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CHEM 1515.001 - 006 Exam V John V. Gelder May 6, 2002 Name ________________________ TA's Name ________________________ Section _______ INSTRUCTIONS: 1. This examination consists of a total of 11 different pages. The last four pages include a periodic table; many useful equations and constants; a table of vapor pressures for water; a solubility table; a table of equilibrium constants for acids and bases; and a table of thermodynamic values. All work should be done in this booklet. 2. PRINT your name, TA's name and your lab section number now in the space at the top of this sheet. DO NOT SEPARATE THESE PAGES. 3. Answer all questions that you can and whenever called for show your work clearly. Your method of solving problems should pattern the approach used in lecture. You do not have to show your work for the multiple choice or short answer questions. 4. No credit will be awarded if your work is not shown in 6. 5. Point values are shown next to the problem number. 6. Budget your time for each of the questions. Some problems may have a low point value yet be very challenging. If you do not recognize the solution to a question quickly, skip it, and return to the question after completing the easier problems. 7. Look through the exam before beginning; plan your work; then begin. 8. R e l a x and do well. SCORES Page 2 Page 3 Page 4 Page 5 Page 6 Page 7 TOTAL _____ (18) _____ (16) _____ (16) _____ (18) _____ (20) _____ (12) ______ (100) CHEM 1515 EXAM V (9) PAGE 2 1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as, (l)iquid, (s)olid or (aq)ueous. Soluble ionic compounds should be written in the form of their component ions. a) Zn(NO3)2(aq) b) Na2CO 3(s) c) (CH3)2NH(aq) Na2S(aq) → ZnS(s) + 2NO3-(aq) + 2+(aq) + + HCl(aq) → CO 2(g) + H2O (l) + + 2Cl-(aq) + 2Na+(aq) HCN(aq) → (CH3)2NH2+(aq) + CN-(aq) (4) 2a. Write the ionic and net ionic chemical equation for 1a), 1b) or 1c). Ionic equation 2Na + (aq) + CO3 2-(g) + 2H+ (aq) + 2Cl-(aq) → CO 2 (g) + H2 O (l) + 2Cl-(aq) + 2Na+ (aq) Net Ionic equation CO 3 2- (g) + 2H+ (aq) → CO 2 (g) + H2 O (l) (5) 3. Account for the following observations about NH3 and NF3. In your answer, use appropriate principles of intermolecular forces. Your answer must include reference to both substances. a) NH3 has a normal boiling point of –33.4 ˚C where as NF3 has a boiling point of –128.8 ˚C. NH3 is polar and can hydrogen bond. Hydrogen-bonding is a strong intermolecular attractive force and causes the boiling point of NH3 to be high. NF3 is polar but can not hydrogen bond. The dipole-dipole forces are not as strong an intermolecular attractive force so the boiling point of NF3 is much lower compared to NH3. CHEM 1515 EXAM V PAGE 3 (16) 4. 0.10 M NaF 0.10 M MgCl2 0.10 M C2H5OH 0.10 M CH3COOH (HC2H3O2) Answer the following questions, which refer to the 100 mL samples of aqueous solutions at 25 ˚C in the stoppered flasks shown above (a) Which solution has the lowest electrical conductivity? Explain. The lowest electrical conductivity has the fewest ions… C2H5OH. All other substances above form ions in solution. (b) Which solution has the lowest freezing point? Explain. MgCl2 has he lowest freezing point. Since MgCl2 produces the most ions in solution