79 -7.29
32
By plotting these values on my graph I can find the gradient using: M = Y2 - Y1
X2 - X1
M = (-8.8) – (-7.4) } (3.13 x 10-3) – (2.94 x 10-3)
M = –5263.16 M= –Ea/R –Ea = M x R–Ea = (–5263.16) x (8.314 x 10-3)–Ea = -43.75789474
E∴ a = +43.76 kJmol-1
This shows that the activation energy for the decay of each mole of vitamin C is +43.76 kJmol-1.
Aeration
In my hypothesis I predicted that at increased levels of oxygen, the rate of reaction would
increase as there would be higher levels of oxygen available to oxidise the ascorbic acid to
dehydroascorbic acid.
Once again I can measure the initial rate of reaction by drawing a tangent on my graph and
calculating the gradient of the line:
M= (5.40 x 10 -3 ) – (7.2 x 10 -3 ) = - 2.00 x 10-6 moldm-3s-1
900 – 0
It is clear to see from this that my hypothesis was correct and the rate of reaction did increase
under the presence of oxygen. By comparing this result to my graph showing how the rate of
reaction varies with temperature I can see that the increase in rate for aeration is similar to the
prospective value for heating to 51.5˚C.
Catalysts
In my hypothesis I predicted that the addition of a solution of copper (II) sulphate catalyst
would have a significant affect on the rate of my reaction, but that the iron (II) chloride and iron
(III) sulphate catalysts would not. This can clearly be seen to be true from my results as there is
a significantly lower concentration of vitamin C left after exposure to Cu2+ ions (5.604 x10-3
moldm-3) than when exposed to the other two catalysts. To really illustrate this clearly I have
drawn a bar graph of catalysts against vitamin C concentration.
In addition to this experiment, I also investigated the effect of changing the concentration of the
copper (II) sulphate catalyst on the decomposition of vitamin C. I would predict that as the
concentration increases the concentration of vitamin C would decrease. Once again, in order to
analyse this successfully I have plotted the graph for this reaction. The smooth negative curve
obtained shows that this is the case.
33
Conclusion In conclusion it can be seen that, as expected, varying the temperature of the reaction does
increase the rate of decomposition. This is due to the increased kinetic energy of the molecules,
which causes more bonds to break and the vitamin C molecule to decompose. Despite this, my
results did not quite match the general theory which states that an increase of 10°C will double
the rate. I have calculated that in reality my rate increased by a factor of 1.64. This is because
the enzyme ascorbic acid oxidase is already present in orange juice originally, but is denatured
in the process of heating. As this enzyme speeds up the oxidation of vitamin C, this denaturing
can, to a certain extent, help to preserve the vitamin.
My investigation also allowed me to calculate the activation energy of the vitamin C molecules.
At higher temperatures a larger proportion of molecules would have had this energy available
to make successful collisions and so that also accounts for an increased rate in decomposition.
By investigating the effect of different temperatures over time I was able to use Arrhenius’
equation to calculate the activation energy of my reaction. My calculations gave the activation
energy for the decay of vitamin C to be +43.76 kJmol-1. Through research I have found another
independent experiment which also calculated the activation enthalpy to +43.8kJmol-1. [20]
Although I cannot comment on the reliability of this experiment, the fact that another
investigation got the same result as me makes my result seem more reliable.
By analysing how aeration affects the rate of vitamin c decomposition over time I was able to
find that the rate for this reaction is - 2.00 x 10-6 moldm-3s-1. This is a faster rate of decrease than
that found for 46°C showing that aeration does increase the rate of reaction. In fact, by
comparing my aeration result to my graph showing how the rate of reaction varies with
temperature I was able to estimate that aeration has a similar effect on vitamin C decomposition
as heating to 51.5˚C. This is because when ascorbic acid decomposes, it is actually undergoing
an oxidation process and therefore the rate of this oxidation increases if there are a larger
number of oxygen molecules available for successful collisions to take place.
My catalyst experiment showed that a copper (II) sulphate catalyst had a significant effect on
the decomposition of vitamin C, but that the iron (II) chloride and iron (III) sulphate catalysts
did not. This is because the decomposition of ascorbic acid is a redox process with an associated
with an electrode potential of +0.06 V. The relevant half equation for the oxidation reaction
involving oxygen has a redox potential of +0.40 V. In order for a transition metal to be a feasible
catalyst, it needed to have a standard electrode potential between these two values. This was
the case for Cu2+/Cu+ (+0.16V) but not either of the other two catalysts. Therefore, in the Cu2+
experiment the ascorbic acid was able to reduce the Cu2+ to Cu+ whilst the Cu+ ions formed then
reduced the O2 to OH-. As the activation energies of these two steps were lower than that of the
direct oxidation of ascorbic acid, then Cu2+ was able to catalyse the process.
