Comparing the enthalpy changes of combustion of different alcohols

Oliver White - Comparing the enthalpy changes of combustion of different alcohols.

Theory

When chemical reactions occur, bonds between atoms are broken and new bonds are made. The amount of energy needed to break a particular bond is called its bond enthalpy(H). We cannot measure the enthalpy of a substance but we can measure the change in enthalpy(∆H) when a reaction occurs.

∆H = Hproducts – Hreactants

Hproducts is the final enthalpy of the system, measured in joules. In a chemical reaction, Hproducts is the enthalpy of the products.

Hreactants is the initial enthalpy of the system, measured in joules. In a chemical reaction, Hreactants is the enthalpy of the reactants.

Enthalpy change in a chemical reaction gives the gives the quantity of energy transferred to or from the surroundings, when the reaction is carried out openly. In an exothermic reaction, the enthalpy of the reacting system decreases. ∆H is negative. In an endothermic reaction, the enthalpy of the reacting systems increases. ∆H is positive.

When chemists talk about enthalpy changes they often refer to the system meaning the reactants and the products of the reaction they are interested in. The system can lose or gain enthalpy depending on the surroundings meaning the rest of the world: the test tube, the air etc.

I will be investigating the enthalpy change of combustion of different alcohols. The standard enthalpy change of combustion (∆HcӨ) is the enthalpy change, when one mole of a substance burns completely in oxygen under standard conditions. ∆HcӨ is always negative.

Fig 1 Enthalpy change diagram showing the combustion of methanol.

Plan

I am investigating the enthalpy change of combustion of several alcohols so that I can find how and why enthalpy change is affected by the molecular structure of the alcohol. The alcohols I will be using are methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol. These will give me clear results as they range from short-chain hydrocarbons to long-chain hydrocarbons, so patterns should be fairly clear in the results and my conclusion.

In chemistry alcohol refers to the complete homologous series, of alcohols of which ethanol is just one. Alcohols consist of an hydroxyl group covalently bonded to an alkyl or cycloalkyl chain. The hydroxyl, is the OH group, is the functional group. Alcohols can be classified as primary, secondary or tertiary depending on the number of carbon atoms attached to the carbon atom with the hydroxyl group. The alcohols I am using are all primary, meaning that they have one carbon atom attached to the carbon with the hydroxyl group.

In chemistry, a homologous series is a series of organic compounds with a similar general formula, possessing similar chemical properties due to the presence of the same functional group, and shows a gradation in physical properties as a result of increase in molecular size and mass.
CnH2n + 1 OH

I will be finding my results by using water to measure the amount of heat used. The experiment isn’t too efficient as energy is absorbed by the water, the air and the copper calorimeter. I will be using a fuel to heat up water as water absorbs a specific amount of energy (4.2J per g). I will then work out the number of moles of fuel burnt. I will then be able to work out the difference between mass of fuel before heating the water, and mass of fuel afterwards and from this the energy given out by one mole of the substance.

Energy transferred = mc∆T
m = mass of water (g)
c = specific heat capacity of water (J g-1K-1)
∆T = Change in temperature (K)

Variables

My independent variables are

- The type of alcohol (methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol)

My dependant variables are

- The specific heating capacity of water
- Mass of water in grams
- Change in temperature of water

My controllable variables are

- The change in temperature of water (15oC) If this wasn’t kept the same then different amounts of energy would be being used up
- The mass of water (100g). If this were changed then all the results would be wrong, because it would take different amounts of energy to heat different volumes of water.
- The distance from the bottom of the calorimeter to the top of the wick. Otherwise different amounts of energy could escape ...

