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Liquids Solids


									   Chapter 11 – Liquids & Solids

Online homework for chapters 11&12 (Set #2)

     Selected End-of-Chapter Problems
         End-of-Chapter Problems (pp 467 - 477)
1    2        3    4    5   6    9     10   13    14

21       22       23   24   25 26a      29 30     31

32       33       34   35   36   37 38      39    41

43       46       53   54   55 59      61,62,63,64,65


84       85       87   89   91   132
             I.   Phase Changes
                  A. Types & Examples

   Solid          Liquid      Melting        ΔH = +
   Liquid         Solid       Freezing       ΔH = -

   Liquid         Gas         Vaporization   ΔH = +
   Gas            Liquid      Condensation   ΔH = -

   Solid          Gas         Sublimation    ΔH = +
   Gas            Solid       Deposition     ΔH = -
I.   Phase Changes
     A. Types & Examples
          I.    Phase Changes
                 B. Physical Characteristics
                 1. Introduction

   Melting Point (mp) is a characteristic physical
    property that can be used to 1) identify an
    unknown and 2) to determine rough purity (mp
    lowering & mp range widen when impure).

   Boiling point (BP) depends upon external pressure
    and can be used to 1) identify an unknown and 2)
    to separate a mixture (distillation – see next pg).

   Sublimation can also be used to purify a substance.
   Distillation can be used for separation of a homogeneous
    mixture of low boiling liquids such as in fingernail polish
    remover – acetone, water & oil.
    I.   Phase Changes   B. Physical Characteristics   2. Melting

   Melting points (mp) or freezing points are characteristic
    of the pure solid & are characteristic of the attractive
    forces holding the solid together.
   As solid heats, molecules vibrate and break lattice
    structure at MP. High MP = strong attraction between
    particles; Low MP = weak attraction between particles.
    Examples :
   (MPs in oC) – Attraction :     NaCl (801) - Very Strong
    H2O = (0) - Moderate ;          Ne = (-249) - Very Weak

   Enthalpy of fusion = ΔHfus = heat absorbed to melt one
    mole of the material; is endothermic (+ in value); large
    value = strong attractive forces.
   ΔHfus H2O = + 6.01 kJ/mol         somewhat large
    I. Phase Changes B. Physical Characteristics 3. Vaporization

   Boiling point (BP) characteristic of pure substance. 1) defn: temperature at
    which the vapor pressure = external pressure. 2) high BP = strong
    attractive forces. 3) Accurate BP more difficult to obtain than mp. Why?

   At BP one breaks apart intermolecular forces to convert liquid to a gas; so,
    BP is a measure of the strength of these forces.

   BP: H2Te = -2 oC     H2Se = - 41 oC     H2S = -60 oC    H2O = +100 oC
    Normal trend: larger MW ---) higher BP. Why is H2O far out of the norm?

   Enthalpy of Vaporization (ΔHvap) = heat absorbed to vaporize 1 mol of the
    material; is endothermic & +; is also a measure of attractive forces; high
    value = strong attractive forces.

   ΔHvap H2O = + 40.7 kJ/mol. Note: reverse (condensation) is an
    exothermic process. This explains why steam at 100 oC produces more pain
    than liquid water at 100 oC.
    I. Phase Changes B. Physical Characteristics 3. Vaporization

   Older refrigerators cool by vaporizing freon, CCl2F2 (MW=121). How many
    kg of freon must be vaporized to freeze 5.3x102 g of H2O ?
    ΔHfus H2O = 6.01 kJ/mol                ΔHvap CCl2F2 = 17 kJ/mol

a) First calculate heat removed to freeze 530 g H2O at 0o:

      530 g       1 mol H2O        (– 6.01 kJ)            = -1.8x102 kJ
                   18.0 g           1 mol H2O

b) Now calculate kg of freon vaporized by absorbing 1.8x102 kJ:

1.8x102 kJ    1mol CCl2F2   121 g CCl2F2      1 kg    =      1.3 kg CCl2F2
                17 kJ       1mol CCl2F2      1000g
I . Phase Changes       B. Physical Characteristics 4. Heating Curve:
       (a) Plot of T vs Heat added. Note plateau at a phase change
       (b) ΔHTotal = ΔHa + ΔHb + ΔHc + ΔHd + ΔHe

Heating Solid (a) =
Melting Point (b) =                   MP ΔHfus (T Constant)
Heating Liquid (c) =
Boiling Point (d) =                           BP      ΔHvap (T Constant)
Heating Gas (e) =

                                          Gas (e)

Temp                                      Liquid (c)

                                          Solid (a)
                       Heat Added
         I.    Phase Changes
               B. Physical Characteristics
               5. Heating Calculation

   Calculating Heat During Phase Change: No temperature
    change - all energy goes into melting or boiling.
       ΔH =        ΔH(vap or fus) x  moles compound

   Calculating Heat Between Phase Changes (heat s, l or g):
    Temperature does change as energy absorbed.

