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CHEM 1515.001 - 006 Exam I John I. Gelder February 5, 2002 Name ________________________ TA's Name ________________________ Section _______ INSTRUCTIONS: 1. This examination consists of a total of 8 different pages. The last three pages include a periodic table, a table of vapor pressures for water, a solubility table 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 4a, 4d and 8. 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 TOTAL _____ (23) _____ (35) _____ (18) _____ (24) ______ (100) CHEM 1515 EXAM I (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. + 2H2O(l) → a) 2K(s) b) Na2CO3(aq) c) Ba(OH) 2 · 8H2O (s) + 2K + (aq) + 2OH– (aq) + H2(g) Ca(NO3)2(aq) → 2Na+(aq) + 2NO3–(aq) + CaCO3(s) + 2NH 4SCN (s) → Ba 2+(aq) + 10H2O (l) + 2NH3(aq) + 2SCN– (aq) (4) 2a. Write the ionic and net ionic chemical equation for 1a) or 1b). Ionic equation 2Na + (aq) + CO3 2-(aq) + Ca2+ (aq) + 2NO3 – (aq) → 2Na + (aq) + 2NO3 – (aq) + CaCO3 (s) Net Ionic equation CO 3 2- (aq) + Ca2+ (aq) → CaCO 3 (s) (10) 3. Predict whether the entropy change in the system is positive or negative for each of the following processes. a) CH3OH(l) + O2(g) → HCO2H(l) + H2O(l) The phases are both liquid, while the reactants are a liquid and a gas. There are equal numbers of moles, and the liquid phase is more ordered compared to the gas phase, so the products are more ordered compared to the reactants and ∆S is –. b) 2SO3(g) → 2SO2(g) + O2(g) The reactants and products are all in the same phase, but there are more moles of products compared to reactants. So there are more possible arrangements in the products compared to the reactants and ∆ S is +. CHEM 1515 EXAM I PAGE 3 4. Methanol, CH3OH, can be synthesized by reacting hydrogen gas with carbon monoxide, according to the chemical equation, CO(g) + 2H2(g) → CH3OH(l) (18) a) Calculate ∆H˚rxn, ∆S˚rxn and ∆G˚rxn. CO(g) + 2H2(g) C(s) + 2H2 (g) → CH3OH (l) ° kJ ∆Hf mol -110.5 -0 –239 J 198 131 127 S° . K mol ° kJ ∆Gf mol -137 -0 –166 ° ° ° ∆Hrxn = Σ∆Hf (products) - Σ ∆H f (reactants) ° ° ° = [∆H f (CH3OH (l)) ] – [∆H f (CO(g)) + 2∆H f (H 2 (g) ) ] = (-239 kJ) - [(-110.5 kJ) + 2(0 kJ)] = –128 kJ ° ∆S rxn = Σ S°(products) - Σ S°(reactants) = [S˚(CH3 OH (l) ) ] – [S˚(CO(g) )+ 2S˚(H2 (g) )] J J J = [(127 K )] - [ (198 K ) + 2(131 K )] = –333 J/K ° ° ° ∆Grxn = Σ∆Gf (products) - Σ ∆G f (reactants) ° ° ° = [∆G f (CH3OH (l)) ] – [∆G f (CO(g)) + 2∆G f (H 2 (g) ) ] = (-166 kJ) - [(-137 kJ) + 2(0 kJ)] = –29 kJ (4) b) Which factor, the change in enthalpy, ∆H˚, or the change in entropy, ∆S˚, provides the principal driving force for the reaction at 298 K? Explain. ∆ H is the principal driving force, because the sign of ∆ H in the reaction is negative. A –∆ ∆H ∆ S does not favor spontaneity. favors spontaneity. ∆ S is also negative. –∆ ∆ S a spontaneous reaction has a negative ∆ G. In this equation spontaneity is Since ∆ G = ∆ H – T∆ ∆ ∆ S. In the reaction in part a only the sign of ∆ H favors spontaneity. favored by a –∆ H and a +∆ (5) c) For the reaction, how is the value of the standard free energy change, ∆G˚, affected by an increase in temperature? Explain. ∆ S, since ∆ H is negative and ∆ S is negative as the temperature Using the equation ∆ G = ∆ H – T∆ ∆ ∆ S is more positive, making ∆ G˚ increases the T∆ S term is becoming more negative, and –T∆ more positive and less spontaneous. (8) d) Calculate the ∆S˚f for CH3OH(l). + 1 2 O2(g) J S° . 