This is a preview of the whole essay

My dependant variables are

- The specific heating capacity of water
- Mass of water in grams
- Change in temperature of water

My controllable variables are

Method

I will use the following equipment to carry out my experiement:
- Copper Calorimeter
- 0-50oC thermometer
- 100cm3 meauring cylinder
- Spirit burners containing:
methanol
ethanol
propan-1-ol
butan-1-ol
pentan-1-ol
- A balance
- Bunsen burner
- Draught shielding (aluminium tubing)
- Stand
- Clamp

1 Put 100 cm3 of cold water into a copper calorimeter and measure its temperature.

2 Weigh the spirit burner (with cap on).

3 Support the calorimeter over the spirit burner containing the liquid fuel you are going to test. Arrange the draught exclusion system and measure the distance from the wick to the calorimeter.

4 Light the wick.

5 Use the thermometer to stir the water all the time as it is being heated. Keep heating until the water has increased by 15oC.

6 Extinguish the burner and keep stirring the water and record the highest temperature reached.

7 Weigh the burner (with cap on) to se what mass of fuel has been burnt.

Use this experiment for the rest of your alcohols. Put the recorded information into a suitable table and then work out the enthalpy change of combustion of the alcohols used.
Energy transferred = mc∆T (mass of water x 4.2((specific heat capacity of water)) x temperature rise)

To work out the enthalpy change of combustion of one mole you need to work out the heat given out by the fuel.

Mass of water x 4.2 x temperature rise = X

Work out the number of moles of fuel that has been burnt.

Mass (X) = Y
Mr

Work out the energy given out by 1 mole of the substance.

Heat given out by fuel (X)
Number of moles (Y)

This gives you the answer in J/mol-1, so to get the answer in KJ/mol-1, divide by 1000.

Risk Assessment

I will be using spirit burners so I must be careful that I do not spill fuel anywhere, because it is highly flammable and is a fire hazard. I will wear safety goggles so that I do not get anything in my eyes and I will also wear a lab coat. I must be careful with the calorimeter after it has been heated as it will be very hot. The draught shield could also be very hot.

References

Fig 1 http://www.teachmetuition.co.uk/Chemistry/Energetics/exothermicdiagram.gif

Web sites used:

http://4college.co.uk/as/df/alcohols.php

http://4college.co.uk/as/df/energy.php

http://4college.co.uk/as/df/enthalpy.php

Text Books Used:

Collins advanced modular sciences - Chemistry AS (pages 84-85, 216-221)

Heinemann for Salters (OCR) - Revise AS Chemistry (pages 20, 66)

Heinemann Salters Advanced chemistry - Chemical Ideas (pages 57-58, 305- 307, 310)

Table 1: Results from fuels burnt

This is a table of results from the fuels I burnt. In my plan I stated that I would use Pentan-1-ol but it wasn’t available. I don’t think this will greatly affect my experiment. Using the information gathered I can calculate the amount of heat absorbed by the water for each alcohol, meaning I will then be able to calculate the enthalpy change of combustion of each alcohol. The formula below will enable me to calculate the heat absorbed by the water.

Energy = Mass x 4.2 x Temp. Change

Mass is equal to the amount of water used in the experiment (100g). 4.2j is the specific heat capacity of water meaning it takes 4.2j of energy to raise 1 gram of water by 1 degree Celsius. I will show the calculating of one alcohol, methanol and put the rest into a table so they can be compared.

Energy = Mass x 4.2 x Temp. Change
Energy = 100 x 4.2 x 17
Energy = 7140 j

This tells me that methanol released 7140 joules of energy when burnt. This same calculation was been applied to all alcohols, except for ethanol and butan-1-ol where 16 oC was the temperature change, giving 6720 joules of energy being released for both alcohols.

These results don’t tell me much about the enthalpy change of combustion of the alcohols, only the energy released. To finish my calculations, I will be using the figures of energy released and will work out the number of moles of fuel that has been burnt by working out the difference between the mass of fuel before the water heated and the mass of fuel afterwards. Then I will work out the energy given out by 1 mole of the substance, leaving me and answer in J/mol-1, meaning I will have to finish by dividing the final calculation by 1000 to get an answer KJ.

