CHP14 - I/. 14.28 No; the reaction only appeared to have...

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Unformatted text preview: I/. 14.28 No; the reaction only appeared to have stopped. The amount of product present was not increasing because the rate of the reverse reaction was equal to the rate of the forward reac— tion, with the result that product was disappearing at the same rate as it was being formed. Because both the forward reaction and the reverse reaction are going on even though con— centrations are not changing, the state. is called dynamic equilibrium (dynamic : moving; equilibrium : not changing). 14.29 The position of a reaction’s equilibrium refers to whether there are more products or more reactants present at equilibrium. When there are more reactants, the equilibrium is said to lie toward the left end of the reaction coordinate; when there are more products, the equi- librium is said to lie toward the right end of the reaction coordinate; when there are signifi- cant amounts of both products and reactants, the equilibrium is said to lie toward the mid— dle of the reaction coordinate. 14.30 Far to the right because, by convention, the left end of the coordinate is taken as the starting point of the reaction and the right end is taken as the 100% point. 14.31 Far to the left. That the reaction appears not to occur means the reaction vessel contains essentially all reactants—in other words, the reaction is still at the starting point. By con— vention, the left end of the reaction coordinate is taken as the starting point. 14.32 That the reaction is reversible and runs in both directions, reactants converting to products and products simultaneously converting to reactants. 14.33 The reaction would produce very little product if the equilibrium lay far to the left, mean- ing the chemist would end up with a vessel containing mostly the reactants she started with, no matter how long she waited. _ 14.34 As the forward reaction proceeds, there is a decrease in the amount of forward—reaction reactants. As the concentration of these reactants decreases, the forward reaction slows dOWn. 14.35 Initially there are no forward-reaction products in the vessel. Because these products are the reverse-reaction reactants, the reverse reaction rate is initially zero. 14.36 As the forward reaction proceeds, there is an increase in the amount of forward—reaction products. Because these products are the reverse—reaction reactants, as their concentration increases, the reverse reaction speeds up. 14.37 The reverse reaction because those products present are the reverse-reaction reactants. At the moment after loading, there are no forward-reaction reactants in the vessel, which means the forward reaction cannot occur. Therefore the reverse reaction must be going faster because the rate of the forward reaction is zero. 14.38 Yes. Once the reverse reaction produces products, these products become forward-reaction reactants. As time passes, the reverse reaction slows down (because reverse-reaction reac— tants are consumed) and the forward reaction speeds up (because forward—reaction reac- tants are being formed). Equilibrium is attained when ratemrward = ratemerse. 14.39 Yes. In the sealed container, the liquid water evaporates until the air in the container is saturated with water vapor. After that point, the rate at which liquid water evaporates is equal to the rate at which water vapor condenses, which means the water level remains constant. 