Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 1. Exercises: 25, 27, 33, 39, 41, 43, 51, 53, 57, 61, 63, 67, 69, 71(a), 73, 75, 79

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1 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 1 Text: Petrucci, Harwood, Herring 8 th Edition Suggest text problems Review questions: 1, 5!11, 13!17, 19!23 Exercises: 25, 27, 33, 39, 41, 43, 51, 53, 57, 61, 63, 67, 69, 71(a), 73, 75, Define vapourization, condensation, vapour pressure, non-volatile liquid and volatile liquid. 2. Define boiling, normal boiling point and explain the effect of external pressure on boiling point. 3. The enthalpy of vaporization of liquid methanol is 38.0 kj@mol!1. The vapour pressure of methanol at 5 EC is 40.0 mmhg. Estimate its normal boiling point. 4. Methane has a normal boiling point of!164 EC and a vapour pressure of 42.8 atm at!100 EC. Calculate the enthalpy of vaporization of liquid methane. 5. Define the term hydrogen bond. Distinguish between the covalent bonding involving hydrogen in water and hydrogen bonding involving water. 6. The electronegativity of chlorine is slightly greater than that of nitrogen. Why is chlorine not usually considered to form hydrogen bonds while nitrogen is known to form strong hydrogen bonds? 7. Describe the difference between a dipole-dipole interaction and a dispersion force. Which interaction is present in every substance? 8. Distinguish between polarizability and polarity. Describe how each influences the intermolecular interactions between a pair of molecules. 9. Determine the lattice energy for MgF 2 (s) from the following data: HE f MgF 2 (s) =!1123 kj@mol!1, enthalpy of sublimation of Mg(s) = 148 kj@mol!1, bond dissociation energy of F 2 (g) = 159 kj@mol!1, I 1 for Mg(g) = 738 kj@mol!1, I 2 for Mg(g) = 1450 kj@mol!1 EA for F(g) =!328 kj@mol!1 10. Pick the compound with the higher boiling point and give your reasoning: (a) CO or N 2 (b) NH 3 or PH 3 (c) CH 3 CH 2 OH or CH 3 OCH 3 (d) CCl 4 or CBr 4 (e) CH 3 CH 2 CH 2 CH 3 or C 4 H 8 (the 4 carbons form a ring) (f) NaBr or HBr

2 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS Consider the group VIIA hydrides,hf, HCl, HBr, HI formula Boiling point/ec HF 19.5 HCl!84.9 HBr!67 HI!35.4 (a) (b) (c) Discuss the hydrides in terms of bond polarity and polarizability. Explain why the boiling points increase from HCl to HI. Explain why the boiling point of HF is much greater than the other hydrides contrary to the trend of increasing boiling points from HCl to HI. 12. Which substance do you expect to have the higher enthalpy of vaporization, CH 3 COCH 3 or CH 3 CH 2 CH 2 CH 3? Explain briefly. 13. Predict which type of solid (ionic, molecular, metallic or network covalent) is each of the following: formula T(fusion) T(boiling) other properties PrI 3 733EC 1377EC liquid conducts electricity OsF 6 32EC 46EC liquid does not conduct electricity BN not measured not measured sublimes at 3000EC 14. Hydrazine, NH 2!NH 2, is a liquid at room temperature while ethyl methyl ether, CH 3 CH 2!O!CH 3, is a gas. Explain in terms of the relevant intermolecular forces. 15. Predict which compound will have the higher melting point and the greater solubility in water. Explain briefly. (a) NaF or KF (b) CaS or KBr

3 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 3 Answers 1. vapourization: Passage of molecules from the liquid state to the gaseous or vapour state. condensation: Conversion of a gas or vapour to a liquid. vapour pressure: The pressure exerted by molecules of a substance in the gaseous state (its vapour) in dynamic equilibrium with its liquid at a given temperature. non-volatile liquid: Liquid with a very low vapour pressure at room temperature e.g. mercury. volatile liquid: Liquid with a high vapour pressure at room temperature e.g. acetone. 2. boiling: State in which bubbles of vapour form throughout a liquid sample and escape with a pressure equal to the pressure over the surface of the liquid. normal boiling point: Temperature at which the vapour pressure of a liquid is equal to a standard atmosphere of pressure. Effect of external pressure on boiling point. As the external pressure over the surface of a liquid increases, the vapour pressure of the liquid required for boiling increases. The temperature of the liquid must increase to provide the higher vapour pressure. Hence boiling point increases. 3. Boiling point = K(65.8EC) kj@mol!1 5. Strong dipole-dipole force in which a hydrogen forms a bridge between two very electronegative atoms allowing a very strong interaction between the dipoles. The H is covalently bonded to one atom and bridges to the lone pair on the second atom in the second molecule or ion. H forms two electron pair (polar covalent) bonds, one to each oxygen atom (short bonds). The oxygen can bridge through its two lone pairs to H s on two other water molecules ( long H-bond). 6. Chlorine doesn t form effective hydrogen bonds due to the large volume of its 3 rd shell lone pairs vs the small volume of nitrogen s 2 nd shell lone pair which allows a close approach to a second molecule containing a hydrogen with a big M + charge. 7. A dipole-dipole attraction is the attraction between the permanent dipoles in polar molecules. Dispersion forces are present in non-polar and polar molecules alike and are the only forces present in non-polar species. They are instantaneous dipole-

