7# ,;8)Z?|||||||}}X}n}n}n}}"}}4}x}n~l ~~*~|X~~ ~~~^~~~~~~Enthalpy of Formation of Magnesium Oxide prelab Reading: Before coming to your discussion section, read: 1. sections 12.1 - 12.4 in Olmstead and Williams. 2. the remainder of this experiment in this manual. Pre-lab assignment to hand in: Prepare written responses to the following questions and submit your work to your instructor at the beginning of your discussion period: 1. If dilute HNO3 were used as an acid instead of dilute HCl, what effect would it have on our results? (Both are strong acids; HNO3 produces H+ and NO3- in solution. Nitrate and chloride are spectator ions in the reactions of Mg with very dilute HNO3 and HCl). 2. Given the following results for a trial of this experimental procedure and the molar enthalpy of formation of water (285.8kJ/mol), calculate HA(o,f)(MgO) from the following data: Mg(s) + 2 H+(aq) F( , )> Mg2+(aq) + H2(g) H1 = - 455 kJ/mol MgO(s) + 2 H+(aq) F( , )> Mg2+(aq) + H2O(l) H2 = - 153 kJ/mol Introduction Thermochemical properties of pure substances are important in many fields of science and technology. For example: a. Enthalpies of formation and of reactions such as combustion are essential for the calculation of bond energies. b. Mining engineers need to know how much fuel will be needed to reduce metals from their ores. c. The heat of combustion of an edible compound is its calorie count (actually kilocalorie* count) for weight watchers. d. The energetics of biochemical processes are essential features of our understanding of how and why these processes occur. In this experiment, you will perform measurements using exactly the same method employed by practicing thermochemists. The goal is to measure the enthalpies of formation of Mg2+(aq) and of MgO(s). The enthalpy of formation of Mg2+(aq) is represented simply by the enthalpy of dissolution of 1 mol of Mg metal in a very large amount of very dilute acid: Mg(s) + 2H+(aq) F( , )> Mg2+(aq) + H2(g) H = HA(o,f) (Mg2+,aq) (A) If the reason for this one-step measurement is not clear to you, remember that HA(o,f) for Mg(s), H+(aq)** , and H2(g) are zero by definition. Therefore, the only species in equation (A) with a nonzero HA(o,f) is Mg2+(aq), so that its enthalpy of formation is simply the enthalpy of this reaction. For the case of MgO, however, the direct enthalpy of formation Mg(s) + 1/2 O2(g) F( , )> MgO(s) H = HA(o,f)(MgO) (B) is difficult to measure. Magnesium metal must be sealed in an heavy-walled vessel (a "bomb") with an excess of oxygen under pressure. If this apparatus is properly set up, an electric fuse can ignite the metal sample to initiate the reaction (This is exactly how a conventional flashbulb operates). However, an appreciable amount of magnesium may react gradually before the ignition, and the reaction may be incomplete (as MgO forms, it may coat some of the metal and prevent its combustion with O2). It is convenient to use the first law of thermodynamics in the form of Hess's law to simplify the measurement of HA(o,f)(MgO). If a chemical reaction can be broken up into the sum of two or more other reactions, the algebraic sum of the enthalpies of these reactions will be the enthalpy of the desired reaction. Consider the following reaction equations: Mg(s) + 2 H+(aq) F( , )> Mg2+(aq) + H2(g) H1 (1) Mg2+(aq) + H2O(1) F( , )> MgO(s) + 2 H+(aq) H2 (2) H2(g) + 1/2 O2(g) F( , )> H2O(1) H3 (3) _____ Mg(s) + 1/2 O2(g) F( , )> MgO(s) H = H1 + H2 + H3 The sum of these three equations is the desired equation; thus HA(o,f)(MgO) = H1 + H2 + H3. H1 is simply the heat of solution of a mole of Mg metal in excess acid; H2 is the negative of the heat of solution of MgO in excess acid; and H3 is the molar enthalpy of formation of water (285.8kJ/mol). Procedure(Work in pairs) a. Using an analytical balance, weigh out 0.480 + 0.020 g of 20-mesh (granulated) magnesium into a weighing bottle; record the weight to + 0.0001 g. The solution will be stirred automatically using a magnetic stirrer. Assemble a calorimeter as shown in Figure 1 on the top of a magnetic stirrer. Use two nested styrofoam cups for extra insulation. Check the demonstration set-up in the lab before you begin. Load it with 100 ml of 0.5 M HCl. Measure the temperature four times over 30-second intervals; add the Mg; measure the temperature over 30-second intervals with vigorous stirring until the temperature reaches a plateau then begins to drop (see Figure 3).  