Lab report - standard enthalpy of combustion PDF

Preview

Summary

Download Lab report - standard enthalpy of combustion PDF

Description

Student Name: Kevin Ma

Investigation: Question summary (HL/SL)

Question Summary The relationship between the enthalpy change of combustion per mole and the number of carbons is simple alcohols belonging to the same homologues series and enthalpy change of combustion 1. Introduction Organic compounds produce a large quantity of energy when combusted. These compounds include alkane, alkene and even alcohol. Alcohols are used as fuels for a range of applications ranging from cooking to fueling cars. They have become a significant component for biofuels. However, what makes a good fuel? A good fuel should able to generate a high amount of heat per unit mass (“Science.” Enotes.com, Enotes.com, https://www.enotes.com/homework-help/what-makes-good-fuel-1153289.) In most cases, fuels are burnt in oxygen. Nonetheless, is there any relationship between the energy released by one alcohol and another alcohol? Therefore, in this investigation, the aim of this experiment is to investigate the relationship between the number of carbon atoms in an alcohol chain and its standard enthalpy change of combustion. As well, I will be only focusing on the complete combustion of straight chain alcohols (Propanol and Ethanol).

2. Exploration 2.1.

Background context

Alcohols undergoes complete combustions in excess oxygen to produce carbon dioxide and oxygen. The following reaction will occur: Ethanol: C2H5OH + 3O2 -> 2CO2 + 3H2O Propanol: C3H7OH + 4.5O2 -> 3CO2 + 4H2O H0: The null hypothesis: The difference in the carbon length has no effect on the standard enthalpy change of combustion Ha: The alternative hypothesis: The difference in the carbon length will affect the standard enthalpy change of combustion; The longer the carbon chain the greater the enthalpy change.

2.2.

Apparatus

For this investigation, the following equipment is needed: 1. 1 clamp stand 2. 1 heat proof mat

1

Student Name: Kevin Ma

Investigation: Question summary

3. 1 calorimeter (including copper beaker) 4. Cotton or wool (for insulation) 5. 1 stirrer 6. 1 electronic balance ( ± 0.001 g ) 7. 1 thermometer ( ± 0.05℃ ¿ 8. 1 measuring cylinder (100 cm3) 9. 1 pack of matchsticks 10. 1 Propanol spirit burner 11. 1 Ethanol spirit burner

2.3. Hazardous chemicals being used or made, or hazardous procedure or equipment

Ethanol

Risk Assessment

Nature of the hazard(s) and source(s) of information

Risks and control measures to reduce the risks



There is a serious risk of liquid catching fire



its vapour may catch fire above 13 °C.



The vapour / air mixture is explosive (from 3.3 to 19% ethanol).



Breathing vapour may result in sleepiness: the concentration in the air should not exceed 5,760 mg m-3.



Use the smallest volume possible



wear eye protection.



Make sure the room is well ventilated.



Check that equipment for extinguishing fires is nearby, eg, damp cloth, bench mat, fire blanket.



Do not use near naked flames



if heating is necessary, use an electrically-heated water bath or hot water from a kettle.

Propanol



Risk of catching fire.



Vapour may catch fire below room temperature.



Can cause eye damage and vapours may cause drowsiness and dizziness.

2

Student Name: Kevin Ma

2.4.

Investigation: Question summary 

Wear eye protection.



Make sure the room is well ventilated or, in a laboratory, use a fume cupboard if possible.



Check ways of putting out any fires.



Do not use near naked flames; if heating is necessary, use an electrically-heated water bath or hot water from a kettle.

