Problem Set #10 Assigned November 8, 2013 Due Friday, November 15, 2013 Please show all work for credit. To Hand in
|
|
- Bethany April Gilbert
- 6 years ago
- Views:
Transcription
1 Problem Set #10 Assigned November 8, 013 Due Friday, November 15, 013 Please show all work or credit To Hand in 1. 1
2 . A least squares it o ln P versus 1/T gives the result 3. Hvaporization = 5.8 kj mol 1. Assuming constant pressure and temperature, and that the surace area o the protein is reduced by 5% due to the hydrophobic interaction: G 4 3 r V M N 3 9 r.5 10 m G 0.5 N 4 A A r 0.73mL/ g 60000g / mol ( N 4 r 0.070N / m / mol (4 ) (.5 10 m) 865kJ / mol A Convert to per mole, determine size per molecule We think this is a reasonable approach, but the value seems high 3 )
3 4. The vapor pressure o an unknown solid is approximately given by ln(p/torr) = (K/T), and the vapor pressure o the liquid phase o the same substance is approximately given by ln(p/torr) = (K/T). a. Calculate H vaporization and H sublimation. b. Calculate H usion. c. Calculate the triple point temperature and pressure. a) Calculate Hvaporization and H sublimation. From Equation (8.16) dln P H sublimation dt RT d ln P d ln P dt d ln P H T 1 dt 1 dt R d d T T For this speciic case H R sublimation sublimation 035 H J mol sublimation Following the same proedure as above, H vaporization R H J mol b. Calculate H usion. vaporization 3 1 H usion H sublimation Hvaporization J mol J mol J mol c. Calculate the triple point temperature and pressure. At the triple point, the vapor pressures o the solid and liquid are equal. Thereore, K K T T K T T tp 73.6 K tp tp Ptp 035 ln Torr P Torr tp tp 3
4 ,, 5. The UV absorbance o a solution o a double-stranded DNA is monitored at 60 nm as a unction o temperature. Data appear in the ollowing table. From the data determine the melting temperature. Temperature (K) Absorbance (60 nm) Plotting the relative absorbance versus temperature yields: Relative Absorbance T m C T [degree Celsius] The plot indicates a melting temperature o approximately C. 6. For the ormation o a sel-complementary duplex DNA rom single strands H = 177. kj mol 1, and T m = 311 K or strand concentrations o M. Calculate the equilibrium constant and Gibbs energy change or duplex ormation at T = 335 K. Assume the enthalpy change or duplex ormation is constant between T = 311 K and T = 335 K. irst calculate equilibrium constant at 311 K, the melting temperature, where = 0.5: K, 0.5 K311 K (1- ) -4 C M Then we can calculate the equilibrium constant at 335 K: K H 1 1 ln K R T T 1 1 ( ) ( ) C ds ( 4 C ds ) C ds 4 M (which means ) C ds ( 4 C ds ) C ds 5 M (which means ) And inally we can calculate the Gibbs energy at 335 K: ( ) ( ( ) 4
5 7. At 47 C, the vapor pressure o ethyl bromide is 10.0Torr and that o ethyl chloride is 40.0 Torr. Assume that the solution is ideal. Assume there is only a trace o liquid present and the mole raction o ethyl chloride in the vapor is 0.80 and then answer these questions: a. What is the total pressure and the mole raction o ethyl chloride in the liquid? b. I there are 5.00 mol o liquid and 3.00 mol o vapor present at the same pressure as in part (a), what is the overall composition o the system? (a) what is the overall composition o the system? ( ) b) We use the lever rule. tot n Z x n y Z tot liq B B vap B B Z x Z x x Z B B A A A A y z y z Z y B B A A A A Thereore, n y Z n Z x tot liq EC EC tot vap EC EC 5 3 we know that x EC = 0.50 and y EC = Z Z 0.50 Z Z EC EB EC Z EC EC 5
6 g o bee heart myoglobin dissolved in 50.0 ml o water at T = 98 K generates suicient osmotic pressure to support a column o solution o height d. I the molar mass o myoglobin is 16.9 kg per mole, calculate d. We set the osmotic pressure equal to the pressure o the column: n solutesrt π g h V We can now solve or the height, h: msolutes RT n M solutesrt solutes h V g V g g J K mol 98 K g mol m 1000 kg m 9.81 m s m 9. Calculate the change in the reezing point o water i g o a protein with molecular weight g mol 1 is dissolved in 100. ml o water K m protein / Mprotein -1 ΔT K msolute K 1860 K g mol m solute g / g mol 100 g 10. 6
7
8 13. ( ) ( ) ( ) Assume we have 5g cholesterol dissolved in 95g Dipalmitoyl phosphatidylcholine. ( ) ( ) ( ) 8
