'Enthalpy of Combustion'.

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        Alcohols are very similar to alkanes and alkenes; they are a family of related compounds – a homologous series. Each member of the series has the O-H group in its molecule. The general formulas for alcohols is Cn H2n+1 OH, where n is the number of carbon atoms. There are six alcohols that I will be using and they are Methanol (CH3OH), Ethanol (CH2H5OH), Propan-1-ol (C3H7OH), Butan-1-ol (C4H9OH), Pentan-1-ol (C5H11OH) and Octan-1-ol (C8H17OH).

        In this experiment I will be burning a range of alcohol’s to heat up a container of water. I will be trying to get reliable results that will tell me how the number of carbon atoms relate to the energy released as heat , otherwise known as the ‘Enthalpy of Combustion’. The combustion process is the making and breaking of bonds. In this experiment, the alcohols will have an exothermic reaction because heat is given out. The breaking of bonds is endothermic and the making of bonds is exothermic. In an exothermic reaction the products are at a lower energy level than the reactants, the difference is the heat energy. The energy is given out when forming the bonds between the new water and the carbon dioxide molecules. The complete combustion of an alcohol is when it reacts with oxygen in the air to form water and carbon dioxide. The basic formula for this reaction is

Heat Transferred =

 Mass of Substance (g) x Temperature Change (°C) x Specific Heat Capacity (J)

        The specific heat capacity is the number of joules required to heat one gram of water by 1°C. Water is used because it is safe, easily found and has a reliable heat capacity of 4.2.

        The bonds that are formed in an exothermic reaction can be of two types. The first could be ionic, where a metal is produced. Ionic bonding involves two electrons transferring from one atom to the other leaving an electrostatic force between them. The other is covalent bonding where atoms share electrons to complete their outer shell. An example being methane where four hydrogen atoms each share an electron with a carbon atom.

        In this investigation, I wish to compare heat produced by the alcohols in terms of:

        Equal masses.

        Equal numbers of molecules, i.e. per mole.

        To calculate the theoretical enthalpy of combustion we need to know the energy requiring in breaking a bond.

Methanol (CH3OH)

CH3OH  +  1½ O2  =  CO2  +  2 H2O

Theoretical Heat of Combustion = 3466 – 2808 = 658 kJ/mole

Ethanol  (C2H5OH)

C2H5OH  +  3 O2  =  2 CO2  + 3H2O

Theoretical Heat of Combustion = 6004 – 4728 = 1276 kJ/mole

Propan-1-ol (C3H7OH)

C3H7OH  +  4½ O2  = 3 CO2  +  4 H2O

Theoretical Heat of Combustion = 8542 – 6648 = 1894 kJ/mole

Butan-1-ol (C4H9OH)

C4H9OH  +  6 O2  = 4 CO2  +  5 H2O

Theoretical Heat of Combustion = 11080 – 8568 = 2512 kJ/mole

Pentan-1-ol (C5H11OH)

C5H11OH  +  7½ O2  = 5 CO2  +  6 H2O

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Theoretical Heat of Combustion = 13618 – 10488 = 3180 kJ/mole

Octan-1-ol (C8H17OH)

C8H17OH + 12 O2  = 8 CO2  +  9 H2O

Theoretical Heat of Combustion = 21232 – 16248 = 4984 kJ/mole

        This data is again reinforced by the results from a preliminary experiment involving Alkanes. Methane, Ethane, Propane and Butane were all burnt using the same method as in this investigation. The results were calculated and are as follows:

        The graph produced from these results has help to support my ...

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