the safety materials and wear goggles when

Hess’ LawPeter Jeschofnig, Ph.D.Version 42-0158-00-01Review the safety materials and wear goggles whenworking with chemicals. Read the entire exercisebefore you begin. Take time to organize the materialsyou will need and set aside a safe work space inwhich to complete the exercise.Experiment Summary:Students will have the opportunity to measuretemperature changes taking place in a calorimeterduring neutralization reactions and use themeasurements to calculate enthalpy of reaction.They will illustrate the validity of Hazy’ Law bycomparing the values of enthalpy of two chemicalreactions.Objectives●● To measure temperature changes taking place in a calorimeter during neutralizationreactionsand use the measurements to calculate enthalpy of reaction.●● To compare the enthalpy of two chemical reactions and use these measured valuesto illustratethe validity of Hess’ Law.MaterialsMaterials From: Label orBox/Bag: Qty Item Description:Student Provides Distilled waterWatchCoffee cupsPaper towelsFrom LabPaq 1 Thermometer – Digital1 Goggles-Safety4 Cup, Styrofoam, 8 oz1 Cylinder-25-mLFrom Experiment BagHess’ Law 2 Ammonia , NH3 (comes as aqueousammonia, NH4OH), – 2 M – 10 mL2 Ammonium chloride, NH4Cl – 2M – 10mL2 Hydrochloric acid, HCl – 2 M – 20 mL2 Pipet, Long Thin Stem2 Sodium hydroxide, NaOH – 2M – 20 mLNote: The packaging and/or materials in this LabPaq may differ slightly from that whichis listedabove. For an exact listing of materials, refer to the Contents List form included in theLabPaq.Discussion and ReviewThermochemistry is the study of the heat energy involved in chemical reactions andchanges of physical state. Nearly all chemical reactions involve the release orabsorption of heat, a form of energy. The burning of any fuel such as gasoline, coal, orwood is an example of a heat-releasing reaction. Heat energy is called thermal energy,and it is always spontaneously transferred from hotter to colder matter.The First Law of Thermodynamics is the Law of Energy Conservation. It states thatthe total energy of the universe must remain constant. Therefore, all energy transferredbetween a system and its surroundings must be accounted for as heat or work.The standard S.I. unit for heat energy is the joule, J. It takes 4.184 joules, theequivalent of 1calorie, to raise the temperature of one gram of water by 1° C. The kilojoule, kJ, iscommonly used in many applications: 1000 joule = 1 kilojoule.When a chemical reaction takes place in a stable environment where the temperatureandpressure remain constant, the system defined by the reactants and products eitherproduces orreleases heat energy.●● If the reacting system releases heat energy to its surroundings, a concurrentincrease insurroundings temperature is observed, and the reaction is exothermic●● If the system absorbs heat energy from its surroundings, a decrease in thesurroundingstemperature is observed, and the reaction is endothermic.●● A measure of the amount of heat given off or absorbed in any chemical reaction iscalled theenthalpy change or heat of reaction, and is given the symbol H.When thermodynamic measurements are carried out at standard-state conditions wherethepressure is constant at 1 atm and the temperature is constant at 25oC, the reactionenthalpy isdesignated as the standard enthalpy change or ΔH°. It is important to havestandardized values because the enthalpy of a reaction can vary with different reactionconditions.The following reaction for the formation of water from its constituents is exothermic:H2(g) + ½ O2(g) à H2O(l); ΔH °f = -286 kJFor every mole of H2O (l) formed at standard-state conditions, 286 kilojoules of heatenergy arereleased. When the standard enthalpy change of reaction describes the formation of 1mol ofcompound directly from its elements in their standard states as in this example, thevalue of ΔH of is called the standard heat of formation.To determine the enthalpy change for a given reaction (ΔH°rxn), the summation of theheats offormation (ΔH° f ) for the reactants are subtracted from the summation of the heats offormation ( ΔH ° f ) for the products.ΔH° rxn = [n ΔH°f (products)] – [n ΔH°f (reactants)]Tables containing the standard heats of formation for a number of compounds areavailable in the appendices of any general chemistry textbook.Hess’s Law states that if a reaction is the sum of two or more other reactions, the ΔHfor theoverall process must be the sum of the ΔH values of the constituent reactions.Enthalpy change (ΔH) is independent of the path that a reaction follows to move fromreactantsto products. It only depends on the relative energy difference between the reactant andproductmolecules at constant pressure. Enthalpy change is referred to as a state function dueto itsindependent of pathway. Since the enthalpy of a substance is not commonlydetermined, thechange in enthalpy when reactants are converted to products is often used to describea chemicalor physical process.The thermal energy absorbed or produced by a chemical process reflects a differencebetweenthe enthalpy between the reactants and products (ΔH). For example, in thedecomposition ofliquid water into its component elements, H 2 (g) and O2 (g), there are two successivechanges.First, the liquid water is vaporized. Second, the water vapor decomposes into itsconstituentelements shown below. The ΔH value for this overall process can be determined byadding theΔH values from the equations for each step as shown below.