Chapter 11. States of Matter Liquids and Solids. Enthalpy of Phase Change

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1 Chapter 11 solids: rigid ordered arrangement or particles fixed volume and shape not compressible particles very close together States of Matter Liquids and Solids liquids: fluid (flow) fixed volume but variable shape very slightly compressible short range order; short range motion gases: fluid least dense form of matter shape and volume are variable highly compressible (lots of empty space) constant, random, chaotic motion Enthalpy of Phase Change Phase Changes: the Big 3 energy associated with change of physical state at constant P sign of ΔH (+ or ) indicates the direction of change units kj/mol H fus: interconversion between solid liquid states solid liquid change is endothermic liquid solid change is exothermic T at which this change occurs is a substance s mp H vap: interconversion between liquid gas states liquid gas change is endothermic gas liquid change is exothermic T at which this change occurs is a substance s bp H sub: interconversion between solid gas states solid gas change is endothermic gas solid change is exothermic T at which this change occurs is a substance s sub. pt

2 Calculations Associated with the Heating or Cooling of Substances: if heat is added or removed and the phase does not change but temperature does: Heating and Cooling Curves plot of how temperature changes during heating or cooling of a substance at constant pressure: use the eqn: q = msδt if heat is added or removed and the temperature remains constant but a change of physical state occurs: use the eqn: q = (mol substance)(δhphase change) for the multistep process: qtot = q1 + q2 + q3 + q4... Look at this as a 5-step process: example: Calculate the heat released when 25.0 g of butane vapor is condensed at its boiling point. For butane, C4H10, H vap = 24.1 kj/mol. example: Calculate the total amount of heat required to convert g H2O (s) at 15 C to H2O (l) at 30 C. For H2O: molar mass = g/mol H fus = 6.01 kj/mol; H vap = 40.7 kj/mol ssolid = 2.09 J/g C; sliquid = 4.18 J/g C Some Points About Phases & Phase Changes... the physical state of substance is dependent on both T and P a diagram that shows completely the relationship between phase, T, and P is called a phase diagram (stay tuned... ) phase changes occur at constant T and P as a phase change is occurring, a dynamic equilibrium exists between 2 phases: rate of forward change = rate of reverse change populations of both phases are constant with time

3 Some Phase Oddities superheated and supercooled liquids: Some Phase Oddities: supercritical fluids a gas may be liquefied by an applied, external pressure (Pext) superheating occurs when a liquid exists at a T above its boiling point; Pext required to liquefy a gas is dependent on T i.e. a liquid exists at conditions consistent with vapor phase critical temperature (Tc) - T above which a gas cannot be liquefied - no matter what Pext is applied OR supercooling occurs when a liquid exists at a T below its freezing point; highest T at which a distinct liquid phase can exist P required to liquefy a gas at its Tc is the critical pressure i.e. a liquid exists at conditions consistent with solid phase supercritical fluid is the phase that exists above a substances critical T and P superheated or supercooled liquids are very unstable! Dependence of Pvap on Temperature: Vapor Pressure, Pvap The vapor pressure of a substance is the pressure exerted by the vapor phase when the liquid & vapor phases coexist in a state of dynamic equilibrium. For a given substance, vapor pressure increases with increasing temperature because the number of particles in the gas phase is greater at higher T s.

4 Vapor Pressure and Temperature: the Clausius - Clapeyron Equation & P #!H vap ln$$ 2!! = R % P1 " & 1 1# $$ '!! T T 2 " % 1 example: Calculate the vapor pressure of ethanol at 40 C if the vapor pressure at 78.3 C is 760 Torr. For ethanol, H vap = 38.6 kj/mol. let: T2 = 40 C; P2 =??? T1 = 78.3 C; P1 = 760 Torr Boiling Point (bp) and Vapor Pressure (Pvap) A liquid boils when its Pvap is equal to the external pressure (Pext) on the surface of the liquid. Phase Diagrams: Plot of P vs. T A diagram that shows completely the relationship between phase, temperature, and pressure for a given substance is its phase diagram. bubbles of vapor form in the interior of the liquid if Pext increases, Pvap must increase in order for the liquid to boil... therefore bp must also increase a substance s normal boiling point is the bp when Pext (and therefore Pvap) = 1 atm = 760 mmhg critical point solid liquid triple point gas

