CHEM 10123/10125, Exam 3 April 4, 2012 (50 minutes) Name (please print) Please box your answers, and remember that significant figures, phases (for chemical equations), and units do count! 1. (18 points) SHOW ALL WORK. Bronze age metallurgists isolated copper by heating copper oxide ore with charcoal (carbon). The key reactions involved are Cu 2 O (s) 2 Cu (s) + ½O 2(g) G = 146.0 kj/mol ½O 2(g) + C (s) CO (g) G = -137.2 kj/mol a) Calculate G for the overall metallurgic reaction described above. Is it spontaneous at room temperature? At high temperature? Explain why or why not, including any pertinent equations. b) Would heating the ore in an oven without charcoal have isolated copper metal as well? Why or why not? [answer: a) Cu 2 O (s) + C (s) 2 Cu (s) + CO (g) G = 8.8 kj/mol At room temperature it would not be spontaneous, because it has a positive G value. G = H T S S is positive (gas is formed from solids), so increasing T would increase the T S factor. Since H is small, this could make it spontaneous at high T. b) In the absence of carbon, the rt reaction has positive G so it is not spontaneous. S is still positive, so heating it would decrease the G value, but since it is so much larger, accessible temperatures would be unlikely to overcome the barrier. Extremely high temperatures would be necessary.] 2. (15 points) SHOW ALL WORK. For the following reaction, calculate the appropriate quantities, using tables in the back. a.) H, in kj/mol 179.2 kj/mol, b.) G at 100 C, in kj/mol 119.4 kj/mol. CaO (s) + CO 2(g) CaCO 3(s) H = ( 1207.6) [( 634.9) + ( 393.5)] = ( 1207.6) ( 1028.4) = 179.2 kj/mol Need S, so prod react = [[91.7] [38.1 + 213.8] = 91.7 251.9 = -160.2 J/mol K G = H T S = ( 179.2) (373.15)( 160.2 x 10 3 ) = 119.4 kj/mol 3. (15 points) Briefly define the following terms: Common ion effect the solubility of an ionic compound is lower in a solution containing one of its constituent ions than in water. An example of LeChatelier s principle in action! 2 nd law of thermodynamics For any spontaneous process, the entropy of the universe increases Voltaic cell is an electrochemical device that produces electric current from a spontaneous chemical reaction 1
4. (14 points) Write the balanced net ionic equations for the following qualitative cation analysis procedures. a. precipitation of PbCl 2 from a solution containing Pb 2+ b. dissolution of Zn(OH) 2 in a solution of NaOH a. Pb 2+ (aq) + 2 Cl (aq) PbCl 2(s) b. Zn(OH) 2(s) + OH (aq) Zn(OH) 4 2 (aq) 5. (24 points) SHOW ALL WORK. Calculate the solubility of silver bromide in 0.435 M KCN. AgBr (s) Ag + (aq) + Br (aq) Ag + (aq) + 2 CN (aq) Ag(CN) 2 (aq) Add to get AgBr (s) + 2 CN (aq) Ag(CN) 2 (aq) + Br (aq) K = (7.7 x 10 13 )(5.6 x 10 8 ) = 4.312 x 10 4 CN Ag(CN) 2 Br I 0.435 M 0 0 C 2x +x +x E 0.435 2x x x 4.312 x 10 4 = [Ag(CN) 2 ][Br ]/[CN ] 2 = x 2 /(0.435 2x) 2 x/(0.435 2x) = 0.020765 x = 0.0090327 0.04153x 1.04153x = 0.0090327 x = 0.00867253 sig fig => 0.0087 M 6. (24 points) SHOW ALL WORK. Balance the redox equation in acidic solution. MnO 4 (aq) + H 2 C 2 O 4(aq) Mn 2+ (aq) + CO 2(g) 6 H + (aq) + 2 MnO 4 (aq) + 5 H 2 C 2 O 4(aq) 2 Mn 2+ (aq) + 8 H 2 O (l) + 10 CO 2(g) 7 (8 points) Predict which substance has the higher entropy. a. NO 2(g) or N 2 O 4(g)? b. pure silicon or silicon containing trace impurities? c. O 2(g) at 0 C or O 2(g) at -50 C? d. SnCl 4(l) or SnCl 4(g)? a. NO 2(g) or N 2 O 4(g)? b. pure silicon or silicon containing trace impurities? c. O 2(g) at 0 C or O 2(g) at -50 C? d. SnCl 4(l) or SnCl 4(g)? 2
