intramolecular forces
DESCRIPTION
Intramolecular Forces. Forces (chemical bonds) within a molecule Typical value: 350 kJ/mol C-C bond. Intermolecular Forces. Forces between molecules Typical value: 20 kJ/mol H-bond 1 kJ/mol van der Waals. Dipole-dipole Forces If present usually dominate intermolecular - PowerPoint PPT PresentationTRANSCRIPT
Intramolecular Forces
Forces (chemical bonds) within a molecule
Typical value: 350 kJ/mol C-C bond
Intermolecular Forces
Forces between molecules
Typical value: 20 kJ/mol H-bond1 kJ/mol van der Waals
Dipole-dipoleForces
If present usuallydominate intermolecularinteractions.
Water Held together by O-H•••H hydrogen bonds
Hydrogen bond donors: O-H, N-H, S-H, X-H
Hydrogen bond acceptors: O, N, X, S
N
O
H
N
O
H
For example:
The amide hyrdrogenbond is the dominateintermolecular forcein proteins.
London Dispersion Forcesor
van der Waals Forces
Gvap = Hvap - T Svap
What determines the B.P. of a liquid?
= 0
at the B.P.
Hvap = T Svap
For simple liquids, all svap values are about
the same. So B.P. is proportional to Hvap
Compound Mass g/mol Hvap kJ/mol B.P.ºC
HF 20 25.2 19.7HCl 36.5 16.2 -85HBr 80.9 19.3 -66HI 128 19.8 -34.6NH3 17 23.3 -33.3H2O 18 40.7 100.0
Methane 16 8.2 -162Ethane 30 15 -89Propane 44 19 -42Butane 58 22 -0.5He 4 .08 -269Ne 20 1.7 -246Ar 40 6.4 -186Xe 131 12.6 -108H2 2 0.9 -253N2 28 5.6 -196O2 32 6.8 -183F2 38 6.6 -188Cl2 71 20.4 -34Br2 160 30.0 58.8
Four General Classifications for Solids
1. Metals Fe, Co, Ag
2. Covalent Network Solids Diamond, SiO2
3. Ionic Solids NaCl ZnS
4. Molecular Solids I2 Sugar
d sin d sin
n = 2 d sin Bragg’sLaw
Determination of Crystal Structuresusing X-Ray diffraction.
Diffraction of any wave will take placewhen you have a grid with a spacing similar to the wavelength of the wave.
X-Rays have wavelengths on the orderof 1 Ångstrom. Typical value 0.71Å
Four General Classifications for Solids
1. Metals Fe, Co, Ag
2. Covalent Network Solids Diamond, SiO2
3. Ionic Solids NaCl ZnS
4. Molecular Solids I2 Sugar
Close Packing of Spheres
Hexagonal Close Packed
Co, Ti, Mg
Body Centered Cubic
Fe, Cr, V
Cubic Close Packed - Face Centered Cubic
Ni, Cu, Au
Efficiency of Close PackingWhat fraction of the volume is occupied?
fv = volume spheres in unit cell / volume of cell
e = r 8
e3 = ( r 8 )3
Cube Volume
Sphere Volume 4 x 4/3r3
fv = ( r 8 )3 4 x 4/3r3 = 0.740
The radius of Ag atom is 1.44 Å. Calculate the density. The Ag structure is fcc (ccp).
e = r 8
e2 + e2 = (4r)2
e2 = 8r2
e = 1.44 8 = 4.07
e3 = 67.4 A3o
The radius of Ag atom is 1.44 Å. Calculate the density. The Ag structure is fcc (ccp).
oo 67.4 A3 x ( 10-8 cm/A)3 = 6.74 x 10-23 cm3
=
Density = 10.6 g/cm3
6.74 x 10-23 cm
4 atoms x 1 mol / 6.022 x 1023 atoms x 107.9g / mol
Close packing ofspheres fills 0.74 %of the available space
The remaining spacecan be allotted to threetypes of holes that occurbetween the spheres.
r = 2 R - R = 1.414R - R = .414
r = 3/2 R - R = 1.225 R - R = .225
rtet = 0.225roct = 0.414rcub = 0.732
Sodium Chloride
Na+ .95 Å
Cl- 1.81Å Table 13.7
Ratio = .95 / 1.81 = .52
rtet = 0.225roct = 0.414rcub = 0.732
Zinc Sulfide
Ratio = Zn+2/S-2 = .35
rtet = 0.225roct = 0.414rcub = 0.732
The Zinc atoms occupy 1/2 of thetetrahedral holes
Glass
Quartz
Molecular Solids C60
Iodine I2
Acetylene HCCH
O
O H
O
O
H O
O H
O
O
H O
O H
O
O
H
Oxalic Acid
O
CH3N
H
G = H - TS = - RT ln K
ln K = -(H/R) /T + S/R
Liquid Gas Kp = P
ln P = -(H/R)(1/T) + S/R
P
Triple point 4.588 torr 0.0098°C
