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Chem123/125 Spring 2003

Chem123/125 Spring 2003 Instructor: C. Chieh (sounds like Jay, Nickname peter) C ybersp a ce C hemis t ry (CaCt) science.uwaterloo.ca/~cchieh/cact/ Office: C2-263 Phone: 888-4567 ext. 5816 e-mail: cchieh@uwaterloo.ca You: Please provide your ID number when sending e-mails.

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Chem123/125 Spring 2003

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  1. Chem123/125Spring 2003 Instructor: C. Chieh (sounds like Jay, Nickname peter) Cyberspace Chemistry (CaCt)science.uwaterloo.ca/~cchieh/cact/ Office: C2-263Phone: 888-4567 ext. 5816e-mail: cchieh@uwaterloo.ca You: Please provide your ID number when sending e-mails. 13. States of matterliquid and solids

  2. Please sit in the front & center The communication is easier and better for both you and the instructor. The better the communication, the better you learn. 13. States of matterliquid and solids

  3. I (student) learn how to learn, learn the basics and techniques of science, and apply them to solve problems to survive, make my plan to fit the class schedule; write CaCt quizzes to test my ability, practice, practice and practice problems solving on suggested questions in part a and b of handout, ask my professor or my TA for help whenever I need, will not wait till I got a failure mark, read all instructions, and distinguish assumptions and facts 13. States of matterliquid and solids Check counseling services for study skills: www.adm.uwaterloo.ca/infocus

  4. Announcements Did you fail CHEM 120 last term? Then you cannot take CHEM 123 this term. You must select another course for this term. CHEM 120 is offered this term from 5:30-6:50 pm in DC 1351 on Mon & Tues. Did you miss the CHEM 120 exam for valid medical reasons? The make-up exam will be held on Friday, January 9th from 2-5 pm. Bring your medical documentation with you to the exam. Contact the course coordinator to find out where the exam is being written. (cbissonn@sciborg.uwaterloo.ca) 13. States of matterliquid and solids

  5. Announcements – cont. Tutorials start in week #2 for even-numbered tutorial sections (i.e. sections 102, 104, 106, etc.) Tutorials start in week #3 for odd-numbered tutorial sections (i.e. sections 101, 103, 105, etc.) CHEM 123L and CHEM 125L: Labs start this week! 13. States of matterliquid and solids

  6. Need to Learn Feeling and longing motive all human endeavor Everything that the human race has done and thought is concerned with the satisfaction of deeply felt needs and the assuagement of pain. Albert Einstein 13. States of matterliquid and solids

  7. 13. States of Matter GasABCD gas laws Liquidtypes of liquid Solidcrystal, glass (amorphous) Other terms related to states: plasma, solutions, mixtures, colloid, liquid crystals, super fluid, supercritical fluid, phase Know properties of your material 13. States of matterliquid and solids

  8. Solids Crystals and amorphous (glass or frozen liquid) Describe the difference between crystals and amorphous solids 13. States of matterliquid and solids

  9. Crystals Common crystals: Diamond, ice, dry ice, quartz, calcite, aragonite, sulfur, phosphorus etc. Crystal properties: Melting point, arrangement of atoms (crystal structures), electrical conduction, hardness, crystal habit etc. Give some unique features of crystals 13. States of matterliquid and solids

  10. Liquid Common liquids: Water, alcohol, acetone, carbon tetrachloride, gasoline, nitrogen, helium, hydrogen, oil, etc. Properties of liquid: Freezing point, boiling point, vapor pressure, viscosity, color, surface tension, motion and arrangement of molecules, dielectric constant etc. Describe the arrangement of molecules in liquid and some common properties of liquid 13. States of matterliquid and solids

  11. Phases Explain the concept of phase, define and exemplify A phase is a distinct and homogeneous state of a system with no visible boundary separating it into parts. Phases: solid, liquid, gas, solution, different solid or liquid Think of some systems and determine the number of phases in them. Milk, orange juice, coke, soup, wine, beer, absolute alcohol, gasoline, gas, natural gas, etc. Ruby imbedded in rocks, many phases 13. States of matterliquid and solids

  12. Phase Transitions - Terms solid deposit freeze melt sublimate condense liquid gas vaporize Explain and name phase transitions 13. States of matterliquid and solids

  13. Phase Transitions – in action E = C Tenergy = heat capacity * change in T thermometer temperature boiling energy melting time 13. States of matterliquid and solids Explain changes as energy is transferred into a system

