Chemical: Thermochemical

Application of the first law of thermodynamics to chemical events called thermochemical, which discusses heat accompanying a chemical reaction. Chemical reactions including isothermal process, when done in the open air, the heat of reactionqp = ΔHAs a result, heat can be calculated from the change in the reaction enthalpyq = ΔHreaksi = Hhasil comments - HpereaksiSo that there is uniformity Womanizer set the default state, the temperature of 25 ° C and a pressure of 1 atm. Thus, thermochemical calculations based on the state standards, for example:AB + CD → AC + BD ΔHo = x kJ mol-1ΔHo is a symbol (notation) reaction enthalpy change in circumstances.Judging from the type of reaction, there are four types of heat, which is as follows.• Heat of formation, is the heat that accompanies the formation of one mole of a compound directly from its elements. For example, ammonia (NH3), must be made of nitrogen and hydrogen gas, so the reaction:½ N2 (g) + 1 ½ H2 (g) → NH3 (g) ΔHo = -46 kJ mol-1Since 1 mol NH3 should it coefficient of the reaction of nitrogen and hydrogen may be written as a fraction. The energy released by 46 kJ known heats of formation of ammonia (ΔHoNH3).• Heat decomposition, (opposite of heats of formation), the heat that accompanies the decomposition of 1 mole of the compound directly into its elements, egHF (g) → ½ H2 (g) + ½ F2 (g) ΔH = +271 kJ mol-1• Heat neutralization, is the heat that accompanies the formation of 1 mol of water from the neutralization reaction (acids and bases), for example:HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) ΔH = 121 kJ mol-1• Heat of reaction, the heat that accompanies a reaction with the simplest coefficients, for example:3H2 (g) + N2 (g) → 2NH3 (g) ΔH = -92 kJOf heat that accompanies a reaction can be determined by laboratory experiments. Measured reagents reacted in the calorimeter, which is a tool that will measure the heat produced or absorbed by the reaction. If the reaction is exothermic, the heat produced will raise the temperature of water in the calorimeter. The amount of heat can be calculated as the temperature rises and the water mass in the device. Conversely, if the reaction is endothermic, then the water temperature will drop so that it can be calculated reaction heat is absorbed.• Shukri S. 1999. Basic Chemistry I, London: ITB PressHeat is the transfer of thermal energy. Heat flow from one part to another or from one system to the other system or due to temperature differences. During pengalirannya we do not know the whole process, for example, the state eventually. Heat is not known during the process. A known quantity in the process is the flow rate Q which is a function of time.Almost all chemical reactions absorb or release energy, usually in the form of heat. Calor (heat) is the transfer of thermal energy between two objects of different temperature. It is often said "heat flow" from the hot object to the cold Bena. Although the heat itself implies the transfer of energy, usually called "heat absorbed" or "heat released" when describing the energy changes that occur during the process.To analyze the energy changes associated with chemical reactions should first define the system, or a certain part of the natural concern to us. For chemists, the system usually includes substances that are involved in the change kiia and physics. For example, in an acid-base neutralization experiments, the system can be a calorimeter containing HCl in which NaOH is added. The rest of the world outside the system is called the (surrounding).