4 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 4 induced dipole forces in which temporary dipoles in two non-polar species attract one another. In a polar species, dispersion forces temporarily strengthen the permanent dipoles leading to stronger interactions. 8. Polarizability refers to the tendency for charge separation to occur in a molecule or atom. The dispersion forces in a sample of molecules, both polar and non!polar, increase in strength with the polarizability of the molecule s electrons. Polarity refers to the degree of charge separation in a polar bond or polar molecule. It is best represented for a molecule by its dipole moment. The more polar the molecule, the greater its dipole moment. 9. lattice energy for MgF 2 (s) =!2962 kj@mol!1 10.(a) T B CO > T B N 2 Both have 14 electrons and 2 atoms and have similar polarizability and dispersion forces but CO is polar and has additional dipole-dipole forces to overcome. (b) T B NH 3 > T B PH 3. PH 3 has more electrons than NH 3 (18 vs 10) and hence has greater polarizability and dispersion forces to overcome. However, NH 3 is more polar and can form hydrogen bonding (N:----H!N). When comparing molecules which are not too different in size, hydrogen bonding is the dominant force to overcome and this will give NH 3 the significantly higher boiling point. (c) (d) (e) (f) T B CH 3 CH 2 OH > T B CH 3 OCH 3. Both have 26 electrons and 9 atoms and hence have similar polarizability and dispersion forces. CH 3 CH 2 OH is more polar (O!H is more polar than O!C) but more importantly CH 3 CH 2 OH can form hydrogen bonding between pairs of molecules and this represents a significantly greater force to overcome in order to boil. T B CBr 4 > T B CCl 4. Since they are both class AX 4 and non-polar, dispersion forces are the only ones to consider. CBr 4 has more electrons (146 vs 74) and is more polarizable and has greater dispersion forces to overcome in order to boil. T B CH 3 CH 2 CH 2 CH 3 > T B C 4 H 8. Both are non-polar with nearly the same #electrons (34 vs 32) but different shapes. CH 3 CH 2 CH 2 CH 3 has an elongated shape and greater polarizability than the more compact ring shape of C 4 H 8. Hence it has greater dispersion forces to overcome in order to boil. T B NaBr > T B HBr. NaBr forms an ionic crystal which always have greater forces to overcome (i.e. cation-anion forces) than the dispersion and dipole-dipole forces of HBr. 11.(a) Bond polarity decreases from HF to HI (as EN decreases) but polarizability increases from HF to HI due to increasing numbers of electrons. (10<18<36<54).

5 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 5 (b) (c) The boiling points increase due to the increase in polarizability and dispersion forces. HF has the highest boiling point because only HF can form strong hydrogen bonds. (HF:---H!F). The valence lone pairs of HCl, HBr and HI are too big. 12. H vap CH 3 COCH 3 [(CH 3 ) 2 C=O] > H vap CH 3 CH 2 CH 2 CH 3. CH 3 CH 2 CH 2 CH 3 has slightly more electrons (34 vs 32) and more atoms (14 vs 10) and hence is more polarizable and will have greater dispersion forces. However the difference in dispersion forces should not be great. CH 3 COCH 3 is polar due to the C=O and it s additional dipole-dipole forces should give it the greater total forces to overcome and the greater enthalpy of vaporization. 13. PrI 3 ionic (cation/anion forces are relatively strong to account for relatively high melting and boiling points. An ionic liquid contains mobile ions which can conduct a current. OsF 6 molecular (low melting and boiling point characteristic of a molecular compound with weak intermolecular forces and a non-conductor.) BN network covalent (very high sublimation temperature indicative of very strong intermolecular forces which are characteristic of network covalent solids.) 14. At a given temperature, the phase of a substance is indicative of the relative strength of intermolecular forces vs the average kinetic energy of the molecules. A gas indicates weaker intermolecular forces than a liquid. The particles of a gas have enough kinetic energy to overcome intermolecular forces but the particles of a liquid do not, they have just enough energy to move around but not enough to escape and form a vapour. CH 3 CH 2!O!CH 3 is a somewhat polar molecule (C!O!C part is V-shaped) with significant non-polar fragments ( CH 3 CH 2 and CH 3 ). It has more electrons than (34 vs 18) stretched over more atoms (12 vs 6) and hence is more polarizable and has greater dispersion forces. However, NH 2!NH 2 is extensively hydrogen-bonded and since the two molecules are not largely different in size, hydrogen bonding give NH 2!NH 2 the greater total intermolecular forces. At room temperature, molecules of both have the same average kinetic energy. However such energy is not enough for molecules of NH 2!NH 2 to totally overcome their intermolecular forces and form a vapour but it does represent sufficient energy for molecules of CH 3 CH 2!O!CH 3 to overcome their weaker intermolecular forces and form a vapour. 15.(a) NaF has higher melting point and should be less soluble in water. K + is a bigger ion than Na + hence the distance between the centres of + and! charge is larger reducing the cation!anion attraction and the lattice energy and melting point of KF vs NaF. The lower lattice energy of KF makes it easier for water molecules to pull ions out of the crystal to dissolve them increasing solubility.

6 Chemistry 1050 Chapter 13 LIQUIDS AND SOLIDS 6 (b) CaS has higher melting point and should be less soluble in water. K + is a bigger ion than Ca 2+ and Br! is bigger than S 2! hence the distance between the centres of + and! charge is larger. The charges on the ions of KBr (+1 and!1) are lower than those of CaS (+2 and!2). Both factors of internuclear distance and ion charges reduce the cation!anion attraction, the lattice energy and melting point of KBr vs CaS. The lower lattice energy of KBr makes it easier for water molecules to pull ions out of the crystal to dissolve them increasing solubility. Developed by Dr. Chris Flinn