Colligative Properties of Solutions

CHAPTER 16. Colligative Properties of Solutions 16-1. The number of moles of KMnO 4 (aq) is ( ) 1 mol KMnO4 moles KMnO 4 (5.25 g KMnO 4 ) 0.0332 mol 158.04 g KMnO 4 molality m 0.0332 mol 0.250 kg 0.133 m 16-2. (a) CH 3 OH(l) is an ganic compound, so i 1 and ( ) 1.00 mol m c im (1) 2.00 m c 0.500 kg (b) AgNO 3 (s) is a strong electrolyte with i 2, so m c (2)(2.00 m) 4.00 m c (c) Ca(ClO 4 ) 2 (s) is a strong electrolyte with i 3, so 16-3. The vap pressure lowering is Using Equation 16.10, we have that m c (3)(2.00 m) 6.00 m c P 1 75.0 Tr 63.9 Tr 11.1 Tr x 2 P 1 P 1 11.1 Tr 75.0 Tr 0.148 The fmula mass of benzene, C 6 H 6 (l), is 78.11 g mol 1 and that of naphthalene, C 10 H 8 (s), is 128.16 g mol 1. Now, let z number of grams of naphthalene in the solution. The mass of benzene is (41.3 z) grams, and so x naph n naph n benz + n naph Solving f z, we get 9.16 grams. z g 128.16 g mol 1 (41.3 z) g 78.11 g mol 1 + z g 128.17 g mol 1 0.148 44

Chapter 16: Colligative Properties of Solutions 45 16-4. The mole fraction of (NH 4 ) 2 SO 4 (aq) is given by x (NH4 ) 2 SO 4 n (NH4 ) 2 SO 4 n (NH4 ) 2 SO 4 + n H2 O Because (NH 4 ) 2 SO 4 (aq) is a strong electrolyte, it dissociates completely into NH + 4 (aq) and SO 2 4 (aq) ions. Assume a one kilogram solution. The number of moles of ions in one kilogram of a 0.70 m (NH 4 ) 2 SO 4 (aq) solution is moles of ions (3)(0.70 mol kg 1 )(1.00 kg) 2.1 mol and the number of moles of water is 1000/18.02 g mol 1 55.49 moles. Therefe, 2.1 mol x (NH4 ) 2 SO 4 2.1 mol + 55.49 mol 0.0365 From Table 15.7, we see that PH 2 O 55.4 Tr at 40 C. Using Equation 16.10, we have P H2 O x (NH4 ) 2 SO 4 PH 2 O (0.0365)(55.4 Tr) 2.0 Tr 16-5. Because NaCl(aq) is a strong electrolyte with i 2, we have ( ) 3.5 g NaCl (2) 58.44 g mol m c 1 NaCl 1.24 m c 0.0965 kg H 2 O Now use Equation 16.13 to write T b K b m c ( 0.513 K m 1 ) c (1.24 mc ) 0.64 Therefe, the nmal boiling point will be 100.64 C. 16-6. We use Equation 16.14 to write T f 5.25 K (1.86 K m 1 c )m c m c 2.82 m c. F CaCl 2 (s), i 3, and so m 0.941 m. 16-7. We use Equation 16.15 in the fm i T f T f K f m 5.49 K 2.96 K (5.07 K m 1 0.499 )(1.00 m) where the freezing point of benzene is 5.49 C. Therefe, i 0.5 1/2, a result suggesting the fmation of dimers, (CH 3 COOH) 2. 16-8. Because CaCl 2 (s) is a strong electrolyte, we use Equation 16.16 with M c 3M to write RTM c (0.083145 L bar mol 1 K 1 )(298 K)(3)(0.10 mol L 1 ) 7.4 bar The value and units of depend upon the value of R used. 16-9. Use Equation 16.16 with M c M to write M RT c 19.0 Tr (62.3637 L Tr mol 1 K 1 )(298 K) 1.02 10 3 mol L 1