9 Extra practice or exam, do not hand in Phase equilibrium 1. Use the vapor pressures or C N given in the ollowing table to estimate the temperature and pressure o the triple point and also the enthalpies o usion, vaporization, and sublimation. Phase T ( C) P (Torr) Solid Solid Liquid Liquid P Torr R ln J mol K ln P Torr H sublimation 3.6 kj mol T T 10.5 K 1.4 K Torr J mol K ln 400 Torr 6 Hvaporization K 5. K kj mol 5.6 kj mol H usion kj mol 1 1 To calculate the triple point temperature, take T T 1.4 K P 100 Torr solid, re solid, re 40. K P 400 Torr liquid, re liquid, re J mol J mol 3-1 Ttp J mol 1.4 K 40. K 1 T P tp tp J mol K ln 400 Torr K, Ttp 40.3 K 100 Torr J mol Torr exp 40 Torr J mol K 1.4 K 40.3 K 9
10 . Use the vapor pressures o Cl given in the ollowing table to calculate the enthalpy o vaporization using a graphical method or a least-squares itting routine. T (K) P (atm) T (K) P (atm) A least squares it o ln P versus 1/T gives the result Hvaporization = 0.3 kj mol It has been suggested that the surace melting o ice plays a role in enabling speed skaters to achieve peak perormance. Carry out the ollowing calculation to test this H usion = 6010 J mol 1, the density o ice is 90 kg m 3, and the density o liquid water is 997 kg m 3. a. What pressure is required to lower the melting temperature by 5.0 C? b. Assume that the width o the skate in contact with the ice has been reduced by sharpening to cm, and that the length o the contact area is 15 cm. I a skater o mass 85 kg is balanced on one skate, what pressure is exerted at the interace o the skate and the ice? 10
11 c. What is the melting point o ice under this pressure? a) d. I the temperature o the ice is 5.0 C, do you expect melting o the ice at the ice skate interace to occur? usion usion 1 1 dp Sm Sm.0 J mol K usion dt M M usion Vm 997 kg m 90 kg m HO, l HO, l Pa K 144 bar K kg kg 3 3 The pressure must be increased by 70 bar to lower the melting point by 5.0ºC. b) F 85 kg 9.81 ms P A m 510 m 7 =. 10 Pa =. 10 bar c) dt 1C T P dp 144 bar usion.0 10 bar 1.5 C ; T m = 1.5ºC d) No, because the lowering o the melting temperature is less than the temperature o the ice. 4. Consider the transition between two orms o solid tin, Sn(s, gray) Sn(s, white). The two phases are in equilibrium at 1 bar and 18 C. The densities or gray and white tin are 5750 and 780 kg m 3 H transition = 8.8 J K 1 mol 1. Calculate the temperature at which the two phases are in equilibrium at 00 bar. In going rom 1 atm, 18 C to 00 atm, and the unknown temperature T G V P S T gray gray gray m G V P S T white white white m At equilibrium gray white m m T gray white S S m m gray white gray white gray white G G 0 V V P S S T M V V P Sn 1 1 gray S transition kg mol 199 x10 Pa kg m 780 kg m 9.8 C J K mol T 8. C white P 11
12 5. A protein has a melting temperature o T m = 335 K. At T = 315 K, UV absorbance determines that the raction o native protein is N = At T = 345. K, N = Assuming a two-state model and assuming also that the enthalpy is constant between T = 315 and 345 K, determine the enthalpy o denaturation. Also, determine the entropy o denaturation at T = 335 K. By DSC, the enthalpy o denaturation was determined to be 51 kj mol 1. Is this denaturation accurately described by the two-state model? We irst calculate the equilibrium constants at 315 K and 345 K: K315 K K345 K D N D N N N N N K 345 K ln R ln J mol K K 315 K H 6. kj mol T T 345 K 315 K 1-1 This result deviates rom the DSC result, indicating that the denaturation process is not accurately described by a two-state model. 6. Suppose a DNA duplex is not sel-complementary in the sense that the two polynucleotide strands composing the double helix are not identical. Call these strands A and B. Call the duplex AB. Consider the association equilibrium o A and B to orm duplex AB Assume the total strand concentration is C and, initially, A and B have equal concentrations; that is, C A,0 = C B,0 = C/. Obtain an expression or the equilibrium constant at a point where the raction o the total strand concentration C that is duplex is deined as. I the strand concentration is M, calculate the equilibrium constant at the melting temperature. 1
13 We make the table o concentrations: C initial C equilibrium AB 0 C A = B C/ C/ C The equilibrium constant at the melting temperature with = 0.5 is given by: CAB C C C C C K A B C And or C = M: K C 110 M Ideal and Real Solutions 1. Predict the ideal solubility o lead in bismuth at 80 C given that its melting point is 37 C and its enthalpy o usion is 5. kj mol 1. 13
