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.ResultsStage OneThese are the results I gained from the titration.Titrant (HCl)Rough12Initial (cmï¿½)121022Final (cmï¿½)22.520.732Titre (cmï¿½)10.510.710Stage TwoThe results gained are as follows:Time (s)Temp for expt.1 (ï¿½C)Temp for expt.2 (ï¿½C)Average temp (ï¿½C)0161616518.21818.101025.124.9251528.527.828.152029.428.929.153029.529.129.34029.529.129.35029.529.129.36029.529.129.37029.529.129.38029.529.129.39029.529.129.310029.529.129.312029.229.129.1514029.028.928.9516028.628.228.418027.927.727.820026.826.926.8522026.126.326.224025.525.725.6Calculations for Stage OneConcentration of NaOHMoles = MassMolar massMoles = 8g(23+16+1) = 40= 0.2Therefore concentration of NaOH is 0.2 mol/dmï¿½.Concentration of diluted bench HClMoles of NaOH in 10cmï¿½ = V ï¿½ C = 10 ï¿½0.2 = 0.002 moles1000 1000Equation NaOH + HCl ï¿½ NaCl + H2OMoles 1 1Vol/cmï¿½ 10 10.4 ï¿½ This figure is the average titre gained from Stage One.Conc. g/dmï¿½ 8 xFrom the equation 1 mole of NaOH reacts with 1 mole of HCl to give 1 mole of NaCl and H2O. However, there are only 0.002 moles of NaOH and therefore there must be 0.002 moles of HCl. The concentration of HCl can be worked out the following equation.Concentration = Moles ï¿½ 1000 = 0.002 ï¿½1000 = 0.19 mol/dmï¿½Volume 10.4Concentration of the bench HClThe dilute bench HCl is diluted by a factor of ten and its concentration was found to be 0.19mol/dmï¿½. The bench HCl should be ten times more concentrated.Bench HCl (dilute) = 0.19 mol/dmï¿½Bench HCl = 0.19 ï¿½ 10 = 1.9 mol/dmï¿½Therefore bench HCl has a concentration of 1.9 mol/dmï¿½.Calculations for Stage TwoIn 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 NeutralisationHeat of neutralisation for this experiment can be represented by the following ionic equation:H3O+ (aq) + OH- (aq) ï¿½ 2H2O (l) (Na+ (aq) and Cl- (aq) are spectator ions)The equation for heat of neutralisation is as follows:Q = M ï¿½ SHC ï¿½ ?TFor 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 ï¿½C.The temperature of NaOH and HCl was 16ï¿½C at room temperature. When HCl was added neutralisation took place – this is an exothermic reaction which produced a maximum temperature of 29.3ï¿½C (which is shown on graph 1).Heat of neutralisation is worked out by adding the heat received by the solution to the heat received by the polystyrene cup. To simplify the calculations I am assuming that polystyrene is an insulator and it only takes a very small amount of the heat of neutralisation.Calculations to work out heat of neutralisationHeat from neutralisation = Heat received by waterQ = M ï¿½ SHC ï¿½ ?TQ = 20 ï¿½ 4.2 ï¿½ 13.3Q = 1111.88 J/moleQ = 1.11 kJ/moleThe moles of water formed is 0.02 which can be worked out by referring to the word equation. For stage two the bench acid was used. It was worked out that 10cm3 of NaOH consists of 0.002 moles for the diluted bench HCl (this is diluted by a factor of ten). Therefore the bench acid on its own must consist of 0.02 moles. By ratio the equation shows that 0.02 moles of NaOH reacts with 0.02 moles of HCl to form 0.02 moles of water.Therefore the heat of neutralisation per mole= Q = 1117.2 = 55860 = -55.86 kJ/molmoles 0.02EvaluationThe 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.
Investigation into Enthalpy of Neutralisation Essay
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