Amount of heat changes during chemical changes can be measured in a calorimeter (measured were temperature). Calorimeter consists of a tube made in such a way that there is no exchange or transfer of heat to the environment around it, or around it. While there it can happen as little as possible, so it can be ignored.Termus bottles can be used as a simple calorimeter, connected or wrapped in foam pastik, but it should be noted that there perukaran between the calorimeter and its contents so menera calorimeter (ie heat is absorbed by the surface of the calorimeter), as expeditiously as possible in accordance with the lessons learned.The amount of heat absorbed by the calorimeter to raise its temperature by 1 ° C is called calorimeter constant. One method used to determine the calorimeter constant is to include a number of "cold water" with Mass moles, and temperature T with a number of "hot water" with the mass of a mole, and the temperature T in the calorimeter were determined tetapannya mixed water temperature is no more of 30. If the calorimeter does not absorb heat from the mixture. Heat given hot water to be equal to the heat absorbed by the cold water. Price constant temperature degan indirect calorimeter can be measured, which can be measured is the change in temperature.• Team Teaching Basic Chemistry I. , 2011. Practical Guidance Basic Chemistry I, Gorontalo: UNGD. Tools and Materialsa. Tool1. CalorimeterServes as a place to measure heat changes during chemical reactions2. Measure GlassesServes to measure the volume of solution3. BeakerServes as a place to accommodate solvents or chemicals4. ThermometerServes to measure the temperature5. Pasteur pipetteServes to shed a small amount of solution6. Turkish bathServes to heat the waterb. Material1. WaterPhysical properties:• Colorless liquid• Boiling point 100 ° C, the melting point of 0 ° CChemical properties:• Molar mass 18.0153 g / mol• Solvents are many kinds of chemical substances• Non-combustible2. EthanolPhysical properties:• Liquid no berwarana• Boiling point -114.3 ° C, a melting point of 78.4 ° CChemical properties:• Molar mass 46.07 g / mol• Soluble in water• Highly flammable and evaporates3. HClPhysical properties:• Liquid colorless to pale yellow• The melting point of -27.32 ° C, the boiling point of 48-110 ° C.Chemical properties:• Molar mass 36.46 g / mol• Soluble in water• Corrosive4. NaOHPhysical properties:• The solid solution or white• The melting point of 318 ° C, boiling point 1390 ° CChemical properties:• Molar mass 39.9971 g / mol• Soluble in water• Non-combustibleE. Work Procedures1. Determination of constant Calorimeter2. Determination of Ethanol Heat Dissolution In Water3. Determination of the heat of neutralization of HCl and NaOHF. Observation Results1. Determination of constants calorimeterT = 30 ° C cold water hot water T = 41 oCV = 20 ml hot water cold water V = 20 mlT mixt T (oC)12345678910 35 oC35 ° C35 ° C34 oC34 oC34 oC34 oC34 oC34 oC33 oC2. Determination of Ethanol Heat Dissolution In Watera) To 18 ml water and 29 ml of ethanolT = 31 oC ethanol water T = 32 oCT T water mixturet T (C) t T (oC)1/211 1/22 32 oC31 oC31 oC31 oC ½11 ½22 ½33 ½4 33 oC34 oC34 oC34 oC34 oC35oC35oC35oC
b) For water 27 ml and 19 ml of ethanolT = 31 oC ethanol water T = 32 oCT T water mixturet T (C) t T (oC)1/211 1/22 31 oC31 oC31 oC31 oC ½11 ½22 ½33 ½4 35 oC35 ° C35 ° C35 ° C35 ° C35oC35oC35oC
c) To 36 ml of water and 14.5 ml of ethanolT = 30 oC ethanol water T = 31 oCT T water mixturet T (C) t T (oC)1/211 1/22 31 oC31 oC31 oC31 oC ½11 ½22 ½33 ½4 35 oC35 ° C35 ° C35 ° C35 ° C35oC35oC35oC
