Ch 10 and 11 unit imf, cmpd characteristics, solutions.notebook. April 06, Ch 10 and 11

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1 Ch 10 and 11 Intermolecular forces Structure and types of bonding in solids and metals Molecular solids Phase changes Solutions solubility, colligative properties 1

2 Ch 10: TYPE OF STRUCTURAL FORCE PROPERTIES EXAMPLE SUBSTANCE UNIT BETWEEN UNITS IONIC ions ionic high melting point NaCl, MgO bonding conducts electricity only when (strong) melted or dissolved usually water soluble insoluble in nonpolar solvents MOLECULAR a) nonpolar dispersion forces low melting point/boiling point H 2, CCl 4 molecules (weak) nonconducting, insoluble in water soluble in nonpolar solvents b) polar dispersion forces higher melting point/boiling point HCl molecules dipole nonconducting NH 3 hydrogen bonding likely to be soluble in water (intermediate) COVALENT atoms covalent bond hard, solid C (diamond) NETWORK (strong) VERY high melting point SiO 2 (glass, SOLIDS nonconductors sand, quartz) insoluble in common solvents Si SiC H 2 O METALLIC cations and metallic bond variable melting points Na mobile electrons insoluble in common solvents Hg malleable, ductile Mg good conductors Fe may react with water 2

3 WEAK > STRONG Dispersion Dipole H bonding Metallic ionic Network Covalent low MM Slightly polar increases with large ions > small ions > high MM > very polar more H atoms low charge > high charge (lattice energy) 3

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6 The Strongest Bond Will Give You the 1) Highest freezing point 2) Highest boiling point 3) Greatest viscosity 4) Smallest vapor pressure 5) Greatest heat of fusion 6) Greatest heat of vaporization 7) Greatest surface tension 6

7 Give the type of intermolecular bonding for each of the following: metallic, ionic, dipole, dispersion, hydrogen, network covalent 1) H 2 O 2) PCl 3 3) Ar 4) C 2 H 6 5) CO 2 6) SO 2 7) CH 3 CH 2 OH 8) KCl 9) SiO 2 10) CH 3 NH 2 11) HCl 12) HF 13) (CH 3 ) 3 N 14) CH 3 OCH 3 15) SF 6 16) PBr 5 17) NH 3 18) (CH 3 ) 4 C 19) HOCH 2 CH 2 OH 20) Cl 2 21) NaCl 22) Na 23) CH 3 CN 24) BCl 3 25) MgO 26) CCl 4 27) SO 3 7

8 1) H 2 Se 8) SO 2 15) AsBr 3 2) Kr 9) O 2 16) CH 3 CH 2 CN 3) SiC 10) AlCl 3 17) SiCl 4 4) KOH 11) HCN 18) C 3 H 8 5) BH 19) 3 Fe 6) HI 13) H 2 CO 20) (CH 3 CH 2 ) 2 NH 7) CH 3 CH 2 OCH 3 14) SO 3 21) HOCH 2 CH 2 CH 2 OH 8

9 1) RATIONALIZE THE DIFFERENCE IN BOILING POINTS FOR EACH OF THE FOLLOWING PAIRS OF SUBSTANCES: A) n pentane 36.2 o C neopentane 9.5 o C B) dimethyl ether 25 o C ethanol 79 o C C) HF 20 o C HCl 85 o C D) HCl 85 o C LiCl 1360 o C E) n pentane 36.2 o C n propane 42 o C F) n propane 42 o C dimethyl ether 25 9

10 2) In each of the following groups of substances, pick the one that has the given property. Justify your answer: A) Highest boiling point:hcl, Ar, or F 2 B) Lowest vapor pressure at 25 o C: Cl 2, Br 2, or I 2 C) Lowest freezing point: N 2, CO, or CO 2 D) Lowest boiling point: CH 4, CH 3 CH 3, or CH 3 CH 2 CH 3 E) Highest boiling point: HF, HCl, or HBr F) Highest freezing point: H 2 O, NaCl, or HF G) Lowest vapor pressure: CH 3 CH 2 CH 3, CH 3 COCH 3, CH 3 CH 2 CH 2 OH 10

11 H) Highest boiling point: CCl 4, CBr 4, or CF 4 I) Lowest freezing point: LiF, Cl 2, HBr J) Greatest viscosity: H 2 S, HF, H 2 O 2 K) Greatest heat of vaporization: H 2 CO, CH 3 CH 3, CH 4 L) Smallest enthalpy of fusion: I 2, CsBr, CaO M) Highest melting point: FeBr 2, MgO, CCl 4, H 2 O 11