Figure 1: Coffee Cup Calorimeter. Use teo nested styrofoam cups. b. Weigh (don't just tare) a clean weighing bottle, then weigh into it a weighing bottle approximately as many moles (+ 2%) of MgO as you weighed out magnesium. (MW of MgO = 40.30) Record the mass accurately. Using a clean calorimeter with 100 ml of 0.5 M HCl, perform the same measurements as for Mg (step a). Make sure all the MgO falls into the acid in the calorimeter; if any sticks to the weighing bottle, reweigh it to determine the mass of MgO actually reacted. Stir the reaction mixture vigorously to make sure all the MgO reacts, as well as to achieve a uniform temperature. Continue your readings until the temperature reaches a plateau then begins to drop c. Determine the heat capacity of the calorimeter by the following procedure. Prepare two calorimeters, each similar to the form illustrated in Figure 2. Compare two thermometers by immersing them together in water at room temperature for 1 minute and reading the temperature of each as nearly as possible to the nearest 0.10 oC. Be careful to avoid parallax in your readings. Always use the same thermometer in the calorimeter in which the temperature change occurs and in all subsequent readings apply any necessary correction to the other, so that the readings of both thermometers will always correspond. To correct for the heat lost to the calorimeter, place 50.0 ml of tap water at room temperature in one calorimeter and 50.0 ml of tap water which has been heated to 15-25oC above room temperature in the other. With the lids and thermometers in place, make careful temperature readings (+ 0.10 oC.) of each at 1-minute intervals for 3 minutes. At the next minute interval, pour the warmer water quickly and as completely as possible into the other calorimeter, and continue the readings for the next 3 minutes. Use the thermometer in the cooler water to record subsequent temperatures. You can extrapolate the temperatures of the separate samples and of the mixture back to the time of mixing by making a graph on which you plot the temperature of each along the ordinate (vertical axis) and the elapsed time for each along the abscissa (horizontal axis). A sample graph is illustrated in Figure 2. (If temperature changes are only slight, you need not make the graph but can, instead, simply extrapolate the data.) REPEAT this determination as a check and average your results.  Figure 2: Typical data on temperature and time that are obtained in the determination of the heat capacity of the calorimeter. Calculate the heat lost by the warmer water and the heat gained by the cooler water (Weight water x t oC x specific heat.). Assume the density of water = 1.0 g/ml, and the specific heat = 4.184 J/g oC. The difference, representing the joules gained by the calorimeter, divided by the temperature increase of the cooler water, gives the water equivalent of the calorimeter in joules/oC. It is unnecessary to express this result on a gram basis. The result simply measures the amount of heat required to raise the styrofoam cups (and the thermometer) 1.0oC. The following example will illustrate how the heat capacity of the calorimeter should be determined. The heat capacity of the calorimeter will be used as a correction factor in subsequent calculations. ______________________________________________________________________________ Example Water equivalent of the calorimeter data: Temperature of 50 ml of warmer water 37.9 oC Temperature of 50 ml of cooler water 20.9 oC Temperature after mixing 29.1 oC Heat lost by warm water: (50 g x 8.8 oC x 4.184 J/g oC) = 1841 J Heat gained by cool water: (50 g x 8.2 oC x 4.184 J/g oC) = 1715 J Heat lost to calorimeter 126 J Water equivalent of this calorimeter: (126 Joule/8.2 C) = 15. J/oC _____________________________________________________________________________ Data Interpretation: a. Make graphs for each reaction as shown in Figure 3, below. If you make the time of mixing as 0 min, the y-intercept will be equal to Tmax. Determine the corrected temperature rise for each. Use the specific heat (4.07 J/g oC) of 0.4 M HCl solution (the mean molarity during the reaction) and its density (1.01 g/cm3) to calculate the heat evolved: Heat evolved = [(specific heat of solution) x (mass of solution) + Cpc] (Tmax - Ti)  Figure 3: Graphical method of calculating corrected temperature rise: (Tmax - Ti). Cpc represents the heat capacity of the calorimeter vessel. For this experiment the heat capacities of Mg and MgO are negligible.b. For each reaction, calculate the heat evolved per mole of Mg or MgO: H = F(heat evolved,moles of sample) The molar heat of dissolving Mg represents reaction (1) and is HA(o,f)(Mg2+,aq)***. Is it exothermic or endothermic? Be sure to indicate the correct sign for HA(o,f)(Mg2+,aq). The molar heat of dissolving MgO represents the reverse of reaction (2). Use the correct sign for the molar heat of dissolving MgO and change the sign to yield H2. Finally, calculate HA(o,f)(MgO) by summing the enthalpies for reactions (1) through (3), keeping the correct sign in each case. Report a. Include your partner's name in your report. b. Submit the calculations necessary to determine the heat capacity of the calorimeter. It is not necessary to include your graphs. c. Submit the calculations requested in the Data Interpretation section of this experiment. It is not necessary to include your graphs. d. Questions: 1. Calculate your percent error by comparing your value of HA(o,f)(MgO) with a tabulated value. 2. Suppose the literature value for HA(o,f)(Mg2+,aq) is -455kJ/mol. Devise a calorimeter method for analyzing the Mg content of an alloy containing Mg and an inert metal. Outline the method and predict the heat of solution of 1.00 g of an alloy that is 25% Mg by mass. * 1 kcal = 4184 J ** H+ is the only aqueous ion for which HA(o,f) is zero by definition.  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