3. Determination of the heat of neutralization of HCl and NaOHT = 31 ° C T HCl NaOH = 31 oCt T (oC)½11 ½22 ½33 ½44 ½5 4040th40th40th40th39th39th39th39th39thG. Calculation and Discussiona. Calculation1. Determination of constants calorimeterDik: Va: Volume of cold water: 20 mlVa: Volume of hot water: 20 mlρ water: 1 g / mlWater S: J / g KTA1: 30 ° C + 273 = 303 KTA2: 41 ° C + 273 = 314 KTable temperature water mixturet (min) T (K)12345678910 308308308307307307307307307306DIT: calorimeter constant (K) .....?Completion:• Calculate the mass of cold waterMA1 = VA1= 20 gr• Calculate the temperature of the mixtureTcamp = ΣT / n= 308 +308 +308 +307 +307 +307 +307 +307 +307 +30610 K= 307210K= 307.2 K• Calculate the change in temperature of the cold waterΔT1 = Tcamp - TA1= 307.2 K - 303 K = 4.2 K• Calculate the change in temperature of the hot waterΔT1 = TA2 - Tcamp= 314 K - 307.2 K = 6.8 K• Calculate the heat absorbed by the cold waterq1 = MA1 x Sair x ΔT1= 20 grams x 4.2 J / g · K x 4.2 K= 352.8 Joule• Calculate the heat released hot waterq1 = Ma2 x Sair x ΔT2= 20 grams x 4.2 J / g · K x 6.8 K= 571.2 Joule• Calculate the heat received by the calorimeterq3 = q2 - q1= 571.2 J - 352.8 J= 218.4 J• Calculate the calorimeter constantQ3/ΔT2 = 218.4 K = J4, 2 K = 52 J / K• Graph hubungana between the temperature interval2. Determination of the heat of ethanol dissolving in watera. To a mixture of 18 ml water with 29 ml of ethanolDik: ρair = 1 g / mlρetanol = 0.79 g / mlS water = 4.2 J / g · KS ethanol = 1.92 J / g · KWater T = 304.25 KT = 304 K ethanolDIT: dissolution enthalpy (ΔH)?Adjustment: - Calculating the mass of water - Calculate the mass of ethanolMa = Va x ρair M ethanol = Vetanol xρetanol= 18 ml x 1 g / ml = 29 ml x 0.79 g / ml= 18 g = 22.91 g• Calculate the temperature of the mixtureTcamp = ΣT / n = 2458/8 = 307.25 KCamp ΔT1 = T - T water = 307.25 to 304.25 = 3 KΔT2 camp = T - T ethanol = 307.25 to 304 = 3.25 K• Heat absorbed water (qa) - Heat is absorbed ethanol (qe)qa = Ma x S x ΔT1 qe = Methanol x S x ΔT2= 18 x 4.2 x 3 = 226.8 J = 22 x 1.92 x 3.25= 226.8 J = 142.95 J• Heat is absorbed calorimeter (QK)QK = K x ΔT2= 52 x 32.5= 1690 J• Heat generated in the solutionQL = qa qe + + QK = 226.8 + 142.96 + 1690 = 2059.76 J• dissolution enthalpy (ΔH)ΔH1 = q1/29/58 = 2059.76 / 0.5 = 4119.52 J• Graph showing the relationship between the temperature intervalT (K)t (Minutes)b. To a mixture of 27 ml water with 19 ml of ethanolDik: ρair = 1 g / mlρetanol = 0.79 g / mlS water = 4.2 J / g · KS ethanol = 1.92 J / g · KT water = 304 KT = 304 K ethanolDIT: dissolution enthalpy (ΔH)?Adjustment: - Calculating the mass of water - Calculate the mass of ethanolMa = Va x ρair M ethanol = Vetanol xρetanol= 27 ml x 1 g / ml = 19 ml x 0.79 g / ml= 27 g = 15.01 g• Calculate the temperature of the mixtureTcamp = ΣT / n = 2464/8 = 308 KCamp ΔT1 = T - T water = 308-304 = 3 KΔT2 camp = T - T ethanol = 308-304 = 3 K• Heat absorbed water (qa) - Heat is absorbed ethanol (qe)qa = Ma x S x ΔT1 qe = Methanol x S x ΔT2= 27 x 4.2 x 4 = 15.01 x 1.92 x 4= 453.6 J = 115.27 J• Heat is absorbed calorimeter (QK)QK = K x ΔT2= 52 x 4= 208 J• Heat generated in the solutionQL = qa qe + + QK = 453.6 + 115.27 + 208 = 776.87 J• dissolution enthalpy (ΔH)ΔH1 = q1/29/58 = 776.87 / 0.5 = 1553.74 J• Graph showing the relationship between the temperature intervalT (K)t (Minutes)c. To a mixture of 36 ml of water with ethanol 14.5 mlDik: ρair = 1 g / mlρetanol = 0.79 g / mlS water = 4.2 