Lab 11

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ENTHALPY OF FORMATION OF AMMONIUM SALTS BACKGROUND INFORMATION There are a very large number of chemical reactions that have never been investigated. There may be good reasons for not performing certain chemical reactions. For example, some reactions may pose great risks to the environment or to living organisms. Other reactions may be too costly in terms of time or money. Studying the chemical reactivity of diamonds would certainly be an expensive undertaking. Fortunately, there are methods that allow us to study certain aspects of chemical reactions without ever having to perform the reactions. Hess’s Law (also known as the Law of Constant Heat Summation) is an example of a method that allows chemists to study the thermodynamics of a chemical reaction without having to perform the reaction itself. Germain Henri Hess discovered this fundamental law of nature in 1840 while studying the heat evolved when acids and bases are mixed with water. Hess’s Law states that the enthalpy change (AH) of individual steps in a process can be added or subtracted to determine the overall enthalpy change of the process. Hess’s Law is a direct consequence of the First Law of Thermodynamics, which requires that the enthalpy change of a reaction be the same regardless of the reaction pathway that leads from reactants to products. Thus, the net amount of heat liberated or absorbed in a chemical process is the same, regardless of whether the process is performed in one step or multiple steps. Because of the additive property of AH, we can use a standard set of reactions from which we can gener- ate any other reaction. All we have to do is add the enthalpies of these standard reactions and we have the enthalpy of our reaction of interest. The standard reactions we use are called formation reactions and the enthalpies are enthalpies of formation symbolized by AHf. A formation reaction has one single product: one mole of the compound. The reactants are elements in their standard states. Thus, on paper, we can decompose any compound to its elements (AH = —AHy and then take those elements and rearrange them into the product compounds (AH = AH?). This leads to the equation 1. AH°= Y AH?- Y, AH} (1) Product Reactants This process for the reaction 3C,H, Cg¢Hg is shown in figure 1. In today’s experiment, you will use Hess’s Law to determine the enthalpy change (AH) of a chemical reaction that cannot be easily measured by experimental methods. This reaction is represented by equa- tion (2). From General Chemistry, 2nd Edition by Department of Chemistry, University of North Carolina at Wilmington. Copyright © 2004 by University of North Carolina at Wilmington. Reprinted by permission of Kendall/Hunt Publishing Company. 115
116 LAB 11: ENTHALPY OF FORMATION OF AMMONIUM SALTS 6C + 3H, 3C,H, Enthalpy CeHe FIGURE 1. Enthalpy of conversion of - acetylene to benzene. %Nz(g) + 2H,(g) + %Clz(g) NH,CI(s) or Nx(g) + 2H,(g) + % O,(g) - NH4,NOs(s) AH, (2) You will determine the value of AH for this reaction by measuring the enthalpy changes of two related chemical reactions. These reactions are represented by equations (3) and (4) below. NH;(aq) + HCl(aq) - NH,Cl(aq) or NHs(aq) + HNO;(aq) - NH;NOs(aq) AH, (3) NH,Cl(aq) 222 5 NH,CI(s) or NH4;NO;(aq) —:° y NH,NO; (s) AH; (4) Equation (2) can be obtained by adding equation (3) to equation (4) along with the enthalpies of forma- tion of the aqueous and ammonia (shown in equations (5) and (6), below). | 1 3 Hy(g) + - Cla(g) — 205 HCl(aq) Ha(g) + > N28) + 5 0:(8) %0 s HNOs(ag) AH; (5) 1 3 7 Na(g) + 5 Ha(g) —=2 5 NHs(aq) AH, (6) We simply add the enthalpies for (3), (5) and (6) together with the reverse of (4) AH}) = AH3 + AH4 + AHS <+ AH6 (7) The enthalpies for (5) and (6) are readily available from the literature. Equation (1) can be obtained by adding equation (2) to the reverse of equation (3). If we reverse equa- tion (3), we must also change the sign of the enthalpy change for the reaction. According to Hess’s Law, if we add the equations for two reactions together, we can also add their enthalpy changes together. Thus, the enthalpy change for reaction (1) is simply the sum of the enthalpy change for reaction (2) and the reverse of reaction (3). AH, = AH, + (-AH;) (8)