14.40 Equilibrium is the state in which the rate of the forward reaction is equal to the rate of the reverse reaction. Because the general form of the rate law is Rate = k[Reactants]”"d”, the definition for equilibrium Rateforward I Ra tereverse kforward[ReaCtantslordfl : kl‘t'versr’ll) rOd uctS] order 14.41 No, because it is not possible to reach equilibrium if both rates increase, which is what the rate meters show. As the reverse reaction proceeds, the reverse—reaction reactant concentra— tions decrease, which means Ratereverse = kreverse[Reactantslmder must decrease. CHAPTER 14: CHEMICAL EQUILIBRIUM 447 /———————7 this is a gas-state reaction, all four species are gases and the balanced equation. Because Km] expressmn: therefore all four appear in the : [ClxlDl eq [A]2 X [1313 14.49 Put product concentration Raise each concentration to the balanced equation. Because therefore all four appear in the K _lA12 ><IBP “1 ‘ ICIxIDI 14.50 The two KEGl values a tor in Problem 14.49; conclude that the Kuq for a reVe reaction. 14.51 Put product concentrations Raise each concentration to I the balanced equation. Because the Keq expression: K 1 [N02]2 e“ [N204] 14.52 Put product con Raise each concentration to a po the balanced equation. Because a Keq express1on: _ [NO]4 x[HZO]6 K _ 4 5 eq [NHsl Xlozl 14.53 You would place some N304 in a container and allow the reaction to proceed to equilibri— um. Then you would determine the concentration of N20,1 and NO2 at equilibrium and substitute those concentration values into the Keq expression shown in the solution to , Problem 14.51. 14.54 No, because Keq depends only on temperature. If y tant B at 25°C and use equilibrium concentrations to calculate a K,eql 10 M reactant A with 7 M reactant B at 25°C also gives IKeq = 30. Only if you to tion at some temperature other than 25°C will your calculated value for Keq change. NO 2 . 1 2 q = l 1 =—e—(0 ULM) :4ex10‘4 [N2]><[Og] 0.25 Mx1.2 M C lies farther to the left because Keq is smaller at 25°C than at 2000°C. the fraction [Products]/ [Reactants]. The smaller the K9 value, ator, the higher the reactant concentration, and the farther K concentrations in the denominator. tance’s stoichiometric coefficient in pecies are gases and s in the numerator, reactant a power equal to the subs this is a gas—state reaction, all four 5 K8q expressmn: 11 Problem 14.48 is denomina— or in 14.49. From this you can f the Keq for the forward hat is numerator i 14.48 is numerat l to the inverse 0 re inversely related: w what is denominator in rse reaction is equa in the numerator, reactant concentrations in the denominator. a power equal to the substance's stoichiometric coefficient in both species in the reaction are gases, show both of them in centrations in the denominator. s stoichiometric coefficient in gases, show all of them in the centrations in the numerator, reactant con wer equal to the substance ll species in the reaction are ou react 1 M reactant Awith 1 M reac— of, say, 30, then reacting n the reac- 14.55 Ke 14.56 The equilibrium at 250 The Keq value comes from therefore, the larger the denomin to the left the equilibrium lies. 14.57 That the concentration of products reactants. The general expression 5 at equilibrium is much greater than the concentration of hows that the numerator (product concentration) must be much larger than the denominator (reactant concentration) if Keg > 103. 1458 N0; Keq : 2.05 means the equilibrium lies near the middle of the reaction coordinate. Only for reactions whose equilibria lie extremely far to the right (K8L1 2> 105) may you use a single arrow to the right, indicating that the reaction goes essentially to completion. a.“ ...... m.turrWuW‘wu—.WWMMW_M‘~....MMWMW. m...“MMWMWWM.“new”.-... . . 