LBS 172

Monday Jan 8th, 2006

Gases

  • Hardest state of matter  to see
  • Easiest state to understand at a molecular level
  • Close connection between microscopic behavior and macroscopic properties

Molecular Level / “Real World” Level
*Gas molecules are not in contact (liquid or solid molecules are in contact) / * Gases have VERY low density
* g/ml or g/cm3 (Dgas < Dl or Ds)
* Density of gas in g/L
*Gas molecules move quickly & randomly / * Gases can be used to do WORK
*Gas molecules collide with the walls of their container / *Gases exert pressure
*Gas molecules can collide with each other & these collisions are elastic
* (Total Momentum transfer)
* Gas molecule = “pool ball” not “silly putty” / * simple relationships between pressure (P), volume (V), temperature (T), amount of gas (m)
* Gases mix evenly & completely, very rapidly

Pressure

  • Pressure = force per unit area N/m2 = Pa (Pascal)
  • Pascal is too small of a unit so kPa are used
  • kPa = 1000 Pa (kPa = kilo Pascal)
  • 1 atm = 101.5 kPa
  • Atmospheric Pressure at 0˚ C, sea level
  • US Units (lame) – lbs/in2 = psi
  • 1 atm = 15 psi

Barometer

  • Barometer invented by Torricelli
  • The pressure goes down causing the mercury in the barometer to go up
  • Height of the column (Hg) at atmospheric pressure (0˚ C, sea level) = 760 mm Hg
  • 1 atm = 760 mm Hg = 760 torr

Relationship between Pressure and Volume

  • Demonstration = squeeze & burst a balloon
  • As volume decreases (compression) then pressure increases
  • It becomes harder to compress due to pressure increasing
  • Eventually the balloon burst due to overpressure
  • Volume is proportional 1/P – inverse relationship
  • V = k/P therefore PV=K where K = constant for any gas at a specific T
  • PiVi = k = PfVf therefore PiVi = PfVfBOYLES LAW

Relationship between Volume and Temperature

  • Demonstration = a balloon in liquid N2
  • As temperature decreases, volume decrease
  • V is proportional to T – direct relationship
  • V = kT Therefore V/T = k
  • Vi/Ti = k = Vf/Tf
  • CHARLES LAW  Vi/Ti = Vf/Tf

Relationship between Volume and n where n = number of moles of gas

  • Demonstration = inflate a balloon
  • As n increases, volume increases
  • V is proportional to n – direct relationship
  • V = kn therefore V/n = k and Vi/ni = k = Vf/nf
  • Constant temperature and pressure
  • Vi/ni = Vi/ni

Ideal Gas Law

  • Volume is proportional to 1/P ; Volume is proportional to Temperature ; Volume is proportional to n
  • Volume is proportional to (nT)/P
  • Therefore V = (RnT)/P where R = ideal gas constant
  • IDEAL GAS LAW PV = nRT
  • Ideal = elastic collisions / non-sticky molecules
  • R = 0.08206 (L∙atm)/(mol∙K)
  • Must use Kelvin
  • K = ˚C + 273
  • P x V  N/m2 x m3 = N x m = WORK!
  • Measured as energy
  • Volume accepted by 1 mol of gas at 0˚C, 1 atm(standard temperature & pressure)
  • PV = nRT
  • V = nRT/P = [(1 mol)(.08206 (L∙atm)/(mol∙K))(273 K)] / 1 atm = 22.4 L

Wednesday January 10th, 2007

Example 1

  • 16.2 grams of dry ice is placed in a 3.00 L glass flask which is then stoppered at 25˚C. If the pressure inside the flask exceeds 1300 torr, the glass will explode. BOOM or NOT??
  • R = .08206 (L∙atm)/(mol∙K) M(CO2) = 44.01 g/mol 25 + 73 = 298L
  • 16.2 g CO2 x (1 mol CO2/44.01 g CO2) = .36809 mol CO2
  • P = [(.36809 mol CO2)(.08206 (L∙atm)/(mol∙K))(298K)] / 3 L
  • P = 3 atm x 760 torr / 1 atm = 2280.362 ≈ 2300 torr
  • 2300 > 1300 so….
  • The glass will explode!!!