34
Errors In order to assess the accuracy of my results I will need to find the percentage errors of my
equipment using the following formula:
Percentage error = Standard apparatus error Reading taken from apparatus
25cm3 volumetric pipette
In my experiment a 25cm3 volumetric pipette was used to measure samples of orange juice and potassium iodate (V).
The error on this piece of equipment is ±0.06cm3.
% error = 0.06 x 100 = 0.24% 25
This is a relatively small error and particularly in the case of the orange juice should have had little effect on my final results. The error may however had some small effect on the amount of iodine produced from the potassium iodate (V) which may have reduced the reliability of my procedure. Nevertheless, there was no readily available equipment of greater accuracy that I could have used instead, so I am happy with my choice in this equipment.
5cm3 volumetric pipette
In my experiment a 5cm3 volumetric pipette was used to measure samples of sulphuric acid and potassium iodide.
The error on this piece of equipment is ±0.05cm3.
% error = 0.05 x 100 = 1.00% 5
This is a relatively large error in relation to the results measured, however I am confident that it would have had little to no effect on my results. For the production of iodine in my experiment the potassium iodide was in excess so a slight inaccuracy in the measurement here should have had little effect. Similarly, as the sulphuric acid was not directly involved in my reaction I do not anticipate any error here having any great effect on the reliability of my experiment.
2cm3 volumetric pipette
In my experiment a 2cm3 volumetric pipette was used to measure samples of transition metal catalysts.
The error on this piece of equipment is ±0.01cm3.
% error = 0.01 x 100 = 0.5% 2
This is a relatively large error in relation to the amount of catalyst I was measuring out. As such, it may have had some effect on the reliability of my final results. However, there was no readily available equipment of greater accuracy that I could have used instead, so I am happy with my choice in this equipment.
35
100cm3 volumetric flask
In my experiment a 100cm3 volumetric flask was used to dissolve the transition metal ion catalysts in 100cm3 of distilled water.
The error on this piece of equipment is ±0.08cm3.
% error = 0.08 x 100 = 0.08% 100
This is a very small error that can be considered insignificant and is unlikely to have had any effect on the reliability of my final results.
250cm3 measuring cylinder
In my experiment a 250cm3 measuring cylinder was used to measure a larger sample of orange juice in preparation for aeration.
The error on this piece of equipment is ±0.5cm3.
% error = 0.5 x 100 = 0.5% 100
This may seem to be quite a large error however in relation to the amount measured it is actually rather insignificant. In addition, the exact measurement of orange juice was not necessary at this stage. This measurement was just to ensure a large enough volume of oxygen was being aerated for me to later take 25cm3 samples for titration. I am therefore happy with my choice in this piece of equipment and so not anticipate that it had any effect on the reliability of my results.
50cm3 burette
In my experiment I used a 50cm3 burette to titre the sodium thiosulphate.
The error on this piece of equipment is ±0.05cm3, however as I will be reading twice for each titration the error on each titre is actually ±0.1 cm3.
The % errors for my measurements are:
19°CTime
(s)Average Titre (cm3) Percentage error from burette
(%)600 19.17 0.52
1200 19.13 0.521800 19.20 0.52
46°CTime
(s)Average Titre (cm3) Percentage error from burette
(%)0 19.00 0.53
300 19.23 0.52600 19.47 0.51900 19.70 0.51
1200 19.73 0.511500 19.90 0.501800 20.00 0.50
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53°CTime
(s)Average Titre (cm3) Percentage error from burette
(%)0 19.10 0.52
300 19.37 0.52600 19.73 0.51900 20.70 0.48
1200 20.93 0.481500 21.27 0.471800 21.37 0.47
65°CTime
(s)Average Titre (cm3) Percentage error from burette
(%)0 19.03 0.53
300 19.40 0.52600 20.00 0.50900 20.83 0.48
1200 21.23 0.471500 21.57 0.461800 21.87 0.46
78°CTime
(s)Average Titre (cm3) Percentage error from burette
(%)0 19.13 0.52
300 19.70 0.51600 20.67 0.48900 21.57 0.46
1200 22.33 0.451500 22.93 0.441800 23.27 0.43
AerationTime
(s)Average Titre (cm3) Percentage error from burette
(%)0 19.17 0.52
600 19.33 0.521200 19.80 0.511800 19.70 0.512400 20.17 0.503000 20.27 0.493600 20.23 0.49
Changing catalystsTransition Metal
CatalystAverage Titre (cm3) Percentage error from burette
(%)Cu2+ 20.33 0.49Fe2+ 19.40 0.52Fe3+ 19.60 0.51
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Concentration of copper (II) sulphate catalystAmount of Cu2+
(cm3)Average Titre (cm3) Percentage error from burette
(%)2 20.33 0.494 20.60 0.496 21.07 0.478 21.23 0.47
10 21.33 0.47
The largest percentage error from my burette readings was therefore 0.53%. This is a small enough percentage error that I am happy with my choice in equipment and confident that it had no large effect on my results.
