CHAPTER 15 PRINCIPLES OF CHEMICAL EQUILIBRIUM

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1 CHAPTER 15 PRINCIPLES OF CHEMICAL EQUILIBRIUM PRACTICE EXAMPLES 1A 1B (E) The reation is as follows: Cu (aq) Sn (aq) Cu (aq) Sn (aq) Therefore, the equilibrium exression is as follows: Cu Sn Cu Sn Rearranging and solving for Cu +, the following exression is obtained: 1/ + + 1/ Cu Sn + Cu = + x x x Sn 1.8x 1. (E) The reation is as follows: Fe (aq) Hg (aq) Fe (aq) Hg (aq) Therefore, the equilibrium exression is as follows: x Fe Hg Fe Hg Rearranging and solving for Hg +, the following exression is obtained: Fe Hg Hg M Fe A (E) The examle gives = for the reation N g + H g NH g The reation we are onsidering is one-third of this reation. If we divide the reation by, we should take the ube root of the equilibrium onstant to obtain the value of the 5 equilibrium onstant for the divided reation: B (E) First we reverse the given reation to ut NObgg on the reatant side. The new equilibrium onstant is the inverse of the given one. 1 NO g NO g + O g ' = 1/ (1. 10 ) = Then we double the reation to obtain moles of NObgg as reatant. The equilibrium onstant is then raised to the seond ower. 5 NO g NO g +O g = =

2 A (E) We use the exression = have n gas RT. In this ase, n gas = +1 = and thus we 9 6 = RT = = B (M) We begin by writing the exression. We then substitute P n / V RT =[ onentration] RT for eah ressure. We ollet terms to obtain an exression relating and, into whih we substitute to find the value of. { P( H)} { P( S)} H RT) RT [H [S RT RT P HS)} ([ ] ([S ] ) { ( HS] RT ] ] ([ ) [H S] The same result an be obtained by using RT n 1.10 = = =1.110 RT gas = b g, sine n gas =+1 = But the reation has been reversed and halved. Thus final = A (E) We remember that neither solids, suh as Ca5( PO) OH(s), nor liquids, suh as HO(l), aear in the equilibrium onstant exression. Conentrations of roduts aear in the + Ca 5 HPO numerator, those of reatants in the denominator. = + H B 5A (E) First we write the balaned hemial equation for the reation. Then we write the equilibrium onstant exressions, remembering that gases and solutes in aqueous solution aear in the exression, but ure liquids and ure solids do not. Fe s + H O g Fe O s + H g { PbH g} H = = Beause ngas = =0, = { PbHO g} HO (M) We omute the value of Q. Eah onentration equals the mass bmg of the substane divided by its molar mass (this quotient is the amount of the substane in moles) and further divided by the volume of the ontainer. 667

3 1 mol CO 1 mol H m m.0 g CO.0 g H 1 COH = V V.0.0 Q = = COH O 1 mol CO 1 mol HO m m 8.0 g CO 18.0 g H O V V (In evaluating the exression above, we anelled the equal values of V, and we also anelled the equal values of m.) Beause the value of Q is larger than the value of, the reation will roeed to the left to reah a state of equilibrium. Thus, at equilibrium there will be greater quantities of reatants, and smaller quantities of roduts than there were initially. 5B (M) We omare the value of the reation quotient, Q, to that of Q { P(PCl )}{ P(Cl )} = = = { P(PCl 5} = RT = RT = (61 7) = 1.99 Beause Q, the net reation will roeed to the right, forming roduts and onsuming reatants. 6A (E) Obgg is a reatant. The equilibrium system will shift right, forming rodut in an attemt to onsume some of the added O (g) reatant. Looked at in another way, O is inreased above its equilibrium value by the addition of oxygen. This makes Q smaller than. (The O is in the denominator of the exression.) And the system shifts right to drive Q bak u to, at whih oint equilibrium will have been ahieved.. 6B (M) (a) The osition of an equilibrium mixture is affeted only by hanging the onentration of substanes that aear in the equilibrium onstant exression, = CO. Sine CaO(s) is a ure solid, its onentration does not aear in the equilibrium onstant exression and thus adding extra CaO(s) will have no diret effet on the osition of equilibrium. b g will inrease CO above its equilibrium value. The reation will shift left to alleviate this inrease, ausing some CaCObsg to form. (b) The addition of CO g () Sine CaCO (s) is a ure solid like CaO(s), its onentration does not aear in the equilibrium onstant exression and thus the addition of any solid CaCO to an equilibrium mixture will not have an effet uon the osition of equilibrium. 668