449 CHAPTER 14: CHEMICAL EQUILIBRIUM ntration) if Kt, < 10‘3 14 60 No, because K = 1 5 >< 10‘ means the equilibrium hes far to the left. There would be very little product present when the reaction reaches equihbrlum. K8 q [C0] X lel 0.387 mol —00387M HO: ~00387M 10.0 L I 10.0 L 0.0387 M x 0.0387 M Ke i—t—‘m : 5 3.93 q 0.0613 M ><(0.1839 M)3 - ' ghly '3, indicating equilibrium essential] = 103, in e does not change if the temperature at which a reaction is run does not Change 4 [Cszlxlflzl 2 = 3.59 [Cde [H28] To determine whether or not the reaction i trations to see if the ra s at equilibrium, substitute in the given concen- tio equals 3.59: and any reaction for which K, eq is greater than 105 goes to completion 1081) 3 hardly occurs at all. Thus reaction a (Keq “F: lly all of the reactants converted to prod- eq — 10595) does not happen, with almost none COMPLETE SOLUTIONS The equilibrium molar concentrations are 0'65” “‘01 w 0.0812 M [121: 0275 mm 100344 M [H11 2 M : 0.375 M [P121 = — 8.00 L 8.00 L 8-00 L (0.375 M); _ 50.3 K 2 ,._..7s _ E‘i 0.0812 M x 0.0344 M The equilibrium lies near the middle of the reaction coordinate because 50.3 is roughly halfway between Keq : 10—3, indicating equilibrium essentially all the way to the left (main— ly reactants present), and Keq 1' 103, indicating equilibrium essentially all the way to the right (mainly product present). 14.68 The two values would be the same reaction. " 14.69 The value of K, does not change b changes only w en the temperatur 14.70 When a chemical reaction at equilibrium is disturbe partially undo the disturbance. 14.71 (a) To the right because Cl2 appears on the left side of the equation. Equilibrium shifts right- ward to consume some of the added C12 via the reaction PCl3 + Cl2 —-> PC15. (b) To the left so that the reaction PCl5 ——> PCl3 + Cl2 produces more Cl2 to replace some of the removed C12. (c) To the left so that the reaction (d) To the left so that the reaction PCIS a PCl3 + Cl2 pro of the removed PC13. ' (e) No shift becauSe H2 does not take PC13/ C12 / PCl5 equilibrium. 72 Because when you remove reactants, product to reactants to partially replace the .73 (a) To the left because the reverse reaction, some of the added products and restore equilibrium. (b) Greater than before the disturbance. Some of the added products are converted to reac- tants as the new equilibrium is established, but the net result is an increase in the amount of products (and an increase in the amount of reactants, produced when the reaction shifted left). In other words, the equilibrium shift never consumes all of the substance (or substances) added to disturb the equilibrium. (c) Greater than before the disturbance. See explanation in part (b). 74 (a) You are free to choose any arbitrary needle position for the initial setting. The only restriction is that the forward rate must be identical to the reverse rate because the mixture is at equilibrium: 9 0 Forward Reverse (’0) Because PCl5 appears on the right in the equation, adding it shifts the reaction left. This means the reverse rate increases as the reaction PCI5 ——> PCl3 + Cl2 speeds up to consume some of the added PC15; therefore your reverse meter must read higher than because .lcf/kr :: Kml whenever the reaction is a one-step ecause you have not changed the temperature. Km] e at which the reaction is run changes. d, the reaction shifts in such a way as to PCl5 —r9 PC13 + Cl2 uses up some of the added PC15. duces more PC13 to replace some part in the reaction and therefore does not affect the _ the reaction shifts to the left, converting some of the reactants you took away. Products —-’ Reactants, speeds up to consume CHAPTER 14: CHEM1CAL EQU'ILIBRIUM in (a). Because the speeded—up reaction PC15 —» PC]3 -|- C12 creates more PC]3 + C12, after a few moments the forward rate also increases (only slightly at first): Q 9 Forward Reverse (c) As more and more PC}3 and C12 are formed, the forward rate continues to increase. At the same time, the reverse rate is decreasing because I’Cl5 is being used up. Equilibrium is restored when the increasing forward rate and the decreasing reverse rate meet: 00 Forward Reverse Note that the new equilibrium rate is higher than the initial equilibrium rate of part (a). (d) Left because the initial disturbance was caused by adding a substance that appears on the right in the balanced equation. 