Individual Investigation Essays

Individual Investigation   Essays Individual Investigation   Essay Individual Investigation   Essay Aim: The aim of this experiment is to answer the following question: What is the effect of temperature on the equilibrium constant for the hydrolysis of an ester? The reaction I aim to investigate is a reversible reaction where an ester (an organic compound with RCOOR group) is produced and. Esters are generally insoluble in water but are soluble in other solvents. Esters are formed in an esterification reaction where and carboxylic acid (RCOOH group) and an alcohol (ROH), react to form an ester. Below is the esterification reaction/ester hydrolysis reaction. Alcohol + Carboxylic acid Ester + Water The Equipment and apparatus I will be using in my investigation are as follows: * Safety Goggles * Test Tube x 5 * Test tube stoppers x 5 * Water Bath with thermostatic control * Test tube holder (suitable for use in water baths) * Burette * Clamp stand with clamp fixings * Safety Mat * Funnel * 250ml Beaker * 250ml Conical Flask * 10ml Measuring cylinder * Thermometer * White tile Safety goggles will be worn so that the risk of chemicals coming into contact with the eyes is lessened (see risk assessment). Test tubes are being used as the environment for the reaction to take place, they are glass so can be monitored and can be easily stored in a test tube holder in the water bath. Test tube stoppers will be used so that none of the water or ethanol can evaporate out of the tube, thus giving inaccurate results. Water baths will be used so that the temperature, can be maintained at different temperatures. Test tube holders will be used so that all the test tubes can be held safely without risk of tipping over. A burette will be used in the titration because it will less small amounts of liquid at a time, and also has an accurate scale up the side, so volumes will be easier to obtain. A Clamp stand will be used so that the burette is held in place above the solution of unknown concentration securely. A safety mat will be used so that any spillages will be kept off of the tables. A funnel will be used so that any pouring of liquids will be safer from the risk of spillage. Beakers of 250ml are used to contain liquid yet will have lips on them to aid pouring. Conical flasks will be used in the titration to stop any of the liquid hitting the side. A 10ml measuring cylinder will be used so that the reactants can be measured accurately. A thermometer will be used so that the temperature of the water bath can be verified. Risk assessment The chemicals I will be using are: * Hydrochloric acid (known concentration of 0.1moldm-3-150ml approx. * Sodium Hydroxide Solution (unknown concentration)- 150ml approx. * Sulphuric acid (unknown concentration)- 150ml approx. * Ethanol (95% concentration with 5% water)-500ml approx. * Ethanoic Acid (Glacial Acetic Acid 100%)-500ml approx. * Methyl Orange * Phenolphthalein The Hydrochloric acid I am using is corrosive, so if it comes into contact with skin it will burn. Therefore when I am transferring the chemical from the sealed jar to the burette, I will ensure that safety goggles are on; so that any spillage will not come into contact with the eyes. I will also ensure that I am using a funnel on the burette. This is so that spillages will be less likely. Also when cleaning I will make sure I am using excess water to clean anything where the acid has contacted because it may react violently with water in small quantities. Similarly the sulphuric acid I am using is corrosive, therefore I must be careful when adding it to the test tube in the reaction, to prevent overfilling and spillages. The Sodium Hydroxide I am using is also corrosive, so when pouring into beakers and the burette, I will use a funnel and wear safety glasses. The ethanol I am using is highly flammable so I will keep it away from any naked flames. It is also very toxic if ingested, therefore I must make sure that it will be kept well away from the face at all times. As the ethanol is used in the reaction it I will make sure it is measured out accurately and that no spillages take place. Ethanoic acid is its concentrated form is also corrosive and will cause burns if it comes into contact with the skin or eyes. There I will be wearing safety goggles when using this chemical to add to the test tube or when it is used in the titration. Phenolphthalein is very harmful even in small quantities, it causes water imbalance and