# CALORIMETRY AND HESS LAW: FINDING H o FOR THE COMBUSTION OF MAGNESIUM

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1 Experiment 12J FV 7/16/06 CALORIMETRY AND HESS LAW: FINDING H o FOR THE COMBUSTION OF MAGNESIUM MATERIALS: Styrofoam coffee cup and lid, thermometer, magnetic stirrer, magnetic stir bar, 50-mL and 100- ml graduated cylinders, 150 ml beaker, aluminum slug, magnesium ribbon, magnesium oxide, 1.00 M HCl, 1.00 M NaOH, copper wire. PURPOSE: The purpose of this experiment is to determine the H o for the combustion of magnesium by determining the H o s for reactions which can be combined together according to Hess Law, yielding the H o for the desired reaction. LEARNING OBJECTIVES: By the end of this experiment, the student should be able to demonstrate the following proficiencies: 1. Construct and use a calorimeter. 2. Use a calorimeter to examine heat transfer and determine the specific heat of a metal. 3. Determine the heat capacity of a calorimeter using a reaction with known H o. 4. Determine a set of reactions which, after applying Hess Law, yield the H o for the desired reaction. 5. Devise procedures to address experimental challenges associated with the reactions. 6. Use a spreadsheet program for data manipulation, graphing, and regression analysis. PRE-LAB: Complete the pre-lab questions on page E12J-11 before lab. DISCUSSION: The First Law of Thermodynamics is the Law of Conservation of Energy, which states that energy can neither be created nor destroyed the total energy of the universe is a constant. A fire or a chemical explosion is just the result of converting the chemical energy of the fuel or explosive into thermal and mechanical energy of the products and the surrounding air. Objects near a fire increase in temperature because energy is transferred in a process called heating; the energy exchanged is the amount of heat (q). Objects near an explosion are moved because energy is transferred in a process called doing work; the energy exchanged is the amount of work done (w). Our primary interest here is with heat transfers (the thermo in thermodynamics). These heat transfers can take place between two objects at different temperatures, and they also can accompany physical or chemical changes (i.e., reactions). Calorimetry is a technique for measuring heat transfers; it makes use of an insulated container called a calorimeter. Qualitatively, the process of heat transfer that occurs in a calorimeter is very familiar to us. We all recognize that a cup of hot coffee left in a cool room will get colder, and eventually will become the same temperature as the room. Of course, the thermal energy of the coffee cannot be lost; it is simply moved into the rest of the room. The coffee gets measurably colder, while the room gets warmer, albeit imperceptibly. The air (and the walls, furniture, etc.) of the room serves as a thermal bath to absorb the heat released by the coffee. The same thing happens in a calorimeter. If a hot object, like an aluminum slug, is dropped into a cold sample of water, the water acts the thermal bath and absorbs the heat; the process stops when both water and slug are at the same temperature, a situation called thermal equilibrium. If the whole process is conducted in an insulated container, like a Styrofoam cup, then no heat is lost to the room. In a thermodynamic sense, the hot Al slug is the system, and the water the surroundings. Energy lost by the system is gained by the surroundings. The insulated cup acts as the whole experimental universe, and the total energy within the cup must remain constant, as dictated by the First Law. As long as there is no phase change, a transfer of energy into or out of a sample of matter will result in a temperature change. The magnitude of the temperature change will depend on the kind of material, its amount (i.e., the mass m or number of moles n), and how much energy is transferred. The relation between the amount of energy transferred in the form of heat, q, and the temperature change of the object, D T, can be written as: q = c m D T or q = c m n D T (1) E12J-1

