An Investigation into How the Amount of Heat Produced By Burninga Fuel Depends On the Mass of the Fuel Burnt

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Firuze Naim 11P

Chemistry Coursework

An Investigation into How the Amount of Heat Produced By Burning a Fuel Depends On the Mass of the Fuel Burnt

Aim

My aim is to measure the energy released when burning ethanol.

Introduction

Exothermic reactions release energy in the form of heat.  A rise in the temperature indicates an exothermic reaction.  The energy that is given out due to the breaking of bonds forms new bonds, therefore an exothermic reaction occurs.  When breaking old bonds the energy is not as great as it is when bonds are forming.

Endothermic reactions continuously need heat to be put in so they can form the necessary chemical bonds.  In an endothermic reaction the products tend to be of more use then the reactants.  A fall in the temperature indicates an endothermic reaction.  The breaking of bonds is an endothermic reaction, this is when the existing bonds are supplied with energy to break them.  When forming new bonds the energy is not as great as it is when bonds are broken.

Scientific Knowledge

The following equation was the one used to calculate the amount of energy released by the ethanol:

C₂H₅OH  2CO₂ + 3H₂O

Bond Energies:

                C – C                 347kJ/mol

                C – H                 413kJ/mol

                O – H                 464kJ/mol

                C – O                 360kJ/mol

                C = C                 612kJ/mol

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                H - H                 436kJ/mol

                C = O                 805kJ/mol

                O = O                 498kJ/mol

Bonds Broken:

                1x C – C                 1x +347

                1x C – O                 1x +360

                1x O – H                 1x +464

                5x C – H                 5x +431

                3x O = O                 3x +498

                                Total = +4730

Bonds Made:

                4x C = O                 4x -805                

                6x O – H                 6x -464

                                Total = -6004

ΔH        = 4730 – 6004

        = -1274Kj/mol¯¹

Calculations:

       H   H

                 

H – C – C – OH + 3O = O  2O = ...

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