Liquids and Solids Chapter 16 E-mail: [email protected]@gmail.com Web-site:

Liquids and SolidsChapter 16E-mail: [email protected]

Web-site: http://clas.sa.ucsb.edu/staff/terri/

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Liquids and Solids – ch. 16

Liquids and Solids – ch. 161. Indicate the types of forces present and the type of solid for each of

the following substances:a. CCl4 h. KOH

b. HF i. BH3

c. NH4Br j. H2CO

d. F2 k. SiO2

e. CH3OH l. Ar

f. Ca m. Pg. H2S n. S8

Liquids and Solids – ch. 16London Dispersion ForcesHydrogen Bonding

A particularly strongdipole-dipole

only for moleculeswith hydrogen

bonded to O, N or F

Dipole-DipoleOnly for molecules with

dipole moments aka polar molecules

Liquids and Solids – ch. 162. Using intermolecular forces predict the following:

a. highest melting point F2 or Br2

b. highest melting point HF or HCl

c. highest boiling point HOCH2CH2OH or CH3CH2OH

d. highest boiling point CH3CH3 or CH3CH2CH3

e. highest vapor pressure CH3CH2OH or CH3OCH3

f. highest vapor pressure CCl4 or CH2Cl2

g. highest freezing point MgO or H2O

h. highest surface tension C14H30 or C24H50


Relative strength of forces ⇒ Ionic > >> Hydrogen bonding > Dipole-Dipole > Dispersionfor molecules comparable in size

Dispersion Forces1. Increase with molecular size2. Decrease with the degree of branching ex:

As the attractive forces get stronger…1. Boiling point ↑2. Freezing point ↑3. Melting point ↑4. Heat of fusion ↑5. Heat of vaporization ↑6. Viscosity ↑7. Surface Tension ↑8. Vapor Pressure ↓9. Volatility ↓


Liquids and Solids – ch. 163. Copper has an inter-planar spacing of 1.36 Å. Calculate the

wavelength of the X ray that should be used if θ is 15° (assume n=1).


Liquids and Solids – ch. 164 Types of Unit Cells for Metallic solids

1. Simple Cubic: 1 atom/unit cell

2. Body-centered Cubic: 2

atoms/unit cell

3. Face-centered Cubic or Cubic

Closest Packed: 4 atoms/unit cellabc hexagonal alignment of

atoms

4. Hexagonal Closest Packed: 6

atoms/unit cellabab hexagonal

alignment of atoms


2 Types of closest packing 1. abab – hexagonal closest packed

2. abca – cubic closest packed


Counting Atoms in a Cubic Unit Cell

Atom is shared between 4 unit cells




Unit CellNumber of

atoms/unit cell – (% by volume)

Coordination Number

Edge Length as a function of atomic

radius

Simple Cubic 1 (52.4%) 6 e = 2r

Body Centered Cubic

2 (68%) 8(e = 2.309r)

Face Centered Cubic aka Cubic Closest Packed

4 (74%) 12(e = 2.828r)

Hexagonal Closest Packed 6 (74%) 12 N/A

Liquids and Solids – ch. 164. Zinc crystallizes in a cubic closest packed structure. The radius of a

zinc atom is 135 pm. Calculate the density in g/mL for solid zinc.

Liquids and Solids – ch. 165. Titanium metal has a body-centered cubic unit cell. The density is

4.50 g/cm3. Calculate the atomic radius in angstroms of titanium. (1010Å= 1m)


6. The unit cell in this two-dimensional crystal contains __________ Xs and __________ Os.

a. 1, 1 b. 2, 1 c. 1, 2 d. 4, 1 e. 1, 4


7. Assume the two-dimensional structure of an ionic compound MxAy is

What is the empirical formula of this ionic compound?

Liquids and Solids – ch. 168. Consider the following cubic closest packed structures. For each

compound determine where the ions are in the lattice.

a. NaCl (Na+ = 95 pm, Cl– = 181 pm )

b. ZnS (Zn2+= 74 pm, S2– = 184 pm)

c. CaCl2 (Ca2+ = 99 pm, Cl– = 181 pm)


Ionic SolidsAnions (or the larger ion) will align themselves in the

lattice points of the unit cellCations (or the smaller ion) will occupy the holes created

by the lattice points

In FCC aka CCP lattices there are 3 types of holes:a. Trigonal holes are occupied when the radius of the

cation is less than 22% of the anion – very rareb. Tetrahedral holes (8 available) are occupied when the

radius of the cation is between 22% and 41% of the anion

c. Octahedral holes (4 available) are occupied when the radius of the cation is between 41% and 73% of the anion

Liquids and Solids – ch. 169. Identify the type of doping in each of the following.

a. Gallium doped with tin

b. Antimony doped with germanium

c. A material is made from Al, Ga, and As. The mole fractions of these elements are 0.25, 0.26, and 0.49, respectively.