  14. Phase Transitions - Notations & Energy 1 mole H2O vapor Please consider how energy causes the changes H2O (s)  H2O (l) Hf 6 kJ/mol H2O (s)  H2O (g) Hs 47 H2O (l)  H2O (g) Hv 41 H2O (l)  H2O (s) -Hf -6 H2O (g)  H2O (s) -Hs -47 H2O (g)  H2O (l) -Hv -41 Please consider other phase transitions and their transition energies. 44.0 kJ Water at 298 K Water at 273 K 6.01 kJ 18 g Ice at 273 K Energy level diagram given in Chem120 13. States of matterliquid and solids Explain the meaning of each equation

  15. Phase Transitions-Vapor Pressurean equilibrium depending on Temperature Critical point Vapor pressure mmHg or Pa Vapor pressures of ice and water are the same Vapor pressure of water Vapor pressure of ice Temperature 13. States of matterliquid and solids 1.8: System on slide 8

  16. Phase Transitions-Vapor Pressure Vapor pressure mmHg or Pa Critical point Vapor pressure of water Triple point (ice, water and vapor) Difference between triple point and melting point Vapor pressure of ice Temperature 13. States of matterliquid and solids

  17. Phase Transitions-Vapor Pressure Critical point Vapor pressure mmHg or Pa Variation of bp with altitude 101.3 kPa Vapor pressure of water Boiling point Vapor pressure of ice Temperature 13. States of matterliquid and solids

  18. Phase Transitions-Vapor Pressure Application P Vapor pressure of water Vapor pressure of alcohol Distillation T 13. States of matterliquid and solids

  19. Phase Transitions-Vapor PressureClausius-Clapeyron Equation The Clausius-Clapeyron equation correlate vapor pressure and heat of phase transition: – Hvap ln P = ––––––– + BRT note ln P = 2.303 log P How to get straight-line plots? ln P 1T Explain the meaning of the Clausius-Clapeyron equation 13. States of matterliquid and solids

  20. Phase Transitions-Vapor Pressure Clausius-Clapeyron Equation The vapor pressure of ether is 400 and 760 mmHg at 18 and 35oC. What is the heat of vaporization? P T400 291 (= 273+18)760 308 (= 273+35) P1 – Hvap 1 1ln ––– = ––––––– (–– – –– )P2R T 1 T2 Derived from the C-C equation dPH P–– = –––– (–––)dTRT2 400– Hvap 1 1ln ––– = –––––––––– (––– – –––) 760 8.31 J mol-1 291 308 HLn P = – –––– + BR T - 0.642 = – Hvap*2.283e-5 Hvap = 28721 J mol-1 13. States of matterliquid and solids

  21. Phase Transitions-Vapor Pressure Clausius-Clapeyron Equation The heat of vaporization for water is 40.7 kJ mol-1. Calculate vapor pressure at 300 K. P T/ KP1 300760 373expect U to know R = 8.314 J mol-1 P1 - Hvap 1 1ln ----- = ---------- ( ---- - ---- )P2R T 1 T2 P1 -40700 J mol-1 1 1ln ------- = ------------------- ( ---- - ----- ) = - 3.914 760 8.31 J mol-1 300 373 P1 / 760 = e-3.914 = 0.020 P1 = 0.020*760 mmHg = 15.2 mmHg 13. States of matterliquid and solids when P=800, b.p.=?

  22. Regardingln N = ln 10 log N b(log a a) = log a N log x N = (log x x)y a b = N = x y b(log a a) = log a N = y log a x (log a a)b = log a N = y log a x = log x N log a x loga N = log a x log x Nlog e N = log e x log x N ln N = ln 10 log N e a 10 x 2.303 13. States of matterliquid and solids

  23. Facts about H2O; m.p. = 272 Kb.p. = 373 KHsub = Hf + Hv Practice Problems • Use a calculator to evaluate the vapor pressure of water at 272, 273, and 360 K. (Use data from lecture material)DH / R = 41000/8.3142=4931P at 272 K = 748 Pa; P at 273 K = 799 Pa, P at 360K = 62846 Pa P at 373 = 101.3 k Pa • Use a spreadsheet to plot the vapor pressure of ice for temperature between 253 and 275 K. DH/R = 47000/8.3142 = 5653; T P /Pa250 91253 119255 141273 610274 658 13. States of matterliquid and solids

  24. Phase Transitions-Phase Diagrams of H2O Critical point Vapor pressure 1 atm W Vapor pressure of water I V b.p. Vapor pressure of ice 100oC Temperature 13. States of matterliquid and solids Draw and explain the phase diagram of water using data on Cact

  25. A Student’s Question For water, the phase diagram says that at 20o C and 1 atm, the water exists as only one phase (liquid), but it still has a vapour pressure. It seems to me that at one atm. water will be 2 phases (vapour and liquid) at any temperature between 0 and 100 degrees celcius. I have the same question about ice too. 13. States of matterliquid and solids