14 . The vapour pressure o -propanol is kpa at C, but it ell to 49.6 kpa when 8.69 g o an involatile organic compound was dissolved in 50 g o -propanol. Calculate the molar mass o the compound. 3. The addition o 5.00 g o a compound to 50 g o naphthalene lowered the reezing point o the solvent by K. Calculate the molar mass o the compound. 4. The osmotic pressure o an aqueous solution at 88 K is 99.0 kpa. Calculate the reezing point o the solution. 14
15 5. The molar mass o an enzyme was determined by dissolving it in water, measuring the osmotic pressure at 0 C, and extrapolating the data to zero concentration. The ollowing data were obtained: c/(mg cm 3 ) h/cm Calculate the molar mass o the enzyme. 6. a 15
16 7. a 16
17 8. a 17
18 9. A and B orm an ideal solution. At a total pressure o bar, y A = and x A = Using this inormation, calculate the vapor pressure o pure A and o pure B. P x P y P * total A a B total P x * a A * B * A * B P bar total ybptotal x 0.63 bar x * A B * * * a B A A 0.450P P P P P P P A yp P P P y * B * * * A B A bar The heat o usion o water is J mol 1 at its normal melting point o K. Calculate the reezing point depression constant K. K K RM T solvent usion H usion J mol 1.86 K kg mol J mol K kg mol K
19 1. 19
Problem Set #10 Assigned November 8, 2013 Due Friday, November 15, 2013 Please show all work for credit To Hand in
Problem Set #10 Assigned November 8, 2013 Due Friday, November 15, 2013 Please show all work for credit To Hand in 1. 2. 1 3. 4. The vapor pressure of an unknown solid is approximately given by ln(p/torr)
More informationPX-III Chem 1411 Chaps 11 & 12 Ebbing
PX-III Chem 1411 Chaps 11 & 12 Ebbing 1. What is the name for the following phase change? I 2 (s) I 2 (g) A) melting B) condensation C) sublimation D) freezing E) vaporization 2. Which of the following
More informationDATA THAT YOU MAY USE UNITS Conventional Volume ml or cm 3 = cm 3 or 10-3 dm 3 Liter (L) = dm 3 Pressure atm = 760 torr = Pa CONSTANTS
DATA THAT YOU MAY USE UNITS Conventional S.I. Volume ml or cm 3 = cm 3 or 0-3 dm 3 Liter (L) = dm 3 Pressure atm = 760 torr =.03 0 5 Pa torr = 33.3 Pa Temperature C 0 C = 73.5 K PV L-atm =.03 0 5 dm 3
More information5. Internal energy: The total energy with a system.
CBSE sample papers, Question papers, Notes or Class 6 to CAPTER 6 TERMODYNAMICS Brie Summary o the chapter:. Thermodynamics: Science which deals with study o dierent orms o energy and quantitative relationship..
More information5. Internal energy: The total energy with a system.
CAPTER 6 TERMODYNAMICS Brie Summary o the chapter:. Thermodynamics: Science which deals with study o dierent orms o energy and quantitative relationship.. System & Surroundings: The part o universe or
More informationThere are eight problems on the exam. Do all of the problems. Show your work
CHM 3400 Fundamentals o Physical Chemistry Final Exam April 23, 2012 There are eight problems on the exam. Do all o the problems. Show your work R = 0.08206 L. atm/mole. K N A = 6.022 x 10 23 R = 0.08314
More informationEquations: q trans = 2 mkt h 2. , Q = q N, Q = qn N! , < P > = kt P = , C v = < E > V 2. e 1 e h /kt vib = h k = h k, rot = h2.
Constants: R = 8.314 J mol -1 K -1 = 0.08206 L atm mol -1 K -1 k B = 0.697 cm -1 /K = 1.38 x 10-23 J/K 1 a.m.u. = 1.672 x 10-27 kg 1 atm = 1.0133 x 10 5 Nm -2 = 760 Torr h = 6.626 x 10-34 Js For H 2 O
More informationThere are five problems on the exam. Do all of the problems. Show your work
CHM 3400 Fundamentals of Physical Chemistry Second Hour Exam March 8, 2017 There are five problems on the exam. Do all of the problems. Show your work R = 0.08206 L atm/mole K N A = 6.022 x 10 23 R = 0.08314
More informationSoluble: A solute that dissolves in a specific solvent. Insoluble: A solute that will not dissolve in a specific solvent. "Like Dissolves Like"
Solutions Homogeneous Mixtures Solutions: Mixtures that contain two or more substances called the solute and the solvent where the solute dissolves in the solvent so the solute and solvent are not distinguishable
More informationCHEMISTRY Topic #2: Thermochemistry and Electrochemistry What Makes Reactions Go? Fall 2018 Dr. Susan Findlay See Exercises in Topic 8
CHEMISTRY 2000 Topic #2: Thermochemistry and Electrochemistry What Makes Reactions Go? Fall 208 Dr. Susan Findlay See Exercises in Topic 8 Vapour Pressure of Pure Substances When you leave wet dishes on
More informationChapter 12 Intermolecular Forces of Attraction
Chapter 12 Intermolecular Forces of Attraction Intermolecular Forces Attractive or Repulsive Forces between molecules. Molecule - - - - - - Molecule Intramolecular Forces bonding forces within the molecule.