is a strong laxative. Therefore this substance must be kept off of tables, hands or any equipment that is likely to be handled. When doing a acid-base titration, the reaction is exothermic so therefore I must make sure nobody touches the beaker when the two chemicals are mixing. Also the water baths will be set so that there is an element heating the water. This must not be touched as it can cause burns. Also the water may too hot to touch so, the water must be touched as little as possible. Another reason for not touching the water too much is that if water gets onto the floor there is a risk of people falling over; therefore this risk should be minimized Variables Although these do not affect the value of the reaction constant I will still aim to control some variables. The variables I am aiming to control are that of concentration and that of total pressure. As all reactants and products are in the liquid phase, keeping the total pressure constant will be just the atmospheric pressure. To control the variable of concentration I will measure accurately different reactants each time so that the same concentration and amounts will be obtained each experiment. I will also be controlling the chain length of the acid and alcohol I use in the reaction. If I change the chain length of the reactants different products are going to be produced and may affect the equilibrium constant of the reaction. For this reason I will use ethanol and ethanoic acid, as they are easily obtainable and they produce an ester that is also a colourless liquid. I will be varying temperature as feel this is feasible to do, by using a water bath. Also because concentration does not actually affect the value for Kc, it will just shift the equilibrium position, away from the part of the reaction with the highest concentrations. The temperature ranges I will be using for this experiment will be in divisions of five degrees Kelvin from 293K, 298K, 303K , 308K , 313K , 318K , 323K. In terms of degrees Celsius this equal to 20oC, 25oC, 30oC, 35oC, 40oC, 45oC, 50oC. I am using this range because I feel I will be able to notice a marked difference in the equilibrium constant. Also I do not want to use any value of temperature above 60oC as above this temperature ethanol will boil, and will be in the gaseous state, separated from this rest of the reaction (condensed on the sides of the test tube) and therefore unable to take part in it. Also if I go above 100oC then the water would boil and the water bath would then not be as effective at keeping temperature constant, as well as damaging the heating element. Other ways I will ensure that I will make my experiment fair, accurate and able to produce repeatable data will now be explained. 🙂 I will make sure that the water bath I am using is set to that temperature for about 15minutes before the experiment, so that the water is an even temperature throughout. 🙂 I will also store the chemicals that I will be using in the water bath so that the reaction starts at the desired temperature, and does not have to be raised to that temperature. 🙂 I will use sterilised equipment throughout so that there is no chance of contamination, as this could affect the reaction. 🙂 I will also use the same equipment all the way through to ensure that no other contaminants can enter the mixtures. After each experiment I will clean out all equipment with both distilled water and the substance that is to be placed in the equipment after cleaning. 🙂 I will always use the same initial volumes and concentration of the alcohol and carboxylic acid. This is so that the 🙂 I will also use the same amount of concentrated sulphuric acid catalyst for each experiment. This amount will be four drops from a small pipette. This is so that the overall concentration of acid will be easier to work out if the drops are constant. 🙂 When titrating my sample I will make sure I use the same concentration of sodium Hydroxide solution. This amount will be 0.4molsdm-3. This will be done so that the same concentration will be used in calculations throughout, my analysis, and will make the spotting of patterns easier, if the sodium Hydroxide solution is of the same concentration all the way through. 🙂 I will use the same amount of phenolphthalein in the sample to measure the colour. This amount will be 6 drops from a small pipette. This will be done because if different amounts were added there may be a colour change at a different point to when I expect one, so making the amount of indicator constant should eliminate this. 