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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 Collection1) Temperature change by using 3.3g of sodium hydrogencarbonateMass of the container on which the sample was weighed = 11.

48gMass of the container and the crystals = 14.98gMass of the container after the crystals were added = 11.70gMass of the crystals that did not react = 00.20gMass of the crystals that reacted = 03.30gTime (s)Temperature ( � C)023.

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03023.06023.09018.012015.015015.018015.021015.024015.527016.030016.02) Temperature change by using 1.88g of sodium carbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 13.48gMass of the container after the crystals were added = 13.60gMass of the crystals that did not react = 00.12gMass of the crystals that reacted = 01.88gTime (s)Temperature ( � C)023.03023.06023.09029.012030.015030.018030.021030.024029.527029.030029.03) Temperature change by using 5.66g of sodium hydrogencarbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 18.48gMass of the container after the crystals were added = 12.82gMass of the crystals that did not react = 01.34gMass of the crystals that reacted = 05.66gTime (s)Temperature ( � C)023.03023.06023.09021.012021.015021.018021.021021.024022.527022.030022.04) Temperature change by using 3.85g of sodium carbonateMass of the container on which the sample was weighed = 11.48gMass of the container and the crystals = 15.48gMass of the container after the crystals were added = 11.63gMass of the crystals that did not react = 00.15gMass of the crystals that reacted = 03.85gTime (s)Temperature ( � C)023.03023.06023.09028.512028.515028.518028.521028.524028.027028.030028.0UNCERTAINITIES+/- 0.01g : Digital weighing scale+/- 0.01s : Stop watch+/- 0.05cm3 : Measuring cylinder+/- 0.05�C : ThermometerOBSERVATIONS- When hydrochloric acid is added to sodium carbonate, some effervescence (bubbles appear) is observed because of the liberation of carbon dioxide gas. It is completely soluble and in the process it starts getting warmer.- The same thing happens when sodium hydrogencarbonate is added to hydrochloric acid except for the fact it cools down instead of getting warmer.CHEMICALS (QUALITATIVE DATA)1) Hydrochloric acid- It is colorless and odorless. It is a monoprotic acid, that is, it produces 1 hydrogen ion when completely dissolved in water. The molarity of hydrochloric acid used in this experiment is 2M.2) Sodium carbonate- Sodium Carbonate is a white, crystalline compound soluble in water (absorbing moisture from the air) but insoluble in alcohol. It forms a strongly alkaline water solution. It is also known as soda ash3) Sodium hydrogencarbonate- sodium bicarbonate or sodium hydrogen carbonate, chemical compound, NaHCO3, a white crystalline or granular powder, commonly known as bicarbonate of soda or baking soda. It is soluble in water and very slightly soluble in alcohol.DATA PROCESSING AND PRESENTATIONFinding the enthalpy of reaction for the following equation:2NaHCO3 Na2CO3 + H2O +CO2a) Using 1.88g of sodium carbonate and 3.3g of sodium hydrogencarbonateEnthalpy cycle for the reaction?H12NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)?H2 ?H3Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc?TSo, Q1 = 25g * 4.18 * (15-23)Q1 = 25* 4.18 * -8Therefore, Q1 = -836 J= -0.836KJCalculating the number of moles present in 3.3g of NaHCO3Number of moles = mass(g)/ molar massMass (g) = 3.3gMolar mass = 23 + 1 + 12 +(16*3)= 23 + 1 + 12 + 48= 84 g/molnumber of moles present in 