    ΔH = grams compound x specific heat (J/goC) x ΔT
    ΔH = g x s x ΔT
      I . Phase Changes        B. Physical Characteristics
                          5. Heating Calculation

How much heat must be added to 1.0 mol (18 g) of ice at
   – 20 0C to convert it to steam at 140. oC?
           ΔHfus = 6.01 kJ/mol         ΔHvap = 40.7 kJ/mol
 si (.00211) sw (.00418) ss (0.00200) are specific heat capacities in
                      kJ/(goC) for ice, water, steam
a)   Ice (-20 to 0 oC)    qa = 18g x si x (20oC) =  0.76
b)   Melting Ice          qb = ΔHfus x 1.0mol =     6.01
c)   Water (0 to 100oC)   qc = 18g x sw x (100oC) = 7.52
d)   Vaporizing           qd = ΔHvap x 1.0mol =     40.7
e)   Steam (100 to 140oC) qe = 18g x ss x (40oC) =   1.4
     ΔHtotal = qa + qb + qc + qd + qe
     ΔHtotal = 56 kJ
I . Phase Changes        B. Physical Characteristics
5. Summary of previous calculation – 1 mol H2O from -20 to 140 oC

  ΔHtotal = qa + qb + qc + qd + qe
  ΔHtotal = 56 kJ
           I . Phase Changes   B. Physical Characteristics
                  6. Vapor Pressure (VP)

   VP of a liquid = partial
    pressure of vapor over the
    liquid at equilibrium.

   We used VP of water in wet
    gases in chapter 5

   Can calculate ΔHvap from VP
    laws – Clausius Clapeyron
    equation (pg 427-430).

   Can  measure       VP    with
    barometer as illustrated
      I . Phase Changes             B. Physical Characteristics
                  7. Triple Point & Critical Temperature

   Melting Point = temp. at which solid converts to liquid state -
    usually a small range of ± about 0.5 oC; (~ independent of pressure).

   Boiling Point = temperature at which vapor pressure of liquid
    equals the external pressure - usually a small range.

   Triple Point = temperature & pressure at which solid, liquid
    and gas all coexist in equilibrium.

   Critical Point (Critical Temperature) = temperature & pressure
    above which the liquid state no longer exists.

   MP, BP, TP, CP = characteristic physical properties.

   For Water: TP = 0.010 oC & 0.0060 atm;     CP = 374 oC & 218 atm
        I . Phase Changes
               B. Physical Characteristics
               7. Critical Temperature

   Above the CP there is only one state = supercritical fluid
   Vapor Pressure at CP = critical pressure
                    II. Phase Diagrams (PD)

   Phase diagrams are graphical ways to illustrate and
    summarize the physical states of a pure substance under
    varied conditions of temperature and pressure.

   The solid lines represent the boundaries between the phases.

   Can get qualitative and quantitative information from PD.

   Can quickly get mp, bp, tp, cp & phases at any temperature
    and pressure from a PD.
II. Phase Diagrams   (General Example)
II. Phase Diagrams - H2O Example
                            II. Phase Diagrams
   Can sketch out the rough phase diagram if have TP and CP.
    - Y axis = Pressure    &    X axis = Temperature
    - Place TP and CP on the diagram
    - Run a solid line from CP through TP. Add second solid line from TP up.
    - Solid = upper left quadrant; Liquid = upper quadrant more to the
    right; Gas = lower portion. Note: above CP no clear boundaries exist
    = SCF.

   Can draw a very crude phase diagram from only TP.

   Note the line between the solid and the liquid phases (mp):
         is ~ straight up if solid and liquid have same densities
         is slanted little to left if solid has less density than liquid
         is slanted to right if solid has greater density than liquid
        These are generalities.
                      III. Intermolecular Forces
                            A. Hydrogen Bonds

   Many properties of liquids, solids
    & gases can be explained on basis
    of weak attractive forces between
    molecules:      hydrogen bonds,
    permanent     dipole   forces,   &
    induced dipole forces.

   These are called intermolecular
    forces (between molecules) & are
    much weaker than covalent bonds.