14.75 At a high temperature because this is an endothermic reaction (AISrxn positive). Therefore heat is a “reactant”: 2 H20(g) + Heat : 2 H2@) + 02(g). Adding a substance that appears on the left in the equation (heat in this case) shifts the reaction right, forming greater amounts of H2 and 02. 14.76 (a) About the middle of the reaction coordinate, with significant amounts of both product and reactants present, because 0.105 is about midway between KEq = 10—3 (mainly reac— tants present) and Keq = 103 (mainly product present). (b) No, because with Keq = 0.105 there are significant amounts of the reactants present at equilibrium. (You need a reaction having a mnch larger KEq if you want to isolate pure product.) (c) You would drive the reaction to completion by causing it to shift continuously to the right. 14.77 (a) To the left, because K3 is smaller at the higher temperature. The smaller the Keq = [Productsl/[Reactantsl1 value, the smaller the numerator. That the product concentration decreases tells you the reaction has shifted left. (b) Exothermic, because increasing the temperature (adding heat) causes the reaction to shift to the left. Heat must therefore be a product in the reaction. (c) CO(g) + 2 H2(g) : CH30H(3) + Heat 14.78 (a) Endothermic, because increasing the temperature (adding heat) causes the reaction to shift to the right. Heat must therefore be a “reactant.” (b) Increase because more product is produced, increasing the numerator in Keq = [Products}/ [Reactants]. (c) Heat + N204(g) 2 2 N02(g) 14.79 Exothermic, because the reaction shifts left when the reaction mixture is heated. Therefore heat must be a product, Cdiammm : Cgmphite + Heat. 14.80 An endothermic reaction means heat is a "reactant": Reactants + Heat .7: Products. To drive the reaction to the right, therefore, you add heat by increasing the temperature at which the reaction is run. 14.81 An exothermic reaction means heat is a product: Reactants :2 Products + Heat. (a) Heating means adding to the right side of the equation, shifting the reaction left. This causes [Products] to decrease and [Reactants] to increase. Therefore Keq : [Products] / [Reactants] decreases. MW 452 COMPLETE SOLUTIONS ——————m (b) Cooling means removing from the right side of the equation, shifting the reaction right. This causes [Products] to increase and [Reactants] to decrease. Therefore K,eq : [Products]/ [Reactants] increases. 14.82 An endothermic reaction means heat is a reactant: Reactants + Heat : Products. (a) Heating means adding to the left: side of the equation, shifting the reaction right. This causes [Products] to increase and [Reactants] to decrease. Therefore Km] 2 [Products] / [Reactants] increases. (b) Cooling means removing from the left side of the equation, shifting the reaction left. This causes [Products] to decrease and [Reactants] to increase. Therefore Km : [Products]/[Reactants] decreases. I 14.83 Decreased. The negative value for AEM tells you this is an exothermic reaction. Thus heat is a product, CO(g) + 2 112(3) : CH3OH(3) + Heat, and adding it to the reaction mixture by raising the temperature shifts the reaction left. [HZ] and [CO] increase, and [CH3OH] decreases. 14.84 Cooling slows down the rate of the reaction. It therefore takes longer to form the product. 14.85 Because it speeds up the forward and reverse reactions by the same amount, a catalyst has no effect on either the position of equilibrium or the value of Km. 14.86 A catalyst decreases the time it takes for a reaction to reach equilibrium by providing an alternative mechanism that has a lower activation energy. A lower activation energy means that more molecules have sufficient energy to react, and the forward and reverse reaction rates both increase. 14.87 (a) That K9q = [NO]2/ [N2] X [02] is higher at the higher temperature means that at the higher temperature [NO] must be higher and [N2] X [02] must be lower. Therefore the reaction shifts to the right. _ (b) Endothermic. The fact that heating shifts the reaction to the right means heat must be a : "reactant." 