12 TYPES OF SOLIDS: 1) CRYSTALLINE 2) AMORPHOUS glass, nylon, plastic, rubber TYPES OF CRYSTALLINE SOLIDS: 1) METALLIC same kind of atom nearly all crystallize in body centered, face centered, or hexagonal closed packed lattice 2) IONIC face centered or simple cubic NaCl, AlCl 3 3) MOLECULAR I 2, H 2 O, CH 3 OH 4) NETWORK COVALENT SiO 2, SiC, Si, C(d) CRYSTAL STRUCTURE: SIMPLE CUBIC 6 NEIGHBORS (coordination #) AAA 1 atom/unit cell BODY CENTERED CUBIC 8 NEIGHBORS (coordination #) 2 atoms/unit cell CUBIC CLOSED PACKED ( FACE CENTERED) 12 NEIGHBORS (coord. #) ABCABC COVER OCTAHEDRAL HOLES 4 atoms/unit cell HEXAGONAL CLOSED PACKED 12 NEIGHBORS (coordination #) ABAB COVER TETRAHEDRAL HOLES 12

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14 Alloy Mixture of metal elements 2 Types of alloys A) Substitutional some of the solute metal is "replaced" with atoms of a similar size but different metal B) Interstitial Some of the holes in the close packed structure are "filled" with smaller metal atoms What is steel? 14

15 Allotropes Allotrope Covalent network lattice Carbon allotropes: Diamond: Graphite: Fullerenes (Buckyballs): Carbon nanotubes: 15

Silicon vs Silicon dioxide: Silicon pure element Characteristics: Silicon dioxide compound (known as silica) Characteristics: Where is silica found and used?

16 Silicon vs Silicon dioxide: Silicon pure element Characteristics: Silicon dioxide compound (known as silica) Characteristics: Where is silica found and used? A) Glassware B) Ceramics C) Additives = glassware type of ceramics type B 2 O 3 = borosilicate glass K 2 O = eye glass lenses clay + heat firing = ceramics = more resistant to breaking 16

17 The liquid state: A) Liquids possess surface tension B) Capillary action resistance of a liquid surface tension = capillary action C) Viscosity resistance to flow surface tension = viscosity 17

18 Summary Statement #1: 1) Intermolecular forces: between particles dispersion forces, dipole, ionic, hydrogen, network covalent) 2) intramolecular forces: within a particle (polar and nonpolar covalent, ionic) 3) Network covalent: C(d), SiC, SiO 2, Si strongest of the intermolecular forces. 4) the stronger the attraction, the a) higher the boiling point b) the higher the melting point c) the higher the viscosity d) the greater the surface tension e) the lower the vapor pressure 5) The greater the number of electrons, the stronger is the dispersion force due to an increase in polarizability which is the ease at which a particle can become distorted. 6) Dispersion forces exist in all particles. It is the primary attraction between nonpolar molecules. 7) For hydrogen bonding to occur BETWEEN particles, a H atom must be bonded to an O, F, or N in the particle. 8) Hydrogen bonding occurs when the exposed nucleus of the hydrogen in one particle is attracted to a lone pair of electrons of an O, F, or N atom of another particle. 18

19 Summary Statement #2: 1) Attraction between hydrocarbons are dispersion forces 2) triple point is where all 3 phases exist at the same time all 3 phases are in equilibrium with each other 3) Lines on the phase diagram represent equilibrium positions 4) Can't liquify a gas if its temperature is above its critical temperature 5) Dispersion forces are weaker than dipole with similar molar masses. With big differences in molar masses, dispersion can be stronger. 6) Positive melting point slope means that the volume of the solid is smaller than the volume of the liquid which means that the density of the solid is greater than the density of the liquid most materials 7) Negative melting point slope means that the volume of the liquid is smaller than the volume of the solid which means that the density of the liquid is greater than the density of the solid water 8) Metals are composed of cations with a sea of electrons. 9) Hydrogen bonding occurs between molecular compounds only, not ionic. 10) Metals can have high (transition metals) or low( alkali metals) melting points. 11) Clausius Clapeyron Equation: plot of ln(vapor pressure) vs 1/T(Kelvin) gives a straight line with slope m = H vap /R where R=8.314 J/mol*K ln(p 1 /P 2 ) = H vap /R (1/T 2 1/T 1 ) where R=8.314 J/mol*K P = vapor pressure T = Kelvin Temp Hvap = heat of vaporization in Joules the normal boiling point can be used as one point 1 atm for pressure boiling point as its corresponding temperature 12) Dispersion forces increase with more electrons 19

20 Example: Vapor pressure The vapor pressure of water at 25 C is 23.8 torr, and the heat of vaporization of water at 25 C is 43.9 kj/mol. Calculate the vapor pressure of water at 50. C. 20