   H bonds: attractive force between
    positive H on N, O, or F & a N, O, F
    on another molecule (Know This).
   Example: Water
III. Intermolecular Forces       - Note 1) BP generally drop with lower MW
       2) Not so with O-H, N-H, or H-F compounds due to H bonding
                      III. Intermolecular Forces
            B. van der Waals forces (weak & two types)

   1) Permanent Dipole Interactions: Weak attractive forces
    between opposite charged dipoles.          Occurs with polar
    molecules such as H-Cl. For a polar molecule need: a) polar
    bond - (Δ electronegativity ≥ 0.2 See Fig 9.15, pg 346 for EN)
    & b) a non-symmetrical molecular shape.

   2) Temporary Dipole Interactions also called London Forces &
    Induced Dipoles: Weaker forces between temporarily induced
    dipoles. Occurs with all molecules (not ionic or metallic).

   Strengths (Very Rough) in kJ / mol:

    Within Compound            Ionic Bond      1000-100
    Within Molecule            Covalent Bond   1000-100
    Intermolecular (between)   H Bond          40-10
    Intermolecular (between)   Van der Waals   10-0.1
               III. Intermolecular Forces - Summary
   H Bonds:          H–F       H–F
                            ||||||||||    ~ 10-40 kJ/mole
                      - Have “minor” overlap of orbitals
                      - Can occur with N-H O-H F-H
                      containing molecules – any combo of above 3

   Per. Dipole:      I – F∂-    I∂+ – F ~ 0.1-10 kJ/mole

                      - Attraction of opposite partial charges
                      - Occurs with polar-covalent molecules

   Induced Dipole: I – I   I–I
                           |||           ~ 0.1-1 kJ/mole
                    - Weak attraction of temporary dipoles &
                    occurs with most molecules. (~ function of MW)

   Hydrophobic:      Special type of nonpolar interaction
        III. Intermolecular Forces - Summary for Molecules
   Can look at a molecule & determine intermolecular forces:
      H bonds: when have N-H / O-H / F-H bonds.     CH3OH

     Perm. Dipole: Occurs when have a polar molecule
    - Two requirements to have a polar molecule:
      1) one or more polar bonds (Δ electronegativity ≥ 0.2 )
      2) a shape that does not cancel out polarity
    H-Cl      F-I    C≡O         = permanent dipoles
    CO2 (linear) BF3 (trigonal) CCl4 (tetrahedral) = all have polar
      bonds but are nonpolar molecules due to cancellation of
      polarities by shapes

       Induced Dipole (London): Nonpolar interactions due to
        temporary dipoles. Occurs with most molecules (not with
        ionic compounds) Br2 CH4 CI4 CH3OH NaCl
    III. Intermolecular Forces - Notes on Molecular Shapes
- Molecular Polarity: For a molecule to be polar: need 1) polar bonds & 2)
   shape that does not cancel out the dipole.
- You need to be familiar with the shapes for 2, 3 & 4 “things” attached to a
   given atom; “things” = atoms & pairs of non-bonding valence electrons.
   See page 377, Figure 10.4. You do not need to memorize the shapes for
   molecules containing 5 or 6 things. See page 381, Figure 10.8 for these
- Examples: (draw Lewis Structure; note bond polarity; note shape; note symmetry)
   CH4 = symmetrical shapes & tetrahedral = non-polar (true of most CxHy)

   CF4   = symmetrical & tetrahedral = non-polar

   NF3   = non-symmetrical & pyramidal = polar

   BF3   = symmetrical & trigonal = non-polar

   OF2   = non-symmetrical & Bent = polar      (note: IF5 wrong pg 376; should be PF5)
          III. Intermolecular Forces - Examples
       Identify the Intermolecular Forces for Each:

1) CH3 -CH2 - OH    1) Hydrogen & Dipole & Induced Dipole

2) I2 or I-Br       2) Induced Dipole (London)

3) CH3F             3) P. Dipole & Induced Dipole (London)

4) CH3-O-CH3        4) P. Dipole & Induced Dipole (London)

5) CH3NH2           5) Hydrogen & P. Dipole & Induced Dipole

                    6) Induced Dipole   (shape cancels polarity)
6) CH3 - CH3
                   III. Intermolecular Forces
                         C. Miscellaneous
   BP, MP, Vapor Pressure, ΔHfus & ΔHvap of liquids &
    solids can be explained by intermolecular forces.
    Larger BP, MP, ΔHfus & ΔHvap = larger intermol. force.
   Two more physical properties explained are:

    1) Surface Tension – a measure of the energy required
    to extend a liquid surface. Note: More difficult to break
    surface of strong intermolecular forced liquid like H2O
    (H. bond, P. & T. Dipole) compared to CCl4 (only T. Dipole).