Methodology

1. Using an electronic scale weigh and record the mass of the spirit burner containing the alcohol 2. Using a measuring cylinder measure out 100 cm3 of water 3. Add it into the copper beaker of the calorimeter 4. Using a thermometer measure and record the initial temperature of the water 5. Insulate the copper beaker by wrapping it with cotton or wool 6. Cover the calorimetry with the lid 7. Insert the stirrer and the thermometer into the copper beaker 8. Using a clamp stand hold the calorimeter above the table 9. Put the spirit burner under the calorimeter 10. Adjust the position of the calorimeter and make sure the distance between them is kept the same throughout the experiment by measuring it with a ruler 11. Take a matchstick and light the spirit burner 12. Adjust the position of the spirit burner, so the flame is directly heating the copper beaker 13. Constantly stir the beaker while it is burning 14. Ensure minimum heat is lost to the surrounding 15. Prevent the spirit burner from burning further when it has risen by 15 ℃ it

by putting the lid over

16. Using an electronic balance, weigh and record the spirit burner containing the alcohol again 17. Calculate the change in mass and record it 18. Wait for the equipment to cool down before starting another trial 19. Repeat steps 1-18 for at least five times for each alcohol testing e.g. for propanol carry out step 118 five times, for ethanol carry out step 1-18 five times.

3

Student Name: Kevin Ma

Investigation: Question summary

a.Set-up

3. Variables

Independent variable Dependent variable

Description Ethanol and Propanol Change in mass of the alcohol Change in mass of the alcohol = Final mass – Initial mass Same mass of distilled water Distance between the spirit burner and the bottom of the copper beaker should be fixed

Controlled variables

As the distance increases, the heat loss to the surrounding also increases. Hence, less energy reaches the bottom of the copper beaker. This will lead to a low rise in temperature of water, thus a wrong ∆ H would be obtained. Same increments of temperature Lifting the thermometer so it would not touch the bottom of the copper can.

Value N/A N/A

Unit N/A grams

Precision N/A ± 0.00 5

100 8

grams cm

± 0.05 ± 0.05

15 N/A

℃ N/A

± 0.05 N/A

4

Student Name: Kevin Ma

Investigation: Question summary

4. Results 4.1.Raw Results Different alcohol in the same homologues series Ethanol Propanol

Trial 1

Volume of alcohol (grams Trial 2 Trial 3

Initial: 273.78 Final: 272.35 Initial: 282.20 Final: 281.14

Initial: 272.35 Final: 271.08 Initial: 281.09 Final: 280.07

Initial: 271.00 Final: 269.69 Initial: 280.07 Final: 279.45

± 0.005) Trial 4

Trial 5

Initial: 269.61 Final: 267.66 Initial: 279.42 Final: 278.61

Initial: 267.65 Final: 266.57 Initial: 278.61 Final: 277.76

Fig 2

4.2.Processed Results Different alcohol in the same homologues series Ethanol Propanol Fig 3

Trial 1

Chane in volume of alcohol (grams ± 0.005) Trial 2 Trial 3 Trial 4

Trial 5

0.93 1.06

1.27 1.02

1.08 0.85

1.31 0.62

1.95 0.81

Different alcohol in the same homologues series

Enthalpy change of combustion per mole (standard enthalpy change of combustion) (kJ mol-1) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5

Ethanol Propanol Fig 4

-310.669 -355.556

-227.497 -369.499

-220.550 -607.886

-148.164 -465.295

-267.520 -443.339

Average standard enthalpy change: Ethanol: -234.88kJ/mol Propanol: -448.315kJ/mol

4.3.Example Calculations To obtain the result in figure 1, use the equation Change in Mass = Final mass – Initial Mass For example (trial 1 of ethanol): Change in Mass = Initial mass – Final Mass ∴ Change in Mass = Initial mass of ethanol – Final Mass of ethanol ∴ Change in Mass = 273.78 – 272.35 ∴ Change in Mass = 0.93g Then repeat this process for all of the data collected to obtain the results shown in Fig 3.