4 7A (E) We know that a derease in volume or an inrease in ressure of an equilibrium mixture of gases auses a net reation in the diretion roduing the smaller number of moles of gas. In the reation in question, that diretion is to the left: one mole of NObgg is formed when two moles of NObgg ombine. Thus, dereasing the ylinder volume would have the initial effet of doubling both NO and NO. In order to reestablish equilibrium, some NO will then be onverted into N O. Note, however, that the NO onentration will still ultimately end u being higher than it was rior to ressurization. b g b g. 7B (E) In the balaned hemial equation for the hemial reation, n gas =1+11+1=0 As a onsequene, a hange in overall volume or total gas ressure will have no effet on the osition of equilibrium. In the equilibrium onstant exression, the two artial ressures in the numerator will be affeted to exatly the same degree, as will the two artial ressures in the denominator, and, as a result, Q will ontinue to equal. 8A (E) The ited reation is endothermi. Raising the temerature on an equilibrium mixture favors the endothermi reation. Thus, N O g higher temeratures and the amount of NO will be greater at high temeratures than at low ones. b g should deomose more omletely at bgg formed from a given amount of N Obgg 1 o 8B (E) The NH (g) formation reation is H 9A N g + H g NH g, = 6.11 kj/mol. This reation is an exothermi reation. Lowering temerature auses a shift in the diretion of this exothermi reation to the right toward roduts. Thus, the equilibrium NH g be greater at 100 C. b g will (E) We write the exression for and then substitute exressions for molar onentrations [H ] [S ] [HS].78 9B (M) We write the equilibrium onstant exression and solve for NO. NO NO 0.06 =.6110 = NO = = = 0.11 M NO Then we determine the mass of NO resent in.6 L mol N O 9.01 g N O NO mass =.6L = 5. g N O 1 L 1 mol N O 669

5 10A (M) We use the initial-hange-equilibrium setu to establish the amount of eah substane at equilibrium. We then label eah entry in the table in the order of its determination (1 st, nd, rd, th, 5 th ), to better illustrate the tehnique. We know the initial amounts of all substanes (1 st ). There are no roduts at the start. Beause initial + hange = equilibrium, the equilibrium amount ( nd ) of Br g enables us to determine hange ( rd ) for Brbgg. We then use stoihiometry to write other entries ( th ) on the hange line. And finally, we determine the remaining equilibrium amounts (5 th ). Reation: NOBr g NOg + Br g Initial: 1.86 mol (1 st ) 0.00 mol (1 st ) 0.00 mol (1 st ) Change: 0.16 mol ( th ) mol ( th ) mol ( rd ) Equil.: 1.70 mol (5 th ) 0.16 mol (5 th ) 0.08 mol ( nd ) mol [NO] [Br ] [NOBr] Here, n gas =+1 =+1. = RT = = B (M) Use the amounts stated in the roblem to determine the equilibrium onentration for eah substane. Reation: SOg SOg + Og Initial: 0 mol mol mol Changes: mol mol / mol Equil.: mol mol 0.05 mol mol mol 0.05 mol Conentrations: 1.5 L 1.5 L 1.5 L Conentrations: M M M We use these values to omute for the reation and then the relationshi RT n gas = b g (with ngas =+1 =+1) to determine the value of. SO O SO = = =.010 =.010 ( )

6 11A (M) The equilibrium onstant exression is = P{ HO} P{ CO} = 0.1 at 100 C. From the balaned hemial equation, we see that one mole of HObgg is formed for eah mole of CO bgg rodued. Consequently, P{ HO} = P{ CO} and = bp{ CO} exression for P {CO } : P{CO } ( P{CO }) atm. 11B (M) The equation for the reation is g. We solve this NH HS s NH g + H S g, = at 5 C. The two artial ressures do not have to be equal at equilibrium. The only instane in whih they must be equal is when the two gases ome solely from the deomosition of NH HS s In this ase, some of the NHbgg has ome from another soure. We an obtain the ressure of HSbgg by substitution into the equilibrium onstant exression, sine we are given the equilibrium ressure of NH g b g = P{H S} P{NH } = = P{H S} atm NH P{H S} = = 0.16 atm So, P total = P PNH = 0.16 atm atm = atm HS b g. 1A (M) We set u this roblem in the same manner that we have reviously emloyed, namely designating the equilibrium amount of HI as x. (Note that we have used the same multiliers for x as the stoihiometri oeffiients.) Initial: mol 0.00 mol 0 mol Changes: x mol x mol +x mol Equil: b0.150 xg mol b0.00 xg mol x mol x 15.0 x = = x 0.00 x x0.00 x Equation: H g + I g HIg We substitute these terms into the equilibrium onstant exression and solve for x. b gb g h x = x 0.00 x 50. = x + x = x + 50.x 0 = 6.x 17.6 x+1.51 Now we use the quadrati equation to determine the value of x. b b a 17.6 (17.6) x = = = a = 0.50 or