(1) H2O (l) à H2O (g); ΔH 1 = +44 kJ(2) H2O (g) à H2 (g) + ½ O2 (g); ΔH 2 = +242 kJ_______________________________________________________________(1) + (2) H2O (l) à H2 (g) + ½ O2 (g); ΔHnet = +286 kJIn order to determine ΔH for the reaction NH3 + HCl à NH4Cl in this experiment, ΔH rxnfor thefollowing two reactions will be measured:1. NaOH (aq) + HCl (aq) à H2O (l) + NaCl (aq)2. NaOH (aq) + NH4Cl (aq) à NH3 + NaCl + H2O (l)Comparison of the calculated results for different parts of the experiment will verify thegeneralization known as Hess’s Law of Constant Heat Summation. In this case thetarget reaction NH3 + HCl à NH4Cl can also be performed directly and the resultscompared to reactions 1 and 2.A Styrofoam coffee cup calorimeter will be used to measure the amount of heat energyevolvedor absorbed during the chemical reactions of this experiment. A digital thermometer isused tomeasure the change in temperature between the final and initial temperatures of thesolutions.Unfortunately, it is impossible to have perfect insulation and some of the heat energy willbe lost to the surroundings, including to the material from which the calorimeter isconstructed.Calibrating the calorimeter before using it to make measurements on an unknownsystem usually solves the problem of heat losses. A known amount of heat energy froma known process is released into the calorimeter system, and the temperature change ismeasured. A simple calculation is done to determine the amount of heat energy loss,called the heat capacity of the calorimeter or calorimeter constant. For this experiment itassumed that the heat capacity of the calorimeter is insignificant and it is ignored.Another practical problem is that heat energy exchanges do not occur instantaneously;i.e., it takes time for energy to move from a hot object to a cold one. An acceptablesolution to this problem is to obtain a cooling curve for the heat energy exchange inquestion and then extrapolate the data back to the exact time that the exchange began.Below is a sample graph from hypothetical data. Notice that at the time of combining thetwo solutions, their starting temperature is 20 oC. Since the starting temperatures are atroomtemperature no initial temperature adjustment is needed. From 0 to 40 seconds thetemperaturerises rapidly to 34.2oC. The temperature then drops gradually 31.1 oC and will continueto drop.Usually recording the temperature in 20-20 second intervals for 5 minutes is enough toprovide agood cooling curve. Extrapolation of these data backward in time determines what thetemperatureat the time of mixing would have been if the temperature of the reaction had beeninstantaneousand the calorimeter had warmed instantaneously. In this example, the temperature atthe timeof mixing determined by extrapolation is 34.3 oC.Calculations: The equation used to calculate heat gained or lost is:qsolution = (mass of solution) x (specific heat) x ΔTDensity = 1.02 g/mL for all solutions in this experiment;Specific Heat = 4.184 JΔT = Final temperature – Initial temperatureA small amount of heat is lost to the surroundings which in this case is the calorimeter.Thisheat loss can be accounted for by using a calorimeter constant, c, which can bedeterminedexperimentally. However, the amount of heat lost to the calorimeter is so insignificantthat it isoften left off, or simply assumed to be 1 J* ΔT. (q cal = c x ΔT).If a correction was to be made for the heat absorbed by the calorimeter, the heat of thereaction,qrxn , could be determined by taking the negative of the heat gained by the solution, qsoln,plus thatgained by the calorimeter, qcal:qrxn = -(qsoln + qcal)Once the total thermal energy transfer is known, the enthalpy of reaction can bedeterminedusing the following equation:ΔH = qrxn /moles NaOH or HClMoles of NaOH or HCl can be determined from the equation: M = moles/L10 mL = 0.01L; 2M = moles/0.01L = 0.02 molesExercise 1: Hess’ LawProcedurePart 1: Reaction: HCl & NaOH → NaCl + H2O1. Before beginning, set up data tables similar to the Data Tables 1 & 2 in the LabReport Assistantsection.2. Construct a calorimeter from 2 Styrofoam cups: Trim the lip of one cup and use thatcup asthe top of the calorimeter. Make a small hole in the top so a thermometer can beinserted, asshown below. Be careful when inserting the thermometer into the calorimeter since ithas apointed tip that could puncture the lower cup if inserted too forcefully. Place thecalorimeterassembly into an empty coffee cup to help prevent it from tipping over.Figure 2:3. Use a graduated cylinder to accurately measure 10 mL of 2M HCl. Use an emptythin-stempipet to remove or add drops of HCl so that the meniscus level is on the 10 mL mark.Pour the10 mL HCl into the Styrofoam calorimeter. Rinse the thin-step pipet according to thismanual’sinstructions on Use, Disposal, and Cleaning of Common Materials.4. Rinse and dry the graduated cylinder and accurately measure 10 mL of 2M NaOHusing