More informationChapter 12.4 Colligative Properties of Solutions Objectives List and define the colligative properties of solutions. Relate the values of colligative
Chapter 12.4 Colligative Properties of Solutions Objectives List and define the colligative properties of solutions. Relate the values of colligative properties to the concentrations of solutions. Calculate
More informationCHEM N-7 November 2005
CHEM1909 2005-N-7 November 2005 Calcium chloride (3.42 g) is completely dissolved in 200 ml of water at 25.00 ºC in a coffee cup calorimeter. The temperature of the water after dissolution is 27.95 ºC.
More informationPHASE CHEMISTRY AND COLLIGATIVE PROPERTIES
PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES Phase Diagrams Solutions Solution Concentrations Colligative Properties Brown et al., Chapter 10, 385 394, Chapter 11, 423-437 CHEM120 Lecture Series Two : 2013/01
More informationFORMULA SHEET (tear off)
FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9 ) ( F - 32) F = ( 9 / 5 )( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pv = nrt
More informationChemistry 2000 Lecture 12: Temperature dependence of the equilibrium constant
Chemistry 2000 Lecture 12: Temperature dependence of the equilibrium constant Marc R. Roussel February 12, 2019 Marc R. Roussel Temperature dependence of equilibrium February 12, 2019 1 / 15 Temperature
More informationLecture 6. NONELECTROLYTE SOLUTONS
Lecture 6. NONELECTROLYTE SOLUTONS NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRATION UNITS percent
More informationCH302 Spring 2009 Practice Exam 1 (a fairly easy exam to test basic concepts)
CH302 Spring 2009 Practice Exam 1 (a fairly easy exam to test basic concepts) 1) Complete the following statement: We can expect vapor pressure when the molecules of a liquid are held together by intermolecular
More information7 Simple mixtures. Solutions to exercises. Discussion questions. Numerical exercises
7 Simple mixtures Solutions to exercises Discussion questions E7.1(b For a component in an ideal solution, Raoult s law is: p xp. For real solutions, the activity, a, replaces the mole fraction, x, and
More informationSchool of Chemical & Biological Engineering, Konkuk University
School of Chemical & iological Engineering, Konkuk University Lecture 7 Ch. 5 Simple Mixtures Colligative properties Prof. Yo-Sep Min Physical Chemistry I, Spring 2009 Ch. 5-2 he presence of a solute in
More informationCHEMISTRY 110 EXAM 3 NOVEMER 12, 2012 FORM A
CHEMISTRY 110 EXAM 3 NOVEMER 12, 2012 FORM A 1. Consider a balloon filled with 5 L of an ideal gas at 20 C. If the temperature of the balloon is increased by 70 C and the external pressure acting on the
More informationThermodynamic condition for equilibrium between two phases a and b is G a = G b, so that during an equilibrium phase change, G ab = G a G b = 0.
CHAPTER 5 LECTURE NOTES Phases and Solutions Phase diagrams for two one component systems, CO 2 and H 2 O, are shown below. The main items to note are the following: The lines represent equilibria between
More informationCHEMISTRY 110 EXAM 3 Nov. 11, 2013 ORM A!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!" 1. The cylinder shown below is filled with enough N 2 gas at 25 o C to reach a
More informationHomework 01. Phase Changes and Solutions
HW01 - Phase Changes and Solu!ons! This is a preview of the published version of the quiz Started: Jan 16 at 1:pm Quiz Instruc!ons Homework 01 Phase Changes and Solutions Question 1 Given that you have
More informationExam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2
Chemistry 360 Dr. Jean M. Standard Fall 2016 Name KEY Exam 3 Solutions 1.) (14 points) Consider the gas phase decomposition of chlorine dioxide, ClO 2, ClO 2 ( g) ClO ( g) + O ( g). At 200 K and a total
More informationIntermolecular Forces
Intermolecular Forces! When two molecules approach one another, they are attracted to some extent! Polar molecules are attracted through the electrostatic interaction of their dipole moments! Non-polar
More informationx =!b ± b2! 4ac 2a moles particles solution (expt) moles solute dissolved (calculated conc ) i =
Properties of Solution Practice Exam Solutions Name (last) (First) Read all questions before you start. Show all work and explain your answers. Report all numerical answers to the proper number of sig.
More informationthe equilibrium: 2 M where C D is the concentration
Problem Set #11 Assigned November 15, 2013 Due Friday, November 22, 2013 Please show all work for credit Colligative properties: 1. P8.7) Assume that 1-bromobutane and 1-chlorobutane form an ideal solution.
