1. Intermolecular forces (van der Waals forces) London dispersion- instantaneous dipole moment -increases with mass -found between all molecules Dipole-dipole- increases with polarity Hydrogen bonding The stronger the intermolecular forces, the greater the melting and boiling points and viscosity will be. Viscosity- resistance to flow. stronger weaker

2. Cohesion binds molecules together Adhesion binds molecules to surfaces water mercury Cohesion > adhesion Adhesion > cohesion

3. SubstanceMol. massBoiling Point (K) F 2 3885.1 I 2 253.8457.6 Propane C 3 H 8 44231 CH 3 CHO 44294 H 2 O18373 List the substances BaCl 2, H 2, CO, HF, and Ne in order of increasing boiling points. H 2 < Ne < CO < HF < BaCl 2 Nonpolar, nonpolar heavier, polar, H bond, ionic

5. The temperature of a substance is a measure of the average kinetic energy of its particles.

6. The effect of increasing temperature and Activation Energy As the temperature increases, the peak for the most probable KE is reduced, and more significantly, moves to the right to higher values so more particles have the highest KE values. At the higher temperature T2, a greater fraction of particles has the minimum KE to react.

7. Fusion- melting process ∆H fus heat of fusion ∆H fus heat of fusion Vaporization- formation of a vapor ∆H vap heat of vaporization ∆H vap heat of vaporization Vapor pressure increases with temperature until it equals the external pressure over the liquid, then it boils.

8. Liquefaction- condensation of a substance that is normally a gas (at room temp) Liquefaction- condensation of a substance that is normally a gas (at room temp) Melting point- temperature at which VP l =VP s Melting point- temperature at which VP l =VP s

9. Heat Energy (q) Energy transferred due to a temperature difference. Measured in Joules or calories 1 calorie = 4.184 J Q = m∙c∙∆T Heat = mass ∙ specific heat∙ change gained or lost in temp

10. States of Matter Solid Solid definite shape definite shape definite volume definite volume Liquid Liquid takes shape of container takes shape of container definite volume definite volume Gas Gas no definite shape or volume no definite shape or volume takes shape of container takes shape of container Plasma Plasma Made of charged particles Made of charged particles Ex.: stars Ex.: stars Condensed phases Can flow

11. Heat of fusion Heat of vaporization Heating Curve for Water

12. Calculate ∆H for a phase change Calculate the enthalpy change upon converting 10.0g of ice at -25 ◦ C to water vapor at 125 ◦ C under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.06 J/g·K, 4.18 J/g·K, and 2.02 J/g·K, respectively. For H 2 O, ∆H fus = 334 J/g For H 2 O, ∆H fus = 334 J/g ∆H vap = 2260 J/g ∆H vap = 2260 J/g

13. Specific Heat (c) The heat needed to raise the temperature of 1g of a substance 1˚C. The heat needed to raise the temperature of 1g of a substance 1˚C. water’s specific heat depends on it state: water’s specific heat depends on it state: Liquid c w = 4.184 J/g∙˚C Ice c i = 2.06 J/g∙˚C Steam c s = 2.02 J/g∙˚C

14. Calculate the enthalpy change for each stage, then total them. To get the ice to 0 ◦ C: ∆H = (10.0g)(2.09J/g·K)(25 ◦ K)=515J To melt the ice: ∆H = (10.0g)(334 J/g)=3340J To heat the water to 100 ◦ C: ∆H = (10.0g)(4.18J/g·K)(100 ◦ K)=4180J To vaporize the water: ∆H = (10.0g)(2260 J/g)=22600J To heat the water to 125 ◦ C: ∆H = (10.0g)(2.02 J/g·K)(25 ◦ K)=505J

15. ∆H = 515J + 3340J + 4180J + 22600J + 505J = 31,140J = 31.1 kJ = 31,140J = 31.1 kJ

16. Phase Diagram Critical point- beyond this, the gas and liquid phases become indistinguishable. Triple point- all three phases are at equilibrium

18. A piece of an unknown metal with mass 23.8g is heated to 100.0˚C and dropped into 50.0 ml of water at 24.0˚C. The final temperature of the system is 32.5˚C. What is the specific heat of the metal? Metal m=23.8 g ∆T=100.0-32.5=67.5˚CWaterm=50.0g∆T=32.5-24.0=8.5˚CC=4.184J/g˚C

19. Heat gained by the water: Q=(50.0g)(4.184J/g˚C) (8.5˚C) = 1778.2 Joules = 1778.2 Joules Heat gained by water = heat lost by metal 1778.2J = (23.8 g) (c) (67.5˚C) 1778.2J = 1606.5 (c) C=1.1J/g˚C

20. Graham’s Law of Diffusion The relative rates at which two gases (at the same temperature and pressure) will diffuse vary inversely as the square root of the molecular masses of the gases. The relative rates at which two gases (at the same temperature and pressure) will diffuse vary inversely as the square root of the molecular masses of the gases.