🙂 When moving the sample I take from the reaction vessel to be titrated I will move it quickly as possible so that the temperature will not go down, if it does then that would change the amount of acid and make the experiment inaccurate. I will help to maintain accuracy by placing the titration apparatus close to the water bath. I will also have the titration equipment already set up, so the titration could be done immediately. 🙂 I will also graph my results as I get them. This way I can make sure that any patterns can be spotted early and any errors in the investigation will be made clear at an early stage. The thing I will be measuring is the amount of Sodium Hydroxide solution it takes to neutralise. The point at which I know that the solution is neutralised is the point when the phenolphthalein changes colour from colourless to pink. In order to find out this I will use a conical flask on top of a white tile. The colour will be deemed changed when I can no longer shake away the pink colour. This will indicate there are excess hydroxyl ions present in the solution. These ions are present in the Sodium Hydroxide solution and when they come into contact with the hydrogen ions in the sample solution they perform this reaction: H+(aq)+ OH- (aq) H2O(l) The indicator I am using is called phenolphthalein and is a organic indicator which I am using because it changes colour strongly at pH 8. I am using it in a 1% alcoholic solution. In acidic conditions the lactone ring in the phenolphthalein molecule is closed and the solution it is in is colourless. However when all the hydrogen ions are reacted with hydroxyl ions, the phenolphthalein is still colourless, as the hydroxyl ions are used up initially. However when there is one drop more of alkali in the solution the solution turns pink, as there is now an excess of hydroxyl ions. This causes the lactone ring to open, yielding the triphenylcarbinol structure. This structure is red/pink colour.1. 2. From the measurement of Sodium Hydroxide solution I obtain I can work out the final values of the acid, alcohol , water and ester present in the solution. Because the reaction below only has one mole of each reactant/product involved the values of final concentration are easily worked out. Alcohol + Carboxylic acid Ester + Water The final value of acid is equal to the titre obtained, as the titre is the amount that is needed to neutralise the acid in the solution. The amount of alcohol at the end, is the same as the amount of acid as one mole of acid is reacted with one mole of alcohol. If I use 20cm3 initially of acid and alkali, then as the amount of water and ester must be the same, as they have the same Stoichiometric values. The amount of ester and water is equal to the initial amount of alcohol minus the final amount of alcohol, as everything that has reacted must be ester. Therefore if I get a titre of 12 cm3 then the concentration is as follows: Acid: 12 cm3 From this I can work out the value of which I am investigating; the equilibrium constant. The equilibrium constant is worked out with this equation: Kc = Concentration products Concentration reactants In the case of this investigation the value of the equilibrium constant is found by this equation: Kc = [Ester] [Water] [Acid][Alcohol] In order to find the concentration I must first find the moles of each chemical in the reaction and to find that I need to know the moles of acid obtained. In order to find the moles I must multiply the volume of titre I obtain by the concentration of the sodium hydroxide I am using. As the reaction of hydrogen ions with hydroxyl ions is one to one the moles of alkali used is the same as the moles of acid used if the reaction mixture is now neutral. As the concentration of alkali in this case is 0.4moldm-3 to find the moles of acid used I multiply 0.012(this value is in decimetres cubed) by 0.4 this gives the answer 0.0048. As one mole of acid reacts with one mole of alcohol the moles of alcohol must also be the same. In order to find the moles of water and ester formed I must find the difference between the initial moles of acid and the final moles of acid. This is because any substance that is not acid or alcohol must be ester and water, so must account for the difference between i nitial acid value and final value. If my initial acid titre after time = 0 is 25 cm3 then the initial acid volume is: 0.025 x 0.4 = 0.01moles Now the difference between 0.01 and 0.0048 