LAB 11: ENTHALPY OF FORMATION OF AMMONIUM SALTS 117 To determine the enthalpy changes (AH, and AH3) of reactions (2) and (3), you must measure the heat (q) released by each reaction. You will do this using the same techniques and concepts you used in the experiment on calorimetry. (You might want to review the Introduction to Calorimetry.) Recall that the heat exchanged in a process can be calculated from the following equation g=cxmxAT 9) where m is the mass of the substance in grams, the specific heat, ¢, of the substance is expressed in units of J/ge°C, and AT is the temperature change the substance undergoes (Tfipa Tiniial)- Some heat will be lost through the walls and lid of the calorimeter. A correction term C, = 15 J/K will be used to estimate this loss. Our corrected relationship between heat and temperature becomes: g = x m + 15)AT. The specific heat capacities can be found in Table 2. Gwater = (418 J/g.oc)(mwater)(ATwater) (10) The heat gained by the water is equal in magnitude by opposite in sign to the heat lost by the reaction. Therefore, we can calculate ge,cion from the following relationship. Yreaction = —Ywater (1 1) Finally, the enthalpy change of each reaction, AH,.,.ion, can be determined by dividing the heat lost by the reaction (Greaciion) by the number of moles of product formed (either NH,Cl(aq) or NH;NOs(aq)). For reaction (2), this calculation takes the following form. AH, = 4 (12) moles NH,Cl or NH,NO; PROCEDURE Caurion: Wear departmentally approved eye protection. If you are found without the eye protection, penalty will be imposed without warning. Gloves and proper attire must be worn during this experiment. This experiment will be performed in pairs. The write-up, calculations, graphing, etc., must be done totally independently. A series of two experiments will be performed. Assume the calorimeter constant, Cec, to be 157 K. Part A. Determination of the enthalpy of reaction of aqueous ammonia with a strong acid (HCI or HNO;) assigned by the instructor. Part B. Determination of the enthalpy of solution of the ammonium salt of the acid used in Part A (NH4C1 or NH4NO3). In addition to reagents, the equipment needed is: two 6-0z pressed Polystyrene cups (and one cup lid), a 150-mL beaker, two 100-mL graduated cylinders, and one thermometer. Keep the cups together, as though they were glued together. You need 2 cups for the additional insulation.
118 LAB 11: ENTHALPY OF FORMATION OF AMMONIUM SALTS Solution temperature as a function of time will be recorded on the data sheet. In addition to recording the data, a temperature-time plot on graph paper should be prepared; this can be done while the data are being collected. The temperature should be plotted on the y-axis and the time in minutes on the x-axis. The T observed will be between 7 and 14° for Part A and between —1 and —6° for Part B. The scale units on the y-axis should be selected so that all points can be plotted on scale; the scale units on the x-axis should cover the range 0—20 minutes. The procedure is essentially the same for both parts: 1. Rinse one Polystyrene cup and beaker with distilled water and wipe dry with a clean piece of absorbent paper. Nest it in the other cup (see figure 2). Clean a beaker with distilled water and wipe dry with absorbent paper. Place the thermometer through the lid of the cup with the rubber ring over the hole set so that the thermometer will not touch the bottom of the cup when the lid is placed on the cup. Do not clamp the thermometer to the ring stand. They are fragile and will likely break if clamped. 2. Using graduated cylinders, measure solution 1 (see Table 1) into the cup and (except in Part B) mea- sure solution 2 into the beaker. Record the concentrations of the solutions used on the Data Sheet. Note: Calculate the concentration of the ammonium salt—NH,C] or NH,NO;—formed in the reaction of Part A. Determine the mass of salt in 100 mL of solution; weigh out that amount of salt and use in place of solution 2 in Part B. 3. Check the temperature of solution 1. Assume the temperature of the two solutions before mixing is essentially the same. Record the temperature of solution 1 at 1-minute intervals for 5 minutes. 4. After 5 minutes of temperature readings, pour solution 2 into the calorimeter cup. Immediately swirl the cup gently to mix the solution well. Record the time of mixing. (Never use a thermometer as a stirring rod!) Continue to record the temperature at 1-minute intervals for 10 minutes. Do one determination of the temperature change for each part of the experiment. Lid——>» " \ /4— Inner Cup Thermometer <«— Outer Cup FIGURE 2. Nested polystyrene cup.