More informationChemistry 425 September 29, 2010 Exam 1 Solutions
Chemistry 425 September 29, 2010 Exam 1 Solutions Name: Instructions: Please do not start working on the exam until you are told to begin. Check the exam to make sure that it contains exactly 6 different
More informationCHM 1046 FINAL REVIEW
CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART I Chapter Description 6 Thermochemistry 11 States of Matter; Liquids and Solids 12 Solutions 13 Rates of Reactions 18 Thermodynamics and
More informationA) sublimation. B) liquefaction. C) evaporation. D) condensation. E) freezing. 11. Below is a phase diagram for a substance.
PX0411-1112 1. Which of the following statements concerning liquids is incorrect? A) The volume of a liquid changes very little with pressure. B) Liquids are relatively incompressible. C) Liquid molecules
More informationChemistry 360 Spring 2017 Dr. Jean M. Standard April 19, Exam points
Chemistry 360 pring 2017 Dr. Jean M. tandard April 19, 2017 Name Exam 3 100 points Note: You must show your work on problems in order to receive full credit for any answers. You must turn in your equation
More informationChemistry 6A F2007. Dr. J.A. Mack. Freezing Point Depression: 11/16/07. t f = nk f M
Chemistry 6A F2007 Dr. J.A. Mack 11/16/07 11/14/07 Dr. Mack. CSUS 1 Freezing Point Depression: Similarly: The Freezing point of a solution is always lower than the freezing point of the pure solvent of
More informationln( P vap(s) / torr) = T / K ln( P vap(l) / torr) = T / K
Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Number 9 Solutions 1. McQuarrie and Simon, 9-4. Paraphrase: Given expressions
More informationPhysical Chemistry I Exam points
Chemistry 360 Fall 2018 Dr. Jean M. tandard October 17, 2018 Name Physical Chemistry I Exam 2 100 points Note: You must show your work on problems in order to receive full credit for any answers. You must
More informationPhase Diagrams. NC State University
Chemistry 433 Lecture 18 Phase Diagrams NC State University Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function of temperature
More informationMCGILL UNIVERSITY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEM 120 MONDAY MARCH 16, :30PM 8:30PM VERSION NUMBER: 1
MCGILL UNIVERSITY FACULTY OF SCIENCE MIDTERM EXAMINATION CHEM 120 MONDAY MARCH 16, 2009 6:30PM 8:30PM VERSION NUMBER: 1 Instructions: BEFORE YOU BEGIN: Enter your student number and name on the computer
More information8.6 The Thermodynamic Standard State
8.6 The Thermodynamic Standard State The value of H reported for a reaction depends on the number of moles of reactants...or how much matter is contained in the system C 3 H 8 (g) + 5O 2 (g) > 3CO 2 (g)
More information5.4 Liquid Mixtures. G i. + n B. = n A. )+ n B. + RT ln x A. + RT ln x B. G = nrt ( x A. ln x A. Δ mix. + x B S = nr( x A
5.4 Liquid Mixtures Key points 1. The Gibbs energy of mixing of two liquids to form an ideal solution is calculated in the same way as for two perfect gases 2. A regular solution is one in which the entropy
More informationPES 2130 Exam 1/page 1. PES Physics 3 Exam 1. Name: SOLUTIONS Score: / 100
PES 2130 Exam 1/page 1 PES 2130 - Physics 3 Exam 1 Name: SOLUTIONS Score: / 100 Instructions Time allowed or this is exam is 1 hours 15 minutes 10 written problems For written problems: Write all answers
More informationColligative Properties and Phase Diagrams - answers at end
Colligative Properties and Phase Diagrams - answers at end Colligative Properties 1) What is the boiling point of 0.10 M CaCl 2( (aq)? Would the actual boiling point be higher or lower than what you calculated?
More informationChem 260 Quiz - Chapter 4 (11/19/99)
Chem 260 Quiz - Chapter 4 (11/19/99) Name (print) Signature Terms in bold: phase transitions transition temperature phase diagram phase boundaries vapor pressure thermal analysis dynamic equilibrium boiling
More informationChapter 10: CHM 2045 (Dr. Capps)
Phase Diagram Phase diagrams for CO 2 and H 2 O Chapter 13. Solutions and Their Physical Properties Shows pressures and temperatures at which gaseous, liquid, and solid phases can exist. Allows us to predict
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 information2. Match each liquid to its surface tension (in millinewtons per meter, mn*m -1, at 20 C).
1. Using your knowledge of the types of intermolecular forces present in CO 2, CH 3 CN, Ne, and CH 4 gases, assign each gas to its van der Waals a parameter. a ( ) 17.58 3.392 2.253 0.2107 gas 2. Match
More informationProblem Set. Assigned October 18, 2013 Due Friday, October 25, Please show all work for credit
Problem Set Assigned October 18, 2013 Due Friday, October 25, 2013 Please show all work for credit To hand in: Thermodynamic Relations 1. The shells of marine organisms contain CaCO 3 largely in the crystalline
More information6 Physical transformations of pure substances
6 Physical transformations of pure substances E6.b E6.2b E6.3b E6.4b Solutions to exercises Discussion questions Refer to Fig. 6.8. The white lines represent the regions of superheating and supercooling.