14.88 One in which not all the reactants and products are in the same phase. A heterogeneous reaction occurs at the interface between the phases present in the reaction mixture. 14.89 Show only (g) and (ray) species in a Kqu expression. Put product molar concentrations in the numerator, reactant molar concentrations in the denominator. Raise each concentration to a power equal to the substance's stoichiornetric coefficient in the balanced equation. (a) K.., = IHCIIfi/IHzor (b) I<..q = ICOZIB/ICOI‘ (c) K. [PW] >< 1603'] (d) Ker] = [Hzcosl/ [CO] 14.90 Show only (3) and (aq) species in a K6,] expression. Put product molar concentrations in the numerator, reactant molar concentrations in the denominator. Raise each concentration to a power equal to the substance’s stoichiometric coefficient in the balanced equation. (a) K..., = 1/[H212 _ (b) Keq = [130431 >< [H30+]3/[H3ro,] (c) Km] : 1/[Pb2+] x [1"]2 (d)K..q = [H30*12/[Ca2+l >< ICOZI 14.91 The fact that the concentrations of solids and liquids do not change during any hetero« q geneous reaction. 15,432 AgCl(s) precipitates. When the NaCl is added, [Cl‘] increases drastically, causing the -- product [Ag+] >< [Cl'] to exceed KHP. 141-93 (a) Because the sealed vessel keeps the H2(g) from escaping. In an open container, H2(g) would escape, continuously shifting the reaction to the left so that no product could form and equilibrium would never be established. CHAPTER 14: CHEMICAL EQUILIBRIUM WWW-“MMWWMMqumnmm"WWW—N an». .. .. —————_————‘~—‘——_—.—'_—”Tfi 14.99 Lead(II) chloride, because it is the one with the largest Kw. -10 1 r T { 78 T 14.100 (a) 4.49x10 11ml8401+]th108.8715pqOH)3 = 4.80%“) w L 1 {1.101 .lie(OH)-3 L (b) Table 14.2 of the textbook tells you that for any sparingly soluble salt of the form XY3, the cation equilibrium concentration is equal to s, the salt's solubility, and the anion equilibrium concentration is equal to 35. Therefore [Fe3 +1 = s : 4.49 X 10'1UM and [OI-1'] = 3s = 1.35 x 10—9 M. 3.79 o x 4.80 x10 “g 1=e(OH)3 1 gallon 1.; 14.101 (a) Table 14.1 of the textbook gives 3.3 X 10TH for the 25°C KSp for PbCO3. Because this is a 1:1 salt, 5: [KSP = 8.8x10‘14 :1.8><10‘71nol/L (c) 20,000 allonsx —3.84x10‘3 Fe(OH) .g g 3 1.8 x 10‘7_m81 Pbcog X 267.2 g Pbcoi 7 4.8 x 10‘5g7PbCCfi L irriolt-P-bCC): L (c) You know from Table 14.2 of the textbook that for a 1:1 salt the equilibrium con— centration of both cation and anion is equal to 5. Therefore [Pb2+] = [C0321 I 1.8 x 10-7 M. (b) . ’ . :1 ‘5 (d)1.000><106gallon3>< 13 72} x 4 835- 0 ngCO3 = 1.8 ><102 PbCO gallon L g 3 14.102 From Table 14.1 of the textbook, on know KSP = 1.0 X 10—26. Because the dissolution 3:; reaction is SnS(s) T“: Sn2+(aq) + S “(013), you know Ksp I [Sn2+] X [52']. Therefore L. K. ~26 7, . g [827 = 51:) _ 1.0 X WV 2 5.0 X 10,43 M ‘1 [Sn2+] _ 2.00x10’1M 14.103 PC15(g) : PC13(g) + C12(g); Keq : [PCI3] >< [C12]/[PC15] _ [PC13]><[C12] : 5.50).: 10’3M x 0.125 M PCl _ a I 5] Keq 7.50x10“ — 9.17x10’3M NH 2 14.104 (3) Keq = WEE}; ('3) K _ INIng eq—#T—_ [H2] ><[N2] [NH3}2 = Km] ><{H2]3 ><[N2] [NH3]: Keq xIHZP XlNzl : J05 x104) X (0.20 M)3 x 0.20 M 42.4 x 16:3 = 4.9 ><1tT3 M £14,105 (a) To the right. Because the reverse reaction is so much slower than the forward reac- tion, there must be a large buildup of product (AB) before the reverse reaction is going fast enough to match the rate of the forward reaction. The large amount of AB means the equilibrium lies to the right. Mumm-itM.WWW.tWMWWW“-Mh N am WW..." .. mmWWMIWWWWumm“ V.............W..N muhtwm “AWN—m.“ mmmmmmvmwmmv. Wham-M CHAPTER 14: CHEMICAL EQUILIBRIUM 45 (b) Because the forward rate is so much faster, kf must be much larger than kt. There- fore the fraction kf/ k1. = Keq must be a large number, which means the equilibrium lies to the right, just as in part (a). 