Determining the enthalpy change for different chemical reactions Essay Example

Determining the enthalpy change for different chemical reactions Essay I familiarised myself with the Material Safety Data Sheets of toxic substances.PLANNING (A)Enthalpy (H)1 The sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume:Esys is the amount of internal energy, while P and V are respectively pressure and volume of the system.However, to make it simpler, this definition can be shortened. Enthalpy (H) is a measure of heat in the system.To measure the enthalpy we have to first figure out the mass of a substance under a constant pressure and determine the internal energy of the system.The enthalpy change (H)2 is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.Standard conditions3 are used in order to allow experiments that are taken at different locations to come out with the same results. Standard pressure is 1 atmosphere or 1.0135 x 105 pascals. Standard temperature is 25o C. Standard state is the physical state at which an element or a com pound exists at standard conditions.Hypothesis: If the temperature of a given substance is known, we may calculate the enthalpy of this substance.Experiment I Part IPLANNING (B)Requirements:- 1 beaker [250 cm3]- 2 test tubes- thermometer- 60 cm -3 of 2 mol dm-3 hydrochloric acid- solid anhydrous sodium carbonate (Na2CO3) [3.75 g]- balanceProcedure:We were provided with 2 mol dm-3 hydrochloric acid, solid sodium hydrogencarbonate and solid anhydrous sodium carbonate.1. One person in each pair measured 30 cm3 of approximately of 2 mol dm-3 hydrochloric acid into the beaker.2. We took the temperature of the acid and recorded it in table 1.3. We weighted a test tube empty and than again when it contained 2.80 g of anhydrous sodium carbonate.4. We recorded the masses in a table similar to table 1.5. Subsequently we added the weighted portion of Na2CO3 to the acid and stirred the mixture carefully with the thermometer until all the solid has reacted.6. While mixing we recorded the maximum temperature of the solution.DATA COLLECTION2HCl (aq) + Na2CO3 (s)à ¯Ã‚ ¿Ã‚ ½ 2NaCl (aq) + CO2 (g) + H2O (l)Mass of tube + sodium carbonate28.17 gMass of empty test tube25.37 gMass of sodium carbonate used (m)2.80 gTemperature of acid initially21.8 oCTemperature of solution after mixing22.0 oCTemperature change during reaction (?T)0.2 oCTable 1.DATA PROCESSING PRESENTATIONCalculating the enthalpy change:?H = ms?Tm = 2.80 g Na2CO3 + 30.00 g HCl = 32.80 gs = 4.2 J g -1 K -1?T = 0.2 oC = 0.2 K?H = 32.80 g * 4.2 J g -1 K -1 * 0.2 K = 27.55 JCalculating the enthalpy change for 1 mole of Na2CO3:M = 106 um2 = 106 gm1 = 2.80 g106 g 1 mole2.80 g x molesx = 2.80g / 106g * 1 mole= 0.03 mole0.03 mole 27.55 J1 mole x Jx = 27.55J / 0.03mole * 1 mole = 918.33 J?H = 918.33 J = 0.92 kJExperiment I Part IIPLANNING (B)Requirements:- 1 beaker [250 cm3]- 2 test tubes- thermometer- 60 cm -3 of 2 mol dm-3 hydrochloric acid- solid sodium hydrogencarbonate (NaHCO3) [3.75 g]- balanceProcedure:We were p rovided with 2 mol dm-3 hydrochloric acid, solid sodium hydrogencarbonate and solid anhydrous sodium carbonate.1. One person in each pair measured 30 cm3 of approximately of 2 mol dm-3 hydrochloric acid into the beaker.2. We took the temperature of the acid and recorded it in