Chapter 12 Liquids and Solids

1. Chapter 12Liquids and Solids L

2. I. Kinetic molecular Theory of Liquids A. Fluid – they can move from place to place. 1. Translation – movement from place to place. 2. Rotation – molecular tumbling 3. Vibration – bond bending and stretching 4. Liquids contain all three types of motion B. Higher density than gases (mass/volume) 1. 1 mole water vapor at STP = 22.4 L 2. 1 mole of liq. water = 18 mL (1 g = 1 mL= 1 cc) C. Molecules are close enough together to interact with one another. D. Liquids definite volume, but can change shape.

3. E. Molecules have the ability to diffuse 1. Demo F. Surface Tension 1. A force that tends to pull adjacent molecules of a liquids’ surface area to the smallest size. 2. Due to the attraction between polarity being greater than the attraction to other molecules (air, etc.) 3. Water Strider, shape of a drop of water G. Capillary action – polarity of water 1. The attraction of a liquid to the surface of a solid The tube needs to be very narrow 2. Trees , plants, etc.

4. H. Evaporation and Boiling 1. Vaporization – the change of a liquid (or solid) to a gas; includes both evaporation and boiling 2. Evaporation – the escape of particles from the surface of a liquid into a gas below the boiling point 3. Ways to increase rate of evaporation a. higher temperature b. more surface area c. move air which is saturated with vapor away from the liquid 4. Evaporation is a cooling process because the molecules with the highest energy escape 5. Boiling is the change of liquid to bubbles of vapor throughout the liquid

5. II. Kinetic Molecular Theory of Solids A. Molecules locked in place 1. Still have vibration 2. Molecular arrangement a. crystals 1) orderly, geometric, repeating pattern 2) cooled slowly b. amorphous solid 1) particles are arranged randomly 2) glass, rubber, plastic B. Solids have a definite shape and volume C. Solids are generally slightly more dense and less compressible than liquids D. Diffusion is millions of times slower in solids than in liquids

6. E. Melting Point a. temperature at which a solid changes to liquid b. energy used to overcome attractive forces holding particles in position c. crystalline substances have definite melting points; amorphous solids such as glass or plastics have no definite melting point d. supercooled liquid – substance which retains certain liquid properties even at temperatures where it appears to be a solid

7. III. Phase changes A. Vocabulary time – p. 382, Fig. 12-15 1) solid ↔ liquid melting/freezing 2) Liquid ↔ Gas condensation/evaporation (Boiling – rapid evaporation) 3) solid  gas a) sublimation b) dry ice c) wet clothes in winter

9. IV. Structure of water A. Structure 1. 105o angle 2. Polar – pos. & neg. end 3. Hydrogen bond – p. 385 4. Density a. solid (ice) less dense than liquid water b. ice floats

10. V. Phase change diagram T vs. Time A. Constant heat is being put into the system B. Plateaus – phase change – no temp. change C. Molar Heat of Fusion a. 1.44 kcal/mole – for water b. 6.009 kJ/mole – for water ↑ Temp Time 

11. D. Molar Heat of Vaporization 1. 9.7 kcal/mole – water 2. 40.79 kJ/mole - water E. calories 1. 1 cal/g oC 2. C = kcal = 1000 cal = food calories 3. p. 519 calorimeter 4. Specific Heat F. Prob. 12-1, p. 386 G. Practice p. 386 #1 and 2

12. H. Specific Heat 1. Works with the non-plateau areas. 2. Used to calculate when there is NOT a phase change. 3. Each substance has a unique value. 4. Water - 1 cal/goC 5. 1 cal. = the amount of energy needed to raise one g of water, 1 oC. (ΔT) - Chg. in temp. calories = (m)(ΔT)S.H. (S.H. = Specific Heat) 6. If I have 37 g of water and change the temperature from 45oC to 100oC, how many calories will be absorbed? 100oC - 45oC = 55oC cal = (37g)(55oC)(1 cal) = 2,035 cal (goC)

