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The aim of this investigation is to show that heat of neutralisation is an exothermic reaction which produces water. The amount of energy given out for one mole of water is about -57.3 kJ/mol. This needs to proven by this experiment as well.

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Introduction

GCSE Assessment - Investigation into Enthalpy of Neutralisation Introduction - The aim of this investigation is to show that heat of neutralisation is an exothermic reaction which produces water. The amount of energy given out for one mole of water is about -57.3 kJ/mol. This needs to proven by this experiment as well. Results Stage One These are the results I gained from the titration. Titrant (HCl) Rough 1 2 Initial (cm�) 12 10 22 Final (cm�) 22.5 20.7 32 Titre (cm�) 10.5 10.7 10 Stage Two The results gained are as follows: Time (s) Temp for expt.1 (�C) Temp for expt.2 (�C) Average temp (�C) 0 16 16 16 5 18.2 18 18.10 10 25.1 24.9 25 15 28.5 27.8 28.15 20 29.4 28.9 29.15 30 29.5 29.1 29.3 40 29.5 29.1 29.3 50 29.5 29.1 29.3 60 29.5 29.1 29.3 70 29.5 29.1 29.3 80 29.5 29.1 29.3 90 29.5 29.1 29.3 100 29.5 29.1 29.3 120 29.2 29.1 29.15 140 29.0 28.9 28.95 160 28.6 28.2 28.4 180 27.9 27.7 27.8 200 26.8 26.9 26.85 220 ...read more.

Middle

The bench HCl should be ten times more concentrated. Bench HCl (dilute) = 0.19 mol/dm� Bench HCl = 0.19 � 10 = 1.9 mol/dm� Therefore bench HCl has a concentration of 1.9 mol/dm�. Calculations for Stage Two In this stage the heat of neutralisation needs to be worked. Firstly, a graph needs to be plotted with the results from stage two in order to work out the maximum temperature rise. Refer to graph 1. Heat of Neutralisation Heat of neutralisation for this experiment can be represented by the following ionic equation: H3O+ (aq) + OH- (aq) � 2H2O (l) (Na+ (aq) and Cl- (aq) are spectator ions) The equation for heat of neutralisation is as follows: Q = M � SHC � ?T For this experiment an assumption is made that the specific heat capacities of NaOH (aq) and HCl (aq) are the same as that of water, which is 4.2 J/g �C. The temperature of NaOH and HCl was 16�C at room temperature. ...read more.

Conclusion

Therefore the heat of neutralisation per mole = Q = 1117.2 = 55860 = -55.86 kJ/mol moles 0.02 Evaluation The experiment went according to plan and there were no anomalous readings in stage two. The aim of the experiment was completed successfully. The heat of neutralisation (exothermic) in my experiment was -55.86 kJ/mol which was very close to the actual reading of -57.3 kJ/mol. My result was within an accuracy of 2.51%. This loss in accuracy may have been due to heat losses through convection, conduction and radiation. This can be minimised by using a vacuum flask which is shown below: The experiment was also simplified because the heat received by the polystyrene beaker was assumed to be negligible. The experiment could have been modified so that the heat received by the polystyrene beaker was also taken account of. This would have produced an accurate result for the heat of neutralisation. In order to investigate this experiment further I would try different acids (sulphuric acid) and alkalis (sodium chloride) in order to prove that heat of neutralisation works for any strong acid or alkali. Abdul Raja 5L ...read more.

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