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Determine the enthalpy of neutralization for HCl +NaOH, CH3COOH +NaOH, and 1/2H2SO4 +NaOH.

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Introduction

Chemistry lab report Aim- to determine then enthalpy of neutralization for 1. HCl +NaOH 2. CH3COOH +NaOH 3. 1/2H2SO4 +NaOH And state the trends in the enthalpy of neutralization for acids. Hypothesis- The enthalpy of neutralization is the heat produced when an acid and a mole react together to form one mole of water. Strong acids completely dissociate in water. Weak acids on the other hand partially dissociate in water. Thus a certain amount of the energy evolved, in reacting a weak acid with a base, is used up in ionizing the unionized molecules. Hence the enthalpy of neutralization should be more for the stronger acids and lesser for the weak acids. Thus the enthalpy of neutralization will be more for hydrochloric acid and sulphuric acid, which are stronger acids and lesser for ethanoic acid, which is a weak acid. The hydrogen ion from the acid and the hydroxide ion from the base react to form the water molecule. the general equation of the reaction would be- H+ + OH- H2O The energy released when the above reaction takes place is -57.3 KJ/mol. ...read more.

Middle

was added to the hydrochloric acid through the opening on the lid of the insulated calorimeter. The final temperature was recorded. Reaction 3- 1. 25 cm3 of dilute sulphuric acid (1.12M) is taken in a glass beaker using a 25-cm3 pipette. 2. The beaker was placed in an insulated calorimeter and the initial temperature recorded using a mercury thermometer. 3. 50 cm3 of a aqueous NaOH (1.04M) was added to the dilute HCl 4. The apparatus was stirred after the lid was closed 5. Since the reaction is quick the thermometer was kept immersed into reaction mixture as soon as the NaOH (aq) was added to the hydrochloric acid through the small opening on the lid of the insulated calorimeter. The final temperature was recorded. After the following steps had been carried out the in each reaction the following step wee carried out 1. The heat evolved in the reaction was calculated using the equation Heat evolved in the reaction= heat gained by water 2. ...read more.

Conclusion

The literature value for the enthalpy of neutralization for sulphuric acid and hydrochloric acid should be the same that is -57.3 KJ/mol. Hence the percentage of error is 17% and 30% respectively. The error for the enthalpy of neutralization (literature value 50.1) is 5.18%. In fact it is expected that the enthalpy of neutralization is the least for the reaction with ethanoic acid but this was not so. It could be due to the following reasons * Some heat was lost to the surroundings due to which the final temperature was accurate. * The apparatus was not adequately stirred. * Since the reactions were quick it is possible that the thermometer was inserted after the reaction was over and thus the exact final temperature was not recorded. * It is also assumed that the acids would have the same density and specific heat capacity as that of water which is not so. Hence this could cause a shift in the values These errors could be minimized in the following ways * The heat loss could be minimized by carrying out the experiment in a closed chamber to minimize the heat loss * Care should be taken the apparatus was adequately stirred. ...read more.

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