2 (As is always true in thermodynamic calculations, D represents a difference between the final and initial values: DT = T final - T initial. ) The proportionality constants in the equation above are different versions of the heat capacity of the object. In the first equation, the constant c is the specific heat capacity (commonly called just specific heat ). Typical units for the specific heat are Joules per gram per o Celsius, or J g -1 o C -1. Basically, it tells you how many Joules of heat you have to add or remove to change the temperature of one gram of the material by one degree Celsius. The constant c m in the second equation is the molar heat capacity; it relates similar information, except it pertains to warming or cooling one mole of material, rather than one gram. Which equation (and constant) you use depends only on your choice of using mass or moles to describe the quantity, but you must be consistent. Again, either one is specific to the substance. If a number of different substances are all being subjected to the same heat input (as with the walls, air and furniture in the room absorbing heat from the coffee), it is usually necessary to use a generic total heat capacity, c tot. There, it represents the amount of heat transfer required to change the temperature of the whole assembly by one o C q = c tot D T (2) Use of this form will be required in our measurements of the heat of reaction, since the whole calorimeter assembly will involve a Styrofoam cup and its lid, and the dilute acid solution, magnetic stir bar and the end of a thermometer within it. No matter which form of the D T expression is used, the general equation for calorimetry works the same way, since the calorimeter again represents the experimental universe, and we are simply watching heat move from one part of that universe to another. To use the example above, the energy lost by the aluminum slug (q Al ) must be gained by the water (q water ). The total energy of the universe is constant, so q Al + q water = 0 (3) Or, this could be rearranged as q Al = q water (4) The negative sign simply results because one object gains heat energy (the water) while the other (Al) loses it. Although the magnitudes of the energy change are the same, the directions are opposite. Both substances have their own unique values of specific heat, and each has a different mass, so each must be represented separately: q Al = c Al m Al D T Al (5) q water = c water m water D T water (6) When resubstituted into the overall calorimetry expression (Eq. 4), the final result is: q Al = q water or c Al m Al D T Al = c water m water D T water (7) Note that the values of T will be different for each substance, even though their final temperatures, T f, are the same, once thermal equilibrium is reached. When measuring energy changes associated with a chemical reaction, the principle is the same. For example, in an exothermic reaction, the reaction will release heat to the surroundings. An exothermic reaction run in a calorimeter acts just like the aluminum slug and water in the simple heat transfer experiment above. The heat transfer associated with the reaction is q rxn ; that associated with the calorimeter is q cal. The law of conservation of energy must hold, so q rxn + q cal = 0 or q rxn = q cal (8) As described above, the calorimeter is an assembly of several different materials, and so its heat equation q cal is best described using the total heat capacity (Eq. 2). Then q rxn = q cal becomes q rxn = c tot cal D T cal (9) E12J-2

3 Again, the negative sign is required because of the direction of the heat change, and reflects the fact that an exothermic reaction (negative q) results in a positive T (temperature increase). In a calorimetry experiment, a reaction is performed inside a calorimeter and the resulting temperature change of the calorimeter is measured. If we know the total heat capacity of the calorimeter and its contents, and can measure the temperature change when the reaction occurs, then we can determine the amount of heat liberated by the reaction. Alternately, if we know the amount of heat transferred, we can determine the heat capacity of the calorimeter by simply using the equation in reverse. As a final theoretical detail, if the heat of reaction is measured in a constant-pressure calorimeter (e.g., a coffee-cup calorimeter), then q rxn = q p = D H (The subscript p emphasizes that the heat of reaction was measured at constant pressure.) (10) Thus we are able to measure D H, which is a goal of the experiment. The correct determination of T for a reaction requires careful consideration of two key issues: (1) reactions do not occur instantaneously, and (2) calorimeters are not perfectly insulated from their surroundings, thus allowing heat energy to slowly enter or escape from the calorimeter over time. Consider an exothermic reaction which occurs in a hypothetical calorimeter which is perfectly insulated. In this case, whether the reaction occurs instantaneously or not, all of the heat produced by the reaction will remain in the calorimeter, resulting in the same final temperature, and hence the same T. Now consider a hypothetical exothermic reaction which occurs instantaneously, but in a realistic calorimeter which is not perfectly insulated. In this case, the temperature of the calorimeter would diminish over time due to the gradual escape of heat energy to the surroundings. The final temperature Instantaneous reaction, perfectly insulated calorimeter to be used in determining T in this case is 30 actually the maximum temperature reached Tf immediately after reaction occurs, since this 28 temperature change is due exclusively to the 26 heat produced in the reaction, and no Instantaneous reaction, partially insulated calorimeter escaping of heat to the surroundings has 24 occurred yet. These three hypothetical cases T i Non-instantaneous reaction, perfectly insulated calorimeter are all depicted on the following temperature 22 vs. time graph: Temp ( o C) Time (min) For real calorimeter experiments, reactions neither occur instantaneously nor are calorimeters perfectly insulated. Thus, during an exothermic reaction the temperature of the calorimeter increases initially, but never has a chance to reach the correct maximum final temperature since heat is escaping to the surroundings even while the reaction is proceeding toward completion. The best way to account for these experimental difficulties, thereby obtaining a very good estimate of the correct final temperature, is to linearly extrapolate the gradual cooling portion of the temperature vs. time graph back to the initial time of mixing. This is the temperature that the calorimeter would have reached if all the heat energy from the reaction had remained absorbed by the calorimeter. This approach to obtaining the correct final temperature is depicted on the following temperature vs. time graph, and can easily be accomplished using a spreadsheet (see Appendix O of the Plebe Chemistry Lab Manual): 30 T f 28 Non-instantaneous reaction, partially insulated calorimeter (extrapolate cooling portion of curve back to time of mixing) Temp ( o C) T i Time (min) E12J-3