46 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly ( ) 0.550 g One liter of the solution will contain (1000 ml) 11.0 g. Thus we have the 50.0 ml crespondence 1.02 10 3 mol 11.0 g 1 mol 1.08 10 4 g 16-10. We use Equation 16.20 in the notation P total Ptol + x ( benz P benz Ptol ) Substituting in the values given, we have x benz 0.329. Now 16-11. We use Equation 16.21 to write 100.0 Tr 59.2 Tr + x benz (124 Tr) y benz P benz (0.329)(183 Tr) 0.602 P total 100.0 Tr M gas P gas k h 1.00 bar 0.10 M 10 bar M 1 University Science Books, 2011. All rights reserved. www.uscibooks.com

CHAPTER 17. Chemical Kinetics: Rate Laws 17-1. See the text. 17-2. The average rate of production of O 2 (g) is rate of reaction [O 2] t 17-3. The equation f the reaction is (0.28 0.16) 10 2 M (20.0 10.0) min 1.2 10 4 M min 1 N 2 O 5 (g) 2NO 2 (g) + 1 2 O 2(g) Accding to the chemical equation, one half of a mole of O 2 (g) is produced f every mole of N 2 O 5 (g) consumed. Therefe, at t 70.0 minutes 17-4. Write [N 2 O 5 ] [N 2 O 5 ] 0 2[O 2 ] 1.24 10 2 M (2)(0.55 10 2 M) 1.4 10 3 M [NO 2 ] 4[O 2 ] (4)(0.55 10 2 M) 0.022 M (rate of reaction) 0 k [CH 3 N 2 CH 3 ] x 0 where x is the der of the reaction. Dividing the data f run 2 by the data f run 1 gives 3.65 10 4 ( ) 0.913 x 2.42 10 4 0.604 1.51 (1.51) x x 1. You obtain the same result using the data from the other runs. To evaluate k, we use run 2 (arbitrarily) to obtain 17-5. Write k (rate of reaction) 0 [CH 3 N 2 CH 3 ] 0 3.65 10 4 1 mol L 1 s 4.00 10 4 s 1 0.913 mol L 1 (rate of reaction) 0 k [NO] x 0 [Br 2] y 0 47

48 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly Dividing the data f run 2 by the data f run 1 gives 2.25 (1.50) x x 2. Dividing the data f run 3 by run 2 gives 3.00 (3.00) y y 1. Thus the rate law is rate of reaction k [NO] 2 [Br 2 ] We evaluate the rate constant using the data f run 2 (arbitrarily). k (rate of reaction) 0 [NO] 2 0 [Br 2] 0 2.93 10 3 M min 1 (1.50 M) 2 (1.00 M) 1.30 10 3 1 M 2 min and so rate of reaction (1.30 10 3 M 2 min 1 )[NO] 2 [Br 2 ] 17-6. Use Equation 17.19 fraction [A] [A] 0 e kt e (0.015 min 1 )(60 min/1 hr)(1.0 hr) e 0.90 0.41 17-7. From Example 17-4, the rate constant of the reaction is 2.20 10 5 s 1. Using Equation 17.20, the half-life is t 1/2 0.693 3.15 10 4 s 8.75 hr k The number of half-lives in 35 hours is 35/8.75 4.0. Using Equation 17.21, the fraction remaining after 35 hours (4.0 half-lives) is given by 17-8. See the text. [A] [A] 0 fraction remaining ( ) 1 n 2 ( ) 1 4.0 0.0625 2 17-9. Use Equation 17.23 ln N ( ) 0.010N0 ln N 0 N 0 ln(0.010) 0.693 t 2.41 10 4 y Solving f t gives 1.6 10 5 years. 17-10. We use Equation 17.23 with t xt 1/2, where x is the number of half-lives. Realize that 0.010% of its iginal value means that N/N 0 0.000 10. ln N ( ) ( ) 0.693 0.693 ln(0.000 10) t xt 1/2 0.693 x N 0 Solving f x gives 13 half-lives. t 1/2 t 1/2 University Science Books, 2011. All rights reserved. www.uscibooks.com

