Chapter 3 Homework Solutions
|
|
- Barrie Paul Gibson
- 5 years ago
- Views:
Transcription
1 Chapter Hmewrk Slutins. n = ml = 5 C = 98 K = 5 C = 98 K p = atm p = 5 atm. Cpm = (5/)R he entrpy changes fr the heating and cmpressin can e calculated separately and added. Heat at cnstant pressure S = nc ln( / ) = n(5 / ) Rln( / ) pm = (.0 ml)(5 / )(8. J / ml K)ln(98 / 98) =+ 8.0 J / K Cmpress at cnstant temperature. Because = cnstant pv = pv V/ V= p/ p= / 5 = 0.0 S = nr ln( V / V ) = ( ml)(8. J / ml K)ln(0.0) = 40. J / K S = S + S = +8.0 J/K -40. J/K = -. J/K In this particular example the negative entrpy change fr cmpressing the sample is larger than the psitive change fr heating the sample yielding a net negative S.. n = ml = 00 K = 50 K CVm = 7.5 J/ml-K Cpm = CVm + R = = 5.8 J/ml-K q = 0 [Adiaatic] dqrev S = = 0 [fr reversile Adiaatic prcess] w = U = nc ( ) = ( ml)(7.5 J / ml K)(50 K 00 K) = 45 J 4.kJ Vm H = nc ( ) = ( ml)(5.8 J / ml K)(50 K 00 K) = 570 J 5.4kJ pm. =98 K p = ar V = V ml n = 4 g = 0.50ml 8 g (a) Reversile Isthermal Expansin S = S = nr ln( V / V ) = (0.50 ml)(8. J / ml K)ln( V / V ) =+.88 J / K sys S = S =.88 J / K ecause it s reversile and Suniv = 0 surr sys
2 () Irreversile Isthermal Expansin against pex = 0 System: Cannt e calculated frm actual prcess which is irreversile. Hwever Ssys = +.88 J/K (part a) since S is a State Functin Surrundings: Usurr = 0 (since isthermal) wsurr = 0 (since pex = 0) qsurr = Usurr - wsurr = 0 herefre Ssurr = 0 Suniv = Ssys + Ssurr = = +.88 J/K > 0 (as expected fr irreversile prcess) (c) Reversile Adiaatic Expansin (dqrev)sys = (dqrev)surr = 0 herefre Ssys = 0 and Ssurr = 0.4 = 6 C = 5 K vaph = 9.4 kj/ml =.94x0 4 J/ml n = 40 g/9.4 g/ml =.0 ml qsys n vaph (.0 ml)(.94x0 J / ml) Ssys = = = = J / K 5 K 4 q qsys n vaph surr (.0 ml)(.94x0 J / ml) Ssurr = = = = = 76.4 J / K 5 K.5 = 78 C = 5 K vaph = 8.6 kj/ml p = ar = 00 kpa he prcess is a cndensatin. herefre we need: cndh = - vaph = -8.6 kj/ml = -.86x0 4 J/ml ml n = 50 g =.6 ml 46 g q = H = n H = (.6 ml)( 8.6 kj / ml) = 5.8kJ cnd ( liq gas ) gas w = p V V + pv =+ nr = (.6 ml)(8.x0 kj / ml K)(5 K) =+ 9.5kJ 9.5kJ U = q + w = = -6. kj qrev H 5.8kJ 0 J S = = = = 0.58 kj / K = 58 J / K 5K kj Nte that S is negative as expected fr a cndensatin ecause the entrpy f the liquid is lwer than the entrpy f the gas.