More informationChemistry 12 Dr. Kline 7 December 2005 Name
Test 4 first letter of last name Chemistry 12 Dr. Kline 7 December 2005 Name This test consists of a combination of multiple choice and other questions. There should be a total of 22 questions on eight
More informationENERGY ANALYSIS: CLOSED SYSTEM
ENERGY ANALYSIS: CLOSED SYSTEM A closed system can exchange energy with its surroundings through heat and work transer. In other words, work and heat are the orms that energy can be transerred across the
More informationPhase Equilibria I. Introduction. Heat and Phase Changes
Phase Equilibria I 2 Introduction In the previous chapter, it was discussed the thermodynamics principles that are the basis of thermochemistry. It was shown how to calculate the energy involved in any
More informationExam 1A Chem 1142 Spring 2013
Exam 1A Chem 1142 Spring 2013 Name: MULTIPLE CHOICE. [4 pts ea.] Choose the best response on the scantron sheet. [48 pts total.] Q1. What angle do the sp2 hybrid orbitals make with respect to each other?
More informationLECTURE 6 NON ELECTROLYTE SOLUTION
LECTURE 6 NON ELECTROLYTE SOLUTION Ch 45.5 pplied Phy Chem First Sem 2014-15 Ch 45.5 Exam II September 1/3 (Multiple Choice/Problem Solving) Coverage: Second/Third Laws of Thermodynamics Nonelectrolyte
More informationHEMISTRY 110 EXAM 3 April 6, 2011 FORM A When the path is blocked, back up and see more of the way. 1. A 250 L vessel is evacuated and then connected to a 50.0 L bulb with compressed nitrogen. The pressure
More informationCHAPTER 4 Physical Transformations of Pure Substances.
I. Generalities. CHAPTER 4 Physical Transformations of Pure Substances. A. Definitions: 1. A phase of a substance is a form of matter that is uniform throughout in chemical composition and physical state.
More informationUseful Information Provided on Exam 1. Sections Covered on Exam , 10.2, 10.8,
Chem 101B Exam 1 Study Questions Name: Chapters 10(partial), 11 & 12(partial) Review Tuesday 2/7/2017 Due on Exam Thursday 2/9/2017 (Exam 1 date) This is a homework assignment. Please show your work for
More informationChem. 112 spring 2012 Exam 1 7:30am/Odago Wednesday March 7, 2012
Chem. 112 spring 2012 Exam 1 7:0am/Odago Wednesday March 7, 2012 Attempt all the questions and fill in your answers correctly on the scantron provided 1. A particular gas exerts a pressure of 4.6 atm.
More information17.4 Calculating Heats Essential Understanding Heats of reaction can be calculated when it is difficult or
17.4 Calculating Heats of Reaction Essential Understanding Heats of reaction can be calculated when it is difficult or impossible to measure them directly. Lesson Summary Hess s Law Hess s law provides
More informationName: Regents Review Quiz #1 2016
Name: Regents Review Quiz #1 2016 1. Which two particle diagrams represent mixtures of diatomic elements? A) A and B B) A and C C) B and C D) B and D 2. At STP, which physical property of aluminum always
More informationa) 1.3 x 10 3 atm b) 2.44 atm c) 8.35 atm d) 4.21 x 10-3 atm e) 86.5 atm
1. (6 pts) A sample of gas with a volume of 750 ml exerts a pressure of 756 mm Hg at 30.0 0 C. What pressure (atm) will the sample exert when it is compressed to 250 ml and cooled to -25.0 0 C? a) 1.3
More informationCYL Classical Thermodynamics Sample Problems
CYL110 2012-2013 Classical Thermodynamics Sample Problems 1. Dieterici s equation of state for a gas is P (V b) exp(a/rv T ) = RT, where a, b, and R are constants. Determine ( V/ T ), ( T/ P ), and ( P/
More informationPhysical Chemistry Physical chemistry is the branch of chemistry that establishes and develops the principles of Chemistry in terms of the underlying concepts of Physics Physical Chemistry Main book: Atkins
More informationCHEM 1412 Practice Exam 1 - Chapters Zumdahl
CHEM 1412 Practice Exam 1 - Chapters 11 13 Zumdahl Some equations and constants: T = Km P = XP = MRT ln[a]t = kt + ln[a]o 1 / [A]t = kt + 1 / [A]o t1/2 = ln(2) / k t1/2 = 1 / k{a]o Kp = Kc(RT) n ln(k1/k2)
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 informationLast Name or Student ID
10/06/08, Chem433 Exam # 1 Last Name or Student ID 1. (3 pts) 2. (3 pts) 3. (3 pts) 4. (2 pts) 5. (2 pts) 6. (2 pts) 7. (2 pts) 8. (2 pts) 9. (6 pts) 10. (5 pts) 11. (6 pts) 12. (12 pts) 13. (22 pts) 14.