Chapter 17: Chemical Kinetics: Rate Laws 49 17-11. We use Equation 17.27 [ ] t (8.27 10 3 1 15.3 disintegration min 1 g years) ln (0.928)(15.3 disintegration min 1 g 618 years 1 ) 17-12. Let s see whether the reaction is first-der second-der. The following two figures show ln([no 2 ]/M) plotted against t (linear f a first-der reaction) and 1/[NO 2 ]/M 1 plotted against t (linear f a second-der reaction). You can see from the figures that the reaction is second-der. 10 t/s 20 30 40 50 ln([no 2 ]/M) 2.0 2.5 3.0 3.5 1/([NO 2 ]/M) 30 25 20 15 10 5 10 20 30 40 50 t/s We can calculate the value of the rate constant by using the various data sets in Table 17.10 and averaging. F example, using the data f the t 10.0 s run (arbitrarily) we obtain from Equation 17.31 1 0.096 M 1 + k (10.0 s) 0.20 M k 0.54 M 1 s 1. The other data sets give 0.54 M 1 s 1,0.54 M 1 s 1,0.53 M 1 s 1, and 0.53 M 1 s 1. 17-13. We are given P total in Table 17.13, so we must first determine P N2 O 3 at various time intervals. P total P N2 O 3 + P NO + P NO2 Because we start with only N 2 O 3 (g), P NO P NO2. Thus, P NO P NO2 P 0 N 2 O 3 P N2 O 3

50 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly where P 0 N 2 O 3 is the initial value of P N2 O 3. Therefe, we have P total P N2 O 3 + 2 P 0 N 2 O 3 2 P N2 O 3 2 P 0 N 2 O 3 P N2 O 3 P N2 O 3 2 P 0 N 2 O 3 P total where P 0 N 2 O 3 400 Tr. The following two figures show ln(p N2 O 3 /Tr) plotted against t (linear f a first-der reaction) and (1/P N2 O 3 )/10 3 Tr 1 plotted against t (linear f a second-der reaction). You can see from the figures that the reaction is first-der. 6.0 ln(pn2o3/tr) 5.5 5.0 5 10 15 20 t /s 1/(PN2O3 /10 3 Tr) 8 7 6 5 4 3 5 10 15 20 t /s We can calculate the value of the rate constant by using the various data sets in Table 17.13 and averaging. F example, using the data f the t 5.0 s run (arbitrarily) we obtain from Equation 17.18, ln 2 P N 2 O 3 P total P N 2 O 3 ln (800 504) Tr 400 Tr k (5.0 s) k 0.060 s 1. The other runs give 0.060 s 1,0.060 s 1, and 0.060 s 1,0.060 s 1 to two significant figures. 17-14. We use Equation 17.35 with P C 5 H 6 500 Tr and k 7.16 10 5 Tr 1 s 1 : t 1/2 1 kp C 5 H 6 1 (7.16 10 5 Tr 1 s 1 )(500 Tr) 27.9 s University Science Books, 2011. All rights reserved. www.uscibooks.com

CHAPTER 18. Chemical Kinetics: Mechanisms 18-1. See the text. 18-2. Using Figure 18.3 as a guide, we see that the value of the activation energy in the reverse direction is 80 kj mol 1 + 50 kj mol 1 130 kj mol 1. 18-3. The time, t, f an egg to boil is inversely proptional to the value of the rate constant at that temperature. The larger the rate constant, the less time it takes to boil the egg. Using Equation 18.9, we have ln k 100 C k 92.2 C ln t 92.2 C t 100 C 0.29 (42 103 J mol 1 )(373.2 K 365.4 K) (8.3145 J mol 1 K 1 )(373.2 K)(365.4) t 92 t 100 e 0.29 (3.0 min)(1.34) 4.0 min 18-4. The slow step controls the overall rate of reaction, and being an elementary step, we see that 18-5. Step (2) controls the overall rate of reaction rate of reaction k 1 [Fe 2+ ] [HNO 2 ] rate of reaction k 2 [NO] [NO 3 ] We can eliminate [NO 3 ], an intermediate species, using Step (1) [NO 3 ] k 1 k 1 [NO] [O 2 ] Substituting this expression into the rate expression f step 1 yields rate of reaction k 1 k 1 k 2 [NO] 2 [O 2 ] k[no] 2 [O 2 ] which is consistent with the experimentally determined rate law. 51