Example 2

  • 4C3H5O9N3 + heat  12CO2 (g) + 10H2O (g) + 6H2 (g) + O2 (g)
  • 100 grams of nitroglycerine is detonated at 425˚C and at atmospheric pressure. What is the volume of gas produced?
  • M(nitroglycerin) = 227.09 g/mol
  • 425 + 273 = 698 K
  • Road map
  • Grams of nitro  moles of nitro mol of gas  volume of gas
  • 100 g NG x 1 mol / 227.09 gram = .44035 mol nitroglycerine
  • 12 moles CO2 + 10 moles H2O + 6 moles H2+ 1 mole O2 = 29 moles of gas
  • .44035 mol nitro x 29 mol gas / 4 mol nitro = 3.193 mol gas
  • PV = nRT  V = (nRT)/P
  • V = [(3.193 mol)(.08206 (L∙atm)/(mol∙K))(698 K)] / 1 atm
  • The volume of the gas produced is 183 liters.

H2 = Fuel of the Future

  • Problem with safe onboard storage
  • Many materials can chemically absorb large quantities of H2

Example 3

  • Pd (s) + H2 (g)  PdH2 (g)
  • A 2.5 L vessel contains H2 (g) at 582 torr and 35˚C. If 4.60 grams of Pd solid is added to the flask. What is the pressure of the H2 gas remaining in torr?
  • M(Pd) = 106.42 g/mol
  • 582 torr x 1 atm / 760 torr = .7657 atm
  • 35 + 273 = 308 K
  • PV = nRT  n = PV/RT
  • n = [(.7657 atm)(2.5 L)] / [(.08206 (L∙atm)/(mol∙K))(308 K)] = .07575 mol H2
  • 4.60 g Pd x 1 mol / 106.42 g = .04322 mol Pd
  • .04322 mol Pd x 1 mol H2 / 1 mol Pd = .04322 mol H2
  • .07575 mol H2 initial - .04322 mol H final = .03253 mol H2 remaining
  • P = (nRT) / V = [(.03253 mol)(.08206 (L∙atm)/(mol∙K))(308 K)] / 2.5 L
  • P = .32887 atm
  • .32887 atm x 760 torr / 1 atm = 249.9 torr ≈ 250 torr
  • The pressure of the H2 gas remaining is 250 torr.

Friday January 12th, 2007

Determination of Molar Mass of a gas

  • PV = nRT
  • P/RT = n/V
  • n = mass/M where M= molar mass
  • P/RT = (mass/M)/V
  • P/RT = mass / (M x V)
  • P/RT = D/M
  • M = (DRT)/P

Example 1

  • The density of air at 20˚C & 1 atm is 1.275 g/L. Assume that air is only N2, O2. Find the percent of O2 in the air according to this data. M(NO2) = 28 g/mol; M(O2) = 32 g/mol. R = .08206 (L∙atm)/(mol∙K)
  • M = DRT / P
  • M = [(1.275 g/L)(.08206 (L∙atm)/(mol∙K))(293 K)] / 1 atm
  • M = 30.6556 g/mol
  • Let x = fractional percentage of O2
  • Let 1 – x = fractional percentage of N2
  • x M(O2) + (1 – x)M(N2) = 30.6556
  • x(32) + (1 – x)(28) = 30.6556
  • 32x + 28 – 28x = 30.6556
  • 28 + 4x = 30.6556
  • x = 2.6556 / 4 = .6639
  • There is about 66.4% of O2

Dalton’s Law of Partial Pressure

  • The total pressure of a gas mixture is equal to the partial pressures of the individual components
  • Gas A and gas B
  • They mix evenly and completely
  • ntotal = nA + nB
  • PV = nRT
  • PTOTV / RT = (PAV / RT) + (PBV / RT)
  • V & T are the same for both gas A and gas B because they are in the same container and has mixed completely
  • PTOT(V/RT) = PA + PB (V/RT)
  • PTOT = PA + PB
  • Expression for one of the components
  • PA / PTOT = (nART / V) / (nTOTRT / V) = nA / nTOT
  • PA / PTOT = nA / nTOT
  • nA / nTOT = XA where is XA the mole fraction of A (0 ≤ x ≤ 1)
  • PA = XAPTOT