Doping ⇒enhancing the conductivity

of a semiconductor by adding a trace amount of an impurity (aka the dope)

2 Different Types of doping:a. n-type impurity has ⇒ more valence electrons in a valence shell with higher energy than the semiconductorb. p-type impurity has fewer valence electrons in a valence shell with lower energy than the semiconductor

Si has 4 valence electrons

As (the dope) has 5 valence electrons

Si has 4 valence electrons

B (the dope) has 3 valence electrons


M.O. modelillustrating doping

and how it enhancesconductivity

n-type doping p-type doping

Liquids and Solids – ch. 1610. Consider the following phase diagrams for water and carbon

dioxide respectively – which phase is the most dense for each substance? How does pressure affect the MP and BP for each substance?

Liquids and Solids – ch. 16pr

essu

re

temperature

melting

freezing

vaporization

condensation

sublimation

deposition

Liquid

Solid

Gas

Equilibrium lines

Liquids and Solids – ch. 1611. Pure compound Z has a triple point at 18 ° C and 72 torr, a normal

melting point at 21 °C, and a normal boiling point at 87 °C. Which of the following statements regarding compound Z is/are correct?

a. The density of the solid is greater that that of the liquid.

b. Sublimation occurs if starting with a solid at a constant temperature of 17 °C the pressure is decreased until a phase change occurs.

c. Condensation occurs if the temperature is decreased from

55 °C to 13 °C at a constant pressure of 1.00 atm.

Liquids and Solids – ch. 1612. The enthalpy of vaporization for water is 44 kJ/mol. What is the

boiling point if the atmospheric pressure were 0.33 atm?

ln =

Liquids and Solids – ch. 1613. How much heat does is required to take 10 g of ice at -31 °C to

vapor at 155 °C? (Csolid = 2.03 J/g°C , Cliquid = 4.18 J/g°C , Cgas = 2.02 J/g°C, ΔHfus = 6.01 kJ/mol, and ΔHvap = 40.7 kJ/mol)


Heatingcurve

for water

q = mcΔTq = n Δ Hfus

q = n Δ Hvap

Liquids and Solids – ch. 1614. Determine the final temperature if a 25 g cube of ice at -7 °C is

placed in 180 mL of water at 64 °C and allowed to come to equilibrium. (Csolid = 2.03 J/g°C , Cliquid = 4.18 J/g°C , Cgas = 2.02 J/g°C, ΔHfus = 6.01 kJ/mol, and ΔHvap = 40.7 kJ/mol)


You have completed ch.16

Liquids and Solids – ch. 161. Indicate the types of forces present and the type of solid for each of

the following substances:a. CCl4 LDF/Molecular solid⇒b. HF ⇒ HB/DD/LDF/Molecular solidc. NH4Br Ionic/Ionic solid⇒d. F2 LDF/Molecular solid ⇒e. CH3OH HB/DD/LDF/Molecular solid⇒ f. Ca ⇒ metallic/metallic solidg. H2S ⇒ DD/LDF/Molecular solid h. KOH Ionic/Ionic solid⇒i. BH3 LDF/Molecular solid⇒j. H2CO ⇒ DD/LDF/Molecular solid k. SiO2 Covalent Network solid⇒l. Ar LDF/Atomic solid⇒m. P Covalent Network solid⇒n. S8 LDF/Molecular solid

Liquids and Solids – Answer Key

2. Using intermolecular forces predict the following:

a. Br2 b. HF c. HOCH2CH2OH d. CH3CH2CH3 e. CH3OCH3

f. CCl4 g. MgO h. C24H50

3. Copper has an inter-planar spacing of 1.36 Å. Calculate the wavelength of the X ray that should be used if θ is 15° (assume n=1).

nλ=2dsinθ ⇒ (1)(λ)=2(1.36 Å)(sin 15°) ⇒ λ = 0.70 Å or 0.07nm


4. Zinc crystallizes in a cubic closest packed structure. The radius of a zinc atom is 135 pm. Calculate the density in g/mL for solid zinc.