3 .6 m = -4 ml C = 59 K fush = 9.45 kj/ml p = ar = 00 kpa n = 50 g =.6 ml 46 g he prcess is a crystallizatin. herefre we need: crysh = - fush = kj/ml = J/ml q = H = n H = (.6 ml)( 9.45 kj / ml) = 0.8kJ cryst ( sl liq ) 0 w= pv V he vlumes f liquids and slids are very small and can e ignred. U = q + w = = -0.8 kj qrev H 0.8kJ 0 J S = = = = 0.94 kj / K = 94 J / K 59 K kj Nte that S is negative as expected fr a crystallizatin ecause the entrpy f the slid is lwer than the entrpy f the liquid..7 = 00 C = 7 K (actual iling pint) = 80 C = 5 K (nrmal iling pint) 7 K 5 K liquid gas Ssys We must calculate Ssys via a reversile path (see diagram ave). his path is: () Cl the liquid frm t (the nrmal iling pint) () Vaprize the liquid reversily at () Heat the gas ack t H ( = 5) S = S + S + S = C ( l) ln( / ) + + C ( g) ln( / ) vap pm pm 5.7x0 J / ml S = (8.7 J / ml K)ln(5 / 7) + + (5. J / ml K)ln(7 / 5) 5K = 7.64 J / ml K + 0. J / ml K +.9 J / ml K = J / ml K J / ml K
4 Ssurr We will calculate the entrpy change f the surrundings as qsurr/(7). Hwever t determine qsurr we will have t knw vaph(=7 K). We can determine that frm the same prcess (ave). [ ] [ ] [ ] [ ] H ( ) = H+ H + H = C () l + H ( ) + C ( g) vap p m vap p m H ( ) = (8.7 J / ml K) 5K 7 K + H (5 K) + (5. J / ml K) 7K 5K vap = J ml + x J ml + J ml 774 / / 70 / = + 60 J / ml q qsys vaph ( ) surr 60 J / ml Ssurr = = = = 7K = 90.6 J / ml K 90. J / ml K vap Suniv Suniv = Ssys + Ssurr = = +5. J/ml-K Suniv > 0 as expected fr the spntaneus vaprizatn f a superheated liquid..8 n =. mles = 98 K Sm (98) = 9.45 J/ml-K S (98) = nsm (98) = ( ml)(9.45 J/ml-K) = J/K S () = S (98) + S c C pm = a + + a = 9.8 =.5x0 - c = -.6x0 5 c n a + + nc pm S = d = d = na d + n d + nc d [ ln( )] [ ] = na + n nc nc = na ln( / ) + n( ) (a) = 00 C = 7 K 5 ()(.6x0 ) ( ) S = ()(9.8) ln(7 / 98) + ()(.5x0 ) 7 98 (7) (98) =. J / K S () = S (98) + S = = J/ml-K
5 () = 500 C = 77 K 5 ()(.6x0 ) ( ) S = ()(9.8) ln(77 / 98) + ()(.5x0 ) (77) (98) = 8.5 J / ml K S () = S (98) + S = = J/ml-K.9 = 600 ml 00kPa C = 87 K n = 00 g =.6ml p = ar = 00kPa 8 g ar = a = 9.6 J/ml-K and =.5x0 J/ml Cpm a CVm = Cpm R= a R= ( a R) = a' a = a - R = =. J/ml-K nr (.6 ml) (8. kpa L / ml K)(87 K) V = = = 95.4 L p 00 kpa (a) Cl at cnstant pressure = 00 C = 57 K U = nc d = n a d = na n = na ' n ln / [ ] ( ) [ ] [ ] ' ' ln( ) Vm = ml J ml K K K ml x J ml (.6 )(. / )[57 87 ] (.6 )(.5 0 / ) ln(57 / 87) = 4700 J J = 90 J.9kJ H = nc d = n a d = na n = na n [ ] ln ( / ) [ ] [ ln( )] pm = ml J ml K K K ml x J ml (.6 )(9.6 / )[57 87 ] (.6 )(.5 0 / ) ln(57 / 87) = 40 J J = 8470 J 8.5kJ q = H = -8.5 kj (ecause p = cnstant) w = U - q = (-8.5) = +6.6 kj
