2005_1_CH101_midterm_Soln

2005_1_CH101_midterm_Soln - 20050 1h } [1] H2O, NH3, CH4 ^...

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Unformatted text preview: 20050 1h } [1] H2O, NH3, CH4 ^ CH4, NH3, H2O [2] ~ p (a) B (b) K (c) D (o) (d) 0 C t S p 6 ^p ` } t t t i 0 4 p< E , ,0 t -0 < E IH } S (0 125 -~ ^ 5 t (20050 40 20H S ) S t } V t . (50 ). 6 J i 6 ) . H -H , H -H , H - t 5 . (0 5 4 p E 5 0 20 , 0 0 p E 1 ) . ` . (0) t 4 M t . (x ) t 4[ t . 00 < . (x) [3] p t . (0 40 ) (a) CuS + HNO3 CuSO4+NO (b) MnO4-(aq)+Fe2+(aq) Mn2+(aq)+Fe3+(aq) (0 ) (a) 3CuS(s) + 8NO3-(aq) +8H3O+(aq) 3Cu2+(aq)+3SO42-(aq)+8NO(g)+12H2O(l) H 2CuS + 8HNO3 3CuSO4 + 8NO + 4H2O (b) MnO4-(aq)+5Fe2+(aq)+8H3O+(aq) Mn2+(aq)+5Fe3+(aq)+12H2O(l) [4] (a) 1.00 mol 0, t t i D .( R = 0.08206 L atm K-1 mol-1, 1 L atm = 101.325 J) ^ p (constant) 0 250 K h } t T = 293 K 0 . 2.0 atm0 1.0 atmh } t 250 KN t S p (reversibly) 0 o 0 y atm0 L~ p ^ t c e T t 293 K 6 T t ( p^ . ). P-V state diagram 6 T t < E . 5 V3 4 e T (200 ) 0 ) 2 e T A4 T TA = 293 K, PA = 2.0 atm * T t K-1 mol-1)*293K / 2.0atm = 12 L. p ^ p ~ p6 ~ p H 6 Me e] B C 6 6 D M] 6 ,2 e T .H VA = nRTA / PA = (1 mol)*(0.08206 L atm PB = 1.0 atm ,* e ] D, VB = (PA .H TB = 293 K 2 e 6 / PB)*VA = (2.0atm / 1.0atm)*12L = 24 L. , 2 e T TC = 250 K2 ] t , TBVB -1 = TCVC -1 H . H TD = 250 K* ] J 6 PBVB = PCVC * ] 6 VC = [(TB / TC) * VB -1]1/( -1) = [(293K / .~ ^ p6 e Me ] 6 TAV -1 A 250K)*(24L)5/3-1]1/(5/3-1) = 30 L. PC = nRTC / VC = (1 mol)*(0.08206 L atm K-1 mol-1)*250K / 30L = 0.67 atm. A* ] = TDVD -1 , H , DH A~ p 6 t 5. t 5 6 VD = [(TA / TD) * VA -1]1/( -1) = [(293K / 250K)*(12L)5/3-1]1/(5/3-1) = 15 L. PD = nRTD / VD = (1 mol)*(0.08206 L atm K-1 mol-1)*250K / 15L = 1.3 atm. 2 P (atm) 1 2.0 atm, 12 L 1.0 atm, 24 L 1.3 atm, 15 L 10 20 V (L) p ` p ,0,0 0.67 atm, 30 L 30 J4e T t .(100 ) (b) J 6 p< x p 0) 1 ^ t HE = 0, w = wAB + wBC + wCD + wDA = -nRTAln(VB/VA) ncV(TC-TB) + q = -w = 248 J. (150 ) nRTCln(VC/VD) + ncV(TA-TD) = -nR(TA-TC)ln(VB/VA) = -(1mol)*(293K-250K)* (0.08206 L atm K-1 mol)*ln(24L/12L) = 2.445 L atm = -248 J. H ~ ^ p C(g) + 2 Cl2(g) CCl4(g) o [5] e T ) Ho = Hfo(CCl4(g)) - Hfo(C(g)) - 2Hfo(Cl2(g)) H k] 6 H .H Cl2(g)H Cl~ ^ p 6 h Hfo(C(g)) atomization ~ ^ p 6 h J i 6 4 V S 6. ] k ] 6 Hfo(CCl4(g)), Hfo(C(g)), H Hfo(Cl2(g)) = 0. H 6 Hfo(Cl2(g)) H Hfo(CCl4(g)) H Hfo(C(g)) . H Hfo(C(g)) = 716.7 kJ mol-1. Hfo(CCl4(g)) 5 6 y p h i C(s, gr) + 2 Cl2(g) CCl4(g) . H Atomization H C(s, gr) + 2 Cl2(g) C(g) + 4 Cl(g) C(g) + 4 Cl(g) CCl4(g). ^ 6 H2 h H k] H1 ] 6 atomization bond enthalpy h ^ p 6 . 4H i . H, H1 = Hfo(C(g)) + 4 C-Cl H2 = h Hfo(Cl(g)) = 716.7 kJ mol-1 + 4*121.7 kJ mol-1 = 1203.5 kJ mol-1. 4*(-328 kJ mol-1) = -1312 kJ mol-1. Hfo(CCl4(g)) = H1 + H2 = 1203.5 kJ mol-1 1312 kJ mol-1 = -108.5 kJ mol-1. Ho = Hfo(CCl4(g)) - Hfo(C(g)) - 2Hfo(Cl2(g)) = -108.5 kJ mol-1 716.7 kJ mol-1 2*0 kJ mol-1 = -825.2 kJ mol-1. [6] (a) k T 5 14,400 g/mol 3 y b <p 4 ^ pt < E 3 w 0.01 mol L-1 k T t 1 L T t . . 1L <p ^p t 25 C V3 .0 h 0)~ ~ ^p ^ t b 6 t 5 6 J i ? (50 ) b T 6 U t . y p 4 S y p < p x b ] 6 b ] 6 t 5 p 0.010 mol p l p < p . p ` p E ?~ ^ .