5 Phase Diagram of Water Phase Diagram of CO2 Phase Diagram of Sulfur Properties of Liquids: Viscosity Viscosity - resistance to flow less flow (or slower flow rate) corresponds to greater viscosity one way to measure viscosity is to determine the rate at which a steel sphere moves through a liquid sample units of viscosity, poise (P) 1 P = 1 g/cm s

6 Surface Tension: Meniscus Formation & Capillary Action Properties of Liquids: Surface Tension Particles in the interior of the liquid feel forces equally in all directions; particles at the surface feel a net inward force. A measure of the strength of these inward forces is given by the liquid s surface tension: surface tension - the energy required to increase the surface area of a liquid by a unit amount Intramolecular vs. Intermolecular Forces Intramolecular forces - forces within a molecule; chemical bonds breaking an intramolecular force - chemical change ex. HCl (g)! H (g) + Cl (g); H = 431 kj/mol Intermolecular forces - forces between different, discrete molecules breaking an intermolecular force - physical change ex. HCl (l)! HCl (g); H = 16 kj/mol consider the relative strengths of cohesive and adhesive forces

7 Types of Intermolecular Forces We will discuss the following intermolecular forces: ion-dipole forces hydrogen bonds * dipole-dipole forces * London dispersion forces all are electrostatic in nature decreasing strength all are significantly weaker than a covalent or ionic bond * dipole-dipole forces & London dispersion forces together are called van der Waal s forces How are a substance s intermolecular forces and properties related? When comparing 2 compounds, the substance with the stronger collection of intermolecular forces will, in general, have the following properties: higher mp, bp larger H phase change higher critical T and P lower Pvap greater viscosity higher surface tension Ion-Dipole Forces ion-dipole forces exist between ions and polar molecules ex. ionic salt dissolved in a polar solvent Dipole-Dipole Forces dipole-dipole forces exist between neutral, polar molecules NaCl (s) Na + (aq) + Cl (aq) strength of ion-dipole force increases with: increasing ion charge and / or increasing polarity of solvent molecules more detail in Chapter 12

8 For molecules of comparable molar mass, the strength of dipole-dipole forces increases with increasing polarity. substance molar mass dipole moment (μ) bp London Dispersion Forces (LDF) instantaneous, and induced dipoles between particles present in all substances the only intermolecular force in samples of neutral, nonpolar molecules or atoms propane CH3CH2CH3 methyl ether CH3OCH3 methyl chloride CH3Cl acetaldehyde CH3CHO acetonitrile CH3CH 44 g/mol 0.1 D 231 K 46 g/mol 1.3 D 248 K 50 g/mol 1.9 D 249 K 44 g/mol 2.7 D 294 K 41 g/mol 3.9 D 355 K increasing polarity (increasing μ) increasing dipoledipole force strength increasing bp For compounds of comparable polarity, the strength of London Dispersion Forces increases with increasing molar mass. greater molar mass correlates to larger e cloud the larger the electron cloud, the greater the polarizeability of the substance strength of LDF s is greater when the polarizeability of the substance is greater Consider the Halogens: substance molar mass bp F2 38 g/mol 85 K Cl2 71 g/mol 239 K Br2 160 g/mol 332 K I2 254 g/mol 458 K increasing size & molar mass increasing polarizeability increasing strength of London dispersion forces increasing bp

9 Consider the Noble Gases: Influence of Molecular Structure/Shape on LDF s Consider 5 alkanes, all with the chemical formula C5H12: substance molar mass bp He 4.0 g/mol 4.6 K Ne 20.2 g/mol 27.3 K Ar 39.9 g/mol 87.5 K Kr 83.8 g/mol K increasing size & molar mass increasing polarizeability increasing strength of London dispersion forces increasing bp Xe g/mol K Why the observed differences in bp and!hvap? example: Consider the data below for acetonitrile and methyl iodide: substance acetonitrile, CH3CN methyl iodide, CH3I molar mass dipole moment bp 41 g/mol μ = 3.9 D 355 K 142 g/mol μ = K Which substance has the stronger dipole-dipole forces? Which substance has the stronger London dispersion forces? Which substance has the stronger total collection of intermolecular forces? example: Consider the following substances: Br2, Ne, HCl, HBr, N2 Identify the primary intermolecular force present in each substance. Which of these should have the strongest LDF s? Which of these should have the strongest dipole - dipole forces? Put these in order of weakest to strongest intermolecular forces. Put these in order of increasing boiling point.