with an I value of 3, ∆ Tf will be the largest. (c) Above which solution is the pressure of water vapor the greatest? Explain. C2H5OH does not ionize, so there are the fewest particles in solution so it will have the higher vapor pressure. (d) Which solution has the highest pH? Explain. NaF will have the highest pH. F- is a weak base. CHEM 1515 EXAM V PAGE 4 (16) 5a. Define the term 'equilibrium vapor pressure'. (b) If a sample of a liquid is injected into an evacuated container, describe what phase change occurs. (Note the amount of the liquid injected is small compared to the volume of the container.) Since the container is evacuated when some liquid is introduced, some of the liquid will immediately vaporize until the equilibrium vapor pressure of the liquid is reached, or until all of the liquid evaporates. So if the equilibrium vapor pressure is reached, both liquid and vapor are present. Otherwise only vapor. (c) Is the phase change exothermic or endothermic? Explain? Evaporation is endothermic, because to escape from the liquid phase to the vapor phase intermolecular attractive forces must be over come. Although weak these forces must be broken for particles to enter the vapor phase. (d) In terms of the 'pressure exerted by the vapor' and the 'equilibrium vapor pressure' how do we determine what phase(s) are present in the container after adding the sample? If the pressure due to the vapor is equal to the equilbirium vapor pressure both liquid and vapor must be present. If the pressure due to the vapor is less than the equilibrium vapor pressure only vapor is present. The pressure dur to the vapor can never exceed the equilibrium vapor pressure. CHEM 1515 EXAM V PAGE 5 (18) 6. A commercial aqueous solution of ammonia is 28% NH3 by mass and has a density of 0.900 g mL-1. (a) Calculate the molarity of the solution. Assume 100 gram of solution 28 g of NH3 and 72 grams of water. Convert the mass of NH3 to moles, then divide the 100 gram of solution by the density to get mLs of solution. Convert mLs to liters and divide into the moles for the molarity Answer is 14.8 M (b) Calculate the molality of the solution. Use the moles determined in part a and divide by the mass of water in kilograms. Answer is 22.9 molal (c) Describe how you would prepare 1.00 L of a 0.400 M NH3 solution from the more concentrated solution in part a). Using the dilution equation Mc·Vc = Md · Vd The volume of concentrated ammonia needed is 27.0 mL. Add 27 mL of NH3 to about 900 mL of water, be sure all the NH3 is dissolved than add enough water to reach a final volume of 1000 mLs. (10) (d) Calculate the pH of the solution prepared in part c) . The pH of the solution is 11.43. This is a simple weak base calculation where the initial concentration of NH3 is 0.400 M (10) (e) Calculate the pH of the solution prepared by adding 0.250 mol of solid NH4Cl to the solution prepared in part c). This is a common ion/buffer solution. The pH is 9.45. CHEM 1515 EXAM V PAGE 6 (12) 7. For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased. PCl3(g) + Cl2(g) PCl3(g) ä (a) What is the sign of ∆S˚ for the reaction? Explain. ∆ S˚ is positive. In the balanced equation there are more moles of gase in the products compared to the reactants. So the products