Stopwatch
In my experiment I used a stopwatch to measure the time at which to remove the sample from a heated water bath.
The error for the stopwatch is ±0.005s. In addition to this there is human limitation to consider as reaction times in stopping the stopwatch also have an effect. Therefore an additional percentage error of ±0.4s must be included.
The % errors for my measurements are:
Time(s)
Percentage error from stopwatch (x10-4 %)
Percentage error from human reaction (x10-2 %)
Total percentage error (x10-2 %)
300 16.67 13.33 13.50600 8.33 6.67 6.75900 5.56 4.44 4.50
1200 4.17 3.33 3.371500 3.33 2.67 2.701800 2.78 2.22 2.252400 2.08 1.67 1.693000 1.67 1.33 1.353600 1.39 1.11 1.12
The two experiments for which I used a stopwatch were changing temperature and aeration. Both of these experiments used different time scales. Therefore I must add up the percentage errors for their respective times in order to find a compound percentage error from the stopwatch. For the changing temperature experiment this gives 0.3307%. For the aeration experiment this gives 0.1653%.
These are very small percentage errors, and as my times were measured in minutes rather than seconds, I am confident that this error is negligible in relation to my results.
Thermometer
In my experiment I used a thermometer to measure the temperature of my orange juice samples.
The error for the thermometer is ±0.05°C.
The % errors for my measurements are:
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Temperature(°C)
Percentage error (%)
19.0 0.2646.0 0.1153.0 0.0965.0 0.0878.0 0.06
The percentage errors for the thermometer are clearly very small and are therefore unlikely to have had an effect on my results.
Top-pan balance
In my experiment a top-pan balance was used to measure samples of solid transition metal catalysts.
The error on this piece of equipment is ±0.00005g.
The % errors for my measurements are:
Transition Metal Catalyst Mass (g) Percentage error (x10-3 %)Cu2+ 0.7086 7.06Fe2+ 0.5645 8.86Fe3+ 1.3696 3.65
Although an inaccuracy in these measurements would have had a large impact on the reliability
of my experiment it is clear to see from these tiny percentage errors that it was not the case. The
top-pan balance I used was very precise and therefore I can be confident that my measurements
were accurate.
Compound Systematic Percentage Error
Sample Measurement
In order to quote the level of uncertainty associated with my final answer, it is necessary to
combine the errors from my measurements. As all of my sample sizes were uniform at 25cm3,
the error associated with each measurement will be the same. The error will be the sum of the
error for measuring the ascorbic acid (0.24%), the potassium iodate (0.24%), the potassium
iodide (1.00%) and the sulphuric acid (1.00%).
0.24 + 0.24 + 1.00 + 1.00 = 2.48%
Reading Measurement
The compound percentage error for reading measurements is not uniform. Instead, I will
therefore need to combine the errors from the thermometer, the burette and the stopwatch to
find this for changing temperature, the errors from the burette and stopwatch for aeration and
the burette and top-pan balance for catalysts.
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Total Error
The total percentage error is equal to the sum of the sample measurement error and the reading
measurement error described above.