  26. Phase Transitions-Phase Diagrams of CO2 Critical point 73 atm Vapor pressure mmHg or Pa Liquid CO2 Solid CO2 Vapor pressure of liquid CO2 d 5.1 atm Gas CO2 266 K 1 atm Vapor pressure of solid CO2 304 K Temperature 216 K 13. States of matterliquid and solids What are the differences between phase diagrams of H2O and CO2

  27. Phase transition - Phase Diagram of Sulfur Triple point liquid Mono-clinic Rhombic vapor 13. States of matterliquid and solids

  28. Intermolecular Forces Intermolecular forces: forces of interaction among molecules Intermolecular forces strength example van der Waals or London dispersion ~ 0.1 kJ mol-1 Ar, S8, CH4 Dipole-dipole ~ 10 kJ mol-1 CO, NO2 hydrogen bonding ~20-40 kJ mol-1 H2O, HF, NH3 ionic attraction ~100-1000 kJ mol-1 NaCl, KBr Intramolecular: covalent bonding ~100-2000 kJ mol-1 diamond Explain each term and correlate properties such as hardness, m.p., b.p., H, of materials with intermolecular forces in them. 13. States of matterliquid and solids

  29. Intermolecular Forces - explained 13. States of matterliquid and solids Summarize types of intermolecular forces

  30. Intermolecular Forces – London (dispersion) Force Non-symmetric distribution of electrons around nuclei resulting in temporally dipole-dipole interaction among molecules. This force exists among all material, but they are very weak when compared to other interactions. • - - -- 10+ * - - - - - • - - -- 10+ * - - - - - • - - -- - 10+ * - - - - Temporary dipole of Ne (Z = 10) 13. States of matterliquid and solids What gives rise to London dispersion force?

  31. Dipole Moment The product of magnitude of charge on a molecule and the distance between two charges of equal magnitude with opposite sign is equal to dipole moment; D (unit is debye,1 D = 3.34E–30 C m (coulumb.metre); representation Cl+H, a vector ) Dipole moment = charge x distanceSymbol: µ = e– x d = dq * dbond For Cl+H, µ = 1.03 D, dH–Cl = 127.4 pmTwo ways of lookint at H+Cl, dq = 1.03*3.34e–30 C m / 1.274e-12 m = 2.70e-20 C (charge separation by H–Cl )Ionic character = dq / e– = 0.17 = 17% d = 3.44e-30 C m / 1.60e –19 C (e– charge) = 2.15E–11 m = 0.215 pm (+e– by 0.215 pm) mH–Cl = 1.03 DmH–F = 1.9 D, find d and % ionic character for them. 13. States of matterliquid and solids Identify molecules that have dipole moment?

  32. Dipole moment of H2O  Please verify:The dipole moment of individual water molecules measured by Shostak, Ebenstein, and Muenter (1991) is 6.18710–30 C m (or 1.855 D). This quantity is a vector resultant of two dipole moments of due to O–H bonds. The bond angle H–O–H of water is 104.5o. Thus, the dipole moment of a O–H bond is 5.05310–30 C m. The bond length between H and O is 0.10 nm, and the partial charge at the O and the H is therefore q = 5.05310–20 C, 32 % of the charge of an electron (1.602210–19 C). Of course, the dipole moment may also be considered as separation of the electron and positive charge by a distance 0.031 nm. For the water molecule, a dipole moment of 6.18710–30 C m many be considered as separation of charge of electron by 0.039 nm. 13. States of matterliquid and solids

  33. Intermolecular Forces – Hydrogen Bond Attractions between H atom covalently bonded to small electronegative atom (N, O, F) and these electronegative atoms are called H-bond, X-H - - - Y (X, Y = N, O, F) Strange properties of water (high density at 277 K, high m.p. & b.p. to molecules of similar mass, high heat capacity, etc. Lewis structure of H2O H : O : H H-bonds are responsible for many important phenomena. Describe H-bonding 13. States of matterliquid and solids

  34. Intermolecular Forces – H-bond and b.p. Explain unusual properties due to the formation of hydrogen bonds. Identify H-bonding molecules. Apply the hydrogen bond formation ability of molecules to predict their properties. 13. States of matterliquid and solids

  35. Intermolecular Forces – H-bond and DNA The double-helix DNA has two strands of phosphoric-acid and sugar linked bases of Adenine, Guanine Cytosine or Thymine. The A-T and G-C pairs stack on top of each other. 13. States of matterliquid and solids http://www.accessexcellence.org/AB/GG/dna_molecule.html

  36. Intermolecular Forces – Solids 13. States of matterliquid and solids

  37. Arrangement of Atoms in Diamonds Image from WebElements.com 13. States of matterliquid and solids

  38. Types of Solid - Covalent Crystals Diamond (C ), Si, Ge, GaAs, CdS, SiO2 (quartz) etc. consist of network formed by covalent bonds. These are high melting, hard, crystalline crystals. 13. States of matterliquid and solids

