midterm1

midterm1 - Chemistry 6B: General Chemistry 1 U. Miiller...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 6B: General Chemistry 1 U. Miiller University of California, San Diego, Winter 2008 First Midterm exam - 30th January 2008 maximum score: 80 points Student name (print): Student ID: If you have a question. quietly raise your hand and wait until a TA is at your place. If you need more space, continue to write on the back of the paper. Problem 1 (10 points): A rubber ball is dropped from a height of 1 m onto a steel block. The mass of the steel block is half the mass of the rubber ball. Before this experiment, both objects have a temperature of 20.0000°C. After a few times of bouncing up and down, the rubber ball lies motionless on the steel block. Calculate the final temperature of the rubber ball and the steel block, assuming that no air friction has absorbed energy, the sphere and the block have thermally equilibrated with each other but are thermally isolated from the environment. The specific heat capacity of steel is 0.51 J K" g", and the specific heat capacity of r ber is 1.60 J K“ g". M“:st = lijlmx‘ifilgicigf} 3' 3‘ C(rub'mr+5w): “3099 3‘ LG“; +5005 xO.Sl--— = IESS E@ “"5 : A C IBSS_, K T: 20.0000‘1: + 0.003; K: 20.0053 0C @ Problem 2 (6 points): Until the mid-20th century, many people believed that the phenomenon of life contradicts the second law of thermodynamics because living organisms amplify ordered structures. State the second law of thermodynamics and explain why there is no contradiction. Tlu Siconoi (an; 0! 'ali’wvrmodynamfes Slate/s 41w} {HQ ,enn‘ropy of an 130(0de 373112114 Wu unit/(N98) 000% ram: aan'L I4 mac/Les an tequila—mum. Lit/m5 ijan15m§ arc (10* {50b #90! from "flu €nv/ronrne’miir J! dhéir enJroW falls} 7”“? end (ofy of TM? SarroumJ/ng Me, an 2m! (aw . @ a (way; r 6?; I Lté‘pphb Student name (print): Problem 3 (15 points): a) You are heating and evaporating 100 g of water at constant pressure, starting from a temperature of 20°C, and ending up with water vapor at 120°C. Calculate the change in enthalpy, assuming that CP(H20“,) = 4.18 J K" g". and AHVMI-lzom) = 40.? kJ mol". Use the estimation of CP(H20(3}) that is based on the degrees of freedom in the molecules. b) Determine the fraction of this enthalpy that is due to expansion work, assuming that water vapor behaves as an ideal gas. The gas constant is 8.3145 J K" mol". a) AH(ZOOC +IOD”C)= XSOK x == 314% '3 1009’ .. 18.9. F 9*S>M@ my»! to a. ‘OHVAP; 40.75:; >35sz —22(o.l Lg,@ cameo» WC? A H(too“¢—-ofgooc)= 5.55M? XZOK 5-” A H: (DJ/((30% 4106C)"*AHVAP+OHOCI5C*MU%) : 2 6 3‘2' k;@ b) w .= n R Tl— nRT, = Esswfivgéltrsg— 1x (120% 400%) = 92%], (0.399 Problem 4 (10 points):@ ad‘swe's 51"“ {305.9 a) Calculate the residual entropy of 2 mol CC|F3 at T=0 K, assuming that its crystals are in a disordered state. The Boltzmann constant is 1.381 * 10'” J K". b) Experimentally, the residual molar entropy of CCIF3 was determined to be 7.5 J K". What can you say about the arrangement of CCIF3 molecules in the crystal at T=0 K? C L Clarify your statement with a drawing of molecules. l FQC"? w: 2%9wl023@ @ 3- ¥ = 1.3mm”? wheat)?an = 2 Joe =25,“ , 5 K v< :t b)5m(b<Er1}meQ/Os 75% : awé) JD «5M (JitfiorC—1'EM) “5;: K I \q: Ct :DQSU/bafflu are OIISOerfirQJ.