Combustion Of Alcohols Essay, Research Paper
Comparing Energy Release from the Combustion of Alcohols I am going to investigate the energy released by burning alcohol – called the “enthalpy of combustion”. I will be trying to find out how the number of Carbon atoms the alcohol contains effects the enthalpy change that occurs during combustion. The Alcohols are a series of related organic molecules where each member has the -OH (hydroxyl) group in the molecule.
The general formula of an alcohol is: Cn H 2n+1 OH, where n is the number of Carbon atoms.
The similarities in molecular structure make alcohols have physically and chemically similar properties. The table below shows a general increase in melting points and boiling points as the number of carbon atoms increases Table 1 Alcohols formulae and properties.
NameMolecular FormulaStructural FormulaMelting PointBoiling Point
MethanolCH3OH-9465
EthanolC2H5OH-11778
Propanol
(Propan-1-ol)C3H7OH-12697
Butanol
(Butan-1-ol)C4H9OH-90117
Pentanol
(Pentan-1-ol)C5H11OH-79137
(The names like propan-1-ol refer to the position of the -OH group on the carbon chain, the OH groups above are on the first Carbon atom, the “1″ position)
This table shows that I will need to investigate a series of different alcohols in my investigation. Ideally I would need at least 5 alcohols for a good range of results for comparison. The complete combustion of an alcohol involves reaction with Oxygen to produce Carbon Dioxide and Water. The general formula for this reaction is:
Cn H 2n+1 OH + (n+n/2)O2 ? nH2O + nCO2
Balanced equations for each of the available alcohols that I will use are as follows:
Methanol2CH3OH + 3O2 ? 2CO2 + 4H2O
EthanolC2H5OH + 3O2 ?2CO2 + 3H2O
Propanol2C3H7OH + 9O2 ? 6CO2 + 8H2O
ButanolC4H9OH + 6O2 ? 4CO2 + 5H2O
The combustion process involves the breaking and remaking of chemical bonds between the reactant elements. Breaking bonds requires energy input, making bonds releases energy.
In an exothermic reaction the products are at a lower energy level than the reactants, the difference is given off as heat energy. The spark or burning splint used to light the flame usually supplies the activation energy. The combustion reaction is then self-supporting as some of the energy released is used to supply the activation energy of the following reactants. In an endothermic reaction the products are at a higher energy level than the reactants, the difference is the amount of energy taken in from the surroundings. These reactions must be heated. E.g. with a Bunsen burner. I will need to find out the energy values for the alcohol and oxygen reactants, then compare these with the energy values for the carbon dioxide and water products.
Notes for following table:
?H is the energy change, (at standard temperature (298K) and pressure (1 atmosphere))
A mole is 6 x 1023 atoms of the chemical; the mass of one mole is equal to the relative atomic mass in grams. E.g. For ethanol C2H5OH
Relative atomic mass = 24 (two Carbon) + 16 (one oxygen) + 5 (five Hydrogen) = 45.
So one mole is 45g of ethanol Table 2 Calculating predicted results
Molecular formula:
C2H5OH+ 3O2? 2CO2+ 3H2O
Structural formula:
+ O=O
O=O
O=O? O=C=O
O=C=O+ H-O-H
H-O-H
H-O-H
Breaking bonds: kJ/moleMaking bonds: kJ/mole
Five C-H bonds
=5 x 413 Three O=O bonds
=3 x 498 kJ/moleFour O=C bonds
= 4 x 805 Six O-H bonds
=6 x 464
One C-O bond = 358
One O-H bond = 464
One C-C bond = 347
Total ?H = 3234Total ?H = 1494 Total ?H = 3220Total ?H = 2784
Left side of reaction total
= ?H 4728 kJ/mole energy absorbedRight side of reaction total
= ?H -6004 kJ/mole energy released
Difference = ?H -1276 kJ/mole The difference of -1276 kJ per mole of ethanol combusted is energy given out, this net release of energy is called an exothermic reaction: heat is released.
The 4728 kJ/mole required to break the ethanol bonds initially is the activation energy for this reaction, this energy must be put in to start the reaction. E.g. a spark or burning splint supplies this activation energy to start ethanol burning. A chemical reaction that needs an overall input of energy is called an endothermic reaction. I can collect and measure the energy liberated in the combustion reaction by using it to heat a measured sample of water. Knowing the temperature rise of the water, its volume and specific heat capacity of the water, I can calculate the energy released.