is equal to is 0.0052. Therefore the moles of ester and water formed is 0.0052moles. Now I can use these values to work out the equilibrium constant. As the equilibrium constant is found from concentrations and concentration is equal to moles/volume it can be written in this form: [moles/volume]1 [mole/volume]2 [moles/volume] 3[moles/volume]4 As the volume is the same for all of the chemicals it is common to all terms in the equation. I can then therefore cancel it, so that the equilibrium constant can be worked out using just the moles of each chemical. In the case of the example I provided the equilibrium constant would be with the moles of ester and water on top divided by the product of acid and alcohol. 0.0052 x 0.0052 0.0048 x 0.0048 = 1.17 3 S.F. As you can see the equilibrium constant in this case has no units. This is because the values of the units on the top of the equation cancel out those on the bottom to make no units. Investigation method 1. Obtain all equipment and chemicals needed for the experiment as listed previously. 2. In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 Therefore I will dissolve 40g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3. Fill the water bath up with water, and set water bath to the desired temperature. In the case of my preliminary work I will set it to 20oC and 60oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and ethanol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a boiling tube. Then add four drops of concentrated sulphuric acid into the boiling tubes. 6. Then measure out using the pipette 20cm3 of ethanoic acid into the boiling tube. 7. At the moment the reaction mixes take a 5cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place boiling tube bungs on all five of the boiling tubes containing the mixture and shake vigorously. Then place all in the test tube rack in the water bath. 8. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can get exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 9. Take a sample again of the reaction mixture after two days and repeat the titration. 10. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 11. Repeat these steps for all repeats and temperatures. Preliminary results The table below shows the results obtained for three repeats at three temperature intervals: 25oC, 45 oC, 60 oC Experiment (after 3 days) Temperature (Celsius) Start cm3 End cm3 Titre Repeat 1 25 0 9.3 9.3 Repeat 2 25 9.3 19.0 9.7 Repeat 3 25 19 28.8 9.8 Average 9.6 Repeat 1 45 0 10.8 10.8 Repeat 2 45 10.8 20.9 10.1 Repeat 3 45 20.9 32.0 11.1 Average 10.7 Repeat 1 60 0 12 12 Repeat 2 60 12 22 11 Repeat 3 60 22 34.3 12.3 Average 11.8 From these results I can see that there is a general upward trend in the titre as temperature. This does not seem to be highly correlated, but is still a relationship that can be useful in making my prediction. However I feel there is room for improvement which I will explain more below. Modifications to Method From my preliminary results and observations I noticed some flaws that could affect my investigation if they are done in the real experiments. They are listed below with ways to change them and make the experiment more accurate. * One of the main flaws with my experiment is that when using a bunged boiling tube, I had to shake the mixture to homogenise it. The problem this caused is that after a while the reaction mixture turned an orange tint. This was because the H+ in the ethanoic acid reacted with the rubber bung causing it to mix with the reaction. The problem this caused was that some of the acid was used up in reacting with the rubber therefore the correct amount was not available to react with the alcohol. The way I have chosen to overcome this is by changing the equipment holding the reaction. I will use glass 30cm3 sample tubes with screw-in plastic tops. This will make sure that all of the acid is available to react with the alcohol in the reaction. * The second problem I found is that the titres I found were rather small. This means that the alkali solution I was using in the titration was too strong. This is because it had too many hydroxyl ions in a given volume, and therefore a small volume neutralised the acid. A way to combat this is to change the concentration of NaOH in the alkali solution. I will use 0.4 moldm-3. The benefit that this has is too make the titres larger and therefore easier to spot trends and patterns. * Another problem I found when using the water bath is that at 60oC, after 