• 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...


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.


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.


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 ...

    In this case, the reaction is as follows: KOH(aq) + HCl(aq) --> KCl(aq) + H2O(l) (Winnipeg School Division, n.d.) Meaning, the initial temperature will be of the two solutions of the 40.0 cm3 1.00 mol dm3 HCl(aq) and 1.00 mol dm-3 KOH(aq)

  2. IB chemistry revision notes

    Change, * It is a measure of the number of possible ways that a system can be organised. This includes both the particles in the system and the quanta of energy within the system. * For anything spontaneous, S must increase (positive)

  1. Chemistry Extended Essay - Viscosity of Xanthan Gum solutions

    In the context of the experiment that will be performed to measure the effect of viscosity on these reactions, the solution in the reaction will be replaced with solutions containing varying concentration gradients of xanthan gum. The salt solution is maintained at the same concentration, but the water that would

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

    The total percentage deviation from the literature value was unimpressively large but at the same time, it should be taken into consideration that it is not possible to perform such experiments require much more sophisticated instruments that those readily available in the school lab.

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

    done by the bomb during the reaction (Mashkevich, 1995). However, the combustion takes place in a sealed container with constant volume, the work done on or by the system is also zero, so that âcU=0 (5) Considering the combustion of the system for both sample and the cotton fuse, âcUsample + âcUcotton = 0 (6)

  1. The chemistry of atmospheric and water pollution.

    + O. (g) ï Cl .(g) + O2 (g) Ozone is destroyed. The chlorine free radical is free to destroy thousands of more ozone molecules just like this until it reacts with methane. Ozone depletion is more frequent in winter and spring due to more ice particles.

  2. Analysis of the Standard Enthalpy of Combustion for Alcohols

    Apparatus: temperature probe datalogger device 5 cm 25 cm alcohol lamp loggerpro collector on computer heatproof mat 100 cm3 distilled water conical flask clamp clamp Variables: 1. Independent The alcohol used to heat water will be changed, however all alcohols will be primary.

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