Example 2

  • Halothane
  • M = 197.38 g/mol
  • A 500 ml gas vessel is filled w/pure O2 at 25˚C and 450 torr. If 750 mg of halothane is admitted to the gas vessel, what is the total pressure inside the flask? Find out the mole fraction of halothane.
  • 450 torr = PO2
  • 25 + 273 = 298 K
  • 750 mg halothane x 1 g / 1000 mg x 1 mol / 197.38 g = .00388 mol halo
  • Phalo = nhaloRT / V
  • Phalo = [(.00388 mol)(.08206 (L∙atm)/(mol∙K))(298 K)] / .5 L
  • Phalo =.1898 atm x 760 torr / 1 atm = 144.22 torr
  • PTOT = PO2 + Phalo
  • PTOT = 144.22 + 450 = 594.22 torr ≈ 594 torr
  • PA = XAPTOT
  • PA / PTOT = XA
  • XA = 144.22 / 594.22 = .243

Diffusion & Effusion of gases

  • Diffusion = gradual mixing of gases when confined to the same container
  • Effusion = gas release through a small hole
  • The same law governs both diffusion and effusion
  • Gas molecules move randomly due to constant molecular collisions
  • Via P-chem
  • PV = 1/3 ∙ n ∙M∙ (ū2)
  • where ū = average speed of a gas molecule in a sample
  • PV = energy
  • n∙ M = grams  moles x g/mol = grams
  • similar to kinetic energy equation
  • E = ½ mV2

Wednesday January 17th, 2007

Diffusion / Effusion

  • PV = 1/3 ∙ n ∙ M ∙ (ū2)
  • nRT = 1/3 ∙ n ∙ M ∙ (ū2)
  • where R = 8.314 J/(K∙mol)
  • Heavier gas molecules more slowly
  • As molar mass increases, average speed of the gas molecules in the sample decreases
  • M ↑, ū↓
  • Higher temperature = faster molecular motion
  • Gas molecules speeds
  • H2 ū = 4300 mph
  • CH4 ū = 1520 mph
  • CO2 ū =920 mph
  • Gas molecules as a sample have a distribution & speeds

Example 1

  • NH3 (g) + HCl (g)  NH4Cl (s)
  • M = 17 g/mol M = 36 g/mol

  • You’ll see the white smoke more towards the HCl bottle because the NH3 molecules move faster.

Naturally Occurring U

  • 235U and 238U
  • 235U = fissionable <18
  • Good for A-bombs and nuclear power
  • Large reaction
  • Take U and hit it with excess fluorine gas, you make UF6
  • U + F6 UF6 (g) + H2O  UO2 + 6HF
  • 235UF6 and 238UF6
  • Spin in a centrifuge and the faster gas will move to the end of the tube
  • Difference in M is very small
  • Hundreds of thousands of centrifuge to enrich 235U content

Thermodynamics of gases

  • Lower density of gases = compressible / expandable
  • Can easily change P by changing V
  • ∆H = change in enthalpy at constant P for gases
  • ∆E = ∆H
  • Total energy change = heat change
  • ∆E = ∆H + W where W>0 if compression & W<0 if expansion
  • W = - P∆V
  • Therefore ∆E = ∆H - P∆V

Ideal gas  assume all collisions are elastic “non sticky”

  • 8.00 mol gas, 27˚C, 4.00 L  P = 49.2 atm by ideal gas
  • In reality
  • 8.00 mol Argon, 27˚C, 4.00 L  Pactual = 47.3 atm
  • Even argon molecules are sticking with each other
  • 8.00 mol Cl2, 27˚C, 4.00 L  Pactual = 9.5 atm HUGE DIFFERENCE
  • Ideal gas law breaks down at higher P, lower T, higher molar mass
  • In reality, gas molecules are sticky
  • Attractive forces between neighboring molecules = “intermolecular forces”

