When a reaction occurs, it is either exothermic or endothermic. The difference between these two types of reactions is that exothermic reactions release energy to the surroundings, where as endothermic reactions take in energy from the surroundings. On a reaction pathway graph this looks like this:



X being the energy of the reactants, Y being the energy of the products and Z being the activated complex. The change in enthalpy is the difference between the X value and the Y value per mole of substance. For exothermic reactions this value is negative, for endothermic reactions this value is positive. Combustion reactions always have a negative enthalpy value. During the combustion of alcohols, the products are H2O and CO2. The alcohols we burned were primary alcohols, meaning they only had one carbon atom directly attached to the hydroxlyized carbon atom. The alcohols we burned were Ethanol, Butan-1-ol, and Hexan-1-ol. Their balanced equations are shown as such:

Ethanol

C2H5OH + 3O2 -----> 2CO2 + 3H2O

Butan-1-ol

C4H9OH + 6O2 -----> 4CO2 + 5H2O

Hexan-1-ol

C6H13OH + 9O2 ---> 6CO2 + 7H2O

Upon combustion, covalent bonds are being broken through the reactants and formed through the products. These procedures give out and take in energy respectively. This is known as bond enthalpy. The structure and bond enthalpies for all the alcohols we burned is shown below:

Ethanol

   H  H									       O
    |   |									      /   \
H-C-C-O-H 	+	3 [O=O]      ------>     2 [O=C=O]	+	3 [H    H]
    |   |
   H  H

5[C-H] = 5x412 = 2060				4[C=O] = 4x743 = 2972
1[C-C] = 1x348 = 348				6[O-H] = 6x463 = 2778
1[C-O] = 1x360 = 360
1[O-H] = 1x463 = 463
3[O=O] = 3x498 = 1494
Total Energy = 4725					Total Energy = 5750

				    4725 - 5750

				  = -1025J



Butan-1-ol

   H  H H H								  	       O
    |   |   |   |									      /   \
H-C-C-C-C-O-H 	+	6 [O=O]      ------>     4 [O=C=O]	+	5 [H    H]
    |   |   |   |
   H  H H H

9[C-H] = 9x412 =3708				8[C=O] = 8x743 = 5944
3[C-C] = 3x348 = 1044				10[O-H] = 10x463 = 4630
1[C-O] = 1x360 = 360
1[O-H] = 1x463 = 463
6[O=O] = 6x498 = 2988
Total Energy = 8563					Total Energy = 10574

				    8563 - 10574

				  = -2011J





Hexan-1-ol

   H  H H H  H H									       O
    |   |   |   |   |   |									      /   \
H-C-C-C-C-C-C-O-H 	+	9 [O=O]      ------>     6 [O=C=O]	+	7 [H    H]
    |   |   |   |   |   |
   H  H H H  H H

13[C-H] = 13x412 =	5356					12[C=O] = 12x743 = 8916
5[C-C] = 5x348 = 1740					14[O-H] = 14x463 = 6482
1[C-O] = 1x360 = 360
1[O-H] = 1x463 = 463
9[O=O] = 9x498 = 4482
Total Energy = 12401					 Total Energy = 15398

				         12401 - 15398

				           = -2997J

To calculate the change in Enthalpy of the reactions, we need to use the formulae:

Change In Enthalpy = Total Energy / Number of moles

Total Energy = Specific Heat Capacity of Water * Mass Of Water Heated * Change In Temperature

Number Of Moles = Mass Of Fuel Burned (Change In Mass) / Gram Formula Mass Of The Alcohol.