18:00-21:00, 15 April, 2016 (Total Score: 106 points)

Transcription

1 Chemistry II Midterm Exam 18:00-21:00, 15 April, 2016 (Ttal Scre: 106 pints) R = atm L/ml K = J/ ml K = bar L/ml K F = C/ml 1. Please answer the fllwing questins. (ttal 16%) (a) Why are mlal and mlar cncentratins f dilutin aqueus slutins apprximately equal? (3%) (b) Why are ice cubes (fr example, thse yu see in the trays in the freezer f a refrigeratr) cludy inside? (3%) (c) Why des carbn dixide escape frm a beer when the cap is remved? First, please write the name f the prper physical law which can answer this questin. (2%) Then, use it t explain yur answer. (3%) (d) If yu try t purify ethanl frm an ethanl-water mixture by fractinal distillatin, the maximum purity btainable is 95%. Please draw a liquid-vapr phase diagram which is rughly clse t the phase diagram f an ethanl-water mixture (3%) and then use the diagram t explain yur answer (2%). 1

2 2. The melting pint f a fictinal substance X is 28.5 C at 1000 bar. The density f the slid phase f X is 5.6 g/cm 3 and the density f the liquid phase is 6.14 g/cm 3 at 1000 bar. The mlar enthalpy f fusin f X, Hfus is 5.6 kj/ml at 1000 bar. The mlecular mass f X is 70.0 g/ml. (ttal 10%) (a) Please use the Clapeyrn equatin t predict whether the increase f pressure wuld increase the melting pint r decrease. (5%) (b) Please estimate the nrmal melting pint f X. (5%) 3. Cnsider the fllwing data fr naphthalene C10H8 Nrmal melting pint = C ; Nrmal biling pint = C The equilibrium vapr pressure Pvap f naphthalene C10H8 is listed in the fllwing table. Here, it is assumed that the mlar enthalpy f vaprizatin Hvap and the mlar enthalpy f sublimatin, Hsub are independent f temperature. (ttal 13%) Temperature (K) Pvap (kpa) (a) Is there enugh infrmatin t calculate Hvap f C10H8? If yes, please calculate Hvap f C10H8. If nt, please list the additinal required physical prperties f C10H8 t assist this calculatin. (5%) (b) Is there enugh infrmatin t calculate the mlar enthalpy f fusin f C10H8, Hfus If yes, please calculate Hfus f C10H8. If nt, please list the additinal required physical prperties f C10H8 t assist this calculatin. (8%) 4. (a) Use the fllwing thermdynamic data t determine the equilibrium cnstant KP fr the reactin 2 NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) at 350 K. (6%) (b) What is the value f equilibrium cnstant Kc fr this reactin at 350 K? (3%) (c) What is the temperature at which the KP = 1. (3%) substance Hf (kj ml -1 ) CP (J K -1 ml -1 ) S (J K -1 ml -1 ) NaHCO3(s) Na2CO3(s) CO2(g) H2O(g)

3 5. The decmpsitin f ammnium hydrgen sulfide NH4HS(s) NH3(g) + H2S(g) is an endthermic prcess. A 3.08 g f sample f the slid is placed in an evacuated 2.00 L vessel at 24 C. After equilibrium has been established, the ttal pressure inside is bar. Sme slid NH4HS remains in the vessel. (a) What is the KP fr this reactin? (3%) (b) What percentage f the slid remains? (3%) (c) If the vlume f the vessel were dubled at a cnstant temperature, what wuld happened t the amunt f the slid? (increase, decrease, r disappear?) (3%) (d) If the vessel were heated and all the slid disappeared, wuld this system be in equilibrium? And please give a brief explanatin fr yur answer. (3%) 6. The reactin Cl2(g) 2 Cl(g) reaches its equilibrium in a vessel. (3% each, ttal 9 %) (a) What will happen t the value f K if the temperature is raised frm 300 K t 800 K? (b) What will happen t the partial pressure f Cl2 and Cl if the vlume f the reactin vessel is dubled? (c) What will happen t the partial pressure f Cl2 and Cl if sme H2 is intrduced int the reactin vessel? 7. Fr the cmplete redx reactin given here, write the half-reactins and identify the xidizing and reducing agents: (3% each, ttal 6%) (a) 2 Li + H2 2 LiH (b) Cl2 + 2 Br 2 Cl + Br2 8. Calculate the standard emf f a cell that uses Fe Fe 2+ (ande) and Cr Cr 3+ (cathde) half-cell reactins. Write the verall cell reactin that ccurs under standard-state cnditins (4%) 9. What is the equilibrium cnstant fr the fllwing reactin at 298 K? Mg 2+ (aq) + Zn(s) Mg(s) + Zn 2+ (aq) (4%) 10. Given that E = 0.52 V fr the reductin Cu + (aq) + e - Cu(s), calculate (a) E, (b) G, and (c) K fr the fllwing reactin at 298 K: 2 Cu + (aq) Cu 2+ (aq) + Cu(s) (4+3+3%) 3