10 Hydrogen Bonding: Effect of Hydrogen Bonding Interactions: Structure of H2O (s) and Density interaction of a H atom bonded to an electronegative atom (typically N, O, or F) with a nearby small, electronegative atom (typically N, O, or F) hydrogen bonds are stronger than van der Waal s forces special case of dipole-dipole force Effect of Hydrogen Bonding Interactions: Trend in bp s of Hydrides of Group V,VI,VII Elements How many hydrogen bonds can form? Effect on properties Consider water, glycerol, and methanol: Which of these compounds would you predict has the highest bp? Which of these compounds would you predict has the highest vapor pressure at RT?

11 Which of the following substances can hydrogen bond? CH4 CH3F H2NNH2 CH3OH H2S substance molar mass H-bonds? polar or nonpolar? CH4 16 g/mol NO nonpolar 109 K CH3F 33 g/mol NO polar 195 K H2S 34 g/mol NO polar 213 K CH3OH 32 g/mol YES polar 338 K H2NNH2 32 g/mol YES polar 387 K bp Some Critical Temperature and Pressure Data: substance molar mass H-bonds? polar or nonpolar? Tc critical P N2 28 g/mol NO nonpolar 126 K 34 atm O2 32 g/mol NO nonpolar 154 K 50 atm Ar 40 g/mol NO nonpolar 151 K 48 atm CO2 44 g/mol NO nonpolar 304 K 73 atm PH3 34 g/mol NO polar 324 K 65 atm NH3 17 g/mol YES polar 406 K 112 atm H2O 18 g/mol YES polar 648 K 218 atm Identifying Intermolecular Forces Present: interacting molecules or ions NO ions involved? YES polar or nonpolar molecules? polar molecules and ions? NONPOLAR POLAR YES NO London Dispersion Forces only ex. I 2, Ar Any H's bonded to N, O, or F? NO YES Ion-Dipole Forces + LDF ex. KCl in H 2O Ionic Bonding + LDF ex. NaCl, NH4NO3 Dipole-Dipole Forces + London Dispersion Forces ex. H 2S, CH 3Cl H-bonding + LDF ex. NH 3, H 2O, HF

12 Classification of Solids: What particles are present? What are the forces holding particles together as an aggregate unit? What are the typical properties of different types of solids? Compare and Contrast 2 Allotropes of Carbon: Diamond vs. Graphite both covalent-network solids Consideration of Some Solids: Classification and Properties Structures of Crystalline Solids: Crystal Lattices & Unit Cells

13 substances crystallize in one of 14 crystal lattices with one of 7 types of unit cells We will focus on the 4 most common crystal lattices and their corresponding unit cells: simple cubic lattice simple cubic unit cell body centered cubic (bcc) lattice cubic close-packed (ccp) lattice body centered cubic (bcc) unit cell face centered cubic (fcc) unit cell hexagonal close-packed hexagonal prismatic (hcp) lattice unit cell For each of the 4 unit cells, we will identify: For each of the 4 lattices, we will identify: arrangement of particles in each layer arrangement of layers relative to one another coordination number of each particle in each layer coordination number of each particle in 3-D lattice particles positions in the unit cell number of particles per unit cell where particles touch define length of unit cell edge (l) in terms of the radius of the particles (r) volume of the unit cell We will discuss 2 calculations based on unit cells: percent space occupied density

14 Simple (or Primitive) Cubic Unit Cell: Simple Cubic Lattice: particles touching, at right angles to one another unit cell is a cube with a particle on each corner each layer of particles stacks directly on top of previous layer # particles in unit cell = 8(1/8) = 1 particles touch along the unit cell edge length of unit cell edge: l = 2r volume of unit cell = l3 = (2r)3 coordination number of a particle in a lattice: number of particles that one particle touches 5 each layer: 1 4 X 3 2 lattice: 1 4 X What Fraction of a Particle Is In a Unit Cell? Relationship Between Unit Cell Edge Length & Particle Radius:

15 Body-Centered Cubic (bcc) Lattice: particles at right angles to one another but not touching in the same layer the 2nd layer of particles sits in crevices created by the 1st layer of particles AB AB AB repeating layer pattern coordination number = 0 within same layer coordination number = 8 within the lattice What Fraction of a Particle Is In a Unit Cell? Body-Centered Cubic (bcc) Unit Cell: unit cell is a cube with a particle on each corner + a particle in the center # particles in unit cell = 8(1/8) + 1 = 2 particles touch along the body diagonal length of unit cell edge: l = 4r/ 3 volume of unit cell = l3 = (4r/ 3)3 Relationship Between Unit Cell Edge Length & Particle Radius:

16 Cubic Close-Packed (ccp) vs. Hexagonal Close-Packed (hcp) Lattices Close-Packed Lattices close-packing is the most efficient way to pack spherical particles hexagonal symmetry is automatic particles in each layer are arranged the same arrangement of layers is different in close-packed lattices: coordination number in same layer = 6 coordination number in lattice = 12 1st layer: 2nd layer: What are the options for the 3rd layer? Coordination Number of Particles in a Close-Packed Lattice:

17 Coordination Number of Particles in a Close-Packed Lattice: Cubic Close-Packed (ccp) Lattice: close-packed layers 2nd layer covers half of the holes in 1st layer 3rd layer covers the other half of holes in 1st layer ABC ABC repeating layer pattern coordination number = 6 within same layer coordination number = 12 within the lattice Face-Centered Cubic (fcc) Unit Cell: unit cell is a cube with a particle on each corner + a particle on each face # particles in unit cell = 8(1/8) + 6(1/2) = 4 particles touch along the face diagonal length of unit cell edge: l = 4r/ 2 volume of unit cell = l3 = (4r/ 2)3

18 Number of Particles in a Face-Centered Cubic Unit Cell: Hexagonal Close-Packed (hcp) Lattice: close-packed layers 2nd layer covers half of the holes in 1st layer 3rd layer sits directly on top of 1st layer AB AB AB repeating layer pattern coordination number = 6 within same layer coordination number = 12 within the lattice Relationship Between Unit Cell Edge Length & Particle Radius:

19 Hexagonal Prismatic Unit Cell: The 4 Unit Cells: unit cell is a hexagonal tube; hexagonal plate on the top and bottom + 3 particles in the middle # particles in unit cell = 12(1/6) + 2(1/2) + 3 = 6 volume of unit cell = 24 2r3 Hexagonal Prismatic Hexagonal Prismatic Calculations:! What percent of space inside of a given unit cell is occupied by spheres? volume occupied by spheres in cell % space inside unit cell occupied = * 100 total volume of unit cell Calculations Based on Unit Cells: Comparing Lattices and Unit Cells: Calculations:! What percent of space inside of a given unit cell is occupied by spheres? (number of spheres in unit cell)(volume of sphere) = * 10 volume occupied by spheres in cell volume of cube % space inside unit cell occupied = * 100 total volume of unit cell note: volume of sphere = 4/3(!r3) volume of cube =!3 of spheres unit cell)(volume of sphere) number of spheres(number in unit cell = 1 forin simple cubic, 2 for body-centered cubic, or 4 for = * 100! = 2r for simple cubic, 4r/"3 forvolume body-centered cubic, 4r/"2 for face-centered cubic of cube therefore,!3 = (2r)3 for simple cubic, (4r/"3)3 for body-centered cubic, (4r/"2)3 for face note: volume of sphere = 4/3(!r3) volume of cube =!3 number of spheres in unit cell = 1 for simple cubic, 2 for body-centered cubic, or 4 for face-centered cubic 2r forissimple cubic, 4r/"3 for body-centered 4r/"2 face-centered cubic!! =What the density of a substance (basedcubic, on its unitforcell)? therefore,!3 = (2r)3 for simple cubic, (4r/"3)3 for body-centered cubic, (4r/"2)3 for face-centered cubic! mass (number of particles)(mass per particle) density = = volume volume of unit cell What is the density of a substance (based on its unit cell)? mass (number of particles)(mass per particle) density =note: = number of spheres in unit cell = number of particles in unit cell; we are know thinking volume volume of unit cell representing atoms, molecules or ions

20 example: Calculate the percent space occupied in a facecentered cubic unit cell. example: Calculate the density of iron in g/cm 3. In its solid state, iron has a body-centered cubic unit cell. For iron, r = 125 pm. lattice simple cubic bodycentered cubic cubic closepacked hexagonal closepacked A Summary of Key Lattice & Unit Cell Details: layer repeating pattern coordination number of each sphere in lattice unit cell A A A A 6 simple cubic (or primative cubic) AB AB AB 8 bodycentered cubic ABC ABC 12 facecentered cubic AB AB AB 12 hexagonal prismatic number of spheres in unit cell in unit cell spheres touch along... % space occupied in unit cell (and lattice) cell edge,! related to radius of sphere, r 1 cell edge 52%! = 2r 2 body diagonal 4 face diagonal 68%! = 4r/"3 74%! = 4r/" %