Equation:
?H = M S ?T
?H = Energy change, energy released ( J )
M = mass of water ( g )
S = Specific heat capacity of water ( 4.2 J/oC /g )
?T = Change in temperature of water ( oC ) So that I can compare my results with secondary sources and other pupils, the number of moles of fuel will be needed, this means I will have to measure the mass of fuel used. From this I can calculate the energy release per mole. (see results table for formulae)
Method I will assemble this apparatus:
? Measuring cylinder, to measure 50ml of water. Accuracy: to about 1 ml
? Thermometer, to measure temperature of water before and after combustion. Accuracy to about 1oC
? Stop clock, to measure 10 minutes of combustion time. Accuracy: to about 1 second.
? Copper calorimeter, to contain the heated water
? Spirit burner, 5 burners each containing a different alcohol.
Methanol
Ethanol
Propanol
Butanol
Pentanol
? Chemical balance to measure the mass of fuel before and after combustion. Accuracy to about 0.01g.
? Heat proof mat, clamp and stand Repeat each alcohol five times (but there was only time for three sets of each result in the lessons)
Repeat with five different alcohols. (But in the lessons there were only four alcohols available)
Variables to control:
1. Position of OH group in alcohol molecule. Propanol and Butanol are long enough carbon chains to form isomers (same chemical formula but different structural formula) the -OH group needs to be on the end Carbon atom in the alcohols tested. Secondary information indicates that the bond energies between two atoms will vary according to where in the molecule the pair of atoms is. For this reason propan-1-ol and butan-1-ol are used so all molecules used will have the -OH group on the end Carbon atom.
2. Purity of alcohol, the alcohol may be mixed with water, all the alcohols supplied are 99% pure.
3. The volume of water used each time will be 50 ml (= 50g mass). The calculation does take into account the mass of water, but for comparison it should be constant if possible.
4. The temperature of the water should be low to start and for convenience tap water is used straight from the tap. If the start temperature is too high the water might boil and would not show any further temperature increase even though energy is still released.
5. The same copper calorimeter will be used each time. The copper calorimeter has a high thermal conductivity and will improve the conduction of heat from the flame to the water.
6. Height of flame, this will depend on the length of exposed wick, this will be kept constant at 1 cm.
7. Height of calorimeter above the wick will be kept to 3cm in every case.
Pilot Experiment
I made a trial run using methanol to check if my chosen quantities and set up would work.
Results:
Mass of burner at start = 92.33g. Mass of burner finish = 86.96g. Mass of fuel burned = 5.37g
Start temp = 25 oC Finish temp = 95 oC, Temp rise = 70 oC
Following the pilot experiment I made some minor adjustments:
Height of can above wick down to 2cm to improve contact with flame and stir the water, checking for the maximum temperature as it continued to rise by a few degrees after the ten minutes due to conduction of heat through the can. I also found that my intention to do 5 sets of results for each alcohol was too much in the time. Also that four alcohols were available.
Safety
Risk Assessment
? Wear safety glasses.
? Tie back long hair, tuck in school ties, and button up cuffs.
? Avoid any naked flame near alcohol being stored or measured.
? Clean up any alcohol spills immediately.
? Take care not to handle hot apparatus, allow to cool before dismantling and cleaning.
Prediction
So that I can make a prediction for my investigation, I can calculate the theoretical energy release from each alcohol tested.
Table 3 Calculating predicted results
AlcoholBonds brokenOne mole of alcohol
?H input kJ/molebonds made?H output kJ/moleTotal ?H kJ/mole?H Difference
kJ/mole
Methanol CH3OHC-H x3
C-O
O-H
O=O x1.5413
358
464
498 ? ?H= 2808C=O x2
O-H x4805
464
? ?H=3466-658
Ethanol C2H5OHC-C
C-H x5
C-O
O-H
O=O x3347
413
358
464
498 ? ?H= 4728C=O x4
O-H x6805
464
? ?H=6004-1276618
Propanol C3H7OHC-C x2
C-H x7
C-O
O-H
O=O x4.5347
413
358
464
498 ? ?H= 6648C=O x6
O-H x8805
464
? ?H=8542-1894618
Butanol C4H9OHC-C x3
C-H x9
C-O
O-H
O=O x6347
413
358
464
498 ? ?H= 8568C=O x8
O-H x10805
464
? ?H=11080-2512618
The above bond enthalpies show an increase in overall energy release as the alcohols increase in size. This is shown in the graph below. The right hand column shows that these differences are constant as the molecule gets larger. Adding each extra H-C-H to the molecule gives a theoretical 618 kJ/mole increase. This gives a straight line graph. I will need to see if my results support this predicted constant increase in energy released.