7 The first root annot be used beause it would afford a negative amount of H (namely, = ). Thus, we have 0.11= 0.6 mol HI at equilibrium. We hek by substituting the amounts into the exression. (Notie that the volumes anel.) The slight disagreement in the two values (5 omared to 50.) is the result of rounding error. = 1B (D) b g b gb g = =5 NO g NO g and =.6110 at 5 C. In the examle, this reation is onduted in a 0.7 L flask. The effet of moving the mixture to the larger, 10.0 L ontainer is that the reation will be shifted to rodue a greater number of moles of gas. Thus, NObgg will be rodued and NObgg will dissoiate. Consequently, the amount of N O will derease. (a) The equation for the reation is (b) The equilibrium onstant exression, substituting 10.0 L for 0.7 L, follows. F x I NO HG x J [ ] [N x O ] ( x) This an be solved with the quadrati equation, and the sensible result is x = moles. We an attemt the method of suessive aroximations. First, assume that x We obtain: (0.00 0) x (0.00 0) Clearly x is not muh smaller than So, seond, assume x We obtain: x ( ) This assumtion is not valid either. So, third, assume x We obtain: x ( ) Notie that after eah yle the value we obtain for x gets loser to the value obtained from the roots of the equation. The values from the next several yles follow. Cyle th 5 th 6 th 7 th 8 th 9 th 10 th 11 th x value The amount of N O at equilibrium is mol, less than the mol N O equilibrium in the 0.7 L flask, as redited. at 67

8 1A (M) Again we base our solution on the balaned hemial equation Equation: Ag aq + Fe aq Fe aq + Ag s =.98 Initial: 0 M 0 M 1.0 M Changes: +x M +x M x M Equil: x M x M b1.0 xg M + = Fe x x x x x Ag Fe =.98 = 1.0 x.98 = = We use the quadrati formula to obtain a solution. b b a 1.00 (1.00) x = = = = 0.88 M or 0.8 M a A negative root makes no hysial sense. We obtain the equilibrium onentrations from x Ag = Fe = 0.88 M Fe = = 0.71 M 1B (M) We first alulate the value of Q to determine the diretion of the reation. Q + + V Cr = = = = + + V Cr Beause the reation quotient has a smaller value than the equilibrium onstant, a net reation to the right will our. We now set u this solution as we have others, heretofore, based on the balaned hemial equation. V + aq + Cr + aq V + aq + Cr + aq initial M M M M hanges x M x M +x M +x M equil ( x)m ( x)m ( x)m ( x)m b g b g + + V Cr x x x = = =7. 10 = V Cr b xgb xg x If we take the square root of both sides of this exression, we obtain x x x = 0.7 7x whih beomes 8 x = 0.1 and yields 0.00 M. Then the + + equilibrium onentrations are: V = Cr = M 0.00 M = M + + V = Cr = M M = 0.15 M F HG I J 67

9 INTEGRATIVE EXAMPLE A. (E) We will determine the onentration of F6P and the final enthaly by adding the two reations: CH 6 1O6 ATP G6PADP G6P F6P CH 6 1O6 ATP ADPF6P ΔH TOT = 19.7 kj mol kj mol -1 = 16.9 kj mol -1 Sine the overall reation is obtained by adding the two individual reations, then the overall reation equilibrium onstant is the rodut of the two individual values. That is, The equilibrium onentrations of the reatants and roduts is determined as follows: C 6 H 1 O 6 + ATP ADP + F6P Initial Change -x -x +x +x Equil x x x x 178 ADPF6P CH OATP x110 x 110 x x xx x x 10 Exanding and rearranging the above equation yields the following seond-order olynomial: 177 x 0.19 x = 0 Using the quadrati equation to solve for x, we obtain two roots: x = and Only the first one makes hysial sense, beause it is less than the initial value of C 6 H 1 O 6. Therefore, [F6P] eq = During a fever, the body generates heat. Sine the net reation above is exothermi, Le Châtelier's rinile would fore the equilibrium to the left, reduing the amount of F6P generated. 67

10 B. (E) (a) The ideal gas law an be used for this reation, sine we are relating vaor ressure and onentration. Sine = for deomosition of Br to Br (very small), then it an be ignored. 1 nrt mol Latm V 8.7L P 0.89 atm (b) At 1000, there is muh more Br being generated from the deomosition of Br. However, is still rather small, and this deomosition does not notably affet the volume needed. EXERCISES Writing Equilibrium Constants Exressions 1. (E). (E) COF g CO g + CF g = (a) COCF COF + Cu Cu s + Ag aq Cu aq + Ag s = + Ag + SO + + Fe S O aq + Fe aq SO aq + Fe aq = + SO 8 Fe + + (b) () 8 P N NH g + O g N g + 6 H O g = P NH P O (a) P P NH 7 H g + NO g NH g + H O g = P H P NO (b) P 7 N g + Na CO s + C s NaCN s + CO g = N () 6 PHO PHO CO. (E) (a) NO + Zn = (b) = () NO O + Ag = OH CO 675

