Logarithm Review ab = c, (a > 0, a ≠ 1) logac = b log c = log10c Definition ab = c, (a > 0, a ≠ 1) logac = b If a = 10, it is called common logarithm log c = log10c If a = e = 2.718281828459045 ∙ ∙ ∙, it is called natural logarithm ln c = logec Keys on your calculator

x > 0, y > 0 ln(xy) = ln(x) + ln(y) ln(x/y) = ln(x) − ln(y) Properties of Logarithm x > 0, y > 0 ln(xy) = ln(x) + ln(y) ln(x/y) = ln(x) − ln(y) ln(xm) = m ln(x) Also see Appendix I B

Intermolecular Forces Chapter 11 Liquids, Solids and Intermolecular Forces continued

Chemistry, continue on Vapor Pressure

Surface Molecules Figure: 11-20

Figure: 11-24a,b Temperature: T Temperature: T

Figure: 11-21

GA, 760 torr = 1 atm Normal Boiling Point H2O 100 °C

Tibet, 480 torr < 1 atm Normal Boiling Point H2O 85 °C

Figure: 11-21

(a) The Vapor Pressure of Water, Ethanol, and Diethyl Ether as a Function of Temperature. (b) Plots of In(Pvap) versus 1/T for Water, Ethanol, and Diethyl Ether T is in K! 1/T (K−1)

Linear relation: y = kx + C slope: k = tg θ θ x C: intercept

Linear relation: y = kx + C 1/T (K−1) Linear relation: y = kx + C ln P = k(1/T) + C

Figure: 11-17

Heat of vaporization ∆Hvap: energy needed to convert one mole of liquid to gas. Unit: J/mol or kJ/mol. ∆Hvap > 0

slope k < 0 y x

ln (P) 1 ln (P1) 2 ln (P2) 1/T1 1/T2 1/T (K−1)

Clausius-Clapeyron Equation

The vapor pressure of water at 25 °C is 23.8 torr, and the heat of vaporization of water is 43.9 kJ/mol. Calculate the vapor pressure of water at 50 °C. Five: T1, T2, P1, P2, ∆Hvap Four known, calculate the other.

PV = nRT Units in ideal gas law P — atm, V — L, n — mol, T — K Option 1 Chem 1211 R = 0.082 atm · L · mol−1 · K−1 P — Pa, V — m3, n — mol, T — K Option 2 Clausius-Clapeyron equation R = 8.314 J · mol−1 · K−1

Carbon tetrachloride, CCl4, has a vapor pressure of 213 torr at 40 °C and 836 torr at 80 °C. What is the normal boiling point of CCl4? ( Please try to work on this question by yourself. Will review next week)

Liquid potassium has a vapor pressure of 10.00 torr at 443 °C and a vapor pressure of 400.0 torr at 708 °C. Use these data to calculate The heat of vaporization of liquid potassium; The normal boiling point of potassium; The vapor pressure of liquid potassium at 100. °C. ( Please try to work on this question by yourself. Will review next week)

Clausius-Clapeyron Equation

slope k < 0 y x

Linear relation: y = kx + C slope: k = tg θ θ x C: intercept

a > b > 0 > c > d y a d c b x Lines tilt to the right have positive slopes (a and b), left negative (c and d). Steeper line has greater absolute value of slope. In this graph, the order of slopes is a > b > 0 > c > d

What is the order of heat of vaporization for these three substances?

Solids

Glass (SiO2)

Crystal Solid Noncrystal

Basis Crystal structure

The basis may be a single atom or molecule, or a small group of atoms, molecules, or ions. NaCl: 1 Na+ ion and 1 Cl− ion Cu: 1 Cu atom Zn: 2 Zn atoms Diamond: 2 C atoms CO2: 4 CO2 molecules

Use a point to represent the basis: =

Lattice Lattice point:

Unit cell: 2-D, at least a parallelogram Unit cell is the building block of the crystal

How many kinds of 2-D unit cells can we have?

                                  

Extend the concept of unit cell to 3-D, the real crystals.

: 3-D, at least a parallelepiped Figure: 11-32 : 3-D, at least a parallelepiped

How many kinds of 3-D unit cells can we have?

5. rhombohedral (trigonal) 4. tetragonal 1. triclinic 2. monoclinic c c a ≠ b ≠ c α ≠ β ≠ γ a a ≠ b ≠ c a b b The 14 Bravais lattices 7 crystal systems 3. orthorhombic α = β = γ = 90° 5. rhombohedral (trigonal) 4. tetragonal a = b = c 90° ≠ α = β = γ < 120° a = b ≠ c α = β = γ = 90° 7. cubic 6. hexagonal a = b = c α = β = γ = 90° a = b ≠ c α = β = 90° ,γ = 120° γ

Chem 1212: assume a lattice point is a single atom Figure: 11-33 (Simple cubic) Chem 1212: assume a lattice point is a single atom

Information of a cubic unit cell • Size of the cell X-ray diffraction

The Wave Nature of Light

Figure: 11-38C

Information of a cubic unit cell • Size of the cell X-ray diffraction • Size of the atoms Soon • Number of atoms in a cell Now!

B C D A

B C D E F A Number of atoms in a unit cell = ¼ x 4 = 1

Number of Atoms in a Cubic Unit Cell 1 2 4

The body-centered cubic unit cell of a particular crystalline form of iron is 0.28664 nm on each side. Calculate the density (in g/cm3) of this form of iron. d = 7.8753 g/cm3 The body-centered cubic unit cell of a particular crystalline form of an element is 0.28664 nm on each side. The density of this element is 7.8753 g/cm3. Identify the element.

The face-centered cubic unit cell of a particular crystalline form of platinum is 393 pm on each side. Calculate the density (in g/cm3) of this form of platinum. d = 21.4 g/cm3

Closest Packing

a b c

· · · abab · · ·

· · · abab · · · Chapter 11, Figure 11.48 Hexagonal Closest-Packing Crystal Structure

5. rhombohedral (trigonal) 4. tetragonal 1. triclinic 2. monoclinic a ≠ b ≠ c a ≠ b ≠ c α ≠ β ≠ γ The 14 Bravais lattices 7 crystal systems 3. orthorhombic α = β = γ = 90° 5. rhombohedral (trigonal) 4. tetragonal a = b = c 90° ≠ α = β = γ < 120° a = b ≠ c α = β = γ = 90° 7. cubic 6. hexagonal a = b = c α = β = γ = 90° a = b ≠ c α = β = 90° ,γ = 120° γ

· · · abcabc · · ·

abcabc = Cubic Closest Packing e.g. Ag, 1 atoms (1 lattice point) in a unit cell

Information of a unit cell • Size of the cell X-ray diffraction • Size of the atoms Soon Now! • Number of atoms in a cell Now!

L 2r L r Al crystallizes with a face-centered cubic unit cell. Example 11.7, page 494 Al crystallizes with a face-centered cubic unit cell. The radius of a Al atom is 143 pm. Calculate the density of solid Al in g/cm3. d = 2.71 g/cm3 L r 2r L

What about simple cubic?

Simple Cubic r L L = 2r

What about body-centered cubic?

Body centered cubic

Body diagonal D = 4r L D

L F L D L Body diagonal D = 4r Pythagorean theorem 

Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm3. Calculate the edge length of the unit cell and a value for the atomic radius of titanium in pm. L = 328 pm r = 142 pm

100-mL container 50 mL 70 mL 50 % 70 %

1 2 4 prove 52 % 68 % L = 2r 74 % Packing Efficiency: fraction of volume occupied by atoms

Chapter 11, Figure 11.44 The Cubic Crystalline Lattices