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

Enthalpy Change

Extracts from this document...

Introduction

Skill P - Planning * The aim of this practical experiment is to determine the enthalpy change for this reaction by an indirect method based on Hess's Law. Both Calcium Oxide and Calcium Carbonate react readily with 2 mol dm Hydrochloric Acid solutions. The temperature changes during these reactions can be measured and the enthalpy changes calculated. * Chemicals and apparatus: * 250 cm measuring cylinder * 2 mol dm Hydrochloric Acid * 250 cm beaker * 0 - 10 0� C thermometer (graduations to 1� C ) * 2.4 - 2.6 g of Calcium Carbonate * 1.3 - 1.5 g of Calcium Oxide * Procedure 1. Weigh out a weighing bottle containing between 2.4 and 2.6 g of Calcium Carbonate 2. Weigh out a weighing bottle containing between 1.3 - 1.5 g of Calcium Oxide 3. Using the measuring cylinder provided place 50 cm3 of 2 mol dm-3 hydrochloric acid (an excess) into a 250 cm3 glass beaker. 4. Measure the temperature of the acid using the thermometer provided. 5. Add the calcium carbonate/ calcium oxide to the acid. 6. Take the temperature again when the reaction is complete. ...read more.

Middle

4.2 7 H= 1514.1 We have to calculate the molar heat. Therefore, 1 mole of CaO = 40 + 16 = 56 KJ mole CaCO CaCl We can calculate based on Hess's Law. Therefore, Hence, = CaO = -18.19 - (- 56.52) = 38.33 KJ mole (Endothermic reaction) Skill E - Evaluating Evidence A balance was used fact that leads to limited accuracy, because it is open to the air and is susceptible to outside forces. The procedure was good and the mix of reactants was intimate, so we can say that the experiment were suitable. The first reaction (CaCO) was a very slow reaction. Therefore, when reading the temperature we had to check it constantly and pick up the biggest value. The second one (CaO) was a very fast reaction, so when reading the temperature, we had to choose the first biggest value, because after that that point the temperature will decrease and the results will be anomalous. 2.42 g of CaCO and 1.50 g of CaO was taken with a tendency to show variation of 0.01g. This will lead to 0.004% error for CaCO and 0.006% for CaO. ...read more.

Conclusion

A greater proportion of molecules exceed the activation energy at higher temperature. The reaction takes place quicker if the reactant, hydrochloric acid, is more concentrated, because there are more molecules per volume and they will exceed the activation energy. This thing applies to the first reaction which involved CaCO, and which was very slowly because it was in lumps form. So, the first experiment could be considerable improved by turning the CaCOin powdered form. In order to find out the enthalpy change we used Hess's Law which shows that whatever the route from given reactants to products, the overall energy change must be the same. Therefore, each of the two enthalpies calculated from the two experiments count half per cent on the final conclusion. Hence, CaCO 2.42 0.01g 0.4 % error on final conclusion + the temperature change CaO 1.50 g 0.01g 0.6 % error on final conclusion + the temperature change A considerably effect on the validity of the final conclusion have the signs of the heat evolved (whether the reactions are endothermic or exothermic). The first two experiments are exothermic reactions and as the final enthalpy change has a positive value, the reaction is an endothermic one. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Inorganic 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 AS and A Level Inorganic Chemistry essays

  1. Peer reviewed

    Determining the concentration of acid in a given solution

    5 star(s)

    Concentration (mol dm-3) = Moles / Volume (dm3) Concentration of sulfuric acid: 0.00253 / (24.6/1000) = 0.10285 mol dm-3 I will round the concentration of sulfuric acid down to 0.103 mol dm-3 (3 sig figs) Evaluation 7, 8 Procedural Uncertainties There are lots of possible errors that could have occurred whilst I was doing the experiment.

  2. effects Concentration and Temperature on the Rate of Reaction

    Figure 4 shows the change when the temperature is increased by 10K. It shows that there is still a wide range of energies but now there are more particles with a higher kinetic energy. At the higher temperature of 310 Kelvin there are many more molecules which have reached the activation enthalpy level than at 300K.

  1. Lab report Determination of Enthalpy Change of Neutralization

    /1000 = 0.05 Kg Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234) * 840) * 10.2 = 2245.741176 J HNO3 (aq) + NaOH (aq) � NaNO3 (aq) + H2O (l) No of mole of NaOH = 2 * 0.025 = 0.05 mol According

  2. Thermal Decomposition of Calcium Carbonate

    calculate the number of moles in calcium oxide: Number of moles = Mass Mr = 1.58 56.1 = 0.02816399287 moles = 0.028 moles (3 d.p.) Therefore, H1 = - 2.1 0.028 = - 75 J mol-1 Hess' Law states that 'for a given overall chemical reaction, the overall energy is

  1. Free essay

    Determination of the Heat of Formation of Calcium and Calcium Carbonate

    Number of mole of calcium used = = 0.02569 mol Heat would have been evolved by one mole of calcium atoms = = 465.94 kJ mol-1 7. In the calculation, only the mass of calcium is the limiting factors. The concentration and the volume of hydrochloric acid do not involve in the calculation.

  2. The periodic table

    This was the number of grams of an element that combined with 8 g of oxygen (They used this because 8 g of oxygen combine with 1 g hydrogen so 8 g of oxygen was equivalent to 1 g hydrogen.)

  1. Finding Out how much Acid there is in a Solution

    For the first titration, as it was the first to be completed after the rough titration, we still were not certain exactly where the end point fell. As our rough titration turned out to have a slightly greater volume than our final average titre, it is possible that this is

  2. Scientific Practical Techniques

    have done this I did the same thing for pH 4 and for the other unknown PH To improve the measurement and get accurate result, I would suggest

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