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Determination of the enthalpy change of neutralization

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Introduction

Chemistry Laboratory Report 4 Date: 27th October, 2008 Topic: Determination of the enthalpy change of neutralization Objective: To determine the enthalpy change of neutralization between different pairs of acid-base used (thermometric titration) Introduction Two methods are used to determine the concentration of sodium hydroxide solution and the enthalpy change of neutralization. First method: Measure the temperature change of the solution when different volume ratios of acid and base are mixed and reacted. Second method: Measure the temperature change of the solution upon each addition of a specified volume of acid to the base. Results > Method 1: Initial temperature of the solutions = 27.1 oC Volume of NaOH (cm3) 5.0 10.0 15.0 20.0 25.0 30.0 35.0 Volume of acid (cm3) 35.0 30.0 25.0 20.0 15.0 10.0 5.0 1.00M Hydrochloric acid: Final Temperature (oC) 29.3 31.5 33.4 33.8 31.9 30.3 28.6 Temperature change (oC) 2.2 4.4 6.3 6.7 4.8 3.2 1.5 1.00M Ethanoic acid: Final Temperature (oC) 29.2 30.7 32.5 33.5 31.8 29.9 28.6 Temperature change (oC) 2.1 3.6 5.4 6.4 4.7 1.8 1.5 > Method 2: Initial temperature of the solutions = 27.0 oC 1.00M HCl Volume (cm3) ...read more.

Middle

The enthalpy change of neutralization of hydrochloric acid is greater than that of ethanoic acid. As ethanoic acid is a weak acid which partially ionizes in water, some energy given out by neutralization is absorbed to dissociate the unionized acid molecules. Therefore the enthalpy change is smaller than that of hydrochloric acid. > Method 2: a) b) The maximum temperature change is produced when 24cm3 of acid is mixed with 25cm3 of sodium hydroxide solution. no. of moles of acid reacted = 1 x 24/1000 = 0.024 mol concentration of sodium hydroxide solution = 0.02425/1000 = 0.96 M c) Energy change of hydrochloric acid = mc?T = (24+25) x 4.18 x (27.0 - 34.4) = -1.5157 kJ Enthalpy change of neutralization of hydrochloric acid = -1.12024(0.96 x 25/1000) = -63.15 kJmol-1 Energy change of ethanoic acid = mc?T = (24+25) x 4.18 x (27.0 - 34.3) = -1.4952 kJ Enthalpy change of neutralization of hydrochloric acid = -1.12024(0.96 x 25/1000) = -62.299 kJmol-1 d) Also, The heat given out by hydrochloric acid is higher than that of ethanoic acid because some energy is absorbed by the acid during the neutralization to ionize weak ethanoic acid molecule. ...read more.

Conclusion

The heat loss will be very significant as the ratio of acid added to the reaction mixture is small. In order to reduce the heat loss and the percentage error in calculations, the titration must be carried out quickly to obtain more accurate results. -Sources of error > Heat loss to surroundings Heat is lost due to evaporation, convection and conduction. The value of enthalpy change of neutralization calculated will be smaller than the actual value. Better insulation, such as wrapping the polystyrene cup with cotton wool, can be taken to reduce the error. > Some heat may be absorbed by thermometer, stirrer and the polystyrene cup In order to reduce the error, the heat capacities of apparatus should be taken into account in calculations. > The specific heat capacity of solutions is not equal to that of water > Density of the solutions is not equal to that of water Conclusion The enthalpy change of neutralization of HCl is -56.012 kJmol-1 whereas the enthalpy change of neutralization of CH3COOH is -53.504 kJmol-1. The objective of the experiment has been fulfilled as the enthalpy changes of neutralization between different pairs of acid-base were found. ?? ?? ?? ?? P. 1/5 ...read more.

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