Answers to Even-Numbered HW Problems

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1 Answers to Even-Numbered HW Problems Ch 5 Gases 8. A bag of potato chips is a constant pressure container. The volume of the bag increases or decreases in order to keep the internal pressure equal to the external (atmospheric) pressure. The volume of the bag increased because the external pressure decreased. This seems reasonable as atmospheric pressure is lower at higher altitudes than at sea level. We ignored n (moles) as a possibility because the question said to concentrate on external conditions. It is possible that a chemical reaction occurred that would increase the number of gas molecules inside the bag. This would result in a larger volume for the bag of potato chips. The last factor to consider is temperature. During ski season, one would expect the temperature of Lake Tahoe to be colder than Los Angeles. A decrease in T would result in a decrease in the volume of the potato chip bag. This is the exact opposite of what actually happened; so apparently the temperature effect is not dominant. 22. Rigid container: As temperature is increases, the gas molecules move with a faster average velocity. This results in more frequent and more forceful collisions resulting in an increase in pressure. Density = mass/volume; The moles of gas are constant and the volume of the container is constant, so density must be temperature independent (density is constant). Flexible container: The flexible container is a constant pressure container. Therefore, the internal pressure will be unaffected by an increase in temperature. The density of the gas, however, will be affected because the container volume is affected. As T increases, there is an immediate increase in P inside the container. The container expands its volume to reduce the internal pressure back to the external pressure. We have the same mass of gas in a larger volume. Gas density will decrease in the flexible container as T increases. 24. a. Containers ii, iv, vi, and viii have volumes twice that of containers i, iii, v, and vii. Containers iii, iv, vii, and viii have twice the number of molecules present as compared to containers i, ii, v, and vi. The container with the lowest pressure will be the one which has the fewest moles of gas present in the largest volume (containers ii and vi both have the lowest P). The smallest container with the most mol of gas present will have the highest pressure (containers iii and vii both have the highest P). All the other containers (i, iv, v and viii) will have the same pressure between the two extremes. The order is: ii = vi < i = iv = v = viii < iii = vii. b. All have the same average kinetic energy since the temperature is the same in each container. Only temperature determines the average kinetic energy. c. The least dense gas will be container ii since it has the fewest of the lighter Ne atoms present in the largest volume. Container vii has the most dense gas since the largest number of the heavier Ar atoms are present in the smallest volume. To figure out the ordering for the other containers, we will calculate the relative density of each. In the table below, m equals the mass of Ne in container i, V equals the volume of container i, and d equals the density of the gas in container i.

2 Container mass, volume i ii iii iv v vi vii viii m, V m, 2V 2m, V 2m, 2V 2m, V 2m, 2V 4m, V 4m, 2V density mass volume m m = d = d 2 V 2 V 2 m V = 2d 2 m = d 2 V 2 m V = 2d 2 m 2 V = d 4 m V = 4d 4 m 2 V = 2d From the table, the order of gas density is: ii < i = iv = vi < iii = v = viii < vii d. µ rms = (3 RT/M) /2 ; the root mean square velocity only depends on the temperature and the molar mass. Since T is constant, the heavier argon molecules will have the slower root mean square velocity as compared to the neon molecules. The order is: v = vi = vii = viii < i = ii = iii = iv. 26. Statements a, c, and e are true. For statement b, if temperature is constant, then the average kinetic energy will be constant no matter what the identity of the gas (KE ave = 3/2 RT). For statement d, as T increases, the average velocity of the gas molecules increases. When gas molecules are moving faster, the effect of interparticle interactions is minimized. For statement f, the KMT predicts that P is directly related to T at constant V and n. As T increases, the gas molecules move faster on average, resulting in more frequent and more forceful collisions. This leads to an increase in P. 32. a Pa; b. 0.5 times the height of a mercury column K ml g Ar remains L O Molecular formula is C 2 H 2 Cl atm kj 82. V, T, and P are all constant, so n must be constant. Because we have equal mol of gas in each container, gas B molecules must be heavier than gas A molecules. a. Both gas samples have the same number of molecules present (n is constant). b. Since T is constant, KE ave must be the same for both gases (KE ave = 3/2 RT). c. The lighter gas A molecules will have the faster average velocity.

3 d. The heavier gas B molecules do collide more forcefully, but gas A molecules, with the faster average velocity, collide more frequently. The end result is that P is constant between the two containers. 96. At constant T and P, Avogadro s law applies; that is, equal volumes contain equal moles of molecules. In terms of balanced equations, we can say that mole ratios and volume ratios between the various reactants and products will be equal to each other. Br F 2 2 X; Two moles of X must contain two moles of Br and 6 moles of F; X must have the formula BrF 3. Ch 6 Thermochem 0. Products have a lower potential energy than reactants when the bonds in the products are stronger (on average) than in the reactants. This occurs generally in exothermic processes. Products have a higher potential energy than reactants when the reactants have the stronger bonds (on average). This is typified by endothermic reactions. o 4. The zero point for Δ H f values are elements in their standard state. All substances are measured in relationship to this zero point. 34. a. The combustion of gasoline releases heat, so this is an exothermic process. b. H 2 O(g) H 2 O(l); Heat is released when water vapor condenses, so this is an exothermic process. c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process. d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic process. 36. a kj; b kj; c kj; d kj 40. ΔH = ΔE + PΔV; From this equation, ΔH > ΔE when ΔV > 0, ΔH < ΔE when ΔV < 0, and ΔH = ΔE when ΔV = 0. Concentrate on the moles of gaseous products versus the moles of gaseous reactants to predict ΔV for a reaction. a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products so ΔV = 0. For this reaction, ΔH = ΔE. b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products so ΔV < 0 and ΔH < ΔE. c. There are 9 moles of gaseous reactants converting to 0 moles of gaseous products so