More informationChemistry 102 Spring 2019 Discussion #4 Chapters 11 and 12 Student name TA name Section
Chemistry 102 Spring 2019 Discussion #4 Chapters 11 and 12 Student name TA name Section Things you should know when you finish the Discussion hand out: Average molar kinetic energy = E = M u 2 rms 2 =
More informationVapor Pressure is determined primarily from!vaph!vaph depends on the intermolecular forces
What do you remember from last time? What do you remember from last time? You have two containers. one has a total volume of 2 L and one has a total volume of 1 L Into each you place 500 ml of liquid ether
More informationPLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET. c) SeF 4
Chem 130 EXAM 4 Fall 99 PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET QUESTIONS 1-5 MAY HAVE MORE THAN ONE POSSIBLE ANSWER CIRCLE ALL CORRECT RESPONSES TO EACH QUESTION
More informationChapter 13 Properties of Solutions
Chapter 13 Properties of Solutions Learning goals and key skills: Describe how enthalpy and entropy changes affect solution formation. Describe the relationship between intermolecular forces and solubility,
More informationChapter 4: Properties of Pure Substances. Pure Substance. Phases of a Pure Substance. Phase-Change Processes of Pure Substances
Chapter 4: roperties o ure Substances ure Substance A substance that has a ixed chemical composition throughout is called a pure substance such as water, air, and nitrogen A pure substance does not hae
More informationSeptember 28, Possibly Useful Information: 1) ( ) ( ) ( ) ( ) ( ) R = L atm / mol K. 2) ( ) ( ) ( ) h = 6.
Name Student ID # CEMISTRY 122 [Tyvoll] EXAM I September 28, 2007 1 2 3 4 5 Possibly Useful Information: 1) ( ) ( ) ( ) ( ) ( ) R = 0.0821 L atm / mol K 2) ( ) ( ) ( ) h = 6.63 x 10-34 J s 3) ( ) ( ) (
More informationCHEM Exam 3 - March 31, 2017
CHEM 3530 - Exam 3 - March 31, 2017 Constants and Conversion Factors NA = 6.02x10 23 mol -1 R = 8.31 J/mol-K = 8.31 kpa-l/mol-k 1 bar = 100 kpa = 750 torr 1 kpa = 7.50 torr 1 J = 1 kpa-l 1 kcal = 4.18
More informationPhase Equilibrium: Preliminaries
Phase Equilibrium: Preliminaries Phase diagrams for two one component systems, CO 2 and H 2 O, are shown below. The main items to note are the following: The lines represent equilibria between two phases.
More informationB. Correct! Good work. F = C P + 2 = = 2 degrees of freedom. Good try. Hint: Think about the meaning of components and phases.
Physical Chemistry - Problem Drill 06: Phase Equilibrium No. 1 of 10 1. The Gibbs Phase Rule is F = C P + 2, how many degrees of freedom does a system have that has two independent components and two phases?
More informationChapter 13. Ions in aqueous Solutions And Colligative Properties
Chapter 13 Ions in aqueous Solutions And Colligative Properties Compounds in Aqueous Solution Dissociation The separation of ions that occurs when an ionic compound dissolves H2O NaCl (s) Na+ (aq) + Cl-
More informationChemistry II Midterm Exam April 24, 2009
Chemistry II Midterm Exam April 24, 2009 Constants R = 8.314 J / mol K = 0.08314 Lbar / K mol = 8.314 L kpa / K mol F = 9.6485 10 4 C/mol h = 6.63 10-34 J s h = 1.05 10-34 J s k = 1.3806504 10 23 J / K
More informationm m 3 mol Pa = Pa or bar At this pressure the system must also be at approximately 1000 K.
5. PHASES AND SOLUTIONS n Thermodynamics of Vapor Pressure 5.. At equilibrium, G(graphite) G(diamond); i.e., G 2 0. We are given G 2900 J mol. ( G/ P) T V V 2.0 g mol.95 0 6 m 3 mol Holding T constant
More informationColligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure
Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure Learning objectives Describe meaning of colligative property Use Raoult s law to determine vapor pressure of solutions
More informationChem/Biochem 471 Exam 2 11/14/07 Page 1 of 7 Name:
Page 1 of 7 Please leave the exam pages stapled together. The formulas are on a separate sheet. This exam has 5 questions. You must answer at least 4 of the questions. You may answer all 5 questions if
More informationLecture 4-6 Equilibrium
Lecture 4-6 Equilibrium Discontinuity in the free energy, G verses T graph is an indication of phase transition. For one-component system, existing in two phases, the chemical potentials of each of these
More informationLiquids and Solids Chapter 10
Liquids and Solids Chapter 10 Nov 15 9:56 AM Types of Solids Crystalline solids: Solids with highly regular arrangement of their components Amorphous solids: Solids with considerable disorder in their
More informationPhysics 119A Final Examination
First letter of last name Name: Perm #: Email: Physics 119A Final Examination Thursday 10 December, 2009 Question 1 / 25 Question 2 / 25 Question 3 / 15 Question 4 / 20 Question 5 / 15 BONUS Total / 100
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 informationEntropy Changes & Processes
Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3 Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas Variation of Entropy with emperature
More informationANSWERS CIRCLE CORRECT SECTION
CHEMISTRY 162 - EXAM I June 08, 2009 Name: SIGN: RU ID Number Choose the one best answer for each question and write the letter preceding it in the appropriate space on this answer sheet. Only the answer
More informationProblem Set 10 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 10 Solutions 1 Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the triple point occurs at 543 K and 114 torr;
More informationCHE 107 Fall 2017 Exam 1
CHE 107 Fall 2017 Exam 1 Your Name: Your ID: Question #: 1 Fill in the blanks with the letter corresponding to the correct term. Use each term only once. Your response for each one should be a single letter.