52 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly 18-6. Start with (see Example 18-6) where 2 is the catalyzed reaction and 1 the uncatalyzed reaction Assume that A 1 A 2 and subtract to get ln k 2 E a 1 E a2 k 1 RT 2.02 ln k 2 E a 2 RT + ln A 2 ln k 1 E a 1 RT + ln A 1 E RT 18-7. We must first make a table of 1/R and 1/[S] 5.00 10 3 J mol 1 (8.3145 J mol 1 K 1 )(298.2 K) k 2 k 1 e 2.02 7.5 (1/R)/mM 1 s 41.6 27.8 18.9 16.7 15.6 (1/[S])/mM 1 0.40 0.20 0.10 0.067 0.050 A plot of 1/R against 1/[S] is a straight line. 40 (1/R)/mM 1 s 35 30 25 20 0.1 0.2 0.3 0.4 (1/[S])/mM 1 Using the second and third data sets (arbitrarily), we calculate the slope to be slope 18.9 mm 1 s 27.8 mm 1 s 8.9 mm 1 s 0.10 mm 1 0.20 mm 1 0.10 mm 89 s K M 1 R max by Equation 18.32. The intercept is about 12 mm 1 s, so 1 R max 12 mm 1 s and so R max 0.083 mm s 1 and K M 7.4 mm. We used just two data points to determine the slope. A me sophisticated treatment of the linear plot gives R max 0.085 mm s 1 and K M 6.4 mm. University Science Books, 2011. All rights reserved. www.uscibooks.com

CHAPTER 19. Chemical Equilibrium 19-1. From the stoichiometry of the chemical equation, we have ( ) ( )( ) change in moles moles per liter 2 mol SO2 per liter of SO 2 of O 2 reacted 1 mol O 2 [ ] ( ) (3.500 0.075) mol O2 2 mol SO 2 1L 1 mol O 2 Therefe; 6.850 mol L 1 [O 2 ] eq 0.075 M [SO 2 ] eq 8.500 M 6.850 M 1.650 M [SO 3 ] eq 5.575 M + 6.850 M 12.425 M 19-2. See the text. Recall that species in a liquid solid state do not enter the equilibrium-constant expression. 19-3. From the data given, we can construct the following concentration table Concentration 2 SO 2 (g) + O 2 (g) 2SO 3 (g) Initial 3.00 M 6.00 M 1.00 M Change +0.40 M Equilibrium 1.40 M Because the change in [SO 3 ]is+0.40 M, we have that the change in [SO 2 ] to reach equilibrium is 0.40 M, and that of [O 2 ]is 0.20 M. Thus, the final concentration table is Concentration 2 SO 2 (g) + O 2 (g) 2SO 3 (g) Initial 3.00 M 6.00 M 1.00 M Change 0.40 M 0.20 M 0.40 M Equilibrium 2.60 M 5.80 M 1.40 M And so, K c [SO 3] 2 [SO 2 ] 2 [O 2 ] (1.40 M) 2 0.0500 (2.60 M) 2 M 1 (5.80 M) 53

54 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly 19-4. We use Equation 19.18 K p K c (RT) ν gas K c (RT) 1 25 atm 1 K c (25 atm 1 )(0.08206 L atm mol 1 K 1 )(1020 K) 2.1 10 3 M 1 19-5. We set up a concentration table Now write Concentration 2 S(g) + C(s) CS 2 (g) Initial 0.700 M 0M Change 2x +x Equilibrium 0.700 M 2x x K c [CS 2] [S] 2 Put this in the standard fm of a quadratic equation: x 9.40 M 1 (0.700 M 2x) 2 37.6 M 1 x 2 27.3x + 4.61 M 0 Solving f x gives 0.266 M and 0.467 M. We reject the 0.467 M root because [S] would be negative. Therefe [CS 2 ] 0.266 M. 19-6. We set up a pressure table Pressure C(s) + CO 2 (g) 2CO(g) Initial 0.22 atm 0 Change x +2x Equilibrium 0.22 atm x 2x Now write K p 1.90 atm (2x) 2 0.22 atm x 4x 2 0.22 atm x 4x 2 + 1.90 atm x 0.418 atm 2 0 Solving f x gives 0.164 atm and 0.639 atm. From the stoichiometry of the reaction, x must be a positive number, so we reject the negative value. Therefe, P CO2 0.22 atm 0.164 atm 0.06 atm P CO (2)(0.164 atm) 0.33 atm P total P CO2 + P CO 0.06 atm + 0.33 atm 0.39 atm 19-7. (a) The equilibrium-constant expression tells us that P NH3 P HI 18 bar 2 P NH3 P HI, and so P NH3 4.24 bar. The total pressure is (b) In this case, P total P NH3 + P HI 8.5 bar P NH3 P HI (5.0 bar + x)(x) 18 bar 2 University Science Books, 2011. All rights reserved. www.uscibooks.com