To calculate the standard enthalpy change of combustion, use the following equations Q= mc ∆ T 5

Student Name: Kevin Ma

Investigation: Question summary

m = mass c = specific heat capacity ∆ T = change in temperature Mole =

Mass Relative molecular mass

Standard Enthalpy change of combustion =

Enthalpy changeof combustion Mole

For example (trial 1 of ethanol): First, obtain the enthalpy change of combustion: Q=

Mass of water × speific heat capacity of water × change ∈temperature

∴Q=

100 g × 4.18 J g-1K-1 ×15 K

∴ Q =-

6270 J

Then calculate the number of moles of ethanol: Mole =

Mass of ethanol Relative molecular mass ∴ Mole =

∴ Mole =

Mass of ethanol Relative molecular mass 0.93 2∗12.01 + 6∗1.01+ 16

∴ Mole = ∴ Mole =

0.93 46.08 0.02018

Finally calculate the standard enthalpy change of combustion: Standard enthalpy change of combustion =

Enthalpy changeof combustion Mole

∴ Standard enthalpy change of combustion =

−6270 J 0.02018

∴ Standard enthalpy change of combustion = - 310668 J /mol ∴ Standard enthalpy change of combustion = - 310.668 k J /mol Then repeat this process for all of the data collected to obtain the results shown in Fig4 6

Student Name: Kevin Ma

Investigation: Question summary

To calculate the average value, use the following equation: Average =

∑ of all terms number of terms

For example (for ethanol): Average =

∴ Average =

∑ of all terms number of terms

−310.669−227.497 −220.550−148.164 −267.520 5 ∴ Average = -234.88

5. Analysis From looking at the data collected, it strongly supports the alternative hypothesis as the alcohol group with more carbon atoms (propanol) has a more negative standard enthalpy change of combustion (propanol is -448.315kJ/mol while ethanol is -234.88kJ/mol). However, the values deviate with the theoretical value by a considerable margin: propanol is off by -1572.685kJ/mol while ethanol is off by -1,132.12kJ/mol. This could be explained as for the investigation I conducted, the environment was not under control and the conditions are not optimum. Firstly, the lab had a poor ventilation system. This means the oxygen level in the air is insufficient for the alcohol to undergo complete combustion, where instead, it resulted in incomplete combustion where the energy produced is smaller than complete combustion. As a result, more alcohol was burnt to produce sufficient energy to raise the water temperature by 15 ℃ . This explanation is supported by the qualitative data as the color of the flame was orange for abundance, indicating the presence of incomplete combustion. On top of this, the room temperature was low. This means the limited energy produced by this reaction is absorbed by the surroundings (air). As a result, little heat energy was transferred to the calorimetry, and the wrong enthalpy value is calculated.

6. Conclusion The longer the chain, the more energy will be produced, thus, the more negative the standard enthalpy change of combustion is. This could be explained through the structural difference between the two alcohol groups. When an alcohol (CnH2n+1OH) group combust in excess oxygen, the C-H, H-O, and C-C bonds are broken, and H-O (in water) and C=O (in CO2) bonds are formed instead. The amount of energy released by forming these H-O and C=O bonds exceeds the amount of energy it takes to break the C-H and C-C bonds. The more carbon and hydrogen atoms there are in a molecule, the more potential energy there is in the molecule since more H-O and C=O bonds will be formed when one molecule of the fuel combusts. Therefore, the more carbon atoms there are, the greater the energy is released, and the enthalpy change. This could be proved mathematically by calculating standard enthalpy change of combustion using bond enthalpies: For ethanol, the reaction for complete combustion is C2H5OH + 3O2 -> 2CO2 + 3H2O

7

Student Name: Kevin Ma

Investigation: Question summary

Hence the displayed formula of this equation is

Using the average bond enthalpy equation and value, we can calculate the enthalpy changes for this reaction: ∆ H r =∑ ( bondbroken )−∑ ( bondmade ) ∴ ∆ H r=( 5 × 414 +346 + 358 + 463 +3 × 498 ) − (4 × 804+ 6 × 463 ) ∴ ∆ H r=4731 −5994

∴ ∆ H r=−1263 kJ /mol For propanol, the reaction for complete combustion is C3H7OH + 4.5O2 -> 3CO2 + 4H2O Hence the displayed formula of this equation is

Using the average bond enthalpy equation and value we can calculate the enthalpy changes for this reaction: ∴ ∆ H r=( 7 × 414 +3× 346 + 358 + 463 +4.5 × 498 ) − (6× 804 + 8× 463 ) ∴ ∆ H r=6998 – 8528 ∴ ∆ H r=−1530 kJ /mol

