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Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure.

Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure. Key aspects: -critical point -normal boiling point -triple point. SPECIFIC HEAT re-visited. The quantity of heat required to raise the temperature of one

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Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure.

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  1. Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure. Key aspects: -critical point -normal boiling point -triple point

  2. SPECIFIC HEAT re-visited The quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin) q = s x m xT ENTHALPY OF A PHASE CHANGE The heat energy required to undergo a change in phase occurs at constant temperature and is associated with the average change in distance between molecules. For water: Hº fus = 335 J/g or 6.02 kJ/mol Hºvap = 2260 J/g or 40.7 kJ/mol

  3. HEATING - COOLING CURVE Calculate the amount of energy required to convert 50.0 g of ice at 0.0 ºC to steam at 100.0ºC Hvap g 100 - T (oC) 0 - l Hfus s Energy (J) qtotal = q(s) + DHfus + q(l) + DHvap + q(g)

  4. Water and the Changes of State Q. How many kilojoules of energy are needed to change 15.0 g of ice at -5.00oC to steam at 125.0 oC? The first step is to design a pathway: q1 = msDT for ice from -5.0 to 0.0 oC, the specific heat of ice is 4.213 J/g oC q2 = DHfus for ice to liquid at 0.0oC q3 = msDT for liquid 0.0oC to 100.0 oC q4 = DHvap for liquid to steam at 100.0oC q5 = msDT for steam 100.0 to 125.0 oC; the specific heat of steam is 1.900 J/g oC so qT = q1 + q2 + q3 + q4 + q5 The next step is to calculate each q: q1= (15.0 g) (4.213 J/g oC) (0.0 - (-5.0) oC) = 316 J q2 = (335 J / g) (15.0 g) = 5025 J q3= (15.0 g) (4.184 J/g oC) (100.0 - (0.0) oC) = 6276 J q4 = (2260 J / g) (15.0 g) = 33900 J q5= (15.0 g) (1.900 J/g oC) (110 - 100 oC) = 285 J qT = 316 J + 5025 J + 6276 J + 33900 J + 285 J = 45.8 kJ

  5. INTERMOLECULAR FORCES INTRAMOLECULAR > INTERMOLECULAR (covalent, ionic) (van der Waals, etc) “ between atoms” “between molecules” TYPES Neutral Molecules: 1. Dipole-dipole forces 2. London Dispersion 3. Hydrogen bonding Ions: 1. Ion-dipole force

  6. INTERMOLECULAR FORCES STRENGTH: BOILING POINTS AND MELTING POINTS ARE DEPENDENT ON STRENGTH OF INTERMOLECULAR FORCES. STRONG FORCE  HIGH BOILING POINT Type of interaction Approximate Energy (kJ/mol) Intermolecular van der Waals 0.1 to 10 (London, dipole-dipole) Hydrogen bonding 10 to 40 Chemical bonding Ionic 100 to 1000 Covalent 100 to 1000

  7. ION-DIPLE FORCES - between ions and polar molecules - strength is dependent on charge of the ions or polarity of the bonds - usually involved with salts & H20 DIPOLE-DIPOLE FORCES - between neutral polar molecules - weaker force than ion-dipole - positive dipole attracted to negative dipole - molecules should be relatively close together - strength is dependent on polarity of bonds. LONDON DISPERSION FORCES - all molecules and compounds - involves instantaneous dipoles - strength is dependent on Molar Mass (size) - contributes more than dipole-dipole - shape contributes to strength

  8. HYDROGEN BONDING AN INTERMOLECULAR ATTRACTION THAT EXISTS BETWEEN A HYDROGEN ATOM IN A POLAR BOND AND AN UNSHARED ELECTRON PAIR ON A NEARBY ELECTRONEGATIVE SPECIES, USUALLY O, F, and N NOTE: (A special type of dipole-dipole interaction) - stronger than dipole-dipole and London dispersion forces - Accounts for water’s unusual properties - high boiling point - solid Less dense than liquid - universal solvent - high heat capacity

  9. FLOWCHART OF INTERMOLECULAR FORCES Interacting molecules or ions Are polar Are ions Are polar molecules involved? molecules involved? and ionsboth present? Are hydrogen atoms bonded to N, O, or F atoms? London forcesDipole-dipolehydrogen bondingIon-dipoleIonic only (induced forcesforcesBonding dipoles) Examples: Examples: Examples Example: Examples: Ar(l), I2(s) H2S, CH3Cl liquid and solid KBr in NaCl, H2O, NH3, HF H2O NH4NO3 NO NO YES NO YES Yes YES NO Van der Waals forces

  10. PROPERTIES OF LIQUIDS VISCOSITY - The resistance of a liquid to flow - Depends on attractive forces between molecules and structural features which cause greater interaction (entanglement). SURFACE TENSION - The energy required to increase the surface area of a liquid by a unit amount (E/A) - Due to interactions between molecules and the lack of interaction if there are no molecules to interact with.