D = mass/volume ⇒ since cubic closest packed has 4 atoms per unit cell ⇒ mass of unit cell = 4(65.39amu)(1g/6.022x1023amu) = 4.34x10-22g

The volume of the unit cell is edge3 ⇒ e3=(2.828(1.35x10-8cm)) 3 ⇒ 5.57x10-

23cm3

D = (4.34x10-22g)/(5.57x10-23cm3) = 7.8g/cm3

5. Titanium metal has a body-centered cubic unit cell. The density is 4.50 g/cm3. Calculate the atomic radius of titanium. Body centered cubic has 2 atoms per unit cell ⇒ mass of unit cell = 2(47.88amu)(1g/6.022x1023amu) = 1.59x10-22g ⇒ volume = m/d ⇒ volume = (1.59x10-22g)/(4.5g/cm3) = 3.53x10-23 cm3 ⇒ edge = (3.53x10-23 cm3 )1/3 = 3.29x10-8cm or 329pm ⇒ radius = 329pm/2.309 = 142pm

6. The unit cell in the two-dimensional crystal contains 1 X and 1O.

X X X X X

O O O O

X X X X X

O O O O

X X X X X


Repeating Unit there’s 4⇒corners where eachcorner has 1/8 X for a total of ½ Xand the face has½ O simplify ⇒1/2X:1/2 O is 1:1

Repeating Unit there’s 4 corners where⇒each corner has 1/8 O for a total of ½ O and the face has½ X simplify 1/2X:1/2 O is 1:1⇒


7. Assume the two-dimensional structure of an ionic compound MxAy is

Repeating Unit4 corners with 1/8 of acation and 1 cation onthe face or ½ for a totalof 1 cation - all of the anions are on the face4x1/2 is 2 anions

What is the empirical formula of this ionic compound? MX2

Smaller ion is most likely to be the cation

Larger ion is most likely to be the anion


8. Compare the cubic closest packed structures for NaCl (radius of Na+ = 0.66 radius of Cl-), ZnS (radius of Zn2+ = 0.35 radius of S2-) and CaCl2 (radius of Ca2+ = 0.68 radius of Cl-).

NaCl => Na+ will fill all of the octahedral holes

ZnS => Zn2+ will fill all of the ½ of the tetrahedral holes

CaCl2 => Ca2+ will fill ½ of the octahedral holes

9. Identify the type of doping in each of the following.

a. Ga/Sn => n-type b. Si/B => p-type

10. When the solid/liquid line has a positive slope the solid is most dense phase (more typical) and as P ↑ MP ↑ – however if the solid/liquid line has a negative slope the liquid is most dense (rare) and as P ↑ MP↓


11. Drawing the points of the phase diagram you get…

a. Tb. Tc. F

note: (+) slope

760torr

72torr

18°C

21°C

87°C


12. The enthalpy of vaporization for water is 44 kJ/mol. What is the boiling point if the atmospheric pressure were 0.33 atm? Since we know that the BP of water at 1 atm is 100°C we can figure out the BP at all other pressures using the ΔHvap

ln(P1/P2) = (ΔHvap/R)(T2-1 – T1

-1)

ln(0.33atm/1atm) = (44kJ/mol/0.008314kJ/molK)(T2-1 – 373-1K)

T2 = 405K

13. How much heat does is required to take 10 g of ice at -31 °C to vapor at 155 °C? (Csolid = 2.03 J/g°C , Cliquid = 4.18 J/g°C , Cgas = 2.02 J/g°C, ΔHfus = 6.01 kJ/mol, and ΔHvap = 40.7 kJ/mol)

This is a 5 step process => 3 changes in temperature (q=mC Δ T) and 2 phase changes (q=n ΔH)


q1=(10g)(2.03 J/g°C)(+31°C) = 629.3J or 0.629kJ

q2=(10g/18g/mol)(6.01kJ/mol) = 3.34kJ

q3=(10g)(4.18 J/g°C)(+100°C) = 4180J or 4.18kJ

q4=(10g/18g/mol)(40.7 kJ/mol) = 22.6kJ

q5=(10g)(2.02 J/g°C)(+55°C) = 1.11kJ

qtotal = 31.86kJ

14. Determine the final temperature if a 25 g cube of ice at -7 °C is placed in 180 mL of water at 64 °C and allowed to come to equilibrium. (Csolid = 2.03 J/g°C , Cliquid = 4.18 J/g°C , Cgas = 2.02 J/g°C, ΔHfus = 6.01 kJ/mol, and ΔHvap = 40.7 kJ/mol)

Heat is transferred from the hot water into the cold water

-qhot = +qcold


-mCΔT = mCsolid Δ T + nHfus +mCliquidΔ T

-(180g)(4.18J/g°C)(Tf-64°C) =(25g)(2.03J/g°C)(7°C)+(25g/18g/mol)(6010J/mol)+(25g)(4.18J/g°C)(Tf-0°C)

Tf = 46°C

Liquids and Solids Chapter 16 E-mail: [email protected]@gmail.com Web-site:

Documents

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