(20 ) 6 ` b ] 6 t t 5 0.005 mol/liter 5 . J i 6 l 6 p p x U 6. p x p J ` (b) s H) ~ ~ ^ 6 mol L . ~ ^ 6 t 5 6 b] 6 -1 y < p ^ t 6 t 5 (0.5 L)*(0.01 mol L-1) = 0.005 mol. (1 L)*(0.005 mol L-1) = 0.005 mol. 1.5 L H . = (0.010 mol) / (1.5 L) = 0.0067 F = (0.005 mol) / (0.0067 mol L-1) = 0.75 L. 0.75 L. J i 6 [7] H2OH Ethanol, 4 azeotrope (h b ] Sy :H =4:96 y y -H, H - azeotropeh ] 6 yq p x ? (2H) -~ ^ pJ 6 . . (3H). H+H 50% H `b ] 6 t 5 . )H (4% water + 96% ethanolH azeotropic mixture)~ ] [8] You are stranded on a tropical island with no access to chemical literature resources, the internet (or norae-bang). Based on your knowledge of intermolecular forces, and molecular geometries, predict the order of normal boiling points for the series: SnCl2, SnCl4, and TeCl2. List these compounds in decreasing temperature with explanations (5H). Answer: SnCl2, 925 K > TeCl2, 595 K > SnCl4 385 K. SnCl4: ~ ^ p T J 6 t S p E Te-Cl P P ^ p t t o. < p E . ) SnCl20 TeCl2: V (Sn-Cl [9] While cast away on your island, your favorite drink is "Moju," which contains variable amounts of ethyl alcohol and water. Given that ethyl alcohol (specific heat capacity 2.44 JK-1g-1) and water (heat capacity 4.18 JK-1g-1) contribute independently to the heat properties of Moju, address the following: a) Calculate the specific heat capacity of Moju c(Moju) which is 45.0 % by weight in alcohol. Give your answer in JK-1g-1 (2H) Answer: c(Moju) = (2.44 JK-1g-1) (0.450) + (4.18 JK-1g-1) (0.550) c(Moju) = 3.40 JK-1g-1 b) A bartender takes an ice cube of mass 5.00 g from the freezer (-20C) and drops it into 50.0 g of the solution from part (a) with a temperature of +20C. In order to not dilute the drink, the bartender pulls out the ice cube when its temperature reaches 0C (assume none of the ice has melted). Calculate the temperature of "Moju" at this point in C. The heat capacity of ice is 2.40 JK-1g-1.(5H) Answer: i. Ice: (-20C, 5.00 g) Ice: (0C, 5.00 g) ii. Moju: (+20C, 50.0 g) Moju: (0C, 50.0 g) q1 + q2 = 0 (5.00 g)(2.44 JK-1g-1) (0C - (-20C)) + (50.0 g)(3.40 JK-1g-1)(Tf (20C)) = 0 Tf = 18.6C q1 = m1c(ice)T1 q2 = m2c(Moju)T2 [10] Later on the island, you realize that you don't produce enough stomach acid to sufficiently digest your octopus stew. Since you are an adventuresome chemist, you resolve to synthesize HCl, from the pure elements of hydrogen and chlorine. Importantly you want to determine the heat of formation Hf for H2(g) + Cl2(g) HCl(g). Calculate this heat Hf in kJ/mol given the following equations (25C): (5H) 1 2 3 4 H2(g) 2H(g) Cl2(g) 2Cl(g) H(g) H+(g) + eCl(g) + e Cl (g) - (436.0 kJ) (243.4 kJ) (1312.2 kJ) (-348.8 kJ) 5 H+(g) + Cl-(g) HCl(g) (-1395.4 kJ) Answer: H2(g) 2H(g) Cl2(g) 2Cl(g) H + (g) (436.0 kJ)(1/2) (243.4 kJ)(1/2) (-1395.4 kJ) (1312.2 kJ) (-348.8 kJ) Hf = -92.3 kJ/mol + Cl - (g) HCl(g) (g) H(g) H+(g) + eCl(g) + e Cl [11] A mixture of H2 and He at 300 K effuses from a very tiny hole in the vessel that contains it. What is the mole fraction of H2 in the original gas mixture if 3.00 times as many He atoms as H2 molecules escape from the orifice in unit time? If the same mixture is to be separated by a barrier-diffusion process, how many stages are necessary to achieve H2 of 99.9% purity? (15H) [12] At 90oC the vapor pressure of toluene is 0.534 atm and the vapor pressure of benzene is 1.34 atm. Benzene (0.400 mol) is mixed with toluene (0.900 mol) to form an ideal solution. Compute the mole fraction of benzene in the vapor in equilibrium with this solution. (15H) ...
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This note was uploaded on 05/19/2008 for the course MS 4032 taught by Professor Anony during the Spring '08 term at A.T. Still University.

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