More informationPhysical Biochemistry. Kwan Hee Lee, Ph.D. Handong Global University
Physical Biochemistry Kwan Hee Lee, Ph.D. Handong Global University Week 9 CHAPTER 4 Physical Equilibria Basic Concepts Biological organisms are highly inhomogeneous. Different regions of each cell have
More informationChem Midterm 3 April 23, 2009
Chem. 101 - Midterm 3 April 3, 009 Name All work must be shown on the exam for partial credit. Points will be taken off for incorrect or no units and for the incorrect number of significant figures. Only
More informationPaper-II Chapter- TS-equation, Maxwell s equation. z = z(x, y) dz = dx + dz = Mdx + Ndy. dy Now. = 2 z
aper-ii Chapter- S-equation, Maxwell s equation Let heorem: Condition o exact dierential: Where M Hence, z x dz dx and N Q. Derive Maxwell s equations z x z zx, z dx + dz Mdx + Nd z d Now 2 z x M N x x
More information1. Which molecule will have the strongest intermolecular forces? _D. 2. Which molecule will have the weakest intermolecular forces?
Use the following information to answer questions 1-5: 1. Which molecule will have the strongest intermolecular forces? _D 2. Which molecule will have the weakest intermolecular forces? _C 3. What is the
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 information( g mol 1 )( J mol 1 K 1
Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Number 11 Solutions 1. McQuarrie and Simon, 11-27. Paraphrase: If a solution
More informationLet's look at how different properties affect vapor pressure. P =0 P =vapor pressure P =vapor pressure. first all liquid
Let's look at how different properties affect vapor pressure P =0 P =vapor pressure P =vapor pressure Quick Quiz You have two containers. one has a total volume of 2 L and one has a total volume of 1 L
More information8. INTRODUCTION TO STATISTICAL THERMODYNAMICS
n * D n d Fluid z z z FIGURE 8-1. A SYSTEM IS IN EQUILIBRIUM EVEN IF THERE ARE VARIATIONS IN THE NUMBER OF MOLECULES IN A SMALL VOLUME, SO LONG AS THE PROPERTIES ARE UNIFORM ON A MACROSCOPIC SCALE 8. INTRODUCTION
More informationColligative Properties
Colligative Properties! Consider three beakers: " 50.0 g of ice " 50.0 g of ice + 0.15 moles NaCl " 50.0 g of ice + 0.15 moles sugar (sucrose)! What will the freezing temperature of each beaker be? " Beaker
More informationMultiple Choice Identify the letter of the choice that best completes the statement or answers the question.
Chem 102--Exam #2 Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. When water is measured in a plastic graduated cylinder, a reverse meniscus
More informationExam Thermodynamics 12 April 2018
1 Exam Thermodynamics 12 April 2018 Please, hand in your answers to problems 1, 2, 3 and 4 on separate sheets. Put your name and student number on each sheet. The examination time is 12:30 until 15:30.
More informationCHEMISTRY CP Name: Period:
CHEMISTRY CP Name: Period: CHEMISTRY SPRING FINAL REVIEW SHEET NOTE: Below are concepts that we have covered in class throughout the second semester. Questions are organized by chapter/concept to help
More informationREMOVE THIS PAGE PRIOR TO STARTING EXAM.
REMOVE THIS PAGE PRIOR TO STARTING EXAM. 1 2 ANSWER KEY CHEMISTRY F14O3 SECOND EXAM 11/10/99 PROFESSOR J. MORROW PRINT NAME, LAST: FIRST: I.D.# : MAXIMUM POINT VALUE IS IN PARENTHESES 1. (15) 9. (15) 17.
More informationTemperature C. Heat Added (Joules)
Now let s apply the heat stuff to real-world stuff like phase changes and the energy or cost it takes to carry it out. A heating curve...a plot of temperature of a substance vs heat added to a substance.
More information