Chapter 19: Chemical Equilibrium 55 where x is the equilibrium pressure of HI(g). Solving f x gives 2.4 bar and 7.4 bar. We obviously reject the second root, so that P HI 2.4 bar P NH3 5.0 bar + 2.4 bar 7.4 bar P total 7.4 bar + 2.4 bar 9.8 bar 19-8. (a) In this case, the concentration table is Concentration H 2 (g) + Cl 2 (g) 2 HCl(g) Initial 0.500 M 0.500 M 0.500 M Change x x +2x Equilibrium 0.500 M x 0.500 M x 0.500 M + 2x Using the equilibrium-constant expression gives (0.500 M + 2x) 2 2.5 10 4 (0.500 M x) 2 0.500 M + 2x 0.500 M x ±0.0158 giving x 0.244 M 0.256 M. The concentration of HCl(g) must be a positive quantity, so we reject the 0.256 M root. Therefe, [H 2 ] [Cl 2 ] 0.744 M [HCl] 0.012 M As a check, (b) See the text. [HCl] 2 [H 2 ] [Cl 2 ] (0.012 M)2 2.6 10 4 (0.744 M) 2 19-9. The equation SO 2 (g) + 1 2 O 2(g) SO 3 (g) is 1/2 of the reverse of so 2SO 3 (g) 2SO 2 (g) + O 2 (g) K p 1 1.9 atm 1/2 (0.29 atm) 1/2 19-10. (a) Decreasing [CO] will drive the reaction to the right, and so [CO 2 ] decreases. (b) Changing the amount of C(s), a solid, has no effect on the equilibrium. 19-11. The value of Hrxn f the reaction as written is H rxn H f [SO 3] { H f [SO 2] + 1 2 H f [O 2(g)]} ( 395.7 kj mol 1 ) {( 296.8 kj mol 1 ) + 0kJ mol 1 } 98.9 kj mol 1 < 0

56 GENERAL CHEMISTRY, FOURTH EDITION McQuarrie, Rock, and Gallogly Because Hrxn < 0, the reaction is exothermic. Therefe, a decrease in temperature will shift the equilibrium to the right, increasing the production of SO 3 (g). 19-12. The equation f the reaction can be written as Using the data given, the value of K p is K p P SbCl 5 P SbCl3 P Cl2 SbCl 3 (g) + Cl 2 (g) SbCl 5 (g) 0.228 bar 0.777 bar 1 (0.670 bar)(0.438 bar) We set up the concentration table with the initial pressures halved because we have doubled the volume of the container. Pressure SbCl 3 (g) + Cl 2 (g) SbCl 5 (g) Initial 0.335 bar 0.219 bar 0.114 bar Change +x +x x Equilibrium 0.335 bar + x 0.219 bar + x 0.114 bar x Using the equilibrium-constant expression, we have 0.114 bar x 0.777 bar 1 (0.335 bar + x)(0.219 bar + x) 0.777 bar 1 x 2 + 0.430x + 0.057 bar 0.114 bar x 0.777 bar 1 x 2 + 1.430x 0.057 bar 0 The two roots are x 0.0390 bar and 1.88 bar. We reject the second root because [SbCl 3 ] and [Cl 2 ] must be positive quantities. Therefe, P SbCl3 0.335 bar + 0.0390 bar 0.374 bar P Cl2 0.219 bar + 0.0390 bar 0.258 bar P SbCl5 0.114 bar 0.0390 bar 0.075 bar Note that the equilibrium position shifts from right to left, in accd with Le Châtelier s principle. 19-13. In this case, Q P P 2 CO P CO2 (485 Tr)2 150 Tr 1570 Tr and Q P K p 1570 Tr 1.93 Tr > 1 Because Q P /K p > 1, the reaction will proceed from right to left until Q P K p. University Science Books, 2011. All rights reserved. www.uscibooks.com