Comparing the enthalpy changes of combustion of different alcohols.

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Comparing the enthalpy changes of combustion of

different alcohols.

Planning

What are alcohols?

Alcohol is the common family name for the hydrocarbon group alkanols. They are part of a homologous series. At least one of the hydrogen groups in the molecule is replaced by an –OH group.

H        H        H                                 H        H        H          

│        │        │                                │        │        │

H  — 1C  —  2C  —   3C  —        H        →        H  —  1C  —  2C  —  3C  —        H

        │        │        │                                │        │        │

        H        H        H                                H       OH        H

      Propane                                            Propan-2-ol                                        

They are all organic compounds.

The general formulas for the alcohols are:

CnH(2n+1)OH

Where n represents a number.

The first and simplest member of the alkanols family is methanol. Its molecular formula is CH3OH.

You can now see that each member of the alcohol family has a different number of carbons in its structure. They increase by one carbon atom and two hydrogen atoms each time.

All combustion reactions are exothermic which is why all enthalpy of combustion values are negative.

The question that this piece of coursework is going to answer is

“Is there a relationship between the number of carbon atoms in a chain and the enthalpy change of combustion?”

Aim

The aim of this experiment is to prove that the longer the hydrocarbon chain the higher the amount of energy transferred to the water, therefore fewer moles of fuel will be used to achieve a temperature rise of 15°C.

The enthalpy change of combustion is the amount of energy transferred when one mole of a substance burns completely in oxygen (always –ve).

To work out the enthalpy change of combustion of the alcohols the energy output must be measured. The easiest way to do this accurately is to use the thermal energy of combustion to raise the temperature of a substance with a known specific heat capacity, (the amount of energy required to increase the temperature of 1g of a liquid by 1°C). We will be using water as our substance with a known specific heat capacity.

Prediction

I predict that the longer the hydrocarbon chain the more energy will be transferred to the water, therefore fewer moles of fuel will be used to achieve the same temperature rise.

So as the number of carbon atoms increases the enthalpy of combustion will become more negative. I think this because as you add a H-C-H bond you increase the energy required to break the bonds by 826KJ/Mol. This is worked out by the average bond enthalpy of the C-H bond, which is 413kJmol-1 being multiplied by two because there are two extra C-H bonds in the molecule.

Using this I will predict the enthalpy change of combustion value for methanol.

To do this I will use the equation

CH3OH(l) + 1½O2 O(g) → CO2(g)  + 2H2O(l)

In order to make this easier to work out and for people to understand I will multiply everything by 2 so that I don’t have any ½ values in the equation. This gives the new equation of;

2CH3OH(l) + 3O2 O(g) → 2CO2(g)  + 4H2O(l)

On the left hand side of this equation you have

2 x (3 x C-H(413))

2 x (1 x C-O(358))

2 x (1 x O-H(464))

3 x (1 x O=O(498))

In total the total energy required to break the bonds is

(6 x 413 kJmol-1) + (2 x 358 kJmol-1) + (2 x 464 kJmol-1) + (3 x 498 kJmol-1) =

2478 kJmol-1 + 716 kJmol-1 + 928 kJmol-1 + 1494 kJmol-1 = 5616kJmol-1

On the right hand side of the equation

2 x (2 x C=O(-805 kJmol-1))

4 x (2 x O-H(-464 kJmol-1))

In total the total energy required to form the bonds is

(4 x -805 kJmol-1) + (8 x -464 kJmol-1)

-3220 kJmol-1 + -3712 kJmol-1 = -6932 kJmol-1

In the data sheets that I used for these calculations two values are given for the C=O bond. It is mentioned that the value I used is specific to the bond in CO2. The other value is 745kJmol-1 which is for other C=O bonds.

Therefore the expected enthalpy change of combustion for methanol is.

-6932 kJmol-1 + 5616 kJmol-1 = -1316 kJmol-1.

As in this equation there were two moles we must divide this by two. Therefore the answer is

-658 kJmol-1. As I am using different data sources this will not be the same as the figure used in the calculations in the analysis.

As I am using data books for the enthalpy change of combustion it will be an insufficient use of my time to calculate the value using this method for all of the alcohols that I plan to use.

Data obtained from Data sheets from Salters Advanced chemistry 2000, Table 20 bond lengths and bond enthalpies

Alcohols to be used to compare

  • Methanol        =        CH3OH
  • Pentan-1-ol        =        C5H11OH
  • Propan-1-ol         =        C3H7OH
  • Propan-2-ol        =        C3H7OH
  • Butan-1-ol        =        C4H9OH
  • Butan-2-ol        =        C4H9OH

I have chosen these alcohols because I will have a chain containing 1 carbon atom, 3 carbon atoms, 4 carbon atoms and 5 carbon atoms. I have also decided to use structural isomers of two of these, Propanol and Butanol. (Propan-1-ol and Propan-2-ol and Butan-1-ol and Butan-2-ol). This will allow me to compare the enthalpy of combustion of increasing chain length and the structural isomers of some of these alcohols.

Propan-1-ol differs from propan-2-ol by the position of the –OH group. In propan-1-ol the –OH group is on the first carbon, in propan-2-ol the –OH group is on the second carbon.

Alcohol structure

Methanol

H             H

 |            ⁄

H  —        C   —        O

 |        

        H

Propan-1-ol

H        H        H                

         |         |         |          

H   —        1C  —        2C  —3C  —        H

         |         |         |                  

        OH        H        H        

Propan-2-ol

H        H        H                

         |         |         |          

H   —        1C  —        2C  —        3C  —        H

         |         |         |                  

Join now!

        H        OH        H

Butan-1-ol

H        H        H        H        

         |         |         |         |        

H   —        1C  —        2C  —        3C  —        4C  —        H

         |         |         |         |        

        OH        H        H        H        

Butan-2-ol

H        H        H        H        

         |         |         |         |        

H   —        1C  —        2C  —        3C  —        4C  —        H

         |         |         |         |        

        H        OH        H        H        

Pentan-1-ol

H        H        H        H        H

         |         |         |         |         |

H   —        1C  —        2C  —        3C  —        4C  —        5C  —        H

         |         |         |         |         |

        OH        H        H        H        H

Equipment required.

  • Small copper can to act as a calorimeter
  • Thermometer (to 1°C of accuracy)
  • 100cm3 measuring cylinder
  • Spirit burners containing –  Methanol

        Pentan-1-ol

      ...

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