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IB IA: Determination of Heat of Neutralization

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Introduction

NAME: Nur Amira Rozali SUBJECT: Chemistry HL ASSESSMENT CRITERIA: DC, DP, EV INTERNAL ASSESSMENT CHEMISTRY HIGHER LEVEL PRACTICAL 16: DETERMINATION OF HEAT OF NEUTRALISATION Data collection: EXPERIMENT REACTANTS INITIAL TEMPERATURE AVERAGE INITIAL TEMPERATURE,T (x�0.5)�C MAXIMUM TEMPERATURE,T (x�0.5)�C ACID BASE ACID BASE 1 HNO3 NaOH 32.0 31.0 31.50 37.0 2 HNO3 KOH 32.0 30.5 31.25 37.0 3 HCl NaOH 32.0 30.0 31.00 37.0 4 HCl KOH 32.0 30.0 31.00 37.0 5 H2SO4 NaOH 32.0 30.0 31.00 43.0 6 H2SO4 KOH 32.0 30.0 31.00 43.5 TABLE 1 Volume of acid used for each experiment = 50cm3 Volume of base used for each experiment = 50cm3 Specific heat capacity of water = 4.18 kg-1K-1 Concentration of each reactant = 1 M Data processing: EXPERIMENT AVERAGE INITIAL TEMPERATURE, T1, (x�0.5) �C MAXIMUM TEMPERATURE,T2, (x�0.5)�C TEMPERATURE DIFFERENCE, T3, (T1 - T2) �C 1 31.50 37.0 5.50 2 31.25 37.0 5.75 3 31.00 37.0 6.00 4 31.00 37.0 6.00 5 31.00 43.0 12.00 6 31.00 43.5 12.50 TABLE 2 Assumption: The acid and base solutions, each has the same density and specific heat capacity as water, so the mass of "water", m in this case = the total amount of solutions used = 50cm3 + 50cm3 = 100cm3 Specific heat capacity of water, c = 4.18 kJ kg-1 K-1 Temperature change, ? ...read more.

Middle

Experiment 4: Equation for the experiment: HCl(aq) + KOH(aq) � KCl(aq) + H2O(l) Heat evolved = mc? = 0.1 x 4.18 x 6.00 = 2.508 kJ Number of moles of acids used = 1 x 0.05 = 0.05 moles From the equation, 1 mole of HCl reacts with 1 mole of KOH and gives 1 mole of water. Therefore, 0.05 moles of HCl reacts with 0.05 moles of KOH and gives 0.05 moles of water. Therefore, heat of neutralization = 2.508 � 0.05 = -50.16kJmol-1 = -50.2kJmol-1 (3 s.f.) Experiment 5: Equation for the experiment: H2SO4(aq) + 2NaOH(aq) � Na2SO4(aq) + 2H2O(l) Heat evolved = mc? = 0.1 x 4.18 x 12.00 = 5.016 kJ Number of moles of acids used = 1 x 0.05 = 0.05 moles From the equation, 1 mole of H2SO4 reacts with 2 moles of NaOH and gives 2 moles of water. Therefore, 0.05 moles of H2SO4 reacts with 0.1 moles of NaOH and gives 0.1 moles of water. Therefore, heat of neutralization = 5.016 � 0.1 = -50.16kJmol-1 = -50.2kJmol-1 (3 s.f.) Experiment 6: Equation for the experiment: H2SO4(aq) + 2KOH(aq) � K2SO4(aq) + H2O(l) Heat evolved = mc? ...read more.

Conclusion

7. However, enthalpy of neutralization is still approximately 57kJmol-1 because enthalpy change of neutralization is defined in term of the formation of 1 mole of water molecules, not 2 moles of water molecules. 8. The major weakness of this experiment is the lost of significant amount of heat to the surrounding. This causes deviation of the experimental value of enthalpy change of neutralization from the theoretical value. 9. This is unavoidable due to the opening of the polystyrene cup, the pouring of acid into the base solution and the absorption of heat by the polystyrene. 10. Besides, the mixture of acid and base solutions may not be well stirred. 11. Solutions to improve the accuracy of the experiment include: a. Initial temperatures of the solutions are to be taken after several minutes to make sure that the solutions have achieved a consistent temperature. b. Acids should be poured quickly and carefully into the bases solutions to reduce the heat lost. c. Any spilling of the chemical substances should be avoided. d. The mixture of solutions should be stirred throughout the experiment to make sure that the temperature is always consistent. e. The reading of the temperature should be observed from time to time so that a maximum temperature can be obtained. ...read more.

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