My prediction is that as the alcohol molecule gets larger, by increasing the number of Carbon atoms, the energy released will be greater. Also that this increase in energy release will increase by a constant factor for each extra Carbon atom in the molecule.
Figure 5 Graph of predicted pattern of results
Results Table
Table 4 Original data and averages
AlcoholStart temp water (oC)End temp water (oC)Temp rise (oC)Average
(oC)Start mass fuel (g)End mass fuel (g)Fuel used (g)Average
(g)
Methanol2481577.336.241.09
Methanol24805657.37.266.300.961.01
Methanol2483597.056.070.98
Ethanol2496723.752.751.00
Ethanol24957170.33.642.720.921.06
Ethanol2492683.982.721.26
Propanol2495716.905.261.64
Propanol24947070.66.265.071.191.53
Propanol2495716.734.981.75
Butanol2496747.216.340.87
Butanol24997574.07.227.071.151.04
Butanol2497737.276.171.1
Analysis
Table 5 Spreadsheet formulae
FuelStart mass (Ave.) (g)End mass (Ave.) (g)Mass fuel used(g)Mass Water (g)Specific heat capacity of water (J/oC/g)start temp (Ave.) (oC)end temp (Ave.) (oC)temp difference (oC)?H Energy released
(J)relative molecular mass of fuelmoles fuel?H energy per mole fuel
(kJ/mol)
Methanol7.216.20=SUM(B2-C2)50.04.22481.3=SUM(H2-G2)=(E2*F2*I2)32=(D2/K2)=-(1/L2*J2)/1000
Ethanol3.792.73=SUM(B3-C3)50.04.22494.3=SUM(H3-G3)=(E3*F3*I3)46=(D3/K3)=-(1/L3*J3)/1000
Propanol6.635.10=SUM(B4-C4)50.04.22494.6=SUM(H4-G4)=(E4*F4*I4)60=(D4/K4)=-(1/L4*J4)/1000
Butanol7.236.19=SUM(B5-C5)50.04.22498.0=SUM(H5-G5)=(E5*F5*I5)74=(D5/K5)=-(1/L5*J5)/1000
Table 6 Spreadsheet analysis
FuelStart mass (Ave.) (g)End mass (Ave.) (g)Mass fuel used(g)Mass Water (g)Specific heat capacity of water (J/oC/g)start temp (Ave.) (oC)end temp (Ave.) (oC)temp difference (oC)?H Energy released
(J)relative molecular mass of fuelmoles fuel?H energy per mole fuel
(kJ/mol)
Methanol7.216.201.0150.04.22481.357.312033320.0316-381.24
Ethanol3.792.731.0650.04.22494.370.314763460.0230-640.66
Propanol6.635.101.5350.04.22494.670.614826600.0255-581.41
Butanol7.236.191.0450.04.22498.074.015540740.0141-1105.73 Table 7 Results of Energy release per mole
AlcoholRelative molecular massActual energy release
?H (kJ/mol)Predicted energy release
?H (kJ/mol)Difference (kJ/mol)
Methanol32-381.24-658.00276.76
Ethanol46-640.66-1276.00635.34
Propanol60-581.41-1894.001312.59
Butanol74-1105.73-2512.001406.27
Table 8 Results analysis of difference in energy release in alcohol series
AlcoholRelative molecular massActual energy release ?H (kJ/mol)Measured Difference (kJ/mol)Predicted Difference (kJ/mol)
Methanol32-381.24
Ethanol46-640.66259.41618.00
Propanol60-581.41-59.25618.00
Butanol74-1105.73524.32618.00
Analysis conclusions Pattern of Results:
The energy released from a mole of alcohol increases with increased molecular weight. The graph, figure 6, page 9, clearly shows a rising line for energy release with increased molecular weight. The number and range of results is enough to support my conclusions. The shape of my graph fig. 6 shows the same pattern as my predicted graph fig 5, page 6.
