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

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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.

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