21 Heating curve: A) Where is heat of fusion? B) Where is heat of vaporization? C) What happens if gas vapor is heated after vaporization? 21

22 Phase change diagram: A) What is the significance of the critical point? B) How does the H 2 O phase change diagram differ from the phase diagram above? C) What is triple point? D) How is a fire extinguisher filled with CO 2 (l)? 22

23 Ch 11: SOLUTION PROCESS: A) Factors that affect solubility: 1) Nature of reactants: 2) Pressure: 3) Temperature: 23

24 B) Factors that affect the rate at which something dissolves: 1) Surface area 2) Temperature 3) Stirring C) Heat of Solution: ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 ΔH 1 = separation of solute particles => (into individual components) ΔH 2 = separation of solvent particles => (overcome IMF of solvent) ΔH 3 = formation from solute solvent interaction => (allow mixing solute + solvent) If ΔH solution is: ΔH ( ) = overall exothermic ΔH (+) = overall endothermic 24

25 Dissolving of substances: "Likes dissolve likes" Polar liquid = will mix/dissolve polar or ionic solute Non polar liquid = will mix/dissolve NP solute 25

26 VAPOR PRESSURE LOWERING: The addition of nonvolatile solute particles reduces the capacity of the solvent particles to move from the liquid phase to the vapor phase. Raoult's Law (nonvolatile solute) A) Raoult's law = linear equation ( y = mx + b) B) Slope = P 0 solvent C) Mole fraction = mol X/mol total Vapor pressure pressure of vapor when equilibrium has been achieved in a closed container A) Liquid vaporizes in a closed container B) Some gaseous molecules recondense to liquid C) Equilibrium is achieved Rate evap = Rate cond Facts: Volatile = high vapor pressure Low intermolecular forces = high vapor pressure High intermolecular forces = low vapor pressure 26

27 VAPOR PRESSURE LOWERING: 1) Sugar, C 12 H 22 O 11, is a nonvolatile, nonionizing solute in water. Determine the vapor pressure at 25 o C, of a sucrose solution prepared by dissolving 50.0 g of sucrose in 117 g of water. Assume the solution behaves ideally. The vapor pressure of pure water at 25 o C is 23.8 torr. 27

28 B) Volatile Solute: 28

29 D) Deviations from Raoult's Law (2 volatile components) 29

30 2) At 40 o C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider the solution that contains 1.00 mol of heptane and 4.00 mol octane. Calculate the vapor pressure of each component and the total pressure above the solution. 30

31 HENRY'S LAW: The solubility of a gas in a liquid is directly proportional to the pressure above the solution. P = kc Ex: In a pop bottle, P = 5.0 atm over liquid. Atmospheric pressure of CO 2 = 4.0 x 10 4 atm. Calculate the concentration of CO 2 in the bottle when opened if K CO2 = 3.1 x 10 2 mol/l*atm GRAPHS: 1) phase diagram: pure solvent vs solution 31

32 BELL JAR: 32

(A) OSMOSIS: the flow of solvent molecules into a solution through a semipermeable membrane (B) Osmotic pressure 1) The pressure necessary to keep water from flowing across a semipermeable membrane

33 (A) OSMOSIS: the flow of solvent molecules into a solution through a semipermeable membrane (B) Osmotic pressure 1) The pressure necessary to keep water from flowing across a semipermeable membrane 2) Osmotic pressure can be used to characterize solutions and determine molar masses (of large molecules) = MRT a) M = molarity of the solution b) R = gas law constant c) T is the Kelvin temperature 33

34 OSMOTIC PRESSURE: 1) Pepsin is an enzyme present in the human digestive tract. A solution of a g sample of purified pepsin in 30.0 ml of aqueous solution exhibits an osmotic pressure of 8.92 torr at 27 o C. Estimate the molecular mass of pepsin. 34

C) Dialysis: 1) Transfer of solvent molecules as well as small solute molecules and ions D) Isotonic Solutions 1) Solutions that have the same osmotic pressure E) Osmotic Pressure and Living cells:

35 C) Dialysis: 1) Transfer of solvent molecules as well as small solute molecules and ions D) Isotonic Solutions 1) Solutions that have the same osmotic pressure E) Osmotic Pressure and Living cells: 1) Crenation a) Cells placed in a hypertonic solution lose water to the solution and shrink 2) Hemolysis a) Cells placed in a hypotonic solution gain water from the solution and swell, possibly burst. F) Reverse Osmosis: 1) External pressure applied to a solution can cause water to leave the solution a) Concentrates impurites (such as salt) in the remaining solution. b) Pure solvent (such as water) is recovered on the other side of the semipermeable membrane 35