4 11. Cnsider a Daniel cell perating under nn-standard-state cnditins. Suppse that the cell reactin is multiplied by 2. What effect des this have n each f the fllwing quantities in the Nernst equatin: (a) E, (b) E, (c) Q, (d) lnq, and (e) n? (5%) 12. Use the data in the fllwing table, calculate the slubility prduct (Ksp) f AgCl. AgCl(s) Ag + (aq) + Cl (aq) (5%) 4

5 104B Chemistry (II) Midterm Exam Answer 1. (ttal 16%) (a) Dilute cncentratins mlality and mlarity are almst the same because the density f the slutin is almst equal t that f the pure slvent. (3%) (b) As the water freezes, disslved minerals in the water precipitate frm slutin, and disslved gases in the water can als frm minuscule bubbles. The precipitates and tiny bubbles refract light and create an paque appearance. (The answer which cntains either precipitated r bubbles is gd enugh t get the full credit.) (3%) (c) Henry s law (2%) When ne pen a beer, the pressure f a beer bttle decreases suddenly. Carbn dixide escapes frm a beer bttle because gases are less sluble in liquids at lwer pressure. (3%) (d) The liquid-vapr phase diagram f an ethanl-water mixture is clse t the phase diagram drawn belw, where the ethanl-water system frms a lw-biling azetrpe and the distillatin prcess will always cnverge t the azetrpe. (2%) (3%) 2. (ttal 10%) (a) First, calculate the mlar vlumes f the slid phase and the liquid phase f X: V (slid) = 70.0 g ml V (liquid) = 70.0 g ml g cm3 m3 5.6 cm3 = 12.5 = ml ml g cm3 m = 11.4 = cm3 ml ml The Clapeyrn equatin is ( dp ) = H fus dt fus T V (1%) (1%) (1%) Here, H fus f X is equal t 5.6 kj/ml. S, H > 0. V is equal t [V (liquid) 6 m3 m3 m3 V (slid)] = = ml ml ml 5

6 S, V < 0. Since T is the abslute temperature, T>0. Because ( dp ) = H fus < 0, the increase f pressure wuld decrease the melting pint dt fus T V f X. (2%) (b) ( dp dt ) fus = P T = H fus 3. (ttal 13%) J = 5600 ml T V ( K) ( m3 P = ( ) 105 = J T T K m 3 T = K = (5%) ml ) = Because the nrmal melting pint f C10H8 is C, the table f vapr pressure f C10H8 J K m 3 actually cver the equilibrium cnditins f different states, which are listed belw Temperature (K) Pvap (kpa) State slid-gas slid-gas liquid-gas (a) There are tw equilibrium vapr pressures f liquid naphthalene required t fulfill the calculatin Hvap f C10H8 by using the Clausius-Clapeyrn Equatin. One data is frm the abve table, which is Pvap = 8.34 kpa at 404 K, and anther data is the nrmal biling pint f naphthalene, which Pvap = kpa at C ( K). (2%) Insert tw equilibrium vapr pressures f liquid naphthalene int the Clausius-Clapeyrn Equatin: kpa ln = H vap ( 1 1 ) 8.34 kpa K 404 K H vap = ln ( 1 1 J kj ) = ml ml (3%) (b) Because Hsub = Hfus + Hvap, Hvap and Hsub are needed first, in rder t calculate Hfus. (1%) Hvap can be btained in previus questin, which is kj/ml. Then, we need tw equilibrium vapr pressures f slid naphthalene required t calculate Hsub f C10H8 by using the Clausius-Clapeyrn Equatin. There are tw equilibrium data f the slid-gas state in the abve table, which is Pvap = kpa at 292 K, and Pvap = 0.25 kpa at 333K. (2 %) Insert tw equilibrium vapr pressures f slid naphthalene int the Clausius-Clapeyrn 6