Mass of water x 4.2 x temperature rise = X

2) Mass = Y
Mr

3) Heat given out by fuel (X)
Number of moles (Y)

4) Final Answer
1000

I will use methanol again as an example to show my working out and then put all my final readings into a table, to be able to compare.

1) 100 x 4.2 x 17 = 7140 j

2) 0.65
32 [Mr = CH3OH = C bonds (1 x 12) + O bonds (1 x 16) + H bonds (4 x 1)]
= 0.0203 (4 d.p)

3) 7140
0.0203
=351724.1379 J/mol-1

4) 351724.1379
1000
= - 351.72 KJ/mol-1 (2 d.p) [answer is negative because it is an exothermic reaction]

Now I can say that the enthalpy change of combustion of one mole of ethanol is equal to 351.72 KJ. Below is a table for the results of the rest of the calculations of the other alcohols.

Table 2: Final Results

My results show fairly clearly that the enthalpies of combustion increase. I think this is because of the atoms and bonds present in each alcohol. As you go down the homologous series of alcohols, another carbon atom is added and two hydrogen atoms to each alcohol, meaning two more H-C bonds and one C-C bond to break each time. So that when you get down to Butan-1-ol, there will be 3 C-C bonds to break and 9 H-H, requiring more energy to break up these bonds. This in turn also means that more bonds have to be made as there are more atoms.

The main reason that Butan-1-ol has a higher enthalpy of combustion than methanol, because Butan-1-ol has more atoms present in one mole of substance, meaning more bonds to break and re-form. It also has a higher molecular mass. As you go down the homologous series of alcohols, the molecular mass of the alcohols increases, meaning that the higher molecular mass, the higher the enthalpy change of combustion. Methanol has a molecular mass of just 32g, whereas butan-1-ol has a molecular mass of 74g, which is more than double, and the ∆Hc/KJ mol-1 result for butan-1-ol is more than 4 times as much.

You can also say that energy released by the fuels can be the same, but alcohols like butan-1-o-l, don’t need to burn as much fuel as methanol does to achieve this. In my calculations, the energy released by the fuels are quite close, there is only a difference of 420j.

When alcohols are burnt, they produce carbon dioxide and water. Below are the chemical equations for the enthalpy change of combustion for methanol and Butan-1-ol

CH3OH + 1.5O2 → CO2 + 2H2O
C4H9OH+ 6O2 → 4CO2 + 5H2O

Here you can clearly see that one mole of butan-1-ol produces a lot more carbon dioxide and water than methanol does, because it has more bonds to form.

From my results that I have collected I have drawn a graph showing the positive correlation of the enthalpy change of combustion against molecular mass, backing up my point.

Graph 1: Scatter graph of enthalpy change of combustion against molecular mass

The graph shows my results in blue and the results under standard conditions in blue. I received these results from my data sheet. The graph correlates almost perfectly with the standard conditions. It has a very similar rate of increase and what the graph doesn’t show, because the scale is too small, that the second molecular mass result for 60g (propan-2-ol) is also less than its first. The results are quite far off the ones under standard conditions, but the pattern is still the same. So as the molecular mass increases, so does the enthalpy change of combustion.

Evaluation

Firstly there are no real anomalies in my results, but from the graph I drew you can see that they are far from close to the expected values. I have managed to maintain the same pattern as the expected results, an increase in enthalpy as the number of carbons increase. There are many reasons from this ranging from procedural errors to percentage errors, but the main reason was heat loss.

We did use a heat shield to prevent heat loss, but all it was, was aluminium tubing. Obviously it helped prevent heat loss much better than had there not been one, but a lot of heat was still lost through it. Much heat was still lost through the shield by conduction and also the convection of heat through holes in the shield or simply through the top. An obvious way to reduce this would have been to use some sort lid, even if it was simply a foil lid it would have prevented heat escaping as easily.

