Problem Set VIII Liquids, Solids, Intermolecular Forces and Phase Diagrams

Transcription

1 Chem 121 Problem set VIII LUTI - 1 Problem et VIII Liquids, olids, Intermolecular orces and Phase Diagrams 1a) this is a point on the vapour pressure curve 1b) gas 1c) gas to liquid Water C 2 2a) solid to vapour or sublimes, 2b) 5.2 atm at the triple point, 2c) The fusion curve has a positive slope. A verticle line drawn in the direction of increasing pressure from any point on the fusion curve takes carbon dioxide into the solid state; that is an increase in pressure will cause liquid carbon dioxide to solidify. The solid carbon dioxide must therefore be more compact than the liquid and have a greater density. 3a) liquid, b) liquid 4a) liquid to solid to vapor, b) liquid to vapour 5a) no, b) yes

2 Chem 121 Problem set VIII LUTI A solid B liquid C gas D solid & gas E solid, liquid & gas liquid & gas 7c) the pure compound changes from a solid to a liquid. 8. Both molecules will have a similar dispersion force. owever carbon monoxide is polar, nitrogen is nonpolar and so C will also have a dipole-dipole force giving it the higher melting point. The melting points of C and 2 are 199 and 210ºC respectively.

3 Chem 121 Problem set VIII LUTI Again both molecules will have a similar dispersion force but because 2 has the larger dipole moment, it will have the larger dipole-dipole force, giving it the higher boiling point and heat of vapourization. 10. ydrogen bonding only occurs with the small electronegative elements,, and attached to a hydrogen. 3 (- bond), C 3 (- bond) and C 3 2 (- bond) show hydrogen bonding. Br, P 3 and C 3 (only has a C- bond) do not. 11a) Chlorine, 2 is larger and has more electrons than oxygen, 2 and so has the larger dispersion force giving it the higher melting point. The normal melting points of 2 and 2 are 101ºC and 218ºC respectively. 11b) Butane is the larger molecule and so has more electrons and thus has the stronger dispersion force and the higher melting point. The normal melting points of propane and butane are 190ºC and 138ºC respectively. 11c) Again i 4 is larger and has more electrons then C 4 and thus has the stronger dispersion force and the higher melting point. The normal melting points of C 4 and i 4 are 170ºC and 110ºC respectively. 12. Largest vapor pressure at 100 K: i 4 C 4 Ge 4 n 4 ; all four molecules are hydrides of group IV elements and are liquids at 100K, since all molecules are tetrahedral, the bond dipoles will cancel and all are nonpolar so the only intermolecular force is dispersion. mallest molecule, weakest intermolecular forces and highest vapour pressure and this is C 4 ighest boiling point: 2 2 e 2 Te 2 ; all four molecules are hyrides of group VI elements and are bent in shape with two lone pairs of electrons. All will exhibit the dispersion force, with 2 Te being the strongest with the most electrons. 2, 2 e and 2 Te exhibit dipole-dipole intermolecular forces while 2 exhibits hydrogen bonding. In this case the hydrogen bonding of water is stronger than the dispersion of 2 Te. ince 2 shows the strongest intermolecular force it will require the higher temperature for its vapour pressure to reach 1 atm and so will have the highest boiling point. 13a) g is a metallic solid, n 2 is an ionic solid, C 2 and butane are molecular solids. C 2 is a nonpolar molecule (vectors cancel), so it only has the dispersion force (22 electrons) and has the weakest attractive force of the four (bp 78 C). g is a metallic solid. The metallic force can be a strong force, but all metals are solids except g, and it has the weakest attractive force of the metals (bp 357 C). n 2 is an ionic solid ( E = 1.7), with a strong force of attraction and has strongest attractive force of the four (bp 1630 C). C 4 10 is a nonpolar hydrocarbon molecule so has the dispersion force (42 electrons) and has a stronger force of attraction than C 2 (bp 0.5 C). The substance with the highest boiling point will have the strongest force of attraction and this is n. 13b) i 2 is a network covalent solid and the other three are molecular solids. i 2 is a network covalent solid (quartz) and has a very strong force, the strongest of the four (bp 2950 C). P 3 has the dispersion force (18 electrons) and the dipole-dipole force (dipole-dipole force: E (P-) = = 0 so thus the vector is due to the lone pair and is relatively strong). Both forces are relatively weak compared to (bp 88 C). 3 has the dispersion force (10 electrons) and hydrogen bonding which is a strong force of attraction. The dipole moment of 3 is the larger than P 3 and P 3 ( E (-) = = 0.9 and is in the direction of the lone pair so the vectors add). ydrogen bonding is the significant attractive force in ammonia and the attractive force is stronger than P 3. (bp 33 C). P 3 has the dispersion force (66 electrons) and the dipole-dipole force.(dipole-dipole force: E (P-) = = 0.9 however this vector is in the opposite direction to the lone pair and so this molecule has the smallest dipole moment). owever the dispersion force is significant with 66 electrons and the overall force of attraction will be larger than 3 or P 3 (bp 76 C).