J / g · KS ethanol = 1.92 J / g · KT water = 304 KT = 303 K ethanolDIT: dissolution enthalpy (ΔH)?Adjustment: - Calculating the mass of water - Calculate the mass of ethanolMa = Va x ρair M ethanol = Vetanol xρetanol= 36 ml x 1 g / ml = 14.5 ml x 0.79 g / ml= 36 g = 11.495 g• Calculate the temperature of the mixtureTcamp = ΣT / n = 2464/8 = 308 KCamp ΔT1 = T - T water = 308-304 = 4 KΔT2 camp = T - T ethanol = 308-303 = 5 K• Heat absorbed water (qa) - Heat is absorbed ethanol (qe)qa = Ma x S x ΔT1 qe = Methanol x S x ΔT2= 36 x 4.2 x 4 = 226.8 J = 11.495 x 1.92 x 5= 604.8 J = 109.968 J• Heat is absorbed calorimeter (QK)QK = K x ΔT2= 52 x 5= 260 J• Heat generated in the solutionQL = qa qe + + QK = 604.8 + 109.968 + 260 = 974.76 J• dissolution enthalpy (ΔH)ΔH1 = q1/29/58 = 974.76 / 0.5 = 1949.536 J• Graph showing the relationship between the temperature intervalT (K)t (min)search for mole mole ratio of water and ethanol in any mixturea. V water = 18 ml, V = 29 ml ethanol• Changes in temperature initially (ΔTm1)ΔTm1 = Tair + T ethanol = 304.25 + 304 = 304.125 K2 2• Changes in the final temperature (ΔTa1)ΔTa1 = Tcamp - ΔTm1 = 307.25 to 304.125 = 3.125 K• Mol airmr water = water = 1818 grams = 1 mol• Mol etanolmr ethanol = g ethanol = 22.9146 = 0.4 mol• Comparison of moles of water to moles of ethanolwater mol: mol ethanol = 1: 0.4b. V water = 27 ml, V = 19 ml ethanol• Changes in temperature initially (ΔTm1)ΔTm1 = Tair + T ethanol = 304 + 304 = 304 K2 2• Changes in the final temperature (ΔTa1)ΔTa1 = Tcamp - ΔTm1 = 308-304 = 4 K• Mol airmr water = water = 2718 grams = 1.5 mol• Mol etanolmr ethanol = g ethanol = 15.0146 = 0.32 mol• Comparison of moles of water to moles of ethanolmoles of water: ethanol = 1.5 mol: 0.32c. V water = 36 ml, V = 14.5 ml ethanol• Changes in temperature initially (ΔTm1)ΔTm1 = Tair + T ethanol = 304 + 303 = 303.5 K2 2• Changes in the final temperature (ΔTa1)ΔTa1 = Tcamp - ΔTm1 = 308 to 303.5 = 4.5 K• Mol airmr water = water = 3618 g = 2 mol• Mol etanolmr ethanol ethanol = g = 0.24 mol = 11.45546• Comparison of moles of water to moles of ethanolwater mol: mol ethanol = 2: 0.243. Determination of the heat of neutralization of HCl and NaOHDik: ρL: 1.03 g / ml T = 31 oC HCl + 273 = 204 KSL: 3.96 J / g K T NaOH = 31 ° C + 273 = 204 KDIT: Heat neutralization ....?Peny:• initial temperature (Tm)Tm: T HCl + NaOH T = 304 +304 / 2 = 304 K2• The final temperature (Ta)ΣT / n = 3125/10 = 312.5• Changes in the final temperature (ΔTa)ΔTa = Ta - Tm= 312.5 -304 = 8.5 KVL = V V NaOH + HCl= 20 + 20 = 40 ml• Mass solution = VL x ρL= 40 x 1.03 = 4.12 g• Heat is absorbed (q1)q1 = g solution x SL x ΔTa= 41.2 x 3.96 x 8.5 = 1386.79 J• Heat is absorbed calorimeter (q2)q2 = K x q1= 52 x 1386.79 = 72113.08 J• Heat generated reaction (q3)q3 = q1 + q2= 1386.79 J + J = 73499.87 J 72113.08• Heat neutralization = q3 = 73499.87 J / mol 0.05 = 13558.22 J / molmole solutionmol solution = mass solution Na0H mr mr + HCl = 4.12 / (40 + 36) = 4.12 / 76 = 0.05 molb. Discussion1. Determination of constants calorimeterThe first experiment aimed to determine the calorimeter constant with hot water and cold water. Running hot and cold water is mixed with an equal volume of the calorimeter is stirred, and the observed temperature. Based on temperature records were obtained in the experiment, the temperature of a mixture of cold water and hot water ranges between 33 oC - 35 oC.In the experimental determination of the constant of the calorimeter is in getting the temperature increase when adding other ingredients