SIMPLE CALORIMETRY TO FIND THE ENTHALPY OF COMBUSTION OF ALCOHOLS

Authors Avatar

Name: Cheng Cheuk Hei
Class: F6B (17)

EXPERIMENT8: SIMPLE CALORIMETRY TO FIND THE ENTHALPY OF COMBUSTION OF ALCOHOLS

Date: 23/11/2010
I. Objective: 
The purpose of this experiment is to estimate the enthalpy of combustion of an alcohol by using calorimetry.

II. Introduction:

Principle

To work out the enthalpy change of combustion of the alcohols, the energy released must be measured. The simplest way of doing this accurately is to use the heat of combustion of the alcohol to raise the temperature of a known amount of substance with a known specific heat capacity. The rise in temperature of this substance can then used to work out the heat transferred to the substance. Thus, the enthalpy change of combustion of the alcohol can be estimated by assuming all the heat is transferred to the substance.

In this experiment, water is heated by combusting the alcohols as specific heat capacity of water is known as 4.2 J g-1 K-1. The enthalpy change of combustion of an alcohol can be calculated as the temperature rise of water and mass of the alcohol used can be found.

i.e. temperature increased of water, mass of water, specific heat capacity of water, mass of the alcohol used and molar mass of the alcohol are needed to collect to work out the enthalpy of combustion of the alcohol.

Chemicals

Ethanol,  = -1367 kJ mol-1 

Propan-1-ol, Butan-1-ol

Materials and apparatus

Safety spectacles

Join now!

12-oz. aluminium beverage can with cut out top

Bench mat

Digital balance (to 0.001g)

Stand, boss, clamp

Spirit burners

Thermometer

100 cm3 measuring cylinder

III. Data

Combustion of Ethanol:

Combustion of Propan-1-ol:

Combustion of Butan-1-ol:

Calculation:

Experimental enthalpy change of combustion of Ethanol:

 = 1120 kJmol-1

Experimental enthalpy change of combustion of Propan-1-ol:

 = 1510 kJmol-1

Experimental enthalpy change of combustion of Butan-1-ol:

 = 2090 kJmol-1

% error of enthalpy change of combustion of Ethanol:
 x 100% =   x 100% = 81.9%

...

This is a preview of the whole essay