LAB 11: ENTHALPY OF FORMATION OF AMMONIUM SALTS 119 TABLE 1 REAGENTS TO BE USED Part A B Solution 1 50.0 mL acid (2.00M HCI or 2.00M HNO;) 100.0 mL distilled water Solution 2 50.0 mL 2.05M NH; See Note, p. 114 CALCULATIONS Study carefully the calculations for NH Br at the end of this section. 1. Find the temperature change AT from the temperature-time plots by extrapolating the plotted data to the time of mixing, reading the initial and final temperatures, and taking the temperature change for each determination. See figure 3 for an example. Use good graphing technique. Data should take up about 2/3rd of the paper. The extrapolation should not run off the top or the bottom of the page. Graph should have a title and axes should be labeled. Origin does not need to be (0, 0). 2. Use the calorimeter constant, C, of 15 J/K for the calculation. 3. Use the data from Part A and the calorimeter constant to calculate the enthalpy of neutralization of NH; and the assigned acid. 4. Use the data from Part B to compute the enthalpy of solution of the ammonium salt. 5. Use the results for the enthalpy of neutralization and the enthalpy of dissolution, along with the liter- ature values of AHy for NH;(aq) and the assigned acid (given in Table 3) to compute the enthalpy of formation of the ammonium salt. AT =104K 25 T (°C) - FIGURE 3. Plot of temperature-time 0 5 10 15 data showing determination of AT. Time (min)
120 LAB 11: ENTHALPY OF FORMATION OF AMMONIUM SALTS Typical Calculations for NH,Br Part A AT=119K, d=1018g/mL C=3911J/gK, MM. =97.8 g/mol Mass of solution = 100 mL x 1.018 g/mL = 101.8 g Cs (heat capacity of the solution) = 101.8 g x 3.91 J/g K =398 J/K AH,; = —(Cs + C))AT =—-(398 J/K + 15.0 J/K)11.9 K =-4915J =-4915kJ Students mixed 50 mL of 2.00M acid with 50.00 mL of 2.05M base, 1.e., 0.100 mol of the reactants were used. Therefore, the measured AH, is for 0.100 mol of product formed. molar AH, = —-4.915 kJ/ 0.100 mol = —49.2 kJ/mol Part B TABLE 2 MOLECULAR WEIGHTS, HEAT CAPACITIES, AND DENSITIES OF SOLUTIONS Molecular wt. Concentration Sp. heat capacity” density” Solute g mol™ mol L™ Jg' K gmL™ NH,Cl 53.49 1.00 3.93 1.013 NH,NO; 80.04 1.00 3.90 1.029 H,O 18.02 4.18 0.997 At 25°C TABLE 3 ENTHALPIES OF FORMATION AT 25 °C® Reaction AH? kJ mole™ INL(2) + 2Hy(2) + —=2 5 NH;(ag, 2.0M) —80.7 JHa(g) + 3Cly(g) —=2 5 HCl(aq, 2.0M) ~164.7 JHy(2) + 3Nx(2) + 50,(g) —22 5 HNO; (aq, 2.0M) ~206.0 Ha(g) +305(g) = H0() -285.8 * “Selected Values of Chemical Thermodynamic Properties,” Circular 500, National Bureau of Standards, Washington, D.C., 1961
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