4 Chem 121 Problem set VIII LUTI - 4 P P The substance with the largest vapour pressure will have the weakest force of attraction and this is P 3. 13c) All four molecules are hydrides of the Group V elements, and are pyramidal in shape with a lone pair. b 3 has a large dispersion force (54 electrons) and a small dipole-dipole force ( E (-b) = = 0.2 and this vector is in the opposite direction to the lone pair). The dispersion force is the largest of the four molecules (bp 17 C). 3 has a weak dispersion force (10 electrons) and a strong hydrogen bonding force of attraction. The dipole moment of 3 is the larger than P 3 ( E (-) = = 0.9) and is in the direction of the lone pair so the vectors add (bp 33 C). As 3 has a relatively large dispersion force (36 electrons) and a weak dipole-dipole force ( E (-As) = = 0.1 and this vector is in the opposite direction to the lone pair). The dipole-dipole force is slightly smaller than b 3, however the difference is insignificant. The dispersion force is also smaller than b 3 (bp 63 C). P 3 has a weak dispersion force (18 electrons) and a relatively strong dipole-dipole force ( E (P-) = = 0, so the vector is entirely due to the lone pair and the vector is larger than b 3 and As 3 ). P 3 has the weakest force of attraction of the four molecules (bp 88 C). b As The order of increasing strength for the dipole-dipole interaction is b 3 < As 3 < P 3 < 3. owever the dispersion force is always stronger than the dipole-dipole force (except for hydrogen bonding) and the order of increasing strength for the dispersion interaction is 3 < P 3 < As 3 < b 3 (the reverse order of the dipole-dipole force). Ammonia has the strongest intermolecular force of all four molecules as it has hydrogen bonding. The molecule with the lowest boiling point is that with the weakest force of attraction or the weakest dispersion force (with the fewest electrons), and that is P 3. ee the figure (Bp of ydrides) on page d) All four molecules are molecular solids. C 4 is a nonpolar molecule and has a weak dispersion force (10 electrons), (bp 161 C). C 3 C 3 is also nonpolar and has a slightly stronger dispersion force (18 electrons), (bp 89 C ). C 3 has the hydrogen bonding force of attraaction and the dispersion force is starting to become noticable (18 electrons). The hydrogen bonding force of attraction is significantly stronger than the dispersion force in C 4 or C 3 C 3 as both of these molecules are gases and C 3 is a liquid (bp 65 C). C 3 C 2 also has the hydrogen bonding force and the dispersion force (26 electrons). It has a stronger overall force of attraction compared to C 3 due to its larger dispersion component (bp 78 C). The substance with the smallest vapour pressure will have the strongest force of attraction and that is C 3 C 2. 13e) K is an ionic solid and the other three are molecular solids. K is an ionic solid with a very strong force of attraction ( E = 2.2), the strongest of the four entities (bp 771 C). 2 has the dispersion force (10 electrons) and the hydrogen bonding force, which is a strong van der Waals force, but weak compared to the ionic, metallic and network covalent forces of attraction. (bp 100 C) 2 has the dispersion force (18 electrons) and the dipole-dipole force which is weak compared to the hydrogen bonding and dispersion forces ( E (-) = = 0.5 and this vector is in the same direction as the lone pairs). It has a weaker attractive force than water (bp 60 C). C 4 is a nonpolar molecule and only has the dispersion force with 10 electrons and so has the weakest force of attraction of the four entities (bp 161 C). P

5 Chem 121 Problem set VIII LUTI - 5 The substance with the highest boiling point will have the strongest force of attraction and that is K. C 3 C 3 C C2 C2 C2 C f C 3 C 3 has the hydrogen bonding force of attraction and the dispersion force (26 electrons), (bp 17 C). C 3 C 2 C 2 C 2 2 has the hydrogen bonding force of attraction with two - bonds. It is also linear and so can pack close together strengthening the dispersion force (42 electrons), giving this molecule the strongest force of attraction (bp 77 C). C 3 (C 3 )C 3 has no hydrogen bonding force of attraction, instead it has a weak dipole-dipole force ( E (-C) = = 0.9 and this vector is in the same direction as the lone pair). It s dispersion force (34 electrons) will be relatively weak as it is spherical and the molecules cannot pack close together. Thus the overall force of attraction will be weak (bp 3 C). C 3 2 has hydrogen bonding with two - bonds and a small dispersion component (18 electrons) compared to the other three molecules. ote that this molecule has the lowest bp even though it can form two hydrogen bonds; in this case the dispersion force dominates (bp 6 C). The substance with the smallest vapour pressure will have the strongest force of attraction and that is C 3 C 2 C 2 C a)The electronegativity vectors for C 2 will cancel however there will be a net vector toward oxygen in C. E(-C) = = 1.0, E(-C) = = 0. C has the strongest dipole-dipole intermolecular force. C C b) The electronegativity vectors for 3 will cancel, while 2 has a net vector toward the lone pair. E(-) = = has the strongest dipole-dipole intermolecular force. c) The electronegativity vectors for Xe 2 will cancel, while 2 has a net vector between the lone pairs. E(-) = = 0.5, E(-Xe) = = has the strongest dipole-dipole intermolecular force. C3 C3 C3 Xe d) The electronegativity vectors for i 4 will cancel, ( E(-i) = = 0.7), the molecule will not have a dipole moment and will have the weakest dipole-dipole intermolecular forces. 4 is not symmetrical and will have a net electronegativity vector. We will take the lone pair electronegativity vectors as approximately 1.5 (since E(lp-) = lp , from lecture notes, then lp 4 and thus E(lp- ) = ). As E(-) = = 1.5, all the electronegativity vectors are approximately the same and will come close to cancelling. The molecule will have a net electronegativity vector, which will be small but it will have a diploe moment and a small dipole-dipole intermolecular force. 4 has the stronger dipole-dipole intermolecular force.