@ I? a;c\? Student name (print): Problem 5 (5 points): Calculate the rise in entropy of a large volume of an ice-water mixture when it is heated with a IOU-Watt heater for 5 minutes. Remember that 1 Watt equals 1 Jr’s. ioogx5><605 =30 Lg,@ _fl=32zo/09L_ 3 53-— Ho;E @ 7 273.514 ' Problem 6 (10 points): a) Calculate the reaction enthalpy for the oxidation of 1 mol ethanol to acetic acid, using the following mean bond enthalpies: CH: 412 Id mol'i, C=O: 743 kJ mol", 0:0: 498 kJ mol", H-O: 463 kJ mol". Ethanol is an alcohol with two carbon atoms; acetic acid is the corresponding carboxy-acid, also with two carbon atoms. b) (no calculation): Assume that the reaction proceeds in your body, at 37°C. Is the entropy change of the reaction positive or negative? Explain the entropic contribution of all molecules in the reaction. What is the driving force of the reaction? a0 a) H3CHCH2-OHH) + Gigi—"9' HjC~CMOM + H300?) ® () QFC'H O'=O C=O I 2¥O-H® .; _ 3k . x-eesuL Jaw vC—Lr‘iékép 4H (7g=0 })+2 C W 3 2% (:ng 0:0 ma 5(5) >7 500' @ b) A$<O Lach 10196030 (,0 ,3 Map; ‘ I +0 AS‘ Problem7(9 points): The C? $in mirtka (No calculation): Decide for the following three reactions whether AS is larger than zero, smaller than zero, or cquai to zero. Explain your decision. a) 2 H2 + 02 -> 2 H20 (at room temperature) 22me 3(5)>>_S(£) anal L?) “yen/"Ki J5 rCcJQ-Uadi _ b) KOH 4- CO2 -> KI—ICO3 (at room temperature) 5&0 51m 3,5)»5wanpt Macs roam Qt NW- 0) c) H20 (1) 9 H20 (5) (at 0 °C) ALS< O 536W 5(2) >S<¢Dt2floifléfmol :5: WWW @ do a. 50234. CD Student name (print): Problem 8 ( 5 points): The standard Gibbs Free Energy of formation is +267! kJ mol'l for Cyclohexane at 25°C. Write the corresponding thermodynamic equation. explain what the standard Gibbs Free Energy of formation means, and state whether this compound is thermodynamically stable or labile. :6 He TA .5 @ OGIO Wan; firth—3L3 (UQWQal 711C (Lace/racer? +0 (generaé! a (NPOMAA £11?“ :15 Agar-malls . taupe Cmpoawl L3) rllxlrmocflzaaamica% 666mm Problem 9 (10 points): Calculate the temperature at which the reduction of magnesium oxide (MgO) with carbon becomes spontaneous. Based on that result, state whether this process would be useful, to obtain magnesium from magnesium oxide. AHf°(COZ,g) = -393.5 kJ mol", AHf°(MgO,s) = -601.'? kJ mol", Sm°(Mg,s)=32.68 J K". Sm°(Mg0,s)=26.94 J K", Sm°( C02,g)=213.'lr J K", Sm°( C,s)=5.7] K“. aé=4H—Ta§@ @ A 6 = (3’ =94>H=Tog 2 M5(5)+C(s)~—r>2 rear (of (j) * AH :97: "5'; @ AH: ZMH; (I15) “slit ((02) — ZXAH‘Q 0150) flag? (C) e l’ ' a? <9 6 : -— 3:13.; Lg~(2><(~—EOL7 Lg» = 801.6: kg} 6) as = zxsnimp + 51(602) egtgmmjoytsm a) _. :2 V3216§+ 2:375} -2xzéfikfi}? -5775?— : K (4 _ s _ 2(a.esfi@ To, ail _ 301G700 5L : 3W0 M =- $917K @ — A—S_ alasgg. :1) owl!) fim‘aera‘lbtrr F5 Joohiflk 40 19¢ ...
View Full Document

Page1 / 4

midterm1 - Chemistry 6B: General Chemistry 1 U. Miiller...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online