• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

IB IA: Determination of Heat of Neutralization

Extracts from this document...

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Enthalpy Change Design Lab (6/6)How does changing the initial temperature (19C, 25C, 35C, and ...

    and KOH(aq) will be multiple. Part of the calculations involved in determining the molar enthalpy change of a reaction is the mass of the reactants. In this investigation, the mass of the reactants will be determined from the volume of the 1.00 mol dm-3 KOH(aq)

  2. IB chemistry revision notes

    The Equilibrium Constant * for a homogenous reaction. * or * can remain the same when equilibrium shifts. This relies on the changes simply cancelling out. * The magnitude of : o Is related to the position of equilibrium. o When >>1, the reaction goes almost to completion (products favoured)

  1. Thermodynamics: Enthalpy of Neutralization and Calorimetry

    See figure 1 for a picture of the set up. To find the heat capacity the calorimeter is first weighed (note: calorimeter must also be weighed after any new substance is added). Next, 50 mL of distilled water are added inside and have the temperature is taken.

  2. To determine the standard enthalpy of formation of Magnesium Oxide using Hess Law.

    + O2 (g) ? MgO (s) ----------------- ?HMgO = -432.2kJ.mol-1 Error Propagation: Total Error = Random Error + Systematic Error To calculate the total random error percentage, the percentage uncertainty of the smallest reading on each apparatus is added. Also, since the final value of ?HMgO was computed using the

  1. Paper Chromatography on Amino Acids IA

    Paper chromatography involves two stages: a stationary phase and a mobile phase. The stationary phase is the water trapped within the cellulose in the paper and the mobile phase is the butan-1-ol, ethanoic acid and water solvent soaking up the paper through capillary action.

  2. The aim of this experiment is to examine the enthalpy of combustion of the ...

    = 1.50g ( from Figure 1) Molar mass of pentanol is needed for further calculations and it can be calculated in a following way : M (C5H11OH) = 5*Ar (C) + Ar (O) + 12*Ar (H) = 5*12.01 + 16.00 + + 12*1.01 = 88.17 g mol-1 From here the amount of burned methanol can be calculated : n (CH3OH)

  1. Analysis of the Standard Enthalpy of Combustion for Alcohols

    Volume of liquid used Measure 100cm3 of distilled water by using 100 cm3 ± 0.08 cm3 graduated pipette for each trial. If the volume was not exactly 100 cm3 it would directly affect the mass of the water which will affect the q=mcâT value and thus the âH value.

  2. Bomb calorimetry. The goal of this experiment was to use temperature data over ...

    and yield the value for the standard molar heat of formation for the sample. Theory The thermodynamic of combustion: The equation for combustion of naphthalene and benzoic acid are the following equations: C6H5COOH (s) + 7.5 O2 (g) ï® 7 CO2 (g) + 3 H2O (l) + 26.434 kJ/g (1)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work