6 Chem 121 Problem set VIII LUTI - 6 i i 15. The vapour pressure of a liquid is only dependant on temperature and independant of volume. Thus if there is liquid water in the container when the volume is reduced by three its vapour pressure will be 12 torr (the temperature does not change). ince the gas is saturated with water vapour when the volume is V, and when we reduce the volume to one third the original volume, one third of the number of moles of water are necessary to maintain a pressure of 12 torr, the other two thirds will form liquid water. owever the pressure of the gas will increase by a factor of three when we reduce the volume by three. Thus we first need to find the partial pressure of the gas. P total = P gas + P water vapour P gas = = 272 torr Let the initial volume of the container be V so the final pressure of the gas is: Vinitial V Pfinal = Pinitial = 272torr = 816torr V 1 final 3 V this is the pressure of the gas and we need to add to this the vapour pressure of water so P total = P gas + P water vapour = 816 torr + 12 torr = 828 torr. 16. We use the ausius apeyron equation and first we need to find vap for mercury from the data given. P 1 = torr, T 1 = K and P 2 = torr, T 2 = K so torr vap 1 1 ln = torr J mol.k K K and vap = x 10 4 J/mol now the vapour pressure at the normal boiling point of g is 1 atm and thus we need to find the temperature when the vapour pressure is 1760 torr. Using the ausius apeyron equation with P 3 = 760 torr and T 3 =? torr J 1 1 ln mol = 760torr J mol.k T K T 3 = K or C 17. Again we use the ausius apeyron equation and initially we need to find vap for first liquid from the data given: P 1 = 0.50 atm, T 1 = K and P 2 = 1.00 atm, T 2 = K so 100 ln. atm vap 1 1 = 050. atm J mol.k K K and vap = x 10 4 J/mol. ince vap (1) = 2 vap (2), vap (2) = x 10 3 J/mol, and using the ausius apeyron equation to find the normal boiling point of the second liquid with P 1 = 0.50 atm, T 1 = K and P 2 = 1.00 atm, T 2 =? K and vap (2) = x 10 3 J/mol we have atm J mol 1 1 ln = J 0.50atm mol.k K T2 T 2 = K or the normal boiling point of the second liquid is C. 18. Read text on types of crystaline solids, their bonding and properties. Gold is a metal, sugar is a molecular solid, quartz is a covalent network solid and fluorite is an ionic solid.

7 Chem 121 Problem set VIII LUTI i 2 covalent network solid (quartz) contain no molecules and the entire solid is held together with strong covalent bonds (mp 1713 C cristobalite). KBr ionic solid strong ionic bonds. The particles that make up the crystal lattice are positive and negative ions. Each ion is surrounded by neighbours of the opposite charge and there are no molecules (mp 734 C) P 3 molecular solid dispersion and Dipole-dipole forces. The dispersion force is an instantaneous dipole and is dependant on the number of electrons in the molecule. P 3 has 42 electrons and thus this force is relatively significant. The dipole-dipole force or permanent dipole is dependant on the sum of the electronegativity vectors ( E(-P)=1.8 and E(lp-P) 1.8). The VEPR shape of the molecule is tetrahedral and the four vectors will come close to canceling and we would expect the dipole-dipole force to not be very large. The dispersion force would be similar to butane (also 42 electrons) and the mp s are also similar (mp 138 C, butane; 152 C, P 3 ). C covalent network solid (diamond, graphite) contain no molecules and the entire solid is held together with strong covalent bonds (mp 4440 C, diamond; 4492 C, graphite). 2 molecular solid dispersion and the ydrogen bonding attractive forces. The dispersion is relatively small (10 electrons) and the hydrogen bonding force is the largest of all molecules. ydrogen bonding is an extreme form of the permanent dipole, where a highly electronegative atom strips the electrons from a hydrogen, leaving an exposed proton or positive charge which will interact with another electronegaive atom (mp 0 C) U metallic solid Metallic attractive force. The metal atoms occupy regular lattice sites, however the valence electrons occupy molecular orbitals that extend throughout the entire solid. or two atomic orbitals interacting (two metal atoms coming together) two rather widely spaced energy levels result. As more atomic orbitals become available to form molecular orbitals, the resulting energy levels become more closely spaced., finally producing a band of very closely spaced orbitals that extend through the entire solid. ince the force of attraction is not localised, the metal atoms can slide from one lattice site to another, making metals maleable and ductile (mp 1135 C)