8. (10 points) SHOW ALL WORK. The equilibrium constant K p of N 2 O 4(g) 2 NO 2(g) is 0.14 at 25 C. Calculate G rxn. G = RTlnK = (8.314 x 10 3 )(298)ln (0.14) = 4.87 kj/mol => 4.9 kj/mol 9. (22 points) Given the following pair of half-reactions, Cl 2(g) + 2e 2 Cl (aq) E = 1.78 V Mn 2+ (aq) + 2e Mn (s) E = 1.18 V a. Determine E cell for the corresponding galvanic cell. b. Sketch the galvanic cell using these reactions, and label all of the following things: -cathode and anode -electrode material(s) -direction of electron flow -direction of cation flow in salt bridge -direction of anion flow in salt bridge -what ions are dissolved in each solution a) E cell = 1.78 ( 1.18) = 2.96 V. b.) CHEM 10123/10125, Exam 3 April 4, 2012 (50 minutes) Name (please print) Please box your answers, and remember that significant figures, phases (for chemical equations), and units do count! 1. (18 points) SHOW ALL WORK. Bronze age metallurgists isolated copper by heating copper sulfide ore with air (oxygen). The key reactions involved are Cu 2 S (s) 2 Cu (s) + S (s) G = 86.2 kj/mol 3
O 2(g) + S (s) SO 2(g) G = 300.1 kj/mol a) Calculate G for the overall metallurgic reaction described above. Is it spontaneous at room temperature? At high temperature? Explain why or why not, including any pertinent equations. b) Would heating the ore in an oven without oxygen have isolated copper metal as well? Why or why not? [answer: a) Cu 2 S (s) + O 2(g) 2 Cu (s) + SO 2(g) G = 213.9 kj/mol At room temperature it would be spontaneous, because it has a negative G value. G = H T S S is very small but positive (same # of moles of gas, one more mol of solid), so increasing T would decrease G very slightly. So, it would also be spontaneous at high T. b) In the absence of oxygen, the rt reaction has positive G so it is not spontaneous. S is still very slightly positive, so heating it would decrease the G value, but since it is so much larger, accessible temperatures would be unlikely to overcome the barrier. Exceedingly high temperatures would be necessary.] 2. (15 points) SHOW ALL WORK. For the following reaction, calculate the appropriate quantities, using tables in the back. a.) H, in kj/mol 65 kj/mol, b.) G at 100 C, in kj/mol 20.2 kj/mol. 2 KCl (s) + H 2 SO 4(l) K 2 SO 4(s) + 2 HCl (g) H = [( 1437.8) + 2(-92.3)] [2( 436.5) + ( 814)] = ( 1622.4) ( 1250.5) = 65 kj/mol Need S, so S = [(2)(186.9) + (175.6)] [(2)(82.6) + (156.9)] = (549.4) (322.1) = 227.3 J/mol K G = H T S = 64.6 (373.15)(227.3 x 10 3 ) = 64.6 84.816 = 20 kj/mol 3. (15 points) Briefly define the following terms: Selective precipitation a process for separating and/or identifying dissolved ions, where an added reagent forms an insoluble salt with one of the dissolved cations in a solution, but not the others. 