The actual results are drawn with the straight line of best fit. The readings for Propanol are anomalous compared to the other readings as these results are below the expected value and are below the observed trend. The increase in energy release between each alcohol was supposed to be a constant value. The graph in Figure 5 shows that this prediction did not show in my results. The value for Propanol again is anomalous as the actual reading shows the opposite trend to what I predicted. The reason that my readings for energy release increased with increased molecular weight is because as the number of bonds broken then made increases (with increased molecular weight of the alcohol) then there is proportionately more exothermic energy released. The higher molecular weight alcohol molecules have more bonds in them so produce more heat. Evaluation of reliability of data The anomalous differences between observed and predicted results can be due to the following factors:
? The energy used in making the product gases expand is ignored
? The energy output in light energy is not measured.
? The bond energy depends on the other elements and bonds in the molecule. E.g. C-O bond in methanol ?H= 336 kJ/mol but the C-O bond has an average value of ?H = 358 kJ/mol.
? The calculated bond energies only account for bond breaking and making within molecules. They do not account for attractions broken and made between molecules. The calculations assume all substances are gas and the vaporisation energy to convert liquid methanol (l) to gas methanol (g) is ignored. The anomalous readings for Propanol could be due to experimental errors in measurements and readings. Two of the readings for mass used are rather high compared to other results. This suggests an error in measuring and / or recording mass. Also an unfortunate draught from a door or window at that moment could have blown my flame to one side and lost much of the heat from the calorimeter. My evaluation will give ways of avoiding these errors in further investigations. If I now allow for errors in measuring heat output due to limitations of the apparatus and technique, I can see why my predictions are not strictly accurate in numerical terms.
My prediction was based on the fact that as more bonds are broken then remade with increasing alcohol molecule size then the more energy is released in the exothermic reaction.
My overall prediction for this pattern of results is still valid so my results do support my prediction.
The differences between alcohols in my investigation results do not show the predicted constant value.
My Chemistry text book (Hunt & Sykes) gives data book values for “Heat of Combustion”:
Table 9
AlcoholRelative molecular massActual energy release
?H (kJ/mol)Predicted energy release
From bond energies
?H (kJ/mol)Heat of Combustion
(kJ/mol)
Methanol32-381.24-658.00-726
Ethanol46-640.66-1276.00-1366
Propanol60-581.41-1894.00-2017
Butanol74-1105.73-2512.00-2675 This shows there are different ways of calculating energy release depending on the ideas being used. My observed values still do not exactly match the text book data. However, the pattern remains as predicted.
The calculated bond energies only account for bond breaking and making within molecules. They do not account for attractions broken and made between molecules. The calculations assume all substances are gas and the vaporisation energy to convert liquid methanol (l) to gas methanol (g) is ignored. The pattern of results is the same and my conclusions would be the same, which ever way is used. This investigation could be extended by increasing the range of alcohols, to include Pentanol, hexanol, heptanol, octanol and decanol. The predicted and observed trend of increasing exothermic energy should continue with this series. Also the effect of bond position can be investigated by using a range of isomers, e.g. butan-1-ol, butan-2-ol and butan-3-ol. These would show the differences in exothermic energy depending on the structure of the isomer.
Evaluation of method
Sources of inaccuracy and error:
My results were consistently lower than predicted, the reasons for this could be as follows: 1. Incomplete combustion of any of the alcohols would result in the production of some Carbon Monoxide gas, ( CO ) and Carbon soot ( C ) so releasing less exothermic energy. A good supply of air is necessary for complete combustion, so there must not be any constriction of airflow around the apparatus. This is probably not a significant error in my investigation. 2. The mass and specific heat capacity of the copper calorimeter was not taken into account. The calorimeter will have absorbed some of the exothermic energy but not have been measured; this gives a lower reading than expected for energy release. I would need to find out the mass and specific heat capacity of the calorimeter in a repeated investigation. This error will give a lower reading than expected. 3. A lot of heat would not have been absorbed by the calorimeter, as convection currents will have taken heated air up and round the apparatus without it being measured. This error will give a lower reading than expected. An improved investigation will include draught shielding to stop warm air escaping. I could solve these problems by using a “bomb calorimeter” which retains all substances inside a water jacket; to absorb all the energy produced. 4. Measurements 1: Rather than a measuring cylinder, I would choose to use a more accurate pipette, like we used for titration investigations. This will improve accuracy from 1 ml to 0.1 ml 5. Measurements 2: I would choose to use a long mercury in glass thermometer rather than the short spirit thermometer I used, but these have a safety risk in using a hazardous substance like Mercury. This will improve accuracy to 0.1oC. Also digital data logging with a suitable temperature sensor could be used.