are more disordered. (b) What change, if any, will occur in ∆G˚ for the reaction as the temperature is increased? Explain your reasoning in terms of thermodynamic principles. The reaction is endothermic because , ", it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased ". So if ∆ H˚ is positive and ∆ S˚ is positive ∆ G˚ will be ∆ S˚ means the come more negative with increasing temperature. The equation ∆ G˚ = ∆ H˚ – T∆ ∆ reaction with become more spontaneous as T increases. The T∆ S˚ term will eventually be greater than ∆ H˚ and the reaction will be spontaneous. (c) If the volume of the reaction mixture is decreased at constant temperature to half the original volume, what will happened to the number of moles of Cl2 in the reaction vessel? Explain. If the volume is decreased the pressure will increase in the container and the reaction will proceed in a direction to relive the pressure, so it will got from right to left and the amount of Cl2 will decrease. Q is inversely related to the volume, so as volume decreases Q increases. To re-establish equilibrium the reaction must go from R to L and amount of Cl2 will decrease. CHEM 1515 EXAM V PAGE 7 Periodic Table of the Elements IA VIIIA 1 H 1 1.008 3 2 He IIA IIIA IVA VA VIA VIIA 4.00 4 Li Be 2 6.94 9.01 11 12 Na Mg 3 22.99 24.30 19 20 4 5 6 7 5 6 7 8 9 10 B C N O F Ne 10.81 12.01 14.01 16.00 19.00 20.18 13 14 15 16 17 18 VIII IIIB IVB VB VIB VIIB 21 22 K Ca Sc Ti 23 24 25 26 27 IB 28 29 Al Si P S Cl Ar 31 32 33 34 35 IIB 26.98 28.09 30.97 32.06 35.45 39.95 30 36 V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) 87 88 89 104 105 106 Fr Ra Ac (223) 226.0 227.0 (261) (262) (263) 58 Lanthanides 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (145) 150.4 152.0 157.2 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Useful Information L·atm J PV = nRT R = 0.0821 mol·K = 8.314 mol·K vp2 ∆H˚vap 1 1 g lnvp = – R T – T density of H2O = 1.00 3 1 2 1 cm ∆H˚rxn = ∑n(∆Hf˚(products)) – ∑ m(∆Hf˚(reactants)) 6.02 x 1023 ∆S˚rxn = ∑ n(S˚(products)) – ∑ m(S˚(reactants)) ∆G˚rxn = ∑ n(∆Gf˚(products)) – ∑ m(∆Gf˚(reactants)) ∆G˚ = ∆H˚ - T∆S˚ pH = –log[H+] ∆G˚ = –RTlnK x 1,2 = pH + pOH = 14 Kp = Kc(RT)∆n –b± b2 – 4ac for ax2 + bx + c = 0 2a pOH = –log [OH–] Kw = 1.00 x 10–14 CHEM 1515 EXAM V PAGE 8 1 pm = 10-12 m Psolution = χsolventP˚solvent ∆T = ikm ˚C kf(H2O) = 1.86 m edge length (l) = 2r ˚C kb(H2O) = 0.512 m edge length (l) = 2 2· r Temperature (°C) -5 0 5 10 15 20 25 30 35 40 45 Vapor Pressure(mmHg) 3.2 4.6 6.52 9.20 12.8 17.5 23.8 31.8 42.1 55.3 71.9 edge length (l) = Temperature (°C) 50 55 60 65 70 75 80 85 90 95 100 4r 3 Vapor Pressure(mmHg) 92.5 118.0 149.4 187.5 233.7 289.1 355.1 433.6 525.8 633.9 760 Solubility Table Ion NO3– ClO4– Cl– I– SO42– Solubility soluble soluble soluble soluble soluble Exceptions none none except Ag+, Hg22+, *Pb2+ except Ag+, Hg22+, Pb2+ except Ca2+, Ba2+, Sr2+, Hg2+, Pb2+, Ag+ CO32– insoluble except Group IA and NH4 PO43– -OH S2– Na+ NH4+ K+ insoluble insoluble insoluble soluble soluble soluble except Group IA and NH4 except Group IA, *Ca2+, Ba2+, Sr2+ except Group IA, IIA and NH4+ none none none *slightly soluble + + CHEM 1515 EXAM V Name PAGE 9 Formula Ka1 Acetic Ascorbic HC2H3O2 Arsenic H3AsO4 1.8 x 10– 5 8.0 x 10– 3 5.6 x 10– 3 Arsenous