vitamin c chemistry coursework

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



water

Sink 17

Sodium thiosulphate

solution

0.06 moldm-3 Irritant, harmful if swallowed


lab coat



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


lab coat


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

http://hazard.com/msds/mf/baker/baker/files/p5898

http://www.webexhibits.org/causesofcolor/6AC.html

http://www.outreach.canterbury.ac.nz/chemistry/vitaminCiodate.shtml

http://www.outreach.canterbury.ac.nz/chemistry/vitaminCiodate.shtml

http://www.docbrown.info/page07/redox2.htm#6.6

http://www.chemsheets.co.uk/ASPrac208.doc

http://www.chemguide.co.uk/physical/basicrates/arrhenius.html#top

http://www.chemguide.co.uk/physical/basicrates/arrhenius.html#top

http://www.chemhume.co.uk/A2CHEM/Unit%202a/12%20How%20fast/Ch12Howfastc.htm

http://www.chemhume.co.uk/A2CHEM/Unit%202a/12%20How%20fast/Ch12Howfastc.htm

http://www.chemguide.co.uk/physical/catalysis/introduction.html#top

http://www.chemguide.co.uk/physical/catalysis/introduction.html#top

http://www.chemguide.co.uk/physical/basicrates/catalyst.html#top

http://www.chemguide.co.uk/physical/basicrates/catalyst.html#top

http://www.chemguide.co.uk/physical/basicrates/temperature.html#top

http://www.chemguide.co.uk/physical/basicrates/temperature.html#top

http://www.chemguide.co.uk/physical/basicrates/introduction.html#top

http://www.chemguide.co.uk/physical/basicrates/introduction.html#top

http://www.dietobio.com/vegetarisme/en/vit_c.html

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










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.


Time Intervals(min)


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.












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.


Transition Metal Catalyst


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.












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.


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)


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


Time Intervals(min)


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)




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)


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


24


Time Intervals(min)


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)




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


Time(s)


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



25

Time Intervals(min)


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)




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


Time(s)


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



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)




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


Time(s)


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)


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)




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


Time(s)


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



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


Average Titre (cm3)

Moles of sodium thiosulphate

(x10-3)



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


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)



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


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.


In my experiment a 5cm3 volumetric pipette was used to measure samples of sulphuric acid and potassium iodide.


% 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.


In my experiment a 2cm3 volumetric pipette was used to measure samples of transition metal catalysts.


% 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.


% 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.


% 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


(%)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

36

53°CTime


(%)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


(%)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


(%)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


(%)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

37

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.


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.


38

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.


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.

39

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



Burette 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



Burette 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



Burette 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

40

78°CTime

(s)Sample



Burette 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


Burette 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%)


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%)


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.

42

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.

43

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.

44

vitamin c chemistry coursework

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