7 Equatin: ln 0.25 kpa = H sub ( 1 1 ) kpa K 292 K H sub == ln 0.25 ( 1 1 J kj ) = ml ml (3%) H fus = H sub H vap = = kj ml (2%) 4. (ttal 12%) (a) H = ( )+(-393.5)+(-241.8) - 32 (-950.8) = (kj) S = = (J K -1 ) at 350 K, G = = (kj) 分段給分 (3%) KP = - G /RT = exp( /( )) = 分段給分 (3%) (b) KP = Kc(RT) n = = Kc( ) Kc = /846.8 = (3%) (c) KP = 1, G = 0, T = H / S = 406 (K) 或 ln(1/ ) = ( /8.314)((1/T)-(1/350)), T = 406 (K) (3%) 5. (ttal 12%) (a) KP = P(NH3) P(H2S) = (0.732/2) 2 = (3%) (b) ttal amunt f gases, n = /( ) = (ml) the amunt f slid decmpsed = 51.1 ( )/2 = 1.52 (g) the percentage f the slid remains = ( )/3.08 = 50.6% (3%) (c) fr the 4.00-L vessel, the amunt f NH3 (r H2S) = (0.732/2) 4.00/( ) = ml 51.1 ( ) = 3.03 (g) f NH4HS decmpsed, 3.03 < 3.08 mre NH4HS is decmpsed, and there is still sme NH4HS slid in the vessel. (3%) (d) N. The reactin qutient Q may be smaller than equilibrium cnstant K, but there is n mre slid t decmpse int the prducts. (3%) 因題目未提供 NH4HS 相關資料以及最終溫度, 若同學討論更多, 如 NH4HS 可能昇華 融化 等, 只要合理即可得 3 分, 並可斟酌給予額外加分 6. (ttal 9%) (a) This reactin must be endthermic. K will increase. (3%) (b) Bth f P(Cl2) and P(Cl) will decrease. (3%) (c) Intrducing H2 will prduce HCl. Bth f P(Cl2) and P(Cl) will decrease. (3%) 7

8 7. (ttal 6%) (a) Li Li + + e (1%) ; H2 + 2 e 2 H (1%) H2 is the xidizing agent; Li is the reducing agent. (1%) (b) 2Br Br2 + 2e (1%) ; Cl2 + 2e 2Cl (1%) Cl2 is the xidizing agent; Br is the reducing agent. (1%) 8. Ande (xidatin): 3 Fe(s) 3 Fe 2 (aq) 6 e Cathde (reductin): 2 Cr 3 (aq) 6 e 2 Cr(s) Overall: 3 Fe(s) 2 Cr 3 (aq) 3 Fe 2 (aq) 2 Cr(s) E cathde E ande E Cr 3 /Cr E Fe 2 /Fe 0.74 V ( 0.44 V) 0.30 V (4%) 9. E (cell) = E (cath) E (and) = (-0.76) = V V ln K, n ln K n V, K e n V 10. K e [(2)(-1.61V)/0.0257V] = (4%) Ande (xidatin): Cu (aq) Cu 2 (aq) e Cathde (reductin): Cu (aq) e Cu(s) Overall: 2 Cu (aq) Cu 2 (aq) Cu(s) (a) E cathde E ande E Cu /Cu E Cu 2 /Cu 0.52 V 0.15 V 0.37 V (4%) (b) G = -nfe = = (J) = -36 (kj) (3%) (c) K = exp(- G /RT) = exp( /( )) = (3%) 11. (a)unchanged (b) unchanged (c) squared (d) dubled (e) dubled ( 各 1%, 共 5%) 12. Ag(s) Ag (aq) e AgCl(s) e Ag(s) Cl (aq) AgCl(s) Ag (aq) Cl (aq) E = = (V) Ksp = exp(- G /RT) = exp( /( )) = (5%) 8