that hot water. Prior coupled with hot water, the temperature of the water in the calorimeter at 30 oC. And when added to hot water, the water temperature rose to 35 oC. In this experiment the exothermic process occurs because the system releases heat. This can be seen in the data observations that showed reduced system temperatures (Mixed) were initially at 35 oC down slowly to 33 oC.If the calorimeter does not absorb heat from the mixture of water, the heat provided by hot water equal to the heat absorbed by the cold water. But because calorimeter also absorb heat, the heat absorbed by the calorimeter is given by the difference in heat hot water reduced the heat absorbed by the cold water (q3 = q2 - q1). Calorimeter constant prices is obtained by dividing the amount of heat absorbed by the calorimeter (q3) with warming temperature changes in the calorimeter.C = q3ΔTC = constant of the calorimeter (J / K)q = heat absorbed (J)ΔT = change in temperature (K)Based on the calculations, the constant of the calorimeter is 52 J / K2. Determination of Ethanol Heat Dissolution In WaterHeat or thermal dissolution of ethanol can be obtained by mixing these substances into the calorimeter containing cold water, so it will react and there will be a reaction that is accompanied by changes in temperature, and release some heat. Kalornya changes depending on the concentration of initial and final solution formed.In this experiment, the heat generated dissolution, so the temperature of the water mixed with ethanol increases. The temperature increase is due to a mixture of heat dissolving the heat that accompanies ethanol dissolving in water.3. Determination of the heat of neutralization of HCl and NaOHThe whole point of this experiment is to determine the heat of the reaction of HCl and NaOH. First introduced into the calorimeter HCl and recorded temperature, then a solution of NaOH the temperature equal to the temperature of HCl was mixed with HCl. Having observed temperature changes before and after HCl mixed with NaOH.In peercobaan reaction between hydrochloric acid (HCl) and alkaline sodium hydroxide (NaOH) in a salt with water. The reaction can be expressed by the following equation:HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)Which acts as a system in this reaction is HCl and NaOH and acts as the environment is water and, as a solvent medium of the two substances. Solution to the reaction temperature increases, this is because at the time of the release of reaction heat. Heat released by the reaction system (NaOH and HCl) is absorbed by the solvent and other material (Calorimeter). As a result, the temperature rise shown by the increase in the solution temperature. So in the experiment measured the temperature is not the system, but the environment in which the reaction temperature. While the system down to the reaction temperature and reaches a steady state to form NaCl and H2O.H. ConclusionBased on experiments that have been carried out can be concluded several things:• In each experiment, a mixture of the two solutions is always changing temperatures.• Changes in temperature on the determination of the heat of neutralization of HCl and NaOH is relatively larger than the change of heat in other experiments.• Every chemical reaction is always accompanied by changes in heat.• One way to measure the change in heat is to conduct experiments using a calorimeter.I. Possible errors• Less skilled practitioner in the trials• Less skilled practitioner in the temperature reading on the thermometer