6 131 205 K mol ° ∆S rxn = Σ S°(products) - Σ S°(reactants) → CH3OH (l) 127 1 = [S˚(CH3 OH (l) ) ] – [S˚(C(s) )+ 2S˚(H2 (g) ) + 2 S˚(O2 (g) ) ] J J J J 1 = [(127 K )] - [ (6 K ) + 2(131 K )+ 2 (205 K )] = –243 J/K CHEM 1515 EXAM I PAGE 4 (12) 5. A sample of water is in the vacuum flask shown below. The vacuum flask side arm has vacuum rubber tubing with one end attached to the flask and the other end attached to a vacuum pump (can not see in this diagram). When the vacuum pump is on any gas/vapor in the flask is removed. In an experiment the water placed into the flask initially is at room temperature. The rubber tubing is attached to the flask and the vacuum pump, and the pump is turned on. After the pump has been on for several minutes the following two observations are made: 1. the volume of water in the flask has decreased; 2. what water remains has turned to ice. Explain both of these observations. The volume of liquid water decreases because as the vacuum pump removes vapor from above the liquid, particles escape the liquid phase to maintain the equilibrium vapor pressure. Since particles are evaporating and few particles are condensing, the volume of the liquid will decrease. The reason the liquid eventually turns to solid is because evaporation is an endothermic process, and heat is removed from the system (the liquid water). When enough heat is removed the liquid freezes. Evaporation is endothermic because the conversion of water in the liquid phase to water in the vapor phase requires breaking hydrogen-bonding forces. Breaking bonds is always an endothermic process. (6) 6. Give the name or draw the complete Lewis structure (showing all C-H(X) bonds) for each of the following compounds. 4-ethyl-2,3-dimethylhexane 3, 3, 5–trimethylheptane 1,1-dichloro-2,2-difluoroethane CHEM 1515 EXAM I PAGE 5 (12) 7. Draw and name six different structural isomers for C8H18. (NOTE: You may use condensed formulas when representing the different structural isomers.) n–octane 2-methylheptane 3-methylheptane 4-methylheptane 2,3-dimethylhexane 3,3-dimethylhexane CHEM 1515 EXAM I PAGE 6 (12) 8. CH3I has a vapor pressure of 400 mm Hg at 25.3 ˚C. Calculate the temperature that CH3I has a vapor pressure of 40.0 mm Hg. ∆H˚vap for CH3I is 29.2 kJ mol-1. P1 ∆H -∆ ln P 2 = R (T11 - T12) J -29200 mol 400 ln 40 = J 8.314 mol • K ( 298.31 K - T11) 1 – T1 ) 1 2.303 = -3512 K (3.35 x 10-3 – T 1 ) 1 –6.56 x 10-4 = 3.35 x 10-3 – T 1 1 -3 T1 = 250 K T 1 = 4.00 x 10 ln (10) = -3512 K (3.35 x 10-3 CHEM 1515 EXAM I PAGE 7 Periodic Table of the Elements IA VIIIA 1 1 H 1.008 3 2 2 He IIA IIIA IVA VA VIA VIIA 4.00 4 Li Be 6.94 9.01 11 12 3 Na Mg 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 density of H2O = 1.00 3 lnvp1 = – R T2 – T1 cm g density of H2O = 1.00 3 cm ∆H˚rxn = ∑n(∆Hf˚(products)) – ∑ m(∆Hf˚(reactants)) ∆S˚rxn = ∑ n(S˚(products)) – ∑ m(S˚(reactants)) ∆G˚rxn = ∑ n(∆Gf˚(products)) – ∑ m(∆Gf˚(reactants)) ∆G˚ = ∆H˚ - T∆S˚ CHEM 1515 EXAM I PAGE 8 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 Temperature (°C) 50 55 60 65 70 75 80 85 90 95 100 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 I PAGE 9 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 CHEM 1515 EXAM I H 2O → PAGE 10