14.106 Because this is a simple reaction involving a one—step collision between two reactant molecules, the fact that kf >> kr, established in Problem 14.105, means that 15a for the forward reaction is much smaller than Ea for the reverse reaction. The forward reac— tion is faster because it has the smaller activation—energy barrier to overcome. 14.107 (a) The reverse reaction is faster because it is the only reaction going on at that moment. The forward reaction is not happening because initially there are no reactants for it in the container. (b) The balanced equation tells you that if you begin with only NO, the molar amount of 02 formed must be the same as the molar amount of N2 formed. Therefore the 02 concentration must be equal to the N2 concentration—both are 0.2159 M. — fl _ __(U.;9.6__83 M)2 = 0 100 “1 ' [N2]x[02] 0.2159 M x0.2159 M ' 14.108 Ke ={N20421 ‘1 [N02] N 0 {N002 =[ g 2] eq {N02} = [NZO41 = 0248 M = 40496 = 7.04 x10‘1 Keq 0.500 14.109 Keq must change because the temperature has changed. In an endothermic reaction"— Reactants + Heat : Products—~—heat is a reactant, which means adding heat by increasing the temperature shifts the reaction to the right, creating more product. Therefore [Products] in the fraction [Products] / [Reactants] = Kegl increases, [Reactants] decreases, and so Ke increases. 14.110 Show only (3) and (aq) species in a Keq expression. Put product molar concentrations in the numerator, reactant molar concentrations in the denominator. Raise each con- centration to a power equal to the substance's stoichiometric coefficient in the bal— anced equation. (a) Keq = [HC114/[H20F 02mm = [Col/[H2] >< [(30,] (c) Keq : [M1120 >< [Chi/[WP >< [CI—J2 Keq : (e) Keg : (0 Km =[1\112+1><[0H-i2 2 2 q [N2]x[H2] 0.25 M x (0.15 M) (b) To the right so that, by Le Chatelier’s principle, some of the added H2(g) is used up. (c) Because there is no mention of changing the temperature, you must assume tem— perature is held constant as the H2(g) is added. Therefore Keq does not change because only changes in temperature affect the value of Keq. (d) The new coefficients mean the expression for Keg changes to - [NHgl4 eq [Nle X [Hzlb “WM 456 COMPLETE SOLUTIONS To calculate the value of Keq, use the concentrations given at the beginning of the problem, now raised to the new powers: (0.25 M)2 x (0.15 M)" How can this be? There’s been no temperature change, and yet KeLl changes from 9.6 in (a) to 92 here just because the equation was multiplied through by 2. The answer is that any numeric value of Keq is constant for a given temperature but alsofur a given form of the balanced equation. =92 To see why this must be so, think about putting some N205!) and 112(3) together in a reaction vessel held at some constant temperature. (The amounts you use don‘t matter.) The two gases will react, an equilibrium state will be set up after some time has passed, and at that point the concentrations of N2, H2, and NH3 have attained their (constant) equilibrium values. Plug these values into the two ex— pressions for Keq, and you must get different numbers, as you saw when you calcu— lated your answers to (a) and (d). (Chemists do not usually run into any problem with this interesting phenomenon because published K6,i values are by convention taken to be for chemical equations written in their simplest formmin this case N2 + 3 H2 : 2 NH3 rather than 2 N2 + 6 H2 i 4 N113.) (e) A shift to the right means either (1) something was added to the left side of the equation N2 + 3 H2 2 2 NH3 or (2) something was removed from the right side. Cooling means the removal of heat, and therefore (2) is what happened: heat was removed from the right. Being that the right side is the products side, heat must be a product, which means the reaction is exothermic. 