    2) Viscosity – A measure of the resistance of a liquid to
    flow. Example: A concentrated carbohydrate like
    honey (contains many C-O-H bonds) & is very viscous
    due to strong H bonding.
                IV. Attractive Forces & Solids

   Two classes of solids:
    1) Amorphous - with no ordered structure (glass).
    2) Crystalline - with a well defined 3D structure (most).

   Types of Crystalline Solids:
    Covalent – Solids held together by large networks of
               covalent bonds (sand – polymer of SiO2).
    Ionic – Cations & anions held together by opposite
           charge attractions (NaCl).
    Metallic – Metal atoms in sea of electrons (Cu).
    Molecular – Molecules held together by intermolecular
                forces (Ice with mainly H-bonds).
               IV. Attractive Forces & Solids

   Crystalline solids have ordered, repetitive 3D structure
    that we call the crystal lattice.
   Exact crystal structure can be determined using an X-
    Ray Diffraction instrument.
   Smallest box-like unit that makes up the crystal lattice
    is called the unit cell.
   Are several unit cell shapes: Cubic (most simple type; we
    will concentrate on these), Tetragonal, Hexagonal, etc.

   Three types of Cubic Unit Cells are:
      Simple Cubic Unit Cell               (SC)
      Body-Centered Cubic Unit Cell        (BC)
      Face-Centered Cubic Unit Cell        (FC)
- X-Ray Diffraction is used to determine crystal structure. One directs x-rays
at a crystal and records the intensities and angles of the reflected rays.
- One can determine the unit cell type and the position of each atom in a
molecular solid.
- Early work used photographic plates to record results; was computed by
hand; required a year’s worth of calculations.
      IV. Attractive Forces & Solids
   Some Cubic Unit Cells – Note there is 1/8 of an
atom/ion at each corner = 1 atom per eight corners.
Hints for some cubic unit cell HW Problems

                      - SC = atoms touching along side, X
                      - BC =atoms touch in long diagonal D
                      - FC =atoms touch in short diagonal C

                      - If given density (d=m/V) & mass can
                       get volume & then side (X3 = volume).

                      -If given side (X), then can get either D
                       or C and then the atomic radius.

                      - For SC:     X = 2R

                      - For BC:    D2 = 3X2 ;    D = 4R

                      - For FC:     C2=X2+X2 = 2X2 ; C = 4R

                      - For all cubic cells:    X3 = volume.
                 A few typical unit cell calculations
   Example #1: X-ray diffraction found that a metal has a simple cubic unit
    cell. If the volume of the cell is 4.38x10-23 cm3 then calculate the diameter
    of the metal atom in pm.

    X (side of cube) = 2R = Diameter of the atom

    X3 = Volume = 4.38x10-23 cm3

    X = (4.38x10-23 cm3)1/3 = 3.52x10-8 cm

    3.52x10-8 cm    1m         1012 pm    = 352 pm
                   100. cm       1m
                A few typical unit cell calculations
Example #2: X-ray diffraction reveals that Cr forms cubic unit cells with an
   edge of 2.885x10-8 cm. The density of Cr is 7.20 g/cm3. Calculate the type
   of unit cell (SC, BC or FC).

Cell Volume:      (2.885x10-8 cm)3 / cell = 2.401x10-23 cm3 / cell

Mass Cell:        7.20 g Cr   2.401x10-23 cm3      = 1.73x10-22 g Cr
                    cm3           cell                          cell

Mass of 1 Cr Atom: 52.0 g Cr          1 mole            = 8.63x10-23 g Cr
                     mole         6.02x1023 Cr atom                  atom

Atoms per Cell & type of unit cell:

1.73x10-22 g Cr        atom                =    2.00 atoms      Must be BC
           cell       8.63x10-23 g Cr                cell
                 A few typical unit cell calculations

Example #3: X-ray diffraction reveals that Cu forms a
   face centered cubic cell with an edge of 361 pm.
   Calculate the radius of a Cu atom.

Are four Cu radii (R) per short diagonal (C) for the FC;
        C = 4R

 C2 = X2+X2 = 2X2       C2 = (4R)2 = 2X2       4R = √2X2

 R = √2X2 = √ 2(361 pm)2          = 128 pm
      4         4

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