There was not only heat loss but also unburnt fuel, carbon not being used up efficiently and being left on the copper calorimeter as a sooty deposit. The carbon will have left the burner but remained unburnt, meaning that when it came to weighing the burner, the carbon would not have been taken into account. Ultimately this means that the amount of fuel burnt was too high, as the carbon was not measured, meaning more energy waste and fuel loss.

One last thing about procedural errors is the height of the copper calorimeter. I used my judgement to decide the height of the copper calorimeter for each alcohol. Some of the flames from the burner were much larger than others, meaning that on some alcohols the tip would have been touching the calorimeter and on others the calorimeter would have been half way down the flame. I did my best to keep it just below the tip for each alcohol.

There were a lot of percentage errors involved for each apparatus, it’s not a mistake, its just that there will always be a degree of experimental uncertainty. The source of uncertainty is the precision of the instrument being used and this can be quantified. Each measuring instrument is designed to measure to a different level of precision. To reduce this, simply more accurate apparatus can be used.

A reading of 4.35 g means that the mass is closer to 4.35 than 4.34 or 4.36. This means that it is between 4.345 and 5.355. Meaning that the actual result would be +/- 0.005, giving an absolute precision error of 0.01.

The importance of an experimental error will depend on how large the mass of the object being weighed is. If an object is 200g than the error will be much less important than if an object is 0.2g. I will calculate all my percentage errors using the equation below.

Percentage error = error x 100
reading

I will use this equation to calculate the percentage errors for all my measuring apparatus and will use the measuring of water in the measuring cylinder as an example. The measuring cylinder measures 100cm3, which we filled to 100 with an accuracy of +/- 0.5, the absolute precision error being 1.0 accounting for both the plus and the minus.

Percentage error = 1 x 100
100 = 1%

One way to reduce this could have been to use a taller measuring cylinder, giving a higher degree of accuracy. A better way to reduce this would have been to weigh the water. The balance we used would have given the reading to +/-0.005, rather than +/-0.5. This would have resulted in a percentage error of 0.01%, meaning that this result would have been a 100 times better.

When reading the thermometer, I believe there was an accuracy of +/- 0.5oC. Being a thermometer ranging up to 100oC it has given me a percentage error of 1%. An easy way to reduce this would have been to use only a 50oC thermometer, which would have enabled me to read it to an accuracy of +/- 0.05oC, giving me a percentage error of 0.2% rather than 1%.

The last percentage error is when weighing the spirit burners. This percentage error won’t be as high, as the balance is quite accurate. Nevertheless I could only read it to an accuracy of +/-0.005, giving me a percentage error of 0.0047%. This is very low because the mass of the object is more than double all of the other masses I have been working out.

Percentage error = 0.01 x 100
214.581 (rather than 100 for the measuring cylinder)
=0.0047%

The reading is 214.581 because rather than working this out for all 10 times I used the scales, I added all the weights up including before and after and then divided by 10 to get an average.

There are other points to look at which barely affect the experiment, but must still be pointed out, for example did I constantly stir the water the same for each alcohol or even how long I left for the temperature to rise, after extinguishing the heat. I believe I did do these fairly correctly as I did with how quickly I extinguished the flame itself after burning but there may still be errors involved which accumulate to the overall errors when doing this experiment.

I my results were useful in identifying which alcohol has the highest enthalpy change in combustion, and the pattern involved. I believe that they were as accurate as I could do them with the apparatus available and the time given to me to perform this experiment. If I had the opportunity to redo the experiment with the chance to use the improvements I have suggested, I would get better results but they would still be far from the expected results under standard conditions, which is simply due to human error and the fact that I am not able to perform this experiment under standard conditions.

Even though the improvements I stated would improve my final readings and results, overall my experiment succeeded in finding out which alcohol produces the highest enthalpy change of combustion out of the 5 I used and the trend involved as you go down the homologous series.

References

Web sites used: