# Comparing the Enthalpy changes of combustion of different alcohols

Marc Duxbury

Comparing the Enthalpy changes of combustion of different alcohols

Aim:

The aim of this experiment is to find out how the enthalpy change (total energy released when the alcohols are completely combusted in a plentiful supply of air) for 5 different alcohols is affected by the number of carbon atoms in the alcohol and other factors contributing to the molecular structure.

Prediction:

I predict that as the amount of carbon atoms in the alcohol increases, the higher the enthalpy of combustion will be. I have made this prediction, using the values for the enthalpy change of combustion for each alcohol, calculated using bond enthalpies and Hess’ law.

Methanol’s molecular formula is CH3OH. This is the basic structure for all the alcohols, then to make the larger ones an extra carbon is added to the existing carbon each time and the oxygen-hydrogen molecule gets added to the atoms added to the new carbon atom

When methanol combusts in air, it reacts with oxygen molecules to from water and carbon dioxide. The balanced equation fort this is:

CH3OH (l) + 1.5O2 (g)                        CO2 (g) + 2H2O (l)

This means that the bonds broken are; 3 carbon- hydrogen, 1 carbon-oxygen, 1 oxygen-hydrogen and 1.5 oxygen- oxygen (double bond) and the bonds broken are; 2 carbon- oxygen (double bond) and 4 oxygen-hydrogen.

Constructing a Hess’ law cycle will show how these are linked together:

CH3OH (l) + 1.5O2 (g)                        CO2 (g) + 2H2O (l)

C (g) + 4H (g) + 4O (g)

N.B. bond enthalpies are for elements in their gaseous states

If a calculation for the amount of energy needed to break the bonds is made and then the amount of energy given out from bond formation, the resultant energy difference (negative because the reaction is exothermic) is the enthalpy change of combustion.

Average bond enthalpies for elements in their gaseous states (kJmol-1):

Carbon – Carbon (C-C) = +347

Carbon – Hydrogen (C-H) = +413

Oxygen – Hydrogen (O-H) = +464

Carbon – Oxygen (C-O) = +358

Carbon – Oxygen double bond (C=O) = +805

Oxygen – Oxygen double bond (O=O) = +498

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 3(C-H) + E (C-O) + E (O-H) + E 1.5(O=O)

= 3(413) + 358 + 464 + 1.5(498)

= 2808kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 2(C=O) + E 4(O-H)

= 2(805) + 4(464)

= -3466kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 2808 + -3466

= -658kJmol-1

The combustion of methanol gives out –658kJ of energy for every mole of fuel burnt. This is only an approximate result and the actual value in my experiment may be different because, firstly the bond enthalpies are an average and they may vary in different molecules and secondly because the value are worked out assuming that the reactants and products are in a gaseous state, when in practise the water and alcohol are liquids. This means that my values for enthalpy change of combustion are likely to be different, but as long as the same method is used for each alcohol the pattern can be seen. I will now use the same bond enthalpy values to work out the estimation of enthalpy change of combustion for each alcohol.

Ethanol:

C2H5OH (l) + 3O2 (g)                            2CO2 (g) + 3H2O (l)

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 5(C-H) + E (C-C) + E (C-O) + E (O-H) + E 3(O=O)

= 5(413) + 347 + 358 + 464 + 3(498)

= 4728kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 4(C=O) + E 6(O-H)

= 4(805) + 6(464)

= -6004kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

...