3days much of the water in the water bath had evaporated. This was a bad thing because it meant the water bath could not maintain the correct temperature for the reaction and it can damage the water bath. A modification I am making is to take out 60oC from my temperature range. I will then therefore make the temperature range 25oC to 50oC in 5oC intervals. This will make sure I have enough water to maintain the correct water bath temperature. A second addition I can make is to use polystyrene foam and place it on top of the water bath to stop the water evaporating at the higher temperatures. This is because the foam will absorb some of the water and make a layer on top of the water bath to stop evaporation. * Another thing I found when looking at my chemicals was that the ethanol I used was 5% methanol. This could damage the experiment as the differing chain length could give rise to a different value for the Equilibrium constant than what I should actually get. A way to overcome this is too use an alcohol in a more pure form. Therefore I shall use propan-1-ol. This is bottled in near pure form so I believe it is more suitable for my investigation. Therefor the reaction will now be: Propan-1-ol + Ethanoic Acid Water + Propyl Ethanoate CH3CH2CH2OH + CH3COOH H2O + CH3CH2CH2COOCH3 * The last modification I will make to my investigation will be to change the amount of sample I take. I will take 1cm3 instead of 5cm3. This is because I will take 3 samples of each reaction, and I think a smaller sample will produce a titre that is under 50cm3. Final Method 1.Obtain all equipment and chemicals needed for the experiment as listed previously. 2.In order for me to start the investigation I must make up a solution of 1 moldm-3 Sodium Hydroxide solution. This is for the titrations I will do later on in the investigation. I will do this by dissolving one mole of Caustic soda crystals into one litre of distilled water. One mole of Sodium Hydroxide: Na = 23 O = 16 H = 1 Total = 40 x 0.4 moles = 16g Therefore I will dissolve 16g of sodium Hydroxide in one litre of distilled water. I will weigh the crystals on a balance and stir the solution vigorously. As the enthalpy of solvation of NaOH is quite high, initially the solution will get quite hot, therefore I must exercise care when handling it. I will then put a bung on the one litre volumetric flask for use later in the experiment. 3.Fill the water bath up with water and set water bath to the desired temperature. In this case I will set it to 25oC, 30oC, 35oC, 40oC, 45oC and 50oC. I will place a thermometer in the bath to make sure the correct temperature is reached. 4. I will then decant 200cm3 of glacial acetic acid and propan-1-ol into conical flasks, so that I do not contaminate the stock bottles. I will make sure that all equipment is thoroughly cleaned with distilled water and the substance that it will be containing. 5. I will use a filling pipette with a 20cm3 filling tube to put ethanol into a sample tube. Then add four drops of concentrated sulphuric acid into the sample tubes. Then measure out using the pipette 20cm3 of ethanoic acid into the sample tube 6. At the moment the reaction mixes take a 1cm3 sample of the reaction mixture to be titrated immediately to see how much acid there is initially. Then place sample tube lids on all five of the tubes containing the mixture and shake vigorously. Then place all directly into the water bath to make sure the sample tube is almost fully submerged. 7. The titration involves putting 50cm3 of NaOH solution into a burette initially. I then place the sample of reaction in a conical flask upon a white tile. I then add 6 drops of phenolphthalein into the sample. I then take a measurement of the starting value of the burette, and turn the tap so that the NaOH mixes with the sample. I will allow the NaOH to run slowly so I can discover exactly when the colour changes. I will then measure the final titre and find the amount of solution used in neutralising the acid. 8. Take a sample again of the reaction mixture after two days and repeat the titration. 9. Take a third sample of the reaction mixture after three days and repeat the titration. If the titre of the third is the same as the one done after two days the reaction has reached equilibrium and the third titre can be used as data for that temperature. 