  39. Types of Solid - Molecular Crystals Molecular crystals consist of molecules in their solids. When only London dispersion force hold the molecules together such as solid Ne, Ar, CO2, C6H6, CCl4, I2 etc. they are soft with low m.p. and low heat of vaporization. When the molecules have permanent dipole moment, the molecules are together more strongly. They have higher m.p. and heat of vaporization than those with only London dispersion force. Packing of I2 molecules from Web Elements Identify the molecules of I2 13. States of matterliquid and solids

  40. Types of Solid - Sphere Packing Sphere packing is one of the models used to illustrate packing of atoms in metallic crystals. There are two types of sphere packing: Sequence Type ABAB… hcp (hexagonal closest packing) ABCABC... ccp (cubic closest packing) or fcc (face centred cubic) ABC ABCABC.. ABAB.. 13. States of matterliquid and solids

  41. Types of Solid - Metallic Crystals Three common types:ccp hcp bcc A slight distortion of ccp leads to bcc. 13. States of matterliquid and solids

  42. Types of Solid - Metallic Structures 13. States of matterliquid and solids

  43. Types of Solid – H-bonded Crystals In water, some clusters of water molecules held by H-bonds exist. Ice, solid NH3, and molecules with H-bond ability crystallize making the most number of H-bonds. These solids have higher m.p. than the molecular and dipole-moment molecules. Hydrogen bonding leads to the peculiar properties of melting curve for ice, and highest density for water at 4oC. Crystal structure of ice 13. States of matterliquid and solids lowtem.hokudai.ac.jp/~frkw/english/ss2.html

  44. Solids - Unit Cell of Crystal The smallest convenient unit when repeatedly stacked together generate the entire crystal structure. Two choices of unit cell in a 2-D space How many disk does each unit have?Left 1 right one large one small extend to 3-D space 13. States of matterliquid and solids

  45. Solids - Unit Cell & Density A metal X (atomic mass M) crystallizes in a simple cubic crystal structure with one atom per unit cell with the length of the unit cell a in cm, thenits density d g cm-1 is. M / NAd = ------------a3 Mass of an atom / unit cell Volume of a unit cell NA is the Avogadro’s number If the cubic unit cell has n atoms, then n M d = (NAa3 ). 13. States of matterliquid and solids

  46. Solids - Unit Cell & Radius of atom Simple cubic:Cubic edge a = 2 time radius of atoms,a = 2 r Body centre cubic:face diagonal df2 = 2 a2body diagonal, db2 = df2 + a2 = 3 a2 = (4r)2 Face centre cubic:face diagonal df2 = 2 a2 = (4r)2a = 22r 13. States of matterliquid and solids

  47. Solids – Density and Radii of Atoms Copper crystallize in ccp type structure with a density of 8.92 g mL-1. Calculate its atomic radius based on the hard-sphere packing model. Let the cell edge be a and radius be r. Data NA = 6.023e23, atomic mass of Cu 63.5 and a = 22r 4 * 63.5 gd = ------------------ = 8.92 g cm-3 6.02e23 * a3 Thus, a3 = 4*63.5 / (6.02e23 * 8.92) cm3 = 4.73e-23 cm3 And a = 3.617e-8 cm = 22r Thus, r = 1.279e-8 cm or 0.127 nm volume Work on part b problems Wk 2 & 3 13. States of matterliquid and solids

  48. Solids – Tetrahedral Holes Four balls of radii R touching each other forms a tetrahedral configuration, the center of which is called a tetrahedral hole. To evaluate the size of this hole, imagine the largest ball of radius r placed in the hole touching all for large balls. The 4 balls can be placed on alternate corners of a cube, whose edge = aSince the 4 balls touch each other face diagonal = 2R =  2aSince large balls touching small ballbody diagonal = 2(R+r) =  3 a a = edge 2R = fd=  2 a 2(R + r) = bd = 3 a R＋r 3 ------- = -----R 2 r---- = 0.225 R Details of deriving these are described in lecture. 13. States of matterliquid and solids

  49. Another Td picture What is the radius of the largest ball that can be placed in a tetrahedral hole without disturbing the packing of the spheres of radius R? How do you go about to solve this problem? Solve the same problem for an octahedral hole. 13. States of matterliquid and solids

  50. Solids – Octahedral Holes The equatorial plane of the octahedral is a square of edge a.= 2R Place a small ball in the octahedral hole touching all 6 balls. Then Diagonal = 2 (R + r) =  2 a =  2 (2R) Therefore 2(R+r)  2 --------- = ------ =  2 2R 1 r---- = --------- = ________? R 13. States of matterliquid and solids

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