19°C Time
(s)Sample
measurement error (%)
Thermometer error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
600 2.48 0.26 0.52 6.75 3.33900 2.48 0.26 0.52 3.37 3.29
1200 2.48 0.26 0.52 2.25 3.28
46°CTime
(s)Sample
measurement error (%)
Thermometer error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
0 2.48 0.11 0.53 N/A 3.12300 2.48 0.11 0.52 13.50 3.25600 2.48 0.11 0.51 6.75 3.17900 2.48 0.11 0.51 4.50 3.15
1200 2.48 0.11 0.51 3.37 3.131500 2.48 0.11 0.50 2.70 3.121800 2.48 0.11 0.50 2.25 3.11
53°CTime
(s)Sample
measurement error (%)
Thermometer error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
0 2.48 0.09 0.52 N/A 3.09300 2.48 0.09 0.52 13.50 3.23600 2.48 0.09 0.51 6.75 3.15900 2.48 0.09 0.48 4.50 3.10
1200 2.48 0.09 0.48 3.37 3.081500 2.48 0.09 0.47 2.70 3.071800 2.48 0.09 0.47 2.25 3.06
65°CTime
(s)Sample
measurement error (%)
Thermometer error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
0 2.48 0.08 0.53 N/A 3.09300 2.48 0.08 0.52 13.50 3.22600 2.48 0.08 0.50 6.75 3.13900 2.48 0.08 0.48 4.50 3.09
1200 2.48 0.08 0.47 3.37 3.061500 2.48 0.08 0.46 2.70 3.051800 2.48 0.08 0.46 2.25 3.04
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78°CTime
(s)Sample
measurement error (%)
Thermometer error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
0 2.48 0.06 0.52 N/A 3.06300 2.48 0.06 0.51 13.50 3.19600 2.48 0.06 0.48 6.75 3.09900 2.48 0.06 0.46 4.50 3.05
1200 2.48 0.06 0.45 3.37 3.021500 2.48 0.06 0.44 2.70 3.011800 2.48 0.06 0.43 2.25 2.99
AerationTime
(s)Sample
measurement error (%)
Burette error (%)
Stopwatch error (x10-2%)
Total percentage error (%)
0 2.48 0.52 N/A 3.00600 2.48 0.52 6.75 3.07
1200 2.48 0.51 3.37 3.021800 2.48 0.51 2.25 3.012400 2.48 0.5 1.69 3.003000 2.48 0.49 1.35 2.983600 2.48 0.49 1.12 2.98
Changing catalystsTransition
Metal Catalyst
Sample measurement
error (%)
Burette error (%)
Top-pan balance error (x10-3%)
Total percentage error (%)
Cu2+ 2.48 0.49 7.06 2.98Fe2+ 2.48 0.52 8.86 3.01Fe3+ 2.48 0.51 3.65 2.99
Concentration of copper (II) sulphate catalystAmount of
Cu2+
(cm3)
Sample measurement
error (%)
Burette error (%)
Top-pan balance error (x10-3%)
Total percentage error (%)
2 2.48 0.49 7.06 2.984 2.48 0.49 7.06 2.986 2.48 0.47 7.06 2.968 2.48 0.47 7.06 2.96
10 2.48 0.47 7.06 2.96
It is clear to see from these tables that all of my compound percentage errors lie in the range 2.96%- 3.33%. This is a perfectly acceptable level of error for my experiments which I feel could not have been avoided based on the equipment I had available. Therefore, I am confident in the reliability of my results so far.
41
LimitationsIn addition to the percentage errors mentioned above, there were also other limitations placed
on my experiment that may have had a negative on the reliability and accuracy of my final
results. However, as far as possible I tried to reduce this by controlling the limitations as
follows.
Varying laboratory conditions
As my experiment took quite a long time to conduct, it was necessary for me to carry out my
reactions on different days at different times. The temperature with the laboratory I was
working in was not able to be controlled across these days so the conditions under which I
carried out my experiment would have varied slightly.
This variation may have had a small effect on the reactants, however the temperature of the
actual reaction was carefully monitored and therefore the varying conditions outside of the
water bath should not have had a great effect. In order to possibly improve this in future I could
have measured the room temperature throughout the day to note any fluctuations that could
have affected my results.
Leaving samples to stand
During my experiment my reactants were left to stand for long periods of time. This could mean
that reactions within the solutions may have taken place which could have affected the
concentrations of solutions and therefore had an impact on my reaction. For example, if the
potassium iodide was left to stand for too long it turned from colourless to a very pale yellow,
indicating that a reaction within the solution was occurring. This was a limitation which may
have had a large effect on my results, as the reaction within the solution may have impacted the
amount of iodine produce and therefore the size of my final titration. Also, once I opened my
carton of orange juice to withdraw my samples, despite sealing it again between titrations, it is
possible that air exposure led to the vitamin C beginning to decompose early.
In an attempt to reduce the limitation associated with the potassium iodide I made sure only to
put small amounts in a beaker at a time. I could then easily replace my iodide with fresh out of
the bottle at regular intervals. Although attempts were made to control this variable, this
problem potentially had a significant impact on my results and may account for any anomalous
results obtained.