By comparing the enthalpy change of combustion between ethanol and propanol, it is clear that the standard enthalpy change is more negative for propanol (-1530kJ/mol) than ethanol (-1263kJ/mol). In addition, the literature value for the standard enthalpy change of combustion also supports my hypothesis as the literature standard enthalpy change of combustion value for propanol is (-2021kJ/mol) while for ethanol it is only (-1367kJ/mol) Moreover, this could be explained through a theoretical approach. For CO2, the hybridization state is sp hybridized, having two pi bonds in each carbon atom in CO2 molecules. Thus, the CO2 molecule is effectively sheathed in a cylinder of negative charge where the rotation is restricted by the pi bonds (Sykes, Peter. A Guidebook to Mechanism in Organic Chemistry. Pearson, 2019.). While on the other 8

Student Name: Kevin Ma

Investigation: Question summary

hand, the carbon is alcohol is sp3 hybridized leading to free rotation about a carbon – carbon single bond. Hence, the more carbon there are in a compound, the more CO2 molecule will be produced who is more thermodynamically stable. Therefore, the more negative the value is.

7. Evaluation 7.1.Strengths Strength: For this particular investigation, the experiment is repeatable, so I was able to carry out numerous trials for each alcohol. Thus, making the experiment reliable as the data are comparable For each experiment, we are able to repeat at least 5 times. Hence, we are able to find the average and identify the anomalies in the experiments. For each trial, the control variables were kept the same, so the results obtained are comparable The water was constantly stirred by the rod so there is a more equal distribution of heat energy in the water. Hence making the result more reliable as the more accurate temperature of water would be obtained As well, the thermometer was lifted so there were no physical contacts between the thermometer and the bottom of the copper, where it was directly heated by the spirit burner. The copper can would have a higher temperature as it has direct contact with the flame which will result in the standard enthalpy of combustion higher than the actual value. Hence, by lifting the thermometer above the copper can, the results are more reliable.

7.2.Improvements and Weaknesses By comparing my collected results with the literature value in the data booklet, there is an exaggerate difference between them (the literature value for propanol is 450% of the experimental data and the literature value for ethanol is 582% of the experimental data). Hence, the extraneous variables in this investigation greatly affected the result collected. Weakness The room temperature was very low thus increasing the loss of heat to the surrounding.

There was an insufficient supply of oxygen in the lab thus resulting in an incomplete combustion where less energy is released. The distance between the flame and the bottom of the copper can was still too far away thus majority of the heat energy is lost to the surrounding There was a limited number of test variables (ethanol and propanol) to generate a well-supported conclusion.

Improvement Turn on the heater so the temperature of the room is similar to the temperature of the water so more of the heat energy generated from the combustion will be transferred Conduct the experiment in a lab with a better ventilation system and as well do not conduct the experiment at the same time with everyone else Minimize the heat loss to surrounding by decrease the distance between the bottom of the copper can and the flame even further Have a broader range of test variable in the next experiment which will able to generate a bigger data base for a better conclusion. In addition. By having a wider range of data, graphs demonstrating trend can be generated. For example, having more alcohol from the same homologues series but with 9

Student Name: Kevin Ma

Human error – It was hard to monitor the exact moment when the temperature increases by 15 ℃ as human has a delayed reaction time which would result in more grams of alcohol burnt. Hence the standard enthalpy of combustion would be smaller than the supposed value. As well, the eyes are not precise enough to identity the scale on the thermometer, thus making it difficult to obtain the exact point when the temperature rises by 15 ℃ .

Investigation: Question summary

different number of carbon (methanol, butanol, pentanol and hexanol) Use the logger pro program and the wire thermometer (connected with the logger pro program to track temperature). This way the results collected would be more accurate as computer algorithms are more precise than human.

8. Bibliography Sykes, Peter. A Guidebook to Mechanism in Organic Chemistry. Pearson, 2019. “Science.” Enotes.com, Enotes.com, https://www.enotes.com/homework-help/what-makes-good-fuel1153289.

10...