6 nc pm a S = d = n d na [ ln ] n = = na ln( / ) + n = (.6 ml)(9.6 J / ml K)ln(57 / 87) + (.6 ml)(.5x0 J / ml) 57 K 87 K = 4.85 J / K +.94 J / K 9.9 J / K () Cl at cnstant vlume = 00 C = 57 K U = ncvm d.9 kj See part (a) H = nc pm d 8.5kJ See part (a) w = 0 (ecause V = cnstant) q = U = -.9 kj (ecause V = cnstant) ncvm a' S = d = n d na ' [ ln ] n = = na 'ln( / ) + n = (.6 ml)(. J / ml K)ln(57 / 87) + (.6 ml)(.5x0 J / ml) 57 K 87 K = 4.66 J / K +.94 J / K 0.7 J / K
7 (c) Isthermal Cmpressin t p = 5 ar = = 600 C = 87 K U = nc d = Vm 0 H = nc d = pm 0 V p ar Isthermal: pv = pv = = = 0.40 V p 5ar w = nr ln( V / V ) = (.6 ml)(8. J / ml K)(87 K)ln(0.40) = J = 7.5kJ q = U - w = 0 - w = -7.5 kj ( ) S = nr ln V / V = (.6 ml)(8. J / ml K) ln(0.40) = 0.0 J / K.0 (a) CHCHO(g) + O(g) CHCOOH(l) S = [ S ( CH COOH )] [ S ( CH CHO) + S ( O )] r m m m = [( ml)(59.8 J / ml K)] [( ml)(50. J / ml K) + ( ml)(05. J / ml K)] = 86. J / K () Hg(l) + Cl(g) HgCl(s) S = [ S ( HgCl )] [ S ( Hg) + S ( Cl )] r m m m = [( ml)(46.0 J / ml K)] [( ml)(76.0 J / ml K) + ( ml)(. J / ml K)] = 5. J / K. (a) CHCHO(g) + O(g) CHCOOH(l) G = [ G ( CH COOH )] [ G ( CH CHO) + G ( O )] r f m f m f m = [( ml)( 89.9 kj / ml] [( ml)( 8.9 kj / ml) + ( ml)(0 kj / ml)] = 5.0kJ () Hg(l) + Cl(g) HgCl(s) G = [ G ( HgCl )] [ G ( Hg) + G ( Cl )] r f m f m f m = [( ml)( 78.6 kj / ml] [( ml)(0 kj / ml) + ( ml)(0 kj / ml)] = 78.6 kj
8 . 4 HCl(g) + O(g) Cl(g) + HO(l) at = 98 K S = [ S ( Cl ) + S ( H O] [4 S ( HCl) + S ( O )] r m m m m = [( ml)(69.9 J / ml K) + ( ml)(. J / ml K)] [(4 ml)(86.9 J / ml K) + ( ml)(05. J / ml K)] = 66.7 J / K = kj / K H = [ H ( Cl ) + H ( H O] [4 H ( HCl) + H ( O )] r f f f f = [( ml)(0 kj / ml) + ( ml)( 85.8 kj / ml)] [(4 ml)( 9. kj / ml) + ( ml)(0 kj / ml)] = 0.4 kj G = H S = 0.4 kj (98 K)( kj / K) = 9.kJ r r r. = 40 C = K (isthermal) n = 70 g/8 g/ml =.5 ml V = 5 L V = 500 ml = 0.50 L Because = cnstant pv = pv p/ p= V/ V = 5 / 0.50 = 50. G = nr ln( p / p ) = (.5 ml)(8. J / ml K)( K)ln(50) = 5440 J 5.4 kj.4 G = a + a = +560 J = 8.0x0 - J/K = 0 C = 0 K G G dg = Sd + Vdp = d + dp p G herefre: S = p Fr a reactin (prcess) this can e written as: herefre p G S = + = = = = [ a ] S 0 (8.0x0 J / K )(0 K) 48.5 J / K p p.5 p =. atm p = 00 atm m = 0 ml x 78 g/ml = 560 g V = 560 g x ml/0.88 g = 77 ml =.77 L p G = Vdp = V ( p p) [Fr slid r liquid V cnstant] p 0J = = = = L atm 4 G (.77 L)(00 atm atm) 75. L atm.77x0 J 7.7 kj
9 .6 n = 0 ml = 98 K p = atm p = 00 atm nr G Vdp dp nr ln( p / p) p p p = = = p p 5 = (0. ml)(8. J / ml K)(98 K) ln(00 /) =.8x0 J = 8kJ Nte: It is nt surprising that G fr the pressure increase in H(g) is much greater than fr C6H6(l) ecause the vlumes f gases are much greater than the vlumes f liquids (and slids)..7 G = 6.0x0 J/ml = 6.0x0 kpa-l/ml Vm = -5 ml/ml = -.5x0 L/ml At equilirium: G = G + V ( p p ) = 0 G x kpa L ml ( ) / 5 p p = = = x kpa Vm.5x0 L / ml m p = 4.0x0 5 kpa + p = 4.0x0 5 kpa + 00 kpa = 4.0x0 5 kpa x ar/00 kpa = 4000 ar.
REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj
Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)
More informationAP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY
AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither
More informationThermodynamics Partial Outline of Topics
Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)
More informationChapter 17 Free Energy and Thermodynamics
Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall First Law f Thermdynamics
More information" 1 = # $H vap. Chapter 3 Problems
Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius
More informationCHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK
CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature
More informationTHE SECOND LAW Chapter 3 Outline. HW: Questions are below. Solutions are in separate file on the course web site. Sect. Title and Comments Required?
THE SECOND LAW Chapter 3 Outline HW: Questions are below. Solutions are in separate file on the course web site. Sect. Title and Comments Required? 1. The Dispersal of Energy YES 2. Entropy YES We won
More informationEntropy, Free Energy, and Equilibrium
Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs
More informationRecitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol
Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is 0.175 atm and that f O 2 (g) is 0.250 atm in a mixture f the tw gases. a. What is the mle fractin f each
More informationChapter 2 Homework Solutions
. Cl - 3 translations + rotations + vibration Chapter Homework Solutions (a) Um(rigid) = (3/)R + (/)R = (5/)R Hm(rigid) = (7/)R Cpm(rigid) = (7/)R = 9. J/mol-K (b) Um(vib) = (3/)R + (/)R + R = (7/)R Hm(vib)
More informationCHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?
1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.
More informationChemistry 1A Fall 2000
Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur
More informationChemistry 114 First Hour Exam
Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)
More informationThermodynamics and Equilibrium
Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,
More information188 CHAPTER 6 THERMOCHEMISTRY
188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.
More informationChapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w
Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin
More informationThermochemistry Heats of Reaction
hermchemistry Heats f Reactin aa + bb cc + dd hermchemical Semantics q V = Heat f Rxn at [V] = U = Energy (change) f Rxn q P = Heat f Rxn at [P] = H = Enthalpy (change) f Rxn Exthermic rxns: q < 0 Endthermic
More informationWork and Heat Definitions
Wrk and eat Deinitins FL- Surrundings: Everything utside system + q -q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces - eat, q: transer energy resulting rm
More informationGOAL... ability to predict
THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict
More informationExamples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?
NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic
More informationMBN 305 Phase Diagrams & Transformations
MBN 35 Phase Diagrams & ransfrmatins Dr. Ersin Emre Oren Department f Bimedical Engineering Department f Materials Science & Nantechnlgy Engineering OBB University f Ecnmics and echnlgy Ankara - URKEY
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationThermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P
Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry
More informationChapters 29 and 35 Thermochemistry and Chemical Thermodynamics
Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany
More informationAdvanced Chemistry Practice Problems
Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with
More informationCHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25
CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing
More informationUnit 14 Thermochemistry Notes
Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm
More informationCh 10 Practice Problems
Ch 10 Practice Problems 1. Which of the following result(s) in an increase in the entropy of the system? I. (See diagram.) II. Br 2(g) Br 2(l) III. NaBr(s) Na + (aq) + Br (aq) IV. O 2(298 K) O 2(373 K)
More informationChapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes
Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3
More informationChapter 4 Thermodynamics and Equilibrium
Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume
More informationCHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013
CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal
More informationCHM 152 Practice Final
CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the
More informationCHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review
Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system
More informationPhys102 First Major-122 Zero Version Coordinator: Sunaidi Wednesday, March 06, 2013 Page: 1
Crdinatr: Sunaidi Wednesday, March 06, 2013 Page: 1 Q1. An 8.00 m lng wire with a mass f 10.0 g is under a tensin f 25.0 N. A transverse wave fr which the wavelength is 0.100 m, and the amplitude is 3.70
More informationCHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics
CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam
More informationCHAPTER 6 THERMOCHEMISTRY. Questions
CHAPTER 6 THERMOCHEMISTRY Questins 11. Path-dependent functins fr a trip frm Chicag t Denver are thse quantities that depend n the rute taken. One can fly directly frm Chicag t Denver, r ne culd fly frm
More informationUniversity Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:
University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,
More informationALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?
Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S
More informationChemical Equilibrium
0.110/5.60 Fall 005 Lecture #10 age 1 Chemical Equilibrium Ideal Gases Questin: What is the cmsitin f a reacting miture f ideal gases? e.g. ½ N (g, T, ) + 3/ H (g, T, ) = NH 3 (g, T, ) What are N,, and
More informationLecture 16 Thermodynamics II
Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies
More information4F-5 : Performance of an Ideal Gas Cycle 10 pts
4F-5 : Perfrmance f an Cycle 0 pts An ideal gas, initially at 0 C and 00 kpa, underges an internally reversible, cyclic prcess in a clsed system. The gas is first cmpressed adiabatically t 500 kpa, then
More informationChem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points
+2 pints Befre yu begin, make sure that yur exam has all 7 pages. There are 14 required prblems (7 pints each) and tw extra credit prblems (5 pints each). Stay fcused, stay calm. Wrk steadily thrugh yur
More informationCHEMICAL EQUILIBRIUM
14 CHAPTER CHEMICAL EQUILIBRIUM 14.1 The Nature f Chemical Equilibrium 14. The Empirical Law f Mass Actin 14.3 Thermdynamic Descriptin f the Equilibrium State 14.4 The Law f Mass Actin fr Related and Simultaneus
More informationPart One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)
CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationChem 75 February 16, 2017 Exam 2 Solutions
1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml
More informationChem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions
Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces
More informationThermochemistry. Thermochemistry
Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk
More informationLecture 4. The First Law of Thermodynamics
Lecture 4. The First Law f Thermdynamics THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and
More informationBIT Chapters = =
BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride
More informationCHEM 103 Calorimetry and Hess s Law
CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the
More informationChapter 16 - Spontaneity, Entropy, and Free Energy
Chapter 16 - Spontaneity, Entropy, and Free Energy 1 st Law of Thermodynamics - energy can be neither created nor destroyed. Although total energy is constant, the various forms of energy can be interchanged
More informationProblem Set 1 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001
LEEL ROBLEMS rblem Set Slutins. MI ressr Gerbrand Ceder Fall rblem. Gas is heating in a rigid cntainer rm 4 C t 5 C U U( ) U( ) W + Q (First Law) (a) W Since nly wrk is pssible & since the cntainer is
More informationWhen a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q
Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationCHAPTER 6 THERMOCHEMISTRY. Questions
CHAPTER 6 THERMOCHEMISTRY Questins 11. Path-dependent functins fr a trip frm Chicag t Denver are thse quantities that depend n the rute taken. One can fly directly frm Chicag t Denver, r ne culd fly frm
More informationA) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N
Q1. A transverse sinusidal wave travelling n a string is given by: y (x,t) = 0.20 sin (2.5 x 80 t) (SI units). The length f the string is 2.0 m and its mass is 1.5 g. What is the magnitude f the tensin
More informationMore Tutorial at
Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationSolutions to the Extra Problems for Chapter 14
Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +
More information17.1 Ideal Gas Equilibrium Constant Method. + H2O CO + 3 H2 ν i ν i is stoichiometric number is stoichiometric coefficient
17.1 Ideal Gas Equilibrium Cnstant Methd CH 4 + H2O CO + 3 H2 ν i -1-1 1 3 ν i is stichimetric number is stichimetric cefficient ν i (1) dn1 dn = ν1 ν 2 2 Prcess 1. # phases? Methd? (K a methd). 2. Find
More informationCHEM 1001 Problem Set #3: Entropy and Free Energy
CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.