10. Repeat these steps for all repeats and temperatures The esterification reaction that I am performing is a rather slow reaction, therefore an acid catalyst is needed to make the reaction faster. Even with this catalyst the reaction still takes at least three days to reach equilibrium. Equilibrium is reached when the rate of the reaction going forwards is the same as the rate of reaction going backwards. All reactions are in a sense an equilibrium reaction, however some reactions have a small value for enthalpy change of reaction, so the reaction is possible to go in either direction. Another form of this reaction exists where the ester reacts with hydroxyl ions to form an alcohol and the carboxylic acids conjugate base. This reaction goes to completion however the ion produced is not easily measured so this reaction is preferred. Another way to get the reaction to reach equilibrium faster would be to reflux the mixture. However I can not do this as when the mixture cools down the equilibrium would shift again making the heating pointle ss. In the first step, the ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. The proton becomes attached to one of the lone pairs on the oxygen that is double-bonded to the carbon. N.B. curly arrows in some diagrams represent the movement of one or a pair of electrons.4. It is quite misleading to show the positive charge just on the oxygen, because is actual fact the charge is delocalised over the entire double bond with the carbon. Therefore it is more accurate to show the molecules like this: These two structures are not complete representations of the ion as it can occurs in both ways. In this case they are resonant hybrids as both can be used to contribute to the actual structure. The positive charge of the ion attracts the lone pair of electrons on the ethanol and the ethanol gives up its lone pair of electrons to bond to the ion. From this ion a proton is removed from the bottom oxygen and placed onto the next oxygen up. This is not a direct transfer, first the proton is taken in by an unreacted ethanol on its lone pair of electrons, and then it acts as an intermediate to transfer the proton onto the middle oxygen. The overall proton transfer is shown below. Now a water molecule is lost from the ion From this ion the positive charge is now still spread over the carbon. The last hydrogen on the ion is removed by the hydrogensulphate ion that was created at the beginning from the sulphuric acid. Therefore the ester is made along with water and the sulphuric acid is not used up in the reaction, so is the catalyst. The hydrolysis of the ester is the reverse reaction, and will be sped up by the same amount with the use of sulphuric acid. Therefore the catalyst will not actually change the equilibrium constant, it will just help the reaction get to equilibrium quicker. The actual catalyst in this case is the hydroxonium ion (H3O+). The hydrolysis is started by the lone pair of electrons on the oxygen in the carbon-oxygen double bond accepts a proton from the hydroxonium ion. Once again the positive charge is delocalised throughout the double bond and can by draw in a number of resonance hybrids. The lone pair of electrons on the water molecule then attacks the positive charge on this carbon Now a proton on the bottom oxygen is transferred to another lone pair on a water molecule and is eventually placed on the middle oxygen. Now one of the products of the reaction, ethanol is removed from the ion. With the charge spread across the carbon a proton is removed from the ion and placed on a water molecule, restoring the hydroxonium ion, and producing ethanoic acid. In order to find out what way the equilibrium shifts when the temperature changes it is useful to consider Le Chatliers principle. This states that Whatever constraint is placed on a system, the system will move to oppose the change. This change can be adding more of one of the reactants, or changing the temperature or pressure of the system. Initially as there is no ester or water the system will move to produce more ester and water. In the case of temperature, if the temperature increases the system will move to remove the heat. This means that the system will make the two products that cause heat to be removed from the system. The way of the reaction that takes in heat is the endothermic root. To ascertain which way is endothermic I must draw an enthalpy cycle. An enthalpy cycle can be used to find out what the enthalpy change of reaction is by knowing what the enthalpy change of formation of all the reactants and products are. Hesss Law states that the enthalpy change of a react ion will be the