4 ΔV > 0 and ΔH > ΔE g hot water needed 48. T f = 30.3 C 58. C 4 H 4 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(l) ΔH comb = -234 kj C 4 H 8 (g) + 6 O 2 (g) 4 CO 2 (g) + 4 H 2 O(l) ΔH comb = kj H 2 (g) + /2 O 2 (g) H 2 O(l) ΔH comb = -286 kj By convention, H 2 O(l) is produced when enthalpies of combustion are given and, since per mole quantities are given, the combustion reaction refers to mole of that quantity reacting with O 2 (g). Using Hess s Law to solve: C 4 H 4 (g) + 5 O 2 (g) 4 CO 2 (g) + 2 H 2 O(l) ΔH = -234 kj 4 CO 2 (g) + 4 H 2 O(l) C 4 H 8 (g) + 6 O 2 (g) ΔH 2 = -(-2755 kj) 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔH 3 = 2(-286 kj) C 4 H 4 (g) + 2 H 2 (g) C 4 H 8 (g) ΔH = ΔH + ΔH 2 + ΔH 3 = -58 kj 60. ClF(g) + F 2 (g) ClF 3 ΔH = kj 68. a kj; b. -67 kj/mol; c. -37 kj Ch 0 Liquids and Solids 2. C 25 H 52 has the stronger intermolecular forces because it has the higher boiling point. Even though C 25 H 52 is nonpolar, it is so large that its London dispersion forces are much stronger than the sum of the London dispersion and hydrogen bonding interactions found in H 2 O. 8. Equilibrium: There is no change in composition; the vapor pressure is constant. Dynamic: Two processes, vapor liquid and liquid vapor, are both occurring but with equal rates so the composition of the vapor is constant. 20. C 2 H 5 OH(l) C 2 H 5 OH(g) is an endothermic process. Heat is absorbed when liquid ethanol vaporizes; the internal heat from the body provides this heat which results in the cooling of the body.

5 22. The phase change, H 2 O(g) H 2 O(l), releases heat that can cause additional damage. Also steam can be at a temperature greater than 00 C. 28. The typical phase diagram for a substance shows three phases and has a positive sloping solid-liquid equilibrium line (water is atypical). A sketch of the phase diagram for I 2 would look like this: P 90 torr s l g 5 o C T Statements a and e are true. For statement a, the liquid phase is always more dense than the gaseous phase (gases are mostly empty space). For statement e, because the triple point is at 90 torr, the liquid phase cannot exist at any pressure less than 90 torr, no matter what the temperature. For statements b, c, and d, examine the phase diagram to prove to yourself that they are false. 32. Ar exists as individual atoms which are held together in the condensed phases by London dispersion forces. The molecule which will have a boiling point closest to Ar will be a nonpolar substance with about the same molar mass as Ar (39.95 g/mol); this same size nonpolar substance will have about equivalent strength of London dispersion forces. Of the choices, only Cl 2 (70.90 g/mol) and F 2 (38.00 g/mol) are nonpolar. Because F 2 has a molar mass closest to that of Ar, one would expect the boiling point of F 2 to be close to that of Ar. 36. a. CBr 4 ; Largest of these nonpolar molecules so has strongest LD forces. b. F 2 ; Ionic forces in LiF are much stronger than the covalent forces in F 2 and HCl. HCl has dipole forces that the nonpolar F 2 does not exhibit; so F 2 has the weakest intermolecular forces and the lowest freezing point. c. CH 3 CH 2 OH; Can form H bonding interactions unlike the others. d. H 2 O 2 ; H O-O H structure produces stronger H bonding interactions than HF, so has greatest viscosity. e. H 2 CO; H 2 CO is polar so has dipole forces, unlike the other nonpolar covalent compounds. f. I 2 ; I 2 has only LD forces while CsBr and CaO have much stronger ionic forces. I 2 has weakest intermolecular forces so has smallest ΔH fusion.

6 38. A molecule at the surface of a waterdrop is subject to attractions only by molecules below it and to each side. The effect of this uneven pull on the surface molecules tends to draw them into the body of the liquid and causes the droplet to assume the shape that has the minimum surface area, a sphere. 40. CO 2 is a gas at room temperature. As mp and bp increase, the strength of the intermolecular forces also increases. Therefore, the strength of forces is CO 2 < CS 2 < CSe 2. From a structural standpoint this is expected. All three are linear, nonpolar molecules. Thus, only London dispersion forces are present. Since the molecules increase in size from CO 2 < CS 2 < CSe 2, the strength of the intermolecular forces will increase in the same order.