3 rd law of thermodynamics the entropy of a perfect crystal at absolute zero (O K) is zero Galvanic cell -- is an electrochemical device that produces electric current from a spontaneous chemical reaction 4. (14 points) Write the balanced net ionic equations for the following qualitative cation analysis procedures. a. dissolution of Fe(OH) 3 in aqueous HCl b. precipitation of CuS from an acidic solution of Cu 2+ and H 2 S 4
a. Fe(OH) 3(s) + 3 H + (aq) 3 H 2 O (l) + Fe 3+ (aq) b. Cu 2+ (aq) + H 2 S (aq) CuS (s) + 2 H + (aq) 5. (24 points) SHOW ALL WORK. Calculate the solubility of silver chloride in 0.522 M NH 3. AgCl (s) Ag + (aq) + Cl (aq) Ag + + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 (aq) + Add to get AgCl (s) + 2 NH 3(aq) Ag(NH 3 ) 2 (aq) + Cl (aq) K = (1.8 x 10 10 )(1.6 x 10 7 ) = 2.88 x 10 3 NH 3 Ag(CN) 2 Cl I 0.522 M 0 0 C 2x +x +x E 0.522 2x x x 2.88 x 10 3 = [Ag(NH 3 ) + 2 ][Cl ]/[NH 3 ] 2 = x 2 /(0.522 2x) 2 x/(0.522 2x) = 0.053665 x = 0.028013 0.0536653x 1.0536653x = 0.028013 x = 0.026586 sig fig => 0.027 M 6. (24 points) SHOW ALL WORK. Balance the redox equation in acidic solution. HMnO 4(aq) + H 2 SO 3(aq) H 2 SO 4(aq) + Mn 2+ (aq) 4 H + (aq) + 2 HMnO 4(aq) + 5 H 2 SO 3(aq) 2 Mn 2+ (aq) + 3 H 2 O (l) + 5 H 2 SO 4(aq) 7. (8 points) Predict which substance has the higher entropy. a. CO 2(g) or C (s) + O 2(g)? b. O 2(g) or O 3(g)? c. O 2(g) at 1 bar pressure or O 2(g) at 0.01 bar pressure? d. I 2(g) or I 2(s)? a. CO 2(g) or C (s) + O 2(g)? b. O 2(g) or O 3(g)? c. O 2(g) at 1 bar pressure or O 2(g) at 0.01 bar pressure? d. I 2(g) or I 2(s)? 8. (10 points) SHOW ALL WORK. The equilibrium constant K p of FeO (s) + H 2(g) Fe (s) + H 2 O (g) is 0.422 at 700 C. Calculate G rxn. G = RTlnK = (8.314 x 10 3 )(973)ln (0.422) = 6.98 kj/mol 5
9. (22 points) Given the following pair of half-reactions, F 2(g) + 2e 2 F (aq) E = 2.87 V Fe 3+ (aq) + 2e Fe (s) E = 0.036 V a. Determine E cell for the corresponding galvanic cell. b. Sketch the galvanic cell using these reactions, and label all of the following things: -cathode and anode -electrode material(s) -direction of electron flow -direction of cation flow in salt bridge -direction of anion flow in salt bridge -what ions are dissolved in each solution a.) E cell = 2.87 ( 0.036) = 2.91 V b.) 6
USEFUL INFO Selected Thermodynamic Data for Materials at 25 C. Substance H f (kj/mol) S (J/mol K) G f (kj/mol) CaCO 3(s) -1207.6 91.7-1129.1 CaO (s) -634.9 38.1-603.3 CCl 4(l) -95.7 309.9-53.6 CO 2(g) -393.5 213.8-394.4 CO (g) -110.5 197.7-137.2 Cu (s) 0 33.2 0 Cu 2 O (s) -168.6 93.1-146.0 H 2(g) 0 130.7 0 HCl (g) -92.3 186.9-95.3 H 2 O (l) -285.8 70.0-237.1 H 2 O (g) -241.8 188.7-228.6 H 2 O 2(l) -187.8 109.6-120.4 H 2 SO 4(l) -814 156.9-690.0 KCl (s) -436.5 82.6-408.5 K 2 SO 4(s) -1437.8 175.6-1321.4 N 2(g) 0 191.6 0 NH 3(g) -45.9 192.8-16.4 NO (g) 91.3 210.8 87.6 NO 2(g) 33.2 240.1 51.3 O 2(g) 0 205.2 0 SO 3(g) -395.7 256.8-371.1 Selected Equilibrium constants: K sp AgBr = 7.7 x 10 13 K sp AgCl = 1.8 x 10 10 K f Ag(CN) 2 = 5.6 x 10 8 K f Ag(NH 3 ) 2 + = 1.6 x 10 7 K f Zn(OH) 4 2 = 2 x 10 15 7