liquids and solids

Liquids and SolidsChapter 16E-mail: [email protected]: http://clas.sa.ucsb.edu/staff/terri/

mailto:[email protected]

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. KOHb. HF i. BH3

c. NH4Br j. H2COd. F2 k. SiO2

e. CH3OH l. Arf. 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 HClc. highest boiling point HOCH2CH2OH or CH3CH2OHd. 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 H2Oh. highest surface tension C14H30 or C24H50

Liquids and Solids – ch. 16Relative strength of forces ⇒ Ionic > >> Hydrogen bonding > Dipole-Dipole > Dispersion

for 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).


Unit Cell Image Atoms/unit cell (% by volume)

Coordination Number

Edge Length vs. 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 2 (74%) 12 N/A

Liquids and Solids – ch. 162 Types of Closest Packed Solids

1. Hexagonal closest packed

2. Cubic closest packed


Counting Atoms in a Cubic Unit Cell

Atom is shared between 4 unit cells



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)

Liquids and Solids – ch. 166. 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

Liquids and Solids – ch. 167. Assume the following is a two-dimensional structure of an ionic compound MxAy.

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)

Liquids and Solids – ch. 16Ionic Solids

Anions (or the larger ion) will align themselves in the lattice points of the unit cell

Cations (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 tinb. Antimony doped with germaniumc. 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

Liquids and Solids – ch. 16M.O. model

illustrating dopingand how it enhances

conductivity

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 Key2. 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

Liquids and Solids – Answer Key4. 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) = 4.34x10-22gThe 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 OX X X X X O O O OX X X X X

Liquids and Solids – Answer Key

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⇒

Liquids and Solids – Answer Key7. 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 anionsWhat 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

Liquids and Solids – Answer Key8. 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 holesZnS => Zn2+ will fill all of the ½ of the tetrahedral holesCaCl2 => 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↓

Liquids and Solids – Answer Key11. Drawing the points of the phase diagram you get…

a. Tb. Tc. F

note: (+) slope

760torr

72torr

18°C

21°C

87°C

Liquids and Solids – Answer Key12. 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)

Liquids and Solids – Answer Keyq1=(10g)(2.03 J/g°C)(+31°C) = 629.3J or 0.629kJq2=(10g/18g/mol)(6.01kJ/mol) = 3.34kJq3=(10g)(4.18 J/g°C)(+100°C) = 4180J or 4.18kJq4=(10g/18g/mol)(40.7 kJ/mol) = 22.6kJq5=(10g)(2.02 J/g°C)(+55°C) = 1.11kJqtotal = 31.86kJ14. 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

Liquids and Solids – Answer Key-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

Documents