same regardless of the intermediate steps taken. The enthalpy change of formation is the change in enthalpy when one mole of a molecule is formed from its component elements in their standard states and under standard conditions i.e. room temperature 298K and 1atm pressure. The enthalpy change of formation of each chemical is shown below: Water -285.8kJmol-1 Ethanoic Acid -484.5 kJmol-1 Propan-1-ol -302.7 kJmol-1 Propyl Ethanoate -502.7 kJmol-1 5. Notice that all these values are negative this means that energy is given out into the system as energy is given out when bonds are made This is the enthalpy cycle for my reaction: ?Hr? CH3COOH + CH3CH2CH2OH H2O + CH3COOCH2CH2CH3 (-484.5 +-302.7) (-285.8 + -502.7) 5C(s) + 3/2O2(g) + 6H2(g) In order to find the enthalpy change of reaction I must follow the arrows to get to the ester and water. Firstly as I start by going down the first arrow I must take the negative value of it. -484.5 302.7 = -787.2 This value becomes +787.2 as I am going against this arrow. I then add the sum of the esters and waters enthalpy changes of formation to get the enthalpy change of reaction. -285.8 + -502.7 = -788.5 787.2 788.5 = -1.3 kJmol-1 From this I can show that the reaction is exothermic going in the direction of making ester and water. As this value is small it also fits in that this reaction is an equilibrium reaction. From this I can now make a prediction. As an increase in temperature forces the equilibrium to shift to remove heat energy, and the reaction root that is endothermic (removing heat) is making acid and alcohol from ester and water, I predict that as I increase temperature the more acid and alcohol is produced. In terms of the value of Kc I predict that this value decreases as I increase temperature. This is because if more acid is produced it means less ester and water is produced; and as the value for Kc is found by the product concentration divided by reactant concentration, more reactant means lower Kc. 1. Vogels textbook of quantitative chemical analysis 2.chem.ox.ac.uk/vrchemistry/FilmStudio/phenolphthalein/Symbols/equilibrium.jpg 3.ilo.org/public/english/protection/safework/cis/products/icsc/dtasht/_icsc05/icsc0553.htm 4. chemguide.co.uk/physical/catalysis/hydrolyse.html 5. ucdsb.on.ca/tiss/stretton/chem2/data09f.html Below is a list of the main areas of chemistry in my investigation and where they occur in the course. Area Of Investigation Area of Chemistry AS/A2 Enthalpy Cycles Developing Fuels AS Hydrolysis of Esters Whats in a Medicine? AS Acid-Base titrations Whats in a Medicine? AS Equilibrium Constant Engineering Proteins A2 Results Tables The table below shows the readings at each temperature, proving that the reaction has reached equilibrium. The readings for each of the temperatures are taken after one day two three and four. If the third and forth reading are the same then the rate of reaction going forward is the same as the rate of reaction going backwards so there is no net change in the amounts of reactants/products. Temperature (Celsius) Start reading (cm3) End reading (cm3) Titre(cm3) Moles of acid Initial Reading 25 0 22.5 22.5 0.00900 Second Reading 25 0 15.7 15.7 0.00628 Third Reading 25 0 9.9 9.9 0.00396 Final Reading 25 0 9.9 9.9 0.00396 Temperature (OC) Start reading (cm3) End reading (cm3) Titre (cm3) Moles of acid Initial reading 30 0 22.5 22.5 0.00900 Second reading 30 22.5 37.7 15.2 0.00608 Third reading 30 37.7 47.8 10.1 0.00404 Final Reading 30 0 10.1 10.1 0.00404 Initial reading 35 0 22.5 22.5 0.00900 Second reading 35 22.6 37.9 15.3 0.00612 Third reading 35 0 12.3 12.3 0.00412 Forth Reading 35 0 10.3 10.3 0.00412 Final Reading 35 0 10.3 10.3 0.00412 Initial reading 40 0 22.4 22.5 0.00900 Second reading 40 22.4 37.5 15.1 0.00604 Third reading 40 37.5 48.0 10.5 0.00420 Final Reading 40 0 10.5 10.5 0.00420 Initial reading 45 0 22.5 22.5 0.00900 Second reading 45 22.3 37.2 14.9 0.00596 Third reading 45 37.2 47.9 10.7 0.00428 Final Reading 45 0 10.7 10.7 0.00428 Initial reading 50 0 22.5 22.5 0.00900 Second reading 50 22.5 37.5 15.0 0.00600 Third reading 50 37.5 48.4 10.9 0.00436 Final Reading 50 0 10.9 10.9 0.00436 N.B the Results highlighted in red shows that the reaction did not reach equilibrium as fast as all of the others. Although this did not affect the results obtained, it will still be accounted for later on in the evaluation section. Below shows graphs for all of the temperatures showing time from the beginning of reaction to the reaching of equilibrium. The