More informationProcess Engineering Thermodynamics E (4 sp) Exam
Prcess Engineering Thermdynamics 42434 E (4 sp) Exam 9-3-29 ll supprt material is allwed except fr telecmmunicatin devices. 4 questins give max. 3 pints = 7½ + 7½ + 7½ + 7½ pints Belw 6 questins are given,
More informationHeat Effects of Chemical Reactions
eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = -9690
More informationSpontaneous Processes, Entropy and the Second Law of Thermodynamics
Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer
More informationSPONTANEITY, ENTROPY, AND FREE ENERGY
CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and
More informationTHE ANSWER KEY TO THIS EXAM WILL BE POSTED ON BULLETIN BOARD #4 IN THE HALLWAY EAST OF ROOM 1002 GILMAN AND ON THE CHEM 167 WEBSITE.
PROF. JOHN VERKADE SPRING 2005 THIS EXAM CONSISTS OF 3 QUESTIONS ON 9 PAGES CHEM 67 HOUR EXAM II FEBRUARY 28, 2005 SEAT NO. NAME RECIT. INSTR. RECIT. SECT. GRADING PAGE Page 2 Page 3 Page 4 Page 5 Page
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis
More informationHow can standard heats of formation be used to calculate the heat of a reaction?
Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk
More informationKEEP THIS SECTION!!!!!!!!!
KEEP THIS SECTION!!!!!!!!! CHEMISTRY 202 Hour Exam II (Multiple Choice Section) November 1, 2018 Dr. D. DeCoste Name Signature T.A. This exam contains 20 questions on 5 numbered pages. Check now to make
More informationEntropy and Gibbs energy
14 Entrpy and Gibbs energy Answers t wrked examples WE 14.1 Predicting the sign f an entrpy change (n p. 658 in Chemistry 3 ) What will be the sign f the value f S fr: (a) crystallizatin f salt frm a slutin;
More informationYou MUST sign the honor pledge:
CHEM 3411 MWF 9:00AM Fall 2010 Physical Chemistry I Exam #2, Version B (Dated: October 15, 2010) Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only if the correct
More informationCHEM Exam 2 - October 11, INFORMATION PAGE (Use for reference and for scratch paper)
CHEM 5200 - Exam 2 - October 11, 2018 INFORMATION PAGE (Use for reference and for scratch paper) Constants and Conversion Factors: R = 0.082 L-atm/mol-K = 8.31 J/mol-K = 8.31 kpa-l/mol-k 1 L-atm = 101
More information3. Review on Energy Balances
3. Review n Energy Balances Objectives After cmpleting this chapter, students shuld be able t recall the law f cnservatin f energy recall hw t calculate specific enthalpy recall the meaning f heat f frmatin
More informationA. Lattice Enthalpies Combining equations for the first ionization energy and first electron affinity:
[15.1B Energy Cycles Lattice Enthalpy] pg. 1 f 5 CURRICULUM Representative equatins (eg M+(g) M+(aq)) can be used fr enthalpy/energy f hydratin, inizatin, atmizatin, electrn affinity, lattice, cvalent
More informationAP Chemistry Assessment 2
AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND
More informationTypes of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY
CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred
More informationCHEM 305 Solutions for assignment #4
CEM 05 Solutions for assignment #4 5. A heat engine based on a Carnot cycle does.50 kj of work per cycle and has an efficiency of 45.0%. What are q and q C for one cycle? Since the engine does work on
More informationChem 1310 A/B 2005, Professor Williams Practice Exam 3 (chapters 10, 11 and 12) Chapter 10 Thermochemistry
Chem 1310 A/B 2005, Professor Williams Practice Exam 3 (chapters 10, 11 and 12) Chapter 10 Thermochemistry 1. The heat capacity (C P ) is related to the heat absorbed at constant pressure (q P ) and the
More informationUniversity of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2014
Lecture 11 07/18/14 University of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2014 A. he Helmholt Free Energy and Reversible Work he entropy change S provides an absolutely general
More informationWhat determines how matter behaves? Thermodynamics.