Benzoic Boric Butyric acid Carbonic Cyanic Citric H3AsO3 HC7H5O2 H3BO3 HC4H7O2 H2CO3 HCNO H3C6H5O7 6.0 6.5 5.8 1.5 4.3 3.5 7.4 Formic Hydroazoic Hydrocyanic Hydrofluoric Hydrogen chromate ion Hydrogen peroxide Hydrogen selenate ion Hydrogen sulfate ion Hydrogen sulfide Hypobromous Hypochlorous Hypoiodus Iodic Lactic Malonic Oxalic HCHO2 HN3 HCN HF HCrO4– H2O2 HSeO4– HSO4– H2S HBrO HClO HIO HIO3 HC3H5O3 H2C3H2O4 H 2C 2O 4 1.8 1.9 4.9 7.2 3.0 2.4 2.2 1.2 5.7 2.0 3.0 2.0 1.7 1.4 1.5 5.9 Nitrous Phenol Phosphoric Paraperiodic Propionic Pyrophosphoric Selenous Sulfuric Sulfurous Tartaric HNO2 HC6H5O H3PO4 H5IO6 HC3H5O2 H4P2O H2SeO3 H2SO4 H2SO3 H2C4H4O6 E.2 HC6H7O6 10–10 10– 5 10–10 10– 5 10– 7 10– 4 10– 4 x 10– 4 x 10– 5 x 10–10 x 10– 4 x 10– 7 x 10–12 x 10– 2 x 10– 2 x 10– 8 x 10– 9 x 10– 8 x 10–11 x 10– 1 x 10– 4 x 10– 3 x 10– 2 x 10– 4 x x x x x x x 4.5 1.3 x 10–10 7.5 x 10– 3 2.8 x 10– 2 1.3 x 10– 5 3.0 x 10– 2 2.3 x 10– 3 strong acid 1.7 x 10– 2 1.0 x 10–3 Ka2 Ka3 1.0 x 10– 7 3.0 x 10–12 5.6 x 10–11 1.7 x 10– 5 4.0 x 10– 7 1.3 x 10–13 2.0 x 10– 6 6.4 x 10– 5 6.2 x 10– 8 5.3 x 10– 9 4.4 5.3 1.2 6.4 4.6 x x x x x 4.2 x 10–13 10– 3 10– 9 10– 2 10– 8 10– 5 DISSOCIATION CONSTANTS FOR BASES AT 25˚C Name Formula Ammonia Aniline Dimethylamine Ethylamine Hydrazine NH3 C6H5NH2 (CH3)2NH C2H5NH2 H2NNH2 Kb 1.8 4.3 5.4 6.4 1.3 x x x x x 10– 5 10–10 10– 4 10– 4 10– 6 Name Formula Hydroxylamine Methylamine Pyridine Trimethylamine HONH2 CH3NH2 C5H5N (CH3)3N Kb 1.1 4.4 1.7 6.4 x x x x 10– 8 10– 4 10– 9 10– 5 CHEM 1515 EXAM V PAGE 10 Thermodynamic Values (25 ˚C) Substance o ∆H f and State kJ mol Carb o n C(s) (graphite) C(s) (diamond) CO(g) CO2(g) CH4(g) CH3OH(g) CH3OH(l) CH3Cl(g) CHCl3(g) CHCl3(l) H2CO(g) HCOOH(g) HCN(g) C2H2(g) C2H4(g) CH3CHO(g) C2H5OH(l) C2H6(g) C3H6(g) C3H8(g) 0 2 -110.5 -393.5 ? -201 -239 -80.8 -100.8 -131.8 -116 -363 135.1 227 52 -166 -278 -84.7 20.9 -104 Bromine Br2(l) BrCl(g) 0 14.64 o ∆G f kJ mol So Substance J K.mol 0 3 -137 -394 -51 -163 -166 -57.4 6 2 198 214 186 240 127 234 -110 -351 125 209 68 -129 -175 -32.9 62.7 -24 219 249 202 201 219 250 161 229.5 266.9 270 0 -0.96 152. 240 Chlorine Cl 2 (g) Cl2(aq) Cl-(aq) HCl(g) 0 -23 0 7 223 121 -167 -92 -131 -95 57 187 Fluorine F2(g) F-(aq) HF(g) 0 -333 -271 0 -279 -273 203 -14 174 Hydrogen H2(g) H(g) 217 H+(aq) OH-(aq) H2O(l) H2O(g) Magnesium Mg(s) Mg(aq) MgO(s) 0 203 0 -230 0 115 0 -157 131 -242 -229 189 0 -492 -601 0 -456 -569 33 -118 26.9 0 -11 o o So ∆H f and State kJ mol ∆G f kJ mol Oxygen O2(g) O(g) 249 O3(g) 0 232 143 0 161 163 239 Nitrogen N2(g) NCl3(g) NF3(g) NH3(g) NH3(aq) NH2CONH2(aq) NO(g) NO2(g) N2O(g) N2O4(g) N2O5(g) HNO3(aq) HNO3(l) NH4Cl(s) NH4ClO4(s) 0 230 -125 ? ? ? 90 32 82 10 -42 -207 -174 -314 -295 0 271 -83.6 -17 -27 ? 87 52 104 98 134 -111 -81 -201 -89 192 -137 -139 193 111 174 211 240 220 304 178 146 156 95 186 Silver Ag(s) Ag+(aq) Ag(S2O3)3-(aq) AgBr(s) AgCl(s) 0 105.6 -1285.7 -100.4 -127.1 0 77.1 --96.9 -109.8 42.6 72.7 -107.1 96.2 J K.mol 205 Sulfur S(rhombic) SO2(g) SO3(g) H2S(g) 0 -296.8 -395.7 -20.17 0 -300.2 -371.1 -33.0 31.8 248.8 256.3 205.6 Titanium TiCl4(g) TiO2(s) -763 -945 -727 -890 355 50 Aluminum AlCl3(s) -526 -505 184 Barium BaCl2(aq) -872 Ba(OH)2·8H2O(s) -3342 -823 -2793 123 427 Iodine I2(s) HI(g) 0 25.94 0 1.30 116.7 206.3