Contamination
Contamination of pipettes, beakers, boiling tubes and conical flasks may have occurred despite
washing. This would have meant that reactions may have occurred between chemical other than
those I was expecting and so would have had an effect on the results obtained.
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In order to try to remove any possibility of contamination, I made sure to wash every piece of
equipment out after I used it with distilled water. This limitation could have been reduced
further if new pieces of equipment were used for each measurement, however this would have
been impractical and wasteful. In any case, I do not think there would have been any severe
contaminations that could have had a significant effect on my results.
Loss of reactants
In my experiment it was necessary for me to use a pipette to draw up the sample, a boiling tube
to heat the sample and a conical flask to titrate the sample. As a result of these transfers it may
have been that some of the reactants were left behind in the equipment.
I tried to prevent this limitation by washing out each piece of equipment with a small amount of
distilled water between transfer, adding both the solution and the distilled water to the next
piece of equipment.
Inaccurate measurements
Although I took great care to ensure that all my measurements were as accurate as possible, for
example by reading from the bottom of the meniscus, I realise that errors may still have
occurred in my measurements. As the change in titre was so small throughout my reaction,
inaccuracy may have occurred in readings which may have had a large effect on my results and
made them less reliable.
In an attempt to make my results as accurate as possible I made sure to take extra care when
measuring each sample. I also placed a white tile underneath my reaction mixture when
titrating to allow me to view a colour change more easily and therefore determine the end-point
of the reaction. I would have liked to carry out more repeats for each titration so that I had a
greater range of results, which would have made my average more reliable. Unfortunately this
was not possible within the time that I had available.
Reactants not fully reacting
In my experiment I relied on the reaction between potassium iodide and potassium iodate to
produce the iodine that was needed to go on to be reduced by the ascorbic acid. As chemical
reactions are not usually 100% efficient it may have been that the theoretical value of iodine
that I calculated did not accurately reflect what was actually in my solution. If this was the case
and there was less iodine in my reaction mixture than I was expecting, then my value for the
amount of vitamin C would have been higher than was actually true. This would mean that my
results may show that less decay of vitamin C occurred than actually was the case.
Unfortunately, this was not a limitation that I could control in any way, however I am happy
with my choice here as the method I used was more reliable than using iodine directly, as the
iodate solution is more stable that the iodine complex.
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EvaluationAs a result of the errors and limitations previously described there are clearly areas in my
experiment where inaccuracies may have occurred. The main limitations that I feel may have
had an effect on my results were the loss of reactants between transfers and unwanted
reactions such as those in the potassium iodide and orange juice when left to stand. Loss of
reactant or reactants changing composition may have exaggerated the decay and caused slight
variations in my titration.
Another limitation that may have had an effect was successfully calculating the end-point by
determining a clear colour change. Although I used a white-tile to aid in this, it was sometimes
difficult to assess exactly when the end-point had been reached. As a single extra drop from the
burette would make a difference to my final recording it is clear that this would have had some
impact. Small errors such as these may be reflected in my titration results as values fluctuated
by up to 0.3cm3.
On the whole however I am happy that I used the correct apparatus necessary to conduct my
experiment. I feel that this is justified in my percentage error calculations as all of my compound
percentage errors lie in the range 2.96%- 3.33%. This is a relatively low level of error that
should not have had a significant impact on my results and one that I do not think could have
been avoided.
I am very pleased with the calculations I conducted from my results and the conclusions I drew
from them. My calculation of the activation energy for the decay of each mole of vitamin C being
+43.76 kJmol-1 is the same as the another experiment’s value of +43.8 kJmol-1 [20]. This once again
supports the idea that my experiment was sufficiently accurate to draw reliable conclusions. In
addition my experiments were able to prove my hypotheses to be true in relation to aerating the
orange juice and adding a copper (II) sulphate catalyst.
Although overall I am very happy with my experiment there are ways I would have liked to have
improved it had it been possible within the time constraints. In particular I would have liked to
do more repeats of each titration so I could be more confident in the reliability of my averages
and easily identified any anomalies. In addition, if time had allowed I would have liked expand
my experiment so that I could learn more about the decomposition of vitamin C. For example, I
would have liked to use my copper (II) sulphate catalyst at different temperatures in order to
calculate a new activation energy for the decomposition of vitamin C. This would have allowed
me to identify how a catalyst shifted the activation energy line of the Maxwell-Boltzmann
distribution. I would have also liked to explore the effect of pH on vitamin C decomposition by
adding a known concentration of sodium hydroxide and quenching with hydrochloric acid.
These additional experiments would have added extra interest to my investigation.
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