Friday January 19th, 2007

Intermolecular forces

  • Attractive forces between neighboring molecules
  • NOT covalent bonds
  • ∆H = + 930 kJ/mol
  • Takes a large amount of energy to break chemical bonds
  • H2O  H2O (l) OR H2O  H2O (g)
  • Disrupting IMF’s∆H = 41 kJ/mol
  • Solid  liquid has ∆H = 6 kJ/mol
  • ∆Hf < ∆Hvap <∆Hbond breaking
  • Intermolecular forces are very weak
  • Despite being weak, they are critical to properties of nature
  • IMFS
  • Can determine physical state for a substance (s, l, g)
  • Solids have stronger IMFs and gases have weaker IMFs
  • Huge effect on solubility (like dissolves like)
  • Huge effect on chemical reactivity
  • Especially biochemistry
  • Clue to melting point & boiling point

5 types of IMFs

  • Ion – Ion
  • Ion – Dipole
  • Dipole – Dipole
  • London Dispersion Forces (LDFs)
  • Hydrogen bonding

Ion – Ion Force

  • Attractive force between oppositely charged ions in an ionic solid
  • Get an ionic solid when there’s a metal and a nonmetal in the same compound and the ∆x≥ 1.5
  • Strongest type of IMF
  • Example: NaCl = strong because it’s a solid
  • It’s hard to melt, brittle, can’t bear impact very well, & its hard
  • How can NaCl dissolve in water
  • Because of ion – dipole force

Ion – Dipole Force

  • Attractive force between an ion and a polar molecule
  • See the example of water to the right
  • bond geometry must give asymmetric configuration
  • ∆x ≥ 0.5
  • Nonmetal only
  • When NaCl is dissolving in H2O the negative oxygen end is attracted to the Na+ and the Cl- is attracted to the positive hydrogen
  • Ion – dipole forces are responsible for solubility in H2O of ionic compounds
  • Typically won’t encounter this when dealing with pure compounds

Dipole – Dipole Force

  • An attraction between neighboring polar molecules
  • Huge for molecular recognition in biochemical systems and in organic chemistry
  • NH3 in H2O
  • Water has a pyramidal shape & it’s polar. NH3 has a bent shape & it’s polar.
  • H2O and NH3 bonds like this
  • Solids held together by dipole – dipole forces
  • i.e. sugar (C12H22O11) = when sugar is melted no bonds are broken; low melting point; soft/waxy solids
  • How does nonpolar O2 dissolve in polar H2O to help fish to breathe?
  • Because of London Dispersion forces

London Dispersion Forces

  • Attractive forces caused by temporary dipoles
  • O=O ∆x=0 non polar

Monday January 22nd, 2007

London Dispersion Forces

  • Temporary dipole due to circulation of the electrons throughout the molecule
  • This causes a weak attraction between neighboring molecules
  • LDFs = weakest type of IMF
  • LDF are stronger if M is higher
  • More electrons, more likely to create temporary dipoles
  • More space to “spread out”
  • LDFs increase if the molecule is more polarizable (“cushier electron clouds”)

Cl2 / Br2 / I2
At Room Temp. / Gas / Liquid / Solid
  • The elements in the table above have LDFs only (∆x=0, nonpolar)
  • Molecules with greater surface have stronger LDF
  • More area for intermolecular contact
  • Pentane boiling pt = 38˚C
  • 2-methyl butane boiling pt = 32˚C

Compound / Boiling Point
H2Te / -2˚C
H2Se / -42˚C
H2S / -61˚C
H20 / +100˚C WTF; why is it so high?

Hydrogen bonding

  • Why is H2O so special?
  • Hydrogen “bonding” = special type of dipole – dipole interaction between an H atom in an A-H bond (where A = N, O, or F) and an A atom (N, O, F) in a neighboring molecule
  • H2O is a liquid at room T due to H-bonding
  • Solid ice = hydrogen bonded network of stacked H2O molecules

Strongest of the IMFs

  • Ion – ion > ion – dipole > H-bonding > dipole – dipole > LDFs
  • Use this ranking first! THEN break ties with molar mass THEN with molecular shape