5 time values every 5-10 seconds for about one minute, and every seconds for about 2 minutes after that. Be sure to swirl the cup frequently. 5. Remove the slug from the water, empty the water into a sink, and dry off the slug and the inside of the cup and lid. Answer in-lab questions #6 and #7 on pages E12J-10. Part C: Measuring the total heat capacity of the calorimeter and its contents 1. Calorimetry requires that the heat capacity of the calorimeter and its contents be known. In this experiment, the calorimeter will consist of a Styrofoam cup and lid. It will contain a stir bar, thermometer, and a dilute solution. You will need to measure how much the temperature of the calorimeter and its contents changes when they absorb a known amount of heat. To produce a known amount of heat, you will perform the following acid-base neutralization reaction, for which H o is well known: H + (aq) + OH - (aq) H 2 O(l) D H o = kj 2. Place the magnetic stir bar into a clean, dry Styrofoam cup. 3. Rinse a clean graduated cylinder with a few milliliters of 1.00 M HCl and transfer 50.0 ml of 1.00 M HCl into the cup. 4. Place the cup on the magnetic stirrer and begin stirring gently. (Why gently?) 5. After rinsing the graduated cylinder with distilled water and with a few milliliters of 1.00 M NaOH, obtain a 50.0 ml sample of 1.00 M NaOH. 6. Record the molarity and volume of each solution used. 7. Before mixing, measure the temperature of the two solutions. (Rinse and dry the thermometer between readings). Record the average of the two temperatures. This is the initial temperature (T i ). 8. Working quickly, note the time (time = 0 s) when the NaOH solution is added to the Styrofoam cup, then put on the lid, and place the thermometer into the solution. (The thermometer must be supported in some way to prevent the cup from tipping over.) 9. Record the time and temperature every seconds for a total of 8 minutes. The time intervals can be increased as the rate of temperature change decreases. 10. When data collection is completed, rinse the calorimeter with distilled water and dry as completely as possible. Important: For this and all subsequent experiments, use a total solution volume of ml. Part D: Measuring D H o for the reaction of Mg or MgO with HCl Your instructor will assign you a specific reaction to study, and give you either a target mass of reagent or target T to work for. See your instructor to obtain these assignments. For students assigned to study the Mg/acid reaction: 1. In part A, you probably observed that the magnesium metal floated on top of the acid solution as it reacted with it. (Think about why this is a problem.) E12J-5

7 DATA ANALYSIS: Part C: Measuring the total heat capacity of the calorimeter and its contents 1. Construct a plot of temperature versus time with Excel and determine the highest temperature (T f ) that would have been achieved if the calorimeter was perfectly insulated (see DISCUSSION section). Use this final temperature to calculate the temperature change ( T = T f - T i ) of the calorimeter. (Refer to Appendix O in the lab manual for help in determining T f using Excel.) 2. Using the following information and the amounts of reactants used in the experiment, calculate the heat (in Joules) produced by the reaction. H + (aq) + OH - (aq) H 2 O(l) H o = kj 3. Calculate the total heat capacity of the calorimeter (c tot cal ) in J/ o C. Part D: Measuring D H o for the reaction of Mg or MgO with HCl 1. Calculate the number of moles of Mg or MgO that were consumed in the reaction that you studied. 2. Determine the temperature change ( T = T f - T i ) for the reaction you studied by constructing a temperature versus time plot on the graph paper provided. (Your instructor will tell you whether you must reproduce this graph in Excel.) 3. Using the total heat capacity of your calorimeter (c tot cal ) and the temperature change, calculate the heat produced by the reaction that you studied (q rxn = H o ). 4. Report the number of moles of Mg or MgO and your calculated H o (q rxn ) to the class. 5. Create an Excel spreadsheet that lists the results from the entire class (moles of reactant and H o for each of the two reactions studied.) 6. Use Excel to graph the class data for both reactions (two graphs) and use the graphs to determine the average value for H o per mole of magnesium, and the average value for H o per mole of magnesium oxide. (Think about which data you should plot on the y-axis.) 7. Combine these results and the H o for the reaction below to obtain H o for the combustion of magnesium. H 2 (g) + ½ O 2 (g) H 2 O(l) H o = kj E12J-7