table 2.3. We weighted a test tube empty and than again when it contained 3.70 g of sodium hydrogencarbonate.4. We recorded the masses in a table similar to table 2.5. Subsequently we added the weighted portion of NaHCO3 to the acid and stirred the mixture carefully with the thermometer until all the solid has reacted.6. While mixing we recorded the maximum temperature of the solution.DATA COLLECTIONHCl (aq) + NaHCO3 (s)à ¯Ã‚ ¿Ã‚ ½ NaCl (aq) + H2O (l) + CO2 (g)Mass of tube + sodium hydrogencarbonate29.08 gMass of empty test tube25.38 gMass of sodium hydrogencarbonate used (m)3.70 gTemperature of acid initially21.5 oCTemperature of solution after mixing14.0 oCTemperature change during reaction (?T)7.5 oCTable 2 .DATA PROCESSING PRESENTATIONCalculating the enthalpy change:?H = ms?Tm = 3.70 g NaHCO3 + 30.00 g HCl = 33.70 gs = 4.2 J g -1 K -1?T = 7.5 oC = 7.5 K?H = 33.70 g * 4.2 J g -1 K -1 * 7.5 K = 1061.55 JCalculating the enthalpy change for 1 mole of NaHCO3:M = 84 um2 = 84 gm1 = 3.70 g84 g 1 mole3.70 g x molesx = 3.70g / 84g * 1 mole= 0.04 mole0.04 mole 1061.55 J1 mole x Jx = 1061.55J / 0.04mole * 1 mole = 26538.75 J?H = 26538.75 J = 26.54 kJThermal decomposition of sodium hydrogencarbonate to sodium carbonate:2NaHCO3 (s) à ¯Ã‚ ¿Ã‚ ½ Na2CO3 (s) + H2O (l) + CO2 (g)This may be also shown in the form of an enthalpy cycle:2HCl (aq) + 2NaHCO3 (s) 2NaCl (aq) + CO2 (g) + H2O (l)Na2CO3 (s) + H2O (l) + CO2 (g) + 2HCl (aq)The enthalpy change for the decomposition of sodium hydrogencarbonate may be obtained by determining the enthalpy change of reaction between sodium carbonate and hydrochloric acid and that between sodium hydrogencarbonate and hydrochloric acid.?H = H(products) H(reactants)? H = 0.92 kJ 26.54 kJ = 25.62 kJExperiment II Part IPLANNING (B)Requirements:- 1 beaker [250 cm3]- 2 test tubes- thermometer- 60 cm -3 of 2 mol dm-3 hydrochloric acid- solid calcium oxide (CaO) [3 g]- balanceProcedure:We were provided with 2 mol dm-3 hydrochloric acid, solid calcium carbonate and solid calcium oxide.1. One person in each pair measured 30 cm3 of approximately of 2 mol dm-3 hydrochloric acid into the beaker.2. We took the temperature of the acid and recorded it in table 3.3. We weighted a test tube empty and than again when it contained 3.00 g of solid calcium oxide.4. We recorded the masses in a table similar to table 3.5. Subsequently we added the weighted portion of CaO to the acid and stirred the mixture carefully with the thermometer until the solid has reacted.6. While mixing we recorded the maximum temperature of the solution.DATA COLLECTION2HCl (aq) + CaO (s) à ¯Ã‚ ¿Ã‚ ½ CaCl2 (aq) + H2O (l)Mass of tube + calcium oxide27.92 gMass of empty test tube24.92 gMas s of calcium oxide used (m)3.00 gTemperature of acid initially20.0 oCTemperature of solution after mixing36.0 oCTemperature change during reaction (?T)16.0 oCTable 3.DATA PROCESSING PRESENTATIONCalculating the enthalpy change:?H = ms?Tm = 3.00 g CaO + 30.00 g HCl = 33.00 gs = 4.2 J g -1 K -1?T = 16.0 oC = 16.0 K?H = 33.00 g * 4.2 J g -1 K -1 * 16.0 K = 2217.60 JCalculating the enthalpy change for 1 mole of CaO:M = 56 um2 = 56 gm1 = 3.00 g56 g 1 mole3.00 g x molesx = 3.00g / 56g * 1 mole= 0.05 mole0.05 mole 2217.60 J1 mole x Jx = 2217.60J / 0.05mole * 1 mole =44352 J?H = 44352 J = 44.35 kJExperiment II Part