  11. VAPOR PRESSURE The pressure exerted by a vapor in equilibrium with its liquid or solid state. 1. Vapor pressure changes with intermolecular forces and temperature 2. Vapor pressure involves a dynamic equilibrium liquidgas 3. Volatile vs nonvolatile 4. Clausius - Clapeyron equation -relates vapor pressure and liquid temperature

  12. CLAUSIUS - CLAPEYRON EQUATION In general, the higher the temperature, the weaker the intermolecular forces, and therefore the higher the vapor pressure. The non-linear relationship between vapor pressure and temperature is given by the Clausius - Clapeyron equation. In P = (-Hvap/RT) + C : a straight line if lnP vs. 1/T It describes the amount of energy required to vaporize 1 mole of molecules in the liquid state R = 8.31 J/mol K T = Kelvin P = vapor pressure In P2 = -Hvap1 - 1 P1 R T2 T1 Q. The vapor pressure of ethanol at 34.9ºC is 100.0 mmHg and at 78.5ºC it’s 760.0 mmHg. What is the heat of vaporization of ethanol?

  13. CYRSTALLINE SOLIDS Type of solid lattice site Type of force properties of examples particle type between particles solids IONIC positive & electrostatic high M.P. NaCl negative ions attraction nonvolatile Ca(NO3)2 hard & brittle poor conductor POLAR polar dipole-dipole & moderate M.P. Sucrose, MOLECULAR molecules London Dispersion moderate C12H22O11 forces volatility Ice, H2O NONPOLAR Nonpolar London Dispersion low M.P., Argon, Ar, MOLECULAR molecules & forces volatile Dry Ice, CO2 atoms MACRO- atoms covalent bonds extremely high Diamond, C MOLECULAR between atoms M.P. nonvolatile Quartz, SiO2 Covalent- Arranged in Very Hard Network Network Poor conductor METALLIC metal atoms attraction between variable M.P. Cu, Fe outer electrons low volatility Al, W and positive good conductor atomic centers

  14. TYPE OF MELTING POINT HARDNESS ELECTRICAL SOLID OF SOLID & BRITTLENESS CONDUCTIVITY Molecular Low soft & brittle Nonconducting Metallic Variable Variable hardness, conducting malleable Ionic High to very hard & brittle Nonconducting high solid (conducting liquid) Covalent Very high Very hard Usually Network nonconducting

  15. CRYSTALLINE SOLIDS - Composed of crystal lattices - the geometric arrangement of lattice points of a crystal consists of unit cells - smallest unit from which atoms can be stacked in 3-D - unit cells (there are different types; see transparencies) - edge lengths and angles are used to describe the unit cell 3 types of unit cells - primitive (simple) - body center cubic - face centered cubic - metals and salts are usually cubic Ni = FCC Na=BCC NaCl=FCC In FCC corners and face are shared with other units Question: Determine the net number of ions in LiF (FCC) Li+ = 1/4 (Li per edge) (12 edges) = 3 1 (center) (1 center) = 1 F- = 1/8 (per corner) (8 corners) = 1 1/2 (face) (6 faces) = 2

  16. MOLECULAR SOLIDS - frozen noble gases, Ne - needs large number of atoms surrounding center for maximum attraction (see transparency) - close packing arrangement variations: -- hexagonal close-packing structure (ABABABA…); 6-unit cell (hcp) -- cubic close-packed structure (ccp); ABC ABC ABC ... Similar to FCC Coordination number = the number of nearest neighbors; 12 for close-pack structures METALLIC SOLIDS - sea of electrons; delocalized (bonding is non-directional) -many metals are cubic or hexagonal close-packed crystals COVALENT NETWORK -directional covalent bonds -diamond, Si, Ge, gray Sn are tetrahedral, sp3 hybridized, FCC -graphite is hexagonal flat sheets; sp2 hybridized; electrical properties are due to the increased delocalization of the electrons.

  17. X-RAY DIFFRACTION - Determining crystal structure x-ray diffraction used to obtain structure of proteins (1962 Nobel Prize myoglobin) & hemoglobin - Due to order structure - crystals consists of repeating planes - Planes act as reflecting surfaces - X-ray reflecting off surface creates diffraction pattern constructive interference gives more intense (higher amplitude) weaves. - only at certain angles will x-rays stay in-phase - creates light and dark areas on photographs - used to determine type of unit cell and size - if molecular, then can determine position of each atom nl = 2d sin q Bragg’s equation d=distance between atomic planes q =angle of reflection

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