8 QUESTIONS FOR CONSIDERATION: 1. Using available thermodynamic tables of H o f (standard enthalpy of formation), determine the H o values for the magnesium/acid and magnesium oxide/acid reactions and compare with your experimentally determined values. Explain any differences. H f (Mg 2+ ) = kj/mol 2. Is the combustion reaction for magnesium the same as the formation reaction for magnesium oxide? Does this mean that the H o obtained is the same as the H o f for magnesium oxide? Compare your experimental value with that found in the table of thermodynamic properties in your textbook. Determine the percent error. 3. Sketch the temperature vs. time graph that you would expect for an endothermic reaction carried out in a calorimeter similar to the one used in the experiment (assume the reaction is not instantaneous and that the calorimeter is not perfectly insulated). 4. Could the same procedures be applied for determining the H o for the combustion of aluminum metal as were used in this experiment? Show how the analogous reactions would combine together, applying Hess Law, to yield the H o for the combustion reaction. 5. Using thermodynamic tables and your calorimeter heat capacity, predict the T if 1.00 g of aluminum oxide completely reacted with hydrochloric acid. H f (Al 3+ ) = kj/mol 6. If the material absorbing the heat produced by reaction were limited to just the water, what would be the expected heat capacity of your calorimeter? How does the experimentally determined heat capacity compare? 7. In Parts C and D, we assumed that the heat of reaction was equal to the standard enthalpy change. For both Part C and Part D, discuss whether or not this was a good assumption. E12J-8

9 Name Section IN-LAB QUESTIONS Experiment 12J Complete these questions during lab. 1. Write the balanced chemical equation for combustion of one mole of magnesium metal. (The coefficient in front of Mg should be 1, but other coefficients may be fractional coefficients. We are doing this because we want to determine H o per mole of magnesium.) 2. Why would it be difficult to study the combustion of magnesium directly using a simple water-filled calorimeter? 3. Write the balanced molecular, complete ionic, and net ionic equations for the reaction of magnesium metal with hydrochloric acid. (Hint: the Activity Series table may help you determine what the products are in this reaction.) Molecular: Complete ionic: Net ionic: 4. Write the balanced molecular, complete ionic, and net ionic equations for the reaction of solid magnesium oxide with hydrochloric acid. (Hint: Magnesium oxide is a base!) Molecular: Complete ionic: Net ionic: E12J-9

10 5. Combine the two deduced reactions (net ionic equations) and the given reaction so that the 3 reactions add to give the desired reaction (combustion of one mole of magnesium). The given reaction is: H 2 (g) + ½ O 2 (g) H 2 O (l) 6. Fill in the following values from your data of Part B. Aluminum Mass (g) Initial Temp. ( o C) Final Temp. ( o C)* D T ( o C) Water *Use the maximum temperature as the final temperature of the Al and water, unless directed by your instructor to perform the temperature-time extrapolation described in the Discussion section. Use of the maximum temperature is an approximation that works reasonably well in Part B because you do not have to wait for an entire samp le to react. 7. Use the data above and Eq. 7 to calculate the specific heat of aluminum. The specific heat of water is J g -1 o C -1. Have your instructor check your result. 8. Based on the Activity Series, identify two different metals that could be used as weights to keep the magnesium metal submerged in the hydrochloric acid solution which would not react with the acid. Explain your answer. E12J-10

11 Name Section Date PRE-LAB QUESTIONS Experiment 12J 1. Plot the following set of temperature vs. time data using a spreadsheet program, with time on the x-axis and temperature on the y-axis. Extrapolate the linear portion of the graph and determine T f of the reaction for which the data have been recorded. (Refer to the Excel directions in Appendix O.) Include your properly formatted graph with your Pre-lab. T f = T = Time (minutes) Temperature ( C) o Review the following websites: Services/Fires_and_Explosions/Magnesium_Fire/magnesium_fire.html a. List 2 military applications for the reaction studied in this experiment. Mg (s) + 2 O 2 (g) MgO (s) i. ii. b. True or False: Water should be used to put out a Magnesium fire. (Hint: watch the video on the website.) 3. Use Hess s Law to determine H f for CaO: Ca (s) + ½ O 2 (g) CaO (s) Ca (s) + 2 H + (aq) Ca 2+ (aq) + H 2 (g) H = kj 2 H 2 (g) + O 2 (g) 2 H 2 O (l) H =! kj CaO (s) + 2 H + (aq) Ca 2+ (aq) + H 2 O (l) H = kj H f (CaO) = E12J-11

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