Temperature ( C) [not to scale]

Transcription

1 Liquids and Solids Key for Quick Review: Use the phase diagram below to answer the following questions. a. which letter in the diagram corresponds to the region of liquid B b. which letter in the diagram corresponds to the region of solid C c. the normal boiling point is (in ºC) 89ºC d. the normal melting point is (in ºC) ~13ºC e. the temperature at the triple point is (in ºC) 13ºC f. the temperature at the critical point is (in ºC) 183ºC g. the pressure at the critical point is (in atm) 95 atm h. the pressure at the triple point is (in atm) 0.12 atm i. which has the lower density the solid or liquid phase liquid C 1 2 B 4 3 A Temperature ( C) [not to scale] Draw arrows for the following processes in the phase diagram above AND write whether each process is endo (or exo)-thermic: ENDO: EXO: melting, vaporization, sublimation freezing, condensation, deposition Q: If the triple point is above 1.00 atm, what is the normal melting point? Normal boiling point? Answer: They do not exist b/c the neither solid and liquid nor the liquid and vapor can exist in equilibrium at 1.00 atm in this case... (refer to the phase diagram for CO 2 in your text) Describe all the process that occur in following path ( ) through the phase diagram. 1 to 2 - heat the solid until the phase boundary is reached, melt the solid, heat the liquid to pt. 2 2 to 3 - decrease pressure until the phase boundary at which point the liquid reaches equilibrium with the vapor, once all the liquid is converted to vapor, the pressure of the vapor is lowered to reach pt. 3

2 3 to 4 - cool the vapor until the phase boundary at which pt. the solid forms by deposition, once all the vapor is deposited (i.e. becomes solid) then the solid is cooled to pt. 4 Q: A compound is found to have a H vap of 3.46 kj /g and a H sub of 4.60 kj /g, calculate the H fus in kj /g. Explain why you can use this data to provide an accurate result. Think about Hess s Law and state functions. X (s) X (l) H fus =? X (l) X (g) H vap = 3.46 kj/g X (s) X (g) H sub = 4.60 kj/g A: therefore H fus = 1.14 kj /g b/c sub = fus + vap, so H sub = H fus + H vap Q: If you were to eat 1.5 kg of ice at 0.0ºC, your body would have to provide how much energy to (a) melt the ice [for water Hfus = J /g] AND (b) to raise the temperature of the water to 37.0ºC (body temperature) [specific heat of water = J /gºc ]? ind heat required to warm ice from 0.0ºC to 37.0ºC, and add to this the heat required to melt the water. A: 732 kj Review Questions: 1. Identify each of the following statements as true or false (T or ). If the statements are false change them to make them true! a. As molecules get larger, dipole-dipole interactions increase. Dispersion forces are larger for larger molecules with more easily polarized electron clouds. b. All polar molecules have higher boiling points than all nonpolar molecules. Boiling pots. Depend on the total intermolecular forces. Some non-polar molecules (for example waxes) have dispersion forces that are greater in strength than the sum of the dispersion forces plus the dipole-dipole forces in a small polar molecule (like chloroform). c. Ionic solids have low melting points due to their weak electrostatic forces. Generally speaking the electrostatic coulombic forces in present in ionic solids result in relatively high melting points. d. London forces are the only intermolecular forces in covalent networks.

3 Covalent networks are held together by covalent bonding (not intermolecular forces). T e. Covalent network substances are hard, brittle, & have high melting point. f. The reason CH3OH has a higher normal boiling point than CH3CH2CH3 is that the CH3OH has only dipole forces acting between its molecules in the liquid phase whereas CH3CH2CH3 has only London forces acting between its molecules in the liquid phase. Replace the only dipole forces with both dispersion forces and hydrogen bonding. 2. In each of the following pairs, which should have the higher melting point? Explain the reasons for your choice in terms of the types of forces involved and the magnitude of the forces. a. C vs. CH4 C (network covalent as perhaps diamond or graphite... or perhaps large molecules like Buckey balls) vs. methane which is a small non-polar molecule with only dispersion forces b. MgO vs. KCl MgO (with a +2 and -2 ion interacting) should have stronger ionic forces than KCl (with a +1 and -1 ion interacting) c. H2O vs. H2S H 2 O (has weaker dispersion forces but has hydrogen bonding) vs.h 2 S (which has larger dispersion forces and dipole forces too) recall that NH 3, H 2 O, and H have anomalously high boiling points d. C3H8 vs. C12H26 C12H26 (is non-polar but has larger dispersion forces) vs. C3H8 (which also has only dispersion forces, but is smaller) e. CH3OCH3 vs. CH3CH2CH3 CH3OCH3 (is polar so it has dispersion forces and dipole forces) CH3CH2CH3 (which has only dispersion forces, and is comparable in size) f. SiO2 vs. NH4Cl SiO2 (network covalent) vs. NH4Cl (which is ionic but has a large cation and the ions are only +1, -1 in charge, so we could expect it to be a lower melting ionic compound) 3. Of the following possible structural arrangements, identify which best matches the statements below. simple cubic cell face-centered cubic cell body centered cubic cell a. the least stable structure (structure has too much empty space) simple cubic cell

4 b. has highest packing efficiency (only 26% empty space) face centered cubic It is worth mentioning that the reasons why a given metal crystallizes in a particular structure is not well understood. However, the type of crystalline structure does impact many properties of metals: for example metals which pack in the simple cubic structure (like alkali metals) have low melting pts., hardness, and densities. Metals with the fcc structure (or a closely related structure called hcp that has the same packing efficiency) tend to have high melting pts., hardness, and densities. Explain why this is sensible. 4. What is the primary reason for the difference in structure observed in the compounds LiCl, NaCl, and CsCl? Apply the radius ratio rules... the way ions can pack together depends on the relative sizes (and charges) of the ions. 5. Which one of the following DECREASES as the strength of the attractive intermolecular forces INCREASES? a. heat of vaporization b. normal boiling point c. the sublimation temperature of a solid d. the vapor pressure of a liquid e. none of the above 6. Under which of the following conditions will vaporization best occur? a. high mass, large surface area, and high kinetic energy b. weak intermolecular forces, high kinetic energy, and large surface area c. high molecular energy and small surface area d. low kinetic energy, strong intermolecular forces, and large surface area e. small surface area, low kinetic energy, and low molecular mass 7. The boiling points of the halogens increase going from 2 to I2. What intermolecular forces are responsible for this trend? a. permanent dipole b. hydrogen bonding c. ion-ion attraction d. London dispersion forces e. ion-dipole attraction 8. In which of the following processes will energy be evolved as heat? a. sublimation b. vaporization c. crystallization d. melting e. none of these

5 9. Which of the statements a to d is INCORRECT? a. molecular solids have high melting points b. the binding forces in molecular solids include London forces c. ionic solids have high melting points d. ionic solids are insulators e. all of the statements a to d are correct 10. A solid and its melt readily conduct electricity. The crystal also has a luster and is easily deformed. Thus, it is (also explain why it cannot be the others): a. an ionic crystal b. a network covalent solid c. a metallic crystal d. a molecular crystal e. not enough data given 11. A crystal does not conduct electricity, even after melting. It is hard and brittle and melts at a very high temperature. What type of crystal is it? a. an ionic crystal b. a network covalent solid c. a metallic crystal d. a molecular crystal e. not enough data given 12. A crystal does not conduct electricity, yet its melt and aqueous solutions do. It is hard and brittle and melts at a very high temperature. What type of crystal is it? a. an ionic crystal b. a network covalent solid c. a metallic crystal d. a molecular crystal e. not enough data given 13. In which of the following substances would dispersion forces be the only significant factor in determining the boiling point? 1. Cl2 2. H 3. Ne 4. KNO2 5. CCl4 a. 1, 3, 5 b. 1, 2, 3 c. 2, 4 d. 2, 5 e. 3, 4, 5

6 14. Which of the substances below is an example of a COVALENT network solid? a. S8 (s) b. SiO2 (s) c. MgO (s) d. NaCl (s) e. C25H52 (s) NOTE: the word COVALENT was added to this question. Ionic compounds are also network solids and thus the initial wording of the question would have yielded three (3) correct choices. 15. Answer each of the following as true or false and give an explanation of your answer. (a) The role of intermolecular forces is more important in a gas than in a liquid. In the gas phase large distances (on the atomic scale) separate the molecules. (b) The process of melting is always exothermic. No, melting is ENDOthermic. (c) Metals are characterized by their high melting points, brittle nature, and good electrical conductivity. Most metals are not BRITTLE... they are malleable and dutile.. T (d) During a phase change from solid to liquid, a substance s temperature remains the same as more liquid is formed and rises only after all the solid has melted. This is true provided that the heat is added slowly enough and the system is mixed to maintain temperature uniformity. T (e) A liquid with a low vapor pressure at 25ºC has a high molar heat of vaporization. If a small amount of thermal energy could vaporize a substance, that substance would generate a high vapor pressure at room temperature (~ K). 16. Arrange the following substances in the order in which you would expect their boiling points to increase. Provide a description of the types of intermolecular forces which are important for each of the substances. Reasonable predictions might be: O2 (MM = 32.00) nonpolar - LOWEST due to weak dispersion forces only Cl2 (MM = 70.91) nonpolar - but higher dispersion forces b/c Cl 2 is larger NO (MM = 49.00) polar - dispersion and dipole forces CH3CH2OH (MM = 46.08) H-bonding and a relatively long molecule compared to the rest (therefore more dispersion forces too) But note that the b.p. are really: O2( ºC) < NO (-59.9ºC) < Cl 2 (-34.05ºC)< CH3CH2OH (78.5ºC) In other words, the greater polarizability of Cl 2 outweighs the polarity of the ON.