36 CONCENTRATION UNITS: mole of solute 1) Molarity(M) = liters of solution g = (M)(L soln )(MM) moles of solute 2) Molality(m) = kg of solvent g = (m)(kg solvent )(MM) moles of solute 3) Mole Fraction solute = (mol of solute + moles of solvent) g of solute x 100 4) % by mass = g of solution g of solute x ) ppm = g of solution NOTE: grams of solute + grams of solvent = grams of solution # of g equivalents (will not cover) 6) Normality = liters of solution 36

37 1) 12.0 g of NaCl is dissolved in enough water to make 125 ml of solution. If the density of the solution is 1.13 g/ml, what is the molality of the solution, the mass percent, and the mole fraction of each component? 37

38 2) The density of a 15.00% by mass aqueous solution of acetic acid is g/ml. What is (a) the molarity (b) the molality (c) the mole fraction of each component 38

39 3) How would you prepare 500.mL of a solution that is 1.50M NaCl and 2.00M KCl? 39

40 4) What volume of 25.0% NaCl ( d = 1.15 g/ml) should be diluted is order to prepare 75.0 ml of 1.25M solution? 40

41 5) 23.5 g of KCl is dissolved in g of water and the resulting solution has a density of 1.06 g/ml. Find the A) molarity B) molality C) mole fraction of the solute 41

42 6) A 1.37M solution of citric acid (H 3 C 6 H 5 O 7 ) in water has a density of 1.10 g/cm 3. Calculate the mass percent, molality, and mole fraction of the citric acid 42

43 7) How many seconds are equal to 1ppm of a year? 43

44 COLLIGATIVE PROPERTIES: properties that are dependent on the number of solute particles but not their identity. 1) boiling point elevation 2) freezing point depression 3) vapor pressure lowering 4) osmotic pressure 1) Boiling Point Elevation: a) nonvolatile solutes elevate the boiling point of the solvent: ΔT = ik b m solute i) ΔT is the boiling point elevation ( bpt solution bpt solvent ) ii) K b is molal boiling point elevation constant dependent on the solvent iii) m solute is the molality of the solute in the solution iv) i is the van't Hoff factor relating to the # of particles the solute breaks into when dissolved in the solvent. 44

45 2) Freezing Point Depression: a) nonvolatile solutes lower the freezing point of the solvent: ΔT = ik f m solute i) ΔT is the freezing point depression ( fpt solvent fpt solution ) ii) K f is molal freezing point depression constant dependent on the solvent iii) m solute is the molality of the solute in the solution iv) i is the van't Hoff factor relating to the # of particles the solute breaks into when dissolved in the solvent. 45

46 1) What is the freezing point of a 10.0% by mass solution of CH 3 OH in water? 2) What is the freezing point of a 10.0% NaCl by mass solution in water? 46

47 3) A solution containing 4.50 g of a nonelectrolyte dissolved in 125 g of water freezes at o C. Calculate the molar mass of the solute. 47

48 4) An aqueous solution containing 28.8 g of a nonvolatile compound having the composition C 2n H 4n O n in 90.0 g of water boils at o C at 1.00 atm pressure. What is the molecular formula of the compound? 48

49 EXAMPLE: 5 A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing g was dissolved in 15.0 g of benzene, and the freezing point of the solution was 5.26 o C. What is the molar mass of the hormone if the freezing point of benzene is 5.50 o C and its freezing pt depression constant is 5.12 K*kg/mol? 49

50 6) A certain nonvolatile nonelectrolyte contains 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. An aqueous solution containing 5.00% by mass of the solute boils at o C. Determine the molecular formula of the compound. 50

51 PROCEDURE: 1) Put BHT in a capillary tube and determine its melting point 2) g of BHT and g of stearic acid(ch 3 (CH 2 ) 16 CO 2 H) 3) Heat the mixture; stir; cool; and place in a capillary tube 4) Repeat Step 2 using an unknown solid 5) Find the melting points of both mixture Using the following data: Substance PART 1: mpt(c o ) mass(g) BHT stearic acid BHT/stearic acid 51.4 PART 2: BHT unknown BHT/unknown 46.5 A) Calculate the freezing point depression constant for BHT B) Calculate the molar mass of the unknown solid. 51

AP CHEM WKST KEY: SOLIDS, LIQUIDS, SOLUTIONS REVIEW

AP CHEM WKST KEY: SOLIDS, LIQUIDS, SOLUTIONS REVIEW ) The particles of solids and liquids are close enough to each other that the interecular forces can exist without any trouble. In gases, the particles are so far apart that the only time IMF s occur is