4.112 (a) You know from Table 7.1 of the textbook that halides of silver(I) are insoluble in water. Therefore the precipitate must be silver chloride, AgCl. (b) Ksp =IAg+]x[c1—] K. A + 3 33p [ g 1 [Cl‘] Get a value for Ksp from Table 14.1 of the textbook: —10 {Ag+] = fl = 8.0x10‘8M 2.0 x 10 M 13 Because KSP values are calculated with molar concentrations, a conversion is your first step: 7.517gAgGgl-l'30'2' X 1 mol AgC2H3O2 _ 4.50x10fmolAgc2H302 1 L 166.9 gAng-I 1302 L Because this is a 1:1 salt, this is also the concentration of both Ag+ and Cl'. Thus _ KSP 2 [Ag] >< [Cg-1302‘] = (4.50 x 10'2 M) x (4.50 x 10”2 M) r 2.03 x 10‘3 ‘14 Because this is a 1:1 salt, the solubility s is s = lep = 01.1 x 10—10 = 1.0x 10“5 M In 200 ml, therefore, 1.0 x 10—5 mol 321804 1000 mL' X 200.0 ml; = 2.0 x 10'6 mol BaSO4 CHAPTER 14: CHEMICAL EQUILIBRIUM 457 (b) Right; reducing the container voi u me has the effect of increasing the concentration of any gases in the container, and the only gas is the reactant C02. Thus you’ve essentially added to the reactant side, forcing a shift to more product. (c) Left; adding a substance that appears as a product in the equation as written shifts the reaction toward reactant. (d) It depends. If all the solid CaCO3 is removed, the reaction must shift left because you are taking away from the reactant side, and Le Chatelier’s principle says more reactant will be formed. However, if only some of the solid CaCO3 is removed, then the constant—density explanation given for graphite on page 565 of the textbook applies and the reaction does not shift. 14.12] (a) KSP, because the reaction is the dissolution in water of a sparingly soluble salt. [Ca2+] 3.3 x 10"4M (c) The rate constant k, for the reverse reaction. Because K < 1, it must be true that the denominator in the fraction .kf/kr : Keq is larger than the numerator. (d) Ea for the forward reaction. The slower the rate of a reaction, the larger the E,1 value for the reaction. Because kf is smaller {see (c)], the forward reaction is slower and therefore has the larger Ed value. (9) The rate of the reverse reaction because k1. is larger [see (c)]. (f) Li2CO3(s) .2 2 Li+(aq) + C032_(riq); Keq : Ksp : [Li+}2 >< [c0321 (g) Lithium carbonate; because these two salts give the same number of moles of ions when they dissolve (1 mol Ca2+ + 2 mol F' 2 3 mol ions; 2 mol Li+ + 1 mol C032" 2 3 mol ions), their solubilities are directly proportional to their KSP values. Li2CO3 has the larger K51, value and therefore is more soluble. 14.122 (a) The reverse reaction, as it has the smaller E2. (the smaller energy hump to get over). (b) kr > kf, because at a given temperature, the reverse reaction is faster, it has the larg- er rate constant. (c) It reaches equillibrium to the left. The reverse reaction is inherently faster due to its lower Ed. The only way to get the forward reaction to match the rate of the inher- ently faster reverse reaction at equilibrium is to have a. high concentration of reac— tants. 4.123 (a) They are the same. Km.1 is a constant for a reaction at a given temperature. (b) The rate meters for the new equilibrium must be wrong. While they must be equal (after all, we are at equilibrium), they can not be at identical setting to the equilibri- um before the disturbance. Since we added reactant, the meters should settle at some higher value than before the disturbance. 4.124 Equilibrium in this case is when the supply of tapes equals the demand for tapes (at ‘ this point, the number of tapes in inventory will not change). This occurs at $3 per tape / 4 million tapes / week (the point at which the supply and demand curves inter- sect). ,4-125 Somewhere between 7 and 8 minutes. It is between these limits that the concentra— tions of product and reactant stop changing. (b) K... =ica2+i><iF12 K. [1312 = 55. 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CHP14 - I/. 14.28 No; the reaction only appeared to have...

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