• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15

An experiment to determine the enthalpy changes using Hess's law

Extracts from this document...

Introduction

CHEMISTRY LAB REPORT Name: Saurabh Lakshman Grade 11 (IB) 17/02/2006 An experiment to determine the enthalpy changes using Hess's law Background Information The main idea behind this experiment is to find out the temperature difference between the room temperature and the final temperature. Sodium carbonate, sodium hydrogencarbonate and hydrochloric acid were used in this experiment. Sodium carbonate, also known as soda ash is got from the reaction of carbonic acid and sodium hydroxide while sodium hydrogencarbonate (baking soda) is a salt formed by the partial replacement of hydrogen by sodium. Data Collection 1) Temperature change by using 3.3g of sodium hydrogencarbonate Mass of the container on which the sample was weighed = 11.48g Mass of the container and the crystals = 14.98g Mass of the container after the crystals were added = 11.70g Mass of the crystals that did not react = 00.20g Mass of the crystals that reacted = 03.30g Time (s) Temperature ( � C) 0 23.0 30 23.0 60 23.0 90 18.0 120 15.0 150 15.0 180 15.0 210 15.0 240 15.5 270 16.0 300 16.0 2) Temperature change by using 1.88g of sodium carbonate Mass of the container on which the sample was weighed = 11.48g Mass of the container and the crystals = 13.48g Mass of the container after the crystals were ...read more.

Middle

Q1 = 25* 4.18 * -8 Therefore, Q1 = -836 J = -0.836KJ Calculating the number of moles present in 3.3g of NaHCO3 Number of moles = mass(g)/ molar mass Mass (g) = 3.3g Molar mass = 23 + 1 + 12 +(16*3) = 23 + 1 + 12 + 48 = 84 g/mol number of moles present in 3.3g of NaHCO3 = 3.3g/ (84g/mol) = 0.039 moles Calculating the amount of energy given out by 1 mole If 0.039 moles of NaHCO3 give -0.836KJ of energy then 1 mole would give out (-0.836/0.039 = -21.44KJ) of energy Therefore, ?H1 = -21.44 KJ/mol Calculating ?H3 by the above method, that is, the reaction between Na2CO3 and HCl Q2 = Mc?T So, Q2 = 25 * 4.18 * (30-23) Q2 = 25 * 4.18 * 7 Therefore, Q2 = 731.5J = 0.7315KJ Calculating the number of moles present in 1.8g of Na2CO3 Number of moles = mass (g)/ molar mass Mass (g) = 1.8g Molar mass = (23 * 2) + 12 +(16*3) = 46 +12 + 48 = 106 g/mol number of moles present in 1.8g of Na2CO3 = 1.8g/ (106g/mol) = 0.018 moles Calculating the amount of energy given out by 1 mole If 0.018 moles of Na2CO3 give -0.731KJ of energy then 1 mole would give out (-0.7315/0.018 = 40.64KJ) ...read more.

Conclusion

= � -83.52KJ/mol Conclusion The reaction between sodium hydrogencarbonate and HCl is endothermic, that is, heat is being absorbed in the reaction and the reaction between sodium carbonate and HCl is exothermic because temperature is given out to the surroundings. Also, in the second part of the experiment when the volume of HCl is increased and also the masses of sodium hydrogencarbonate and sodium carbonate is increased, the temperature difference in the reaction is less than before when the mass were less. The enthalpy of reaction is also decreased in the second part. Evaluation: Reasons for shortcoming in the answers are as follows: 1) While transferring the HCl from the measuring cylinder to beaker, there is a possibility of leaving out some amount of HCl in the beaker itself. 2) An analogue thermometer was used so the temperature may not have been accurate. 3) Some systematic errors in the equipment might have led to some slight changes in the readings. Solution to the above problems: 1) The first problem stated above is a personal error; So it can only be overcome by practice and improving one's concentration while doing the experiment. 2) The second problem could have been overcome by the use of digital thermometer which is more accurate than an analogue thermometer. ?? ?? ?? ?? 1 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous 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 GCSE Aqueous Chemistry essays

  1. Indirect determination of enthalpy change of decomposition of sodium hydrogen carbonate by thermochemical measurement ...

    This means that the reaction is endothermic and ?H is positive (as in experiment 1). ?H is positive because the initial temperature of the reactants is higher than the final temperature, and ?T is therefore also positive. Estimating values of ?T and ?H Before completing my experiments, I roughly estimated the value of ?T for experiment 1 (NaHCO3 and HCl)

  2. An Experiment to Determine the Enthalpy Change for the Decomposition of Calcium Carbonate.

    and calcium oxide (CaO) I then need to calculate how many moles of calcium carbonate (CaCO3) and calcium oxide (CaO) were reacted and then determine the enthalpy change for 1 mole of calcium carbonate (CaCO3) and calcium oxide (CaO). By using the formula: Moles = Mass Mr Then I can

  1. Application of Hess's Law

    gain during reaction * Spatula for measuring out the required amount of powder * Digital scales (to the nearest hundredth of a gram) for weighing powder * Paper tray to prevent loss of powder when measuring on scales * Thermometer for estimating temperatures * Stirring Rod to stir dissolving powders

  2. Softening hard water with sodium carbonate (Na2CO3)

    A lather will only form when all the calcium ions have been removed from the water, and there are some stearate ions left. Therefore when there wasn't enough CO32- to remove all the Ca2+ from the water scum had to form in order to remove the last remnants of the

  1. Analysing; Enthalpy of Decomposition of Sodium Hydrogencarbonate

    Mass of solid + cup (grams) Mass of solid (grams) NaHCO3 1.240 4.740 3.500 NaHCO3 1.070 4.570 3.500 NaHCO3 1.129 3.629 3.500 Table of results Time Experiment 1 Experiment 2 Experiment 3 Temp. of solid (0C) Temp. of solution (0C) Temp. of solid (0C) Temp. of solution (0C) Temp. of solid (0C) Temp. of solution (0C)

  2. Titrating Sodium hydroxide with an unknown molarity, against hydrochloric acid to find its' molarity.

    * Having at least 3 results which were within 0.1cm3 of each other. Precision Precise - the data points are close together (there can be a random error) 1. There were some procedures that I carried out in order to ensure that this experiment and the results that were obtained

  1. Determine the enthalpy change of the decomposition of sodium hydrogen carbonate by thermochemical measurement ...

    Unfortunaetly, it is impossible to get this maximum/minimum temperature using the thermometer alone. The temperature recorded at four minutes is not 100% accurate as some of the reaction has already been taking place since 3.5 minutes.

  2. Investigation of the carbonate - bicarbonate system

    In rivers, lakes and ocean, buffering involves the presence of carbon dioxide (CO2), carbonic acid (H2CO3), bicarbonate (HCO3-), carbonate (CO3-2) and hydrogen [H+]. The reaction involved is thus; CO2 (aq) + H2O H2CO3 (aq)......................................................1 H2CO3- (aq) H+ (aq) + HCO3-(aq)

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