3.3g of NaHCO3 = 3.3g/ (84g/mol)= 0.039 molesCalculating the amount of energy given out by 1 moleIf 0.039 moles of NaHCO3 give -0.836KJ of energy then1 mole would give out (-0.836/0.039 = -21.44KJ) of energyTherefore, ?H1 = -21.44 KJ/molCalculating ?H3 by the above method, that is, the reaction between Na2CO3 and HClQ2 = Mc?TSo, Q2 = 25 * 4.18 * (30-23)Q2 = 25 * 4.18 * 7Therefore, Q2 = 731.5J= 0.7315KJCalculating the number of moles present in 1.8g of Na2CO3Number of moles = mass (g)/ molar massMass (g) = 1.8gMolar mass = (23 * 2) + 12 +(16*3)= 46 +12 + 48= 106 g/molnumber of moles present in 1.8g of Na2CO3 = 1.8g/ (106g/mol)= 0.018 molesCalculating the amount of energy given out by 1 moleIf 0.018 moles of Na2CO3 give -0.731KJ of energy then1 mole would give out (-0.7315/0.018 = 40.64KJ) of energyTherefore, ?H3 = 40.64 KJ/molIn order to find the enthalpy of reaction for:2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hess’s law which states that 2 ?H1 = ?H2 + ?H3?H2 = 2 ?H1 – ?H3so, 2 ?H1 = 2 * -21.44 KJ/mol= -42.88KJ/mol?H3 = 40.64 KJ/molTherefore, ?H2 = -42.88 – 40.64= -83.52KJ/molb) Calculating the enthalpy change of reaction using 5.66g of sodium hydrogencarbonate, 3.85g of sodium carbonate and 50cm� of HCl.Enthalpy cycle for the reaction?H12NaHCO3(s) + 2 HCl 2NaCl(aq) +2 CO2(g)+ 2 H2O(l)?H2 ?H3Na2CO3(s) + CO2(g) + H2O + 2 HCL(aq)Heat released during the reaction between HCl and NaHCO3 (Q1)= Mc?TSo, Q1 = 50g * 4.18 * (21.5-23.0)Q1 = 50* 4.18 * – 1.5Therefore, Q1 = -331.50 J= -0.3135KJ `Calculating the number of moles present in 5.66g of NaHCO3Number of moles = mass (g)/ molar massMass (g) = 5.66gMolar mass = 23 + 1 + 12 + (16*3)= 23 + 1 + 12 + 48= 84 g/molnumber of moles present in 5.66g of NaHCO3 = 5.66g/ (84g/mol)= 0.0674 molesCalculating the amount of energy given out by 1 moleIf 0.0674 moles of Na2CO3 give -0.3135KJ of energy then1 mole would give out (-0.3135/0.0674 = -4.65KJ) of energyTherefore, ?H1 = -4.65 KJ/molCalculating ?H3 by the above method, that is, the reaction between Na2CO3 and HClQ2 = Mc?TSo, Q2 = 50 * 4.18 * (28.5-23)Q2 = 50 * 4.18 * 5.5Therefore, Q2 = 1149.5J= 1.1495KJCalculating the number of moles present in 3.85g of Na2CO3Number of moles = mass (g)/ molar massMass (g) = 3.85gMolar mass = (23 * 2) + 12 + (16*3)= 46 +12 + 48= 106 g/molnumber of moles present in 1.8g of Na2CO3 = 3.85g/ (106g/mol)= 0.036 molesCalculating the amount of energy given out by 1 moleIf 0.036 moles of Na2CO3 give 1.1495KJ of energy then1 mole would give out (1.1495KJ/0.036 = 31.93KJ) of energyTherefore, ?H3 = 31.93KJ/molIn order to find the enthalpy of reaction for:2NaHCO3 Na2CO3 + H2O+ CO2; we use the Hess’s law which states that 2 ?H1 = ?H2 + ?H3?H2 = 2 ?H1 – ?H3so, 2 ?H1 = 2 * -4.65 KJ/mol= -9.3KJ/mol?H3 = 31.93KJ/molTherefore, ?H2 = -9.3KJ/mol – 31.93KJ/mol= -41.23KJ/molc) Error analysisDigital weighing scale: 1) 0.01/3.3 * 100= �0.3%2) 0.01/5.66 * 100= �0.18%Stop watch : 0.01/300 * 100= � 3.3*10^-3Measuring cylinder : 1) 0.05/25 * 100= �0.2%2) 0.05/50 * 100= �0.1%Thermometer : 0.05/23 * 100= �0.22%Total percentage error = 0.22%+ 0.1% + 3.3*10^-3 + 0.18%+ 0.3%= �0.8033%Accounting for the error ?H2 = � -41.23KJ/mol= � -83.52KJ/molConclusionThe 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 andsodium 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.

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Hess Cycle. (2019, Dec 06). Retrieved from https://paperap.com/paper-on-an-experiment-to-determine-the-enthalpy-changes-using-hesss-law/

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