13. 7. S.H. can also be in J/g.K. a. Water(l) - 4.18 J/g.K b. Water(s) - 2.06 J/g.K c. Water(g) - 1.97 J/g.K 8. You can combine S. H. and molar heats of fusion or vaporization. Just take each section by itself. 9. How much heat energy is needed to boil 55 g of water that is at initially at 25oC? a. Break into 2 parts 1) Heating the water (S.H.) 2) The phase change (molar heat of vaporization) 3) Add them together (4.12 kcal + 29.6 kcal) = 33.72 kcal

14. VI. Phase Diagram – P vs. T A. p. 381, Figure 12-14 B. Triple Point –all 3 phases of matter exist C. Critical point/critical pressure- The temp. above which the substance cannot exist in a liquid state.

15. VII. Equilibrium A. A dynamic condition in which two opposing changes occur at equal rates in a closed system. B. The net result is no change, NOT that there is nothing going on. VIII. Boiling A. Temperature at which the vapor pressure of a liquid is equal to the prevailing atmospheric pressure. B. The conversion of a liquid to a vapor within the liquid as well as at its surface. C. Evaporation – the process by which particles escape from the surface of a nonboiling liquid and enter the gas state.

17. 1. vapor pressure- molecules of the liquid moving and some escaping from the liquid. a. The escaping molecules are a gas and have a “pressure” b. The higher the temp, the greater the kinetic energy, the more molecules that escape, the higher the pressure. 2. atmospheric pressure – a. Gas molecules in the air crash into the liquid. b. Some slow down enough to be attracted to the liquid, and turn into liquid molecules.

18. B. Boiling Point Elevation (Chapter 14) 1. Solute (particles that can be dissolved) are added to pure water. 2. This increases the attraction of the water molecules to the liquid state. 3. Therefore, it is harder for the water molecules evaporate and become a gas. 4. The B.P. (boiling point) increases. 5. The B.P. is higher with a solute than in pure water. a. The B.P. raises .51 oC for every one mole of particles added per liter of water. b. Antifreeze c. Cooking time with noodles

19. IX. Freezing A. Freezing – the physical change of a liquid to a solid by the removal of heat. • F.P. (freezing point) is the temperature at which a solid and liquid are in equilibrium at 1 atm. of pressure. • Freezing Point Depression 1. Solute interferes with the crystal lattice structure. 2. More difficult for the liquid to form the solid state. 3. 1 mole of particles lowers the freezing point by -1.86oC per liter (or kilogram) of water. 4. The freezing point of a solution of water is always lower than pure water. 5. The greater the number of particles, the lower the freezing point.

20. X. Colligative Properties – (14-2) A. Particles (solute) tend to interfere with phase changes. 1. The particles takes up space and throw off the equilibrium that has been established. B. Boiling Point Elevation 1. At 100oC, water has the same number of molecules going into the atmosphere as the atmosphere has molecules becoming a liquid. 2. When particles are added, the same number of atmosphere molecules become a liquid, but LESS molecules of liquid water are at the surface. 3. Therefore, less molecules go from a liquid  gas. 4. To compensate for this, you need to raise the boiling point. (Figure 14-7, p. 437)

21. 5. The greater the number of particles, the higher the B.P. has to be raised to reach equilibrium. C. Freezing point depression 1. The solute again takes up surface area in the solution. 2. The relatively pure ice is giving molecules to the liquid state, but the liquid state has less molecules available to become solid. 3. This imbalance can be compensated for by decreasing the temp. below 0. D. In general, adding anything to water raises the B.P. (or lowers the F.P.)

22. E. It is the NUMBER of particles that is important. 1. Sucrose (sugar) when dissolved stays as a molecule. C12H22O11. – one particle. 2. Salt (NaCl) dissolves into 2 particles: Na+ & Cl- so it will have more of an effect on F.P. and B.P. 3. Which substance would have more effect on the F.P. and B.P.? Sugar, salt or Calcium Chloride? (CaCl2)