Wednesday January 24th, 2007

Intermolecular forces

  • Surface molecules will pack more closely together to compensate for reduced IMFs
  • Liquid surfaces froma “tougher” skin
  • Use soap to disrupt IMFs  lowers surface tension
  • Body oils = long chain hydrocarbons and are nonpolar
  • “Lipophilic” = attracted to nonpolar substances
  • Soap  lipophilic & hydrophilic

Nonpolar end | polar end

  • Another implication of weak IMFs at surface of liquids
  • Molecules at the surface are more prone to leave the liquid and enter the gas phase
  • Vapor pressure = P of gas molecules above the surface of a liquid
  • Boiling point = T at which the vapor pressure of a liquid equals external atmospheric pressure
  • Lower external pressure  lower boiling point
  • Higher external pressure  higher boiling point
  • Relation between P,T, physical state for any substance can be represented visually by a phase diagram

  • As pressure decreases, boiling point decreases
  • As pressure increases, boiling point increases
  • As pressure decreases, melting point increases
  • As pressure increases, melting point decreases
  • Freeze – drying
  • Brew coffee
  • Freeze its solid
  • Pull a vacuum at low T
  • Ice sublimes to vapor phase
  • C = critical point  temperature above which a gas cannot be made to liquefy
  • Molecules are moving too fast to be forced into contact
  • AKA Tc for critical temperature
  • Critical Pressure (Pc) = pressure needed to liquefy gas at Tc
  • Above Tc, Pc = super critical fluid
  • High density fluid
  • Supercritical H2O = flammable
  • Phase diagram for CO2

Friday January 26th, 2007

Brief intro to solid state chem

  • Why?
  • A structural materials
  • Conductivity / insulation (heat & electricity)
  • Semiconductors / superconductors
  • Atomic composition & arrangement at the micro level has a HUGE effect on physical properties

Types of Solids

  • Ionic
  • Covalent
  • Molecular
  • Metallic

Ionic Solids

  • Packed arrangement of cations and anions in contact
  • Held together with ion – ion forces
  • High melting point
  • Very brittle
  • Insulators in the solid state but will conduct electricity when melted

Covalent solids

  • Held together into an extended 3-D network with covalent bonds
  • Example 1: Diamonds (carbon), quartz (SO2), and drill bits (WC)
  • Incredibly high melting points
  • Very hard
  • Covalent bonds stronger than ion – ion

Molecular Solids

  • Held together by IMF of different types
  • Separate molecules interacting through IMF
  • Very low melting point
  • Useless instructural materials
  • Sugar, any organic solid

Metallic Solid

  • Held together by “metallic bonds” microscopic view of a metal
  • Light reflection off of free electrons allows mirrors to be made
  • Heat conduction also occurs in metals
  • Superconductors = special type of metal that can conduct electricity
  • Without resistance – repels all magnetic fields

Two Classes of Solid

  • Crystalline
  • Amorphous

Amorphous

  • Lack a well defined long range atomic order
  • e.g. glass – randomly oriented SiO2 chains

Crystalline

  • have a regular arrangement of atoms with a rigid long range order repeating unit called the “unit cell”
  • unit cells stack in 3 to 4 to give a crystal
  • ≈ 0.7Ǻ to ≈ 0.3Ǻ
  • Ǻ = 1 x 10-10 m
  • Distance is comparable to wavelength of x-rays
  • Crystals will diffract x-rays
  • From the pattern of the diffracted x-rays
  • Get size and shape of the unit cell
  • Get positions of atoms within the unit cell
  • Identity of atom on those positions
  • x-ray diffraction
  • incredibly powerful
  • Cubic unit cells (simplistic)
  • Simple cubit = atoms at corner of a cube
  • Simple cubic structure is rare (e.g. Hg solid)
  • Why?Lots of empty space in structure
  • Only 52.3% space is occupied
  • Body centered cubic (bcc)
  • Atoms at corners AND at center of cell
  • Bcc metals – alkali metals, Fe
  • More common because less empty space
  • 68% of space is occupied
  • Face centered cube
  • Atoms on corners of cube and on all faces of the cube
  • Very common
  • Most efficient packing possible for a single type of sphere
  • 74% of space is occupied
  • Aluminum, gold, copper, nickel, platinum, etc. have this