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Calculate the measurement of enthalpy change per mole for each of the three acids (Hydrochloric acid, Nitric acid and Sulphuric acid) for the neutralisation reactions.

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Introduction

Calculate the measurement of enthalpy change per mole for each of the three acids (Hydrochloric acid, Nitric acid and Sulphuric acid) for the neutralisation reactions The Experiment The experiment is to calculate the measurement of enthalpy change per mole for each of the three acids (Hydrochloric acid, Nitric acid and Sulphuric acid) for the neutralisation reactions. Enthalpy change means the heat absorbed, at constant pressure, when a chemical process occurs in molar amounts by an equation. Enthalpy of neutralisation refers to the energy change when one mole of H + ions reacts with one mole of OH - ions to form water, under standard conditions. _Hn. Safety As the experiment involves acids, it will become hazardous, as when in contact with acids the skin will irritate, however this can be dealt with quickly by taking the acid off. If the acid gets into contact with the eyes it will be difficult to remove the acid, which would then cause serious damage to the eyes. So for this reason I will wear safety goggles at all times during the experiment. Results Tables 25cm of the 1.0 mol dm Hydrochloric acid added to 1.0 mol dm Sodium Hydroxide Volume Of Acid Added / Cm Temperature Of Mixture / C 0 20 5 21.5. ...read more.

Middle

Total volume can be achieved by adding the volume required to neutralise the acid and the actual volume which is 25 cm M = (25.5 + 25) = 50.5, mass C = 4.2 T = (26.5 - 20) = 6.5c= 50.5 x 4.2 x 6.5 = -1378.65 heat given off (joules, j) For 1 mole Moles of HCL = volume/1000 x concentration = 25/1000 x 1.0 = 0.025 moles 1 x 1 = 1 mole = 0.025 x 1 = 0.025 moles Energy change per mole of HCL = -1378.65/0.025 = -55146/1000 = -55KJ Experiment 2 :- 1.0 mol dm Nitric Acid added to 1.0 dm Sodium Hydroxide Equation :- HNO (aq) + NaOH(aq) H O(l) Mole Ratio :- 1:1 Heat Transfer = m x c x _T M = (25.5 + 25) = 50.5 C = 4.2 T = (27.5 - 21.5) = 6 = 50.5 x 4.2 x 6 = -1277.6 J For 1 Mole. Moles of HNO = volume/1000 x concentration = 25/1000 x 1.0 = 0.025 moles 1 x 1 = 1mole = 0.025 x 1 = 0.025 moles Energy change per mole of HNO = -1277.6/0.025 = -51104/1000 = -51KJ Experiment 3:- 0.5 dm , H SO added to 1.0 mol dm NaOH Equation:- 0.5H SO (aq) ...read more.

Conclusion

I also believe the differences may also be due to the chemical behaviours for each acid. The experiment generally went well, but problems aroused when heat was lost from the cup, ss there was no lid on, so the results are not accurate. Also, it was difficult, to keep the air bubbles out, of the burette and pipette. The main improvements to the methods would be to use a cup with a lid on top, with a small hole, to keep the thermometer, this would achieve a more accurate result. To extend this experiment, other alkalis could be introduced. This experiment has produced reasonably reliable results using the simple equipment. The conclusions which can be drawn from this experiment are that when a acid ie HCL, and an alkali ie NaOH, neutralise each other the reaction is relatively exothermic. This can be more clearly seen, by looking at the graphs. For example, for HCL, once the acid and alkali have neutralised at point 25cm, the temperature is 26.5 c, the starting temperature was 20.0 c. So there has been a rise of 6.5 c, until the neutralising point, so the reaction is exothermic. Once, the heat energy given to the mixture is used up, the temperature decreases, end temperature for HCL is 23 c. This example is a reflection of all three experiments. ...read more.

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