What determines hw matter ehaves? hermdynamics.. What determines hw matter ehaves? Final Stale State Equilirium State Gis free energy: Minimum (at specific, ). Classical thermdynamics Macrscpic phenmena.
More informationChapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry
Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1
More informationThe Second Law of Thermodynamics (Chapter 4)
The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made
More informationChem 111 Summer 2013 Key III Whelan
Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general
More informationTuesday, 5:10PM FORM A March 18,
Name Chemistry 153-080 (3150:153-080) EXAM II Multiple-Chice Prtin Instructins: Tuesday, 5:10PM FORM A March 18, 2003 120 1. Each student is respnsible fr fllwing instructins. Read this page carefully.
More informationLecture 4. The Second Law of Thermodynamics
Lecture 4. The Second Law of Thermodynamics LIMITATION OF THE FIRST LAW: -Does not address whether a particular process is spontaneous or not. -Deals only with changes in energy. Consider this examples:
More informationNCERT THERMODYNAMICS SOLUTION
NCERT THERMODYNAMICS SOLUTION 1. Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine
More informationUniversity of Washington Department of Chemistry Chemistry 452/456 Summer Quarter 2008
University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 008 Midterm Examination Key July 5 008 Blue books are required. Answer only the number of questions requested. Chose wisely!
More informationFACULTY OF SCIENCE MID-TERM EXAMINATION 2 MARCH 18, :30 TO 8:30 PM CHEMISTRY 120 GENERAL CHEMISTRY
FACULTY OF SCIENCE MID-TERM EXAMINATION 2 MARCH 18, 2011. 6:30 TO 8:30 PM CHEMISTRY 120 GENERAL CHEMISTRY Examiners: Prof. B. Siwick Prof. A. Mittermaier Dr. A. Fenster Name: Associate Examiner: A. Fenster
More informationFind this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring
More informationChemistry 1105 R11 Fall 2018 Test 3
Chemistry 1105 R11 Fall 2018 Test 3 Friday, November 16, 2018 Name: Time: 1 hour 50 minutes Student #: This test consists of seven pages of questions, the periodic table, and a page containing useful constants
More informationCHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS
CHPTER 6 / HRVEY. CHEMICL B. THERMODYNMICS ND C. MNUPULTING CONSTNTS D. CONSTNTS FOR CHEMICL RECTIONS 1. Precipitatin Reactins 2. cid-base Reactins 3. Cmplexatin Reactins 4. Oxidatin-Reductin Reactins
More informationChemistry 1A, Spring 2008 Midterm Exam III, Version A April 14, 2008 (90 min, closed book)
Chemistry 1A, Spring 2008 Midterm Exam III, Version A April 14, 2008 (90 min, closed book) Name: KEY SID: A Name: 1.) Write your name on every page of this exam. 2.) his exam has 15 multiple-choice questions
More informationCHEM Thermodynamics. Entropy, S
hermodynamics Change in Change in Entropy, S Entropy, S Entropy is the measure of dispersal. he natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal
More informationEdexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.
Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied
More informationChapter 8. Thermochemistry 강의개요. 8.1 Principles of Heat Flow. 2) Magnitude of Heat Flow. 1) State Properties. Basic concepts : study of heat flow
강의개요 Basic concepts : study of heat flow Chapter 8 Thermochemistry Calorimetry : experimental measurement of the magnitude and direction of heat flow Thermochemical Equations Copyright 2005 연세대학교이학계열일반화학및실험
More informationExam 3, Ch 7, 19, 14 November 9, Points
Chem 30 Name Exam 3, Ch 7, 9, 4 November 9, 206 00 Points Please follow the instructions for each section of the exam. Show your work on all mathematical problems. Provide answers with the correct units
More informationCHAPTER PRACTICE PROBLEMS CHEMISTRY
Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin
More informationThermodynamics (XI) Assignment(Solution)
SYLLABUS CUM COM./XI/03 4 hermodynamics (XI) Assignment(Solution) Comprehension ype Questions aragraph for Question -5 For an ideal gas, an illustration of three different paths A, (B + C) and (D + E)
More information