table of results shows all of the repeats I did for all of the temperature ranges. It also shows that moles of acid in the sample. Temperature (degrees Celsius) Beginning measurement End measurement Titre Amount of Acid Moles of Acid 25 0 9.8 9.8 9.8 0.00392 25 9.8 19.7 9.9 9.9 0.00396 25 19.7 29.5 9.8 9.8 0.00392 25 29.5 39.4 9.9 9.9 0.00396 25 0 9.8 9.9 9.9 0.00396 Average titre 9.9 9.9 0.00396 30 0 10.1 10.1 10.1 0.00404 30 10.1 20.2 10.1 10.1 0.00404 30 20.2 30.3 10.1 10.1 0.00404 30 30.3 40.4 10.1 10.1 0.00404 30 0 10.1 10.1 10.1 0.00404 Average titre 10.1 10.1 0.00404 35 0 10.3 10.3 10.3 0.00412 35 10.3 20.6 10.3 10.3 0.00412 35 20.6 30.9 10.3 10.3 0.00412 35 30.9 41.2 10.3 10.3 0.00412 35 0 10.3 10.3 10.3 0.00412 Average Titre 10.3 10.3 0.00412 40 0 10.4 10.4 10.4 0.00416 40 10.4 10.9 10.5 10.5 0.0042 40 20.9 31.5 10.6 10.6 0.00424 40 31.5 42 10.5 10.5 0.0042 40 0 10.5 10.5 10.5 0.0042 Average Titre 10.5 10.5 0.0042 45 0 10.7 10.7 10.7 0.00428 45 10.7 21.4 10.7 10.7 0.00428 45 21.4 32.1 10.7 10.7 0.00428 45 32.1 42.8 10.7 10.7 0.00428 45 0 10.7 10.7 10.7 0.00428 Average titre 10.7 10.7 0.00428 50 0 7.6 7.6 7.6 0.00304 50 7.6 18.6 11 11 0.0044 50 18.6 29.5 10.9 10.9 0.00436 50 29.5 40.4 10.9 10.9 0.00436 50 0 10.9 10.9 10.9 0.00436 N.b.Anaomalous result ignored Average titre 10.9 10.9 0.00436 Analysis of results Now for each repeat for each temperature I will work out the moles of each in the reaction, and hence the equilibrium constant for each repeat. 25oC- Repeat 1 Titre initial = 22.5cm Initial moles = volume x concentration of alkali Initial Moles of acid = 0.00900 Volume of Ethanoic acid in 1cm3 at the end = 9.8 Final moles = volume of acid x concentration of alkali Final moles of acid = 0.00392 As one mole of acid reacts with one mole of alcohol then the amount of alcohol must be the same. Final moles of Propan-1-ol = 0.00392 Final moles of Propyl Ethanoate equals difference between initial acid moles and final acid moles. 0.00900 0.00392 = 0.00508 As one mole of Ester makes one mole of water then the moles of water must be exactly the same as that of ester. From these values the value of the equilibrium constant can be worked out: Kc = 0.00508 x 0.00508 0.00392 x 0.00392 = 1.68 3 S.F. Applying this same method to all my other data I will now produce a table showing each repeat and the equilibrium constant of that experiment. I have recorded this table because I feel it makes it easier for me to draw a graph of the equilibrium constant against the temperature. Temperature (degrees Celsius) Initial Moles of Acid Moles of Acid Equilibrium Constant 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00392 1.68 25 0.00900 0.00396 1.62 25 0.00900 0.00396 1.62 Average 0.00900 0.00396 1.62 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 30 0.00900 0.00404 1.51 Average 0.00900 0.00404 1.51 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 35 0.00900 0.00412 1.40 Average 0.00900 0.00412 1.40 40 0.00900 0.00416 1.35 40 0.00900 0.0042 1.31 40 0.00900 0.00424 1.26 40 0.00900 0.0042 1.31 40 0.00900 0.0042 1.31 Average 0.00900 0.0042 1.31 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 45 0.00900 0.00428 1.22 Average 0.00900 0.00428 1.22 50 0.00900 0.00304 3.84 50 0.00900 0.0044 1.09 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 50 0.00900 0.00436 1.13 Average 0.00900 0.00436 1.13 Anomalous Result is ignored From this graph I can deduce that as the Temperature increased the value of Kc decreases. As I said this in my prediction, it goes some way to proving that my prediction was at least to some extent correct. Another conclusion I can draw from my graphs is that as the temperature increased the equilibrium constant changed exponentially. This is because on the graph it shows a slight curve. The reason for this is that as the temperature rises the reaction favours the production of the reactants. Therefore as the reactants increase the products decrease so the equilibrium constant goes down. There is a constant fractional change in the value of Kc as the temperature increases. The reason that the reaction favours the production of the reactants, ethanoic acid and propan-1-ol is because this direction of reaction takes in energy, i.e. its energy level is higher than the products that it makes. As the temperature increases the system tries to remove the extra heat energy being put in. The way it does this is by making the reaction that removes heat energy go faster. Therefore the reaction constant goes down..