IIPLANNING (B)Requirements:- 1 beaker [250 cm3]- 2 test tubes- thermometer- 60 cm -3 of 2 mol dm-3 hydrochloric acid- solid calcium carbonate (CaCO3) [3.75 g]- balanceProcedure:We were provided with 2 mol dm-3 hydrochloric acid, solid calcium carbonate and solid calcium oxide.1. One person in each pair measured 30 cm3 of approximately of 2 mol dm-3 hydrochloric acid into the beaker.2. We took the temperature of the acid and recorded it in table 4.3. We weighted a test tube empty and than again when it contained 3.00 g of solid calcium carbonate.4. We recorded the masses in a table similar to table 4.5. Subsequently we added the weighted portion of CaCO3 to the acid and stirred the mixture carefully with the thermometer until the solid has reacted.6. While mixing we recorded the maximum temperature of the solution.DATA COLLECTION2HCl (aq) + CaCO3 (s) à ¯Ã‚ ¿Ã‚ ½ CaCl2 (aq) + H2O (l) + CO2 (g)Mass of tube + calcium carbonate27.92 gMass of empty test tube24.92 gMass of calcium carbonate used (m)3.00 gTemperature of acid initially20.0 oCTemperature of solution after mixing22.0 oCTemperature change during reaction (?T)2.0 oCTable 4.DATA PROCESSING PRESENTATIONCalculating the enthalpy change:?H = ms?Tm = 3.00 g CaCO3 + 30.00 g HCl = 33.00 gs = 4.2 J g -1 K -1?T = 2.0 oC = 2.0 K?H = 33.00 g * 4.2 J g -1 K -1 * 2.0 K = 277.20 JCalculating the enthalpy change for 1 mole of CaCO3:M = 100 um2 = 100 gm1 = 3.00 g100 g 1 mole3.00 g x molesx = 3.00g / 100g * 1 mole= 0.03 mole0.03 mole 277.20 J1 mole x Jx = 277.20J / 0.03mole * 1 mole = 9240 J?H = 9240 J = 9.24 kJThermal decomposition of calcium carbonate to calcium oxide:CaCO3 (s) à ¯Ã‚ ¿Ã‚ ½ CaO (s) + CO2 (g)This may be also shown in the form of an enthalpy cycle:2HCl (aq) + CaCO3 (s) CaCl2 (aq) + H2O (l) + CO2 (g)CaO (s) + CO2 (g) + 2HCl (aq)The enthalpy change for the decomposition of calcium carbonate may be obtained by determining the enthalpy change of reaction between calcium oxide and hydrochloric acid and that between calcium carbonate and hydrochloric acid.?H = H(products) H(reactants)?H = 44.35 kJ 9.24 kJ = 35.11 kJCONCLUSION EVALUATIONDetermining the enthalpy change for a chemical reaction allows us to decide whether a given reaction is exothermic or endothermic.If the enthalpy has a negative sign, like in the Experiment I, then the reaction is exothermic. Heat energy is ev olved, so the beaker becomes hotter4.If the sign of enthalpy is positive, then similarly the reaction is endothermic, like in the Experiment II. Heat energy is absorbed and the beaker becomes colder5.The physical properties of reactions (different temperatures of beakers) can be easily distinguished in the real life, even without using any instruments.To evaluate this lab I would suggest using the calorimeter to make the records more reliable than by using thermometer. Room temperature might have had an influence on our results and this was probable the most important source of uncertainty. Masses of substances were measured accurately, although some minute amounts might have been lost while pouring. The pressure remained the same, however little changes may have appeared. We also should pay attention to the amount of gas (CO2) that might have escaped during the experiment. It ought to have been gathered and stored to make the results reliable.