thesame technique in step 2 above. Pour the 10 mL NaOH into another Styrofoam cup andplacethe cup into a second empty coffee cup to prevent it from tipping over.5. Turn on the digital thermometer and place it into the HCl solution. Wait 5 minutes andrecordthe temperature of the solution in Data Table 1.6. Remove the thermometer, rinse the tip with distilled water, dry it with a paper towelandplace it into the NaOH solution. Wait 5 minutes and record the temperature of thesolutionin Data Table 1. Remove the thermometer, rinse the tip with distilled water, and dry itwith apaper towel for future use.7. Pour the contents of one Styrofoam cup into the second one, combining the twosolutions.Quickly place the Styrofoam lid on top of the cup containing the combined solutions andinsertthe thermometer through the hole in the lid. Be careful when inserting the thermometertoensure its pointed tip does not puncture the lower Styrofoam cup.8. Record the temperature every 20 seconds for 5 – 6 minutes and record in Data Table.9. Graph the data points using an Excel spreadsheet; time in seconds on the x-axis andtemperature on the y-axis. The graph should look similar to the sample cooling curvebelow.10. Place a ruler on the declining temperature portion of the curve and extrapolate tothe 0-line.Read the extrapolated temperature where the straight line intersects the 0-time line.Thistemperature represents the final temperature of the mixture. Enter this temperature inDataTable 1.11. Dispose of the solution in the calorimeter by flushing it down the drain with water.Recall thatthe solution results from a neutralization reaction and is simply salt water.12. Rinse all equipment used in preparation for reaction 2. This includes thecalorimeters,graduated cylinders, pipets, etc.Part 2: Reaction 2: NH4Cl + NaOH → NH3 + NaCl + H2O1. Repeat the Procedures from Part 1, but using 10 mL of 2M NH 4Cl and 10 mL of 2 mLof NaOH.2. Dispose of the solution in the calorimeter by flushing it down the drain with water.3. Rinse all equipment used in preparation for reaction 3. This includes the calorimeters,graduated cylinders, pipets, etcPart 3: Reaction: NH3 + HCl → NH4Cl1. Repeat the Procedures from reaction 1, but using 10 mL of 2M NH3 and 10 mL of 2mL of HCl.2. Dispose of the solution in the calorimeter by flushing it down the drain with water.3. Rinse all equipment used in preparation for future experiments. This includes thecalorimeters,graduated cylinders, pipets, etc.Hess’ LawPeter Jeschofnig, Ph.D.Version 42-0158-00-01Lab Report AssistantThis document is not meant to be a substitute for a formal laboratory report. The LabReportAssistant is simply a summary of the experiment’s questions, diagrams if needed, anddata tablesthat should be addressed in a formal lab report. The intent is to facilitate students’writing of labreports by providing this information in an editable file which can be sent to aninstructor.Part 1: Reaction: HCl & NaOH → NaCl + H2OPart 1: Reaction: HCI & NaOH → NaCI +H20Data Table 1: Sample DataInitial Temperature of HCl – oCInitial Temperature NaOH – oCAverage Initial Temperature – oC(extrapolated)Final Temperature of mixtureChange in Temperature of mixture, ΔTData Table 2: Sample DataTime after mixing- secondsTemperature – °C20406080100120140160180200220240260280300Part 2: Reaction 2: NH4Cl + NaOH → NH3 + NaCl + H2OData Table 3:Initial Temperature of NaOH – oCoInitial Temperature NH Cl – CAverage Initial Temperature – oCFinal Temperature of mixture(extrapolated)Change in Temperature of mixture, ΔTData Table 4:Time after mixing- seconds20406080100120140160180200220240260280300Part 3: Reaction : NH3 + HCl → NH4ClData Table 5:Temperature – °CInitial Temperature of HCl – oCoInitial Temperature NH – CAverage Initial Temperature – oCFinal Temperature of mixture(extrapolated)Change in Temperature of mixture, ΔTData Table 6:Time after mixing- seconds20406080100120140160180200220240260280300Temperature – °CQuestionsFor A. through E. See the calculations for the Data Tables above.A. Using the data from your data tables calculate ΔT for all three reactions:B. Calculate the heat loss or gain of the three solution mixtures:C. Use Hess’ Law and ΔH for the first two reactions:NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq)NaOH (aq) + NH4Cl (aq) → NH3 + NaCl + H2O (l)to determine ΔH for this reaction: NH3 + HCl → NH4ClD. Compare the results of step 3 above with the experimental results of theNH3 + HCl → NH4ClE. Use the thermodynamic quantities given below to calculate the theoretical ΔH for thisreaction: NH3 + HCl → NH4Cl●● ΔH°f for NH3 (aq) = – 80.29 kJ/mol●● ΔH°f for HCl (aq) = – 167.2 kJ/mol●● ΔH°f for NH4 (aq) = – 132.5 kJ/mol●● ΔH°f for Cl- (aq) = – 167.2 kJ/molF. What was the ΔH value obtained for NH3 + HCl à NH4Cl from Hess’ Law method?G. What was the ΔH value obtained for NH3 + HCl à NH4Cl experimentally?H. What was the calculated ΔH value obtained for NH3 + HCl à NH4Cl using publishedthermodynamic data?What was the % error of the various methods used? (i.e. comparing the results of theresults of Hess’ Law method and the experimental results to the calculated value?J. Name three examples of the practical application for the use of ΔH values.