11 . (E) (a) P{CH }P{H S} P{CS }P{H } = (b) P 1/ = lo q () = P{ } P{ CO H O} 5. (E) In eah ase we write the equation for the formation reation and then the equilibrium onstant exression,, for that reation. HF 1 1 (a) Hg + Fg HFg = 1/ 1/ H F (b) NH N g + H g NH g = N H () [NO] Ng + Og NOg = [N ] [O ] (d) 1 1 Cl (g) + F (g) ClF (l) [Cl ] [F ] 1/ / 6. (E) In eah ase we write the equation for the formation reation and then the equilibrium onstant exression,, for that reation. (a) P NOCl Ng + Og + Clg NOClg P = 1/ 1/ 1/ P N P O P Cl (b) PClNO Ng + Og + Clg ClNOg P = P N P O P Cl () Ng + Hg NHg PNH P = P N P H (d) Ng + Hg + Clg NHCls P = 1/ 1/ P N P P H Cl n 7. (E) Sine = RT g, it is also true that = (a) SO [Cl ] RT ( 1) (b) () n g RT. = = =.9 10 ( ) = SO Cl - -1 ( 1) 5 HS 0 = = = = 0.9 H NO = = RT = ( ) = NO [O ] RT 676

12 8. (E) = (a) (b) () n g 1 1 R = L atm mol P{NO } +1-1 = (RT) = ( ) 0.11 P{NO } P{CH }P{H } ( ) (RT) (0.15)( ) P{CH } P{H } P{CS } ( ) 8 (RT) ( )( ).6 10 P{HS} P{CH } RT, with 9. (E) The equilibrium reation is HOl HOg with n gas =+1. = n = RT g 1 atm. = P{H O} =.8 mmhg 760 mmhg = = RT = = = RT n g RT gives 10. (E) The equilibrium rxn is CH 6 6l CH 6 6g with n gas =+1. Using = = RT = = 0.15 = P{C H } mmhg P{C 6H 6} = 0.15 atm = 95.0 mmhg 1 atm n g RT, 11. (E) Add one-half of the reversed 1 st reation with the nd reation to obtain the desired reation Ng + Og NOg = NOg + Br g NOBr g = net : N g + O g + Br g NOBr g = = (M) We ombine the several given reations to obtain the net reation. 1 N Og N g O (g) = NO g N O g = N (g) + O (g) NO (g) (.110 ) net : N O g + O g N O g = = ( Net) 6 677

13 1. (M) We ombine the values to obtain the value of for the overall reation, and then onvert this to a value for. CO g +H g COg + H Og = 1. 8 C grahite + O g CO g = 1 10 CO g C grahite + CO g = net: (1.) (110 ) H (g) + O (g) HO(g) (Net) 510 (0.6) 16 n = RT = = =510 RT (M) We ombine the values to obtain the value of for the overall reation, and then onvert this to a value for. 1 NO Clg NOClg + O g = NO g + Cl g NO Cl g = 0. 9 N g + O g NO g = ( ) 9 (0.) ( ) net: N (g) + O (g) + Cl (g) NOCl(g) (Net) (1.110 ) n = RT = = = 10 n RT (E) CO (g) HO(l) HCO (aq) In terms of onentration, = a(h CO )/a(co ) In terms of onentration and artial ressure, HCO 16. (E) Fe(s) O (g) FeO (s) a FeO. Sine ativity of solids and liquids is defined as 1, then the exression a a Fe O 1 simlifies to a O P CO Similarly, in terms of ressure and onentration, 1 P /P P O 678

14 Exerimental Determination of Equilibrium Constants 17. (M) First, we determine the onentration of PCl 5 and of Cl resent initially and at equilibrium, resetively. Then we use the balaned equation to hel us determine the onentration of eah seies resent at equilibrium mol PCl mol Cl PCl 5 = = M Cl = = M initial equil 0.50 L 0.50 L Equation: PCl5g PCl (g) + Clg Initial: M 0 M 0 M Changes: xm +xm +xm Equil: M-xM xm xm M (from above) At equilibrium, [Cl ] = [PCl ] = M and [PCl 5 ] = M xm = M PCl PCl Cl ( M)( M) M = (M) First we determine the artial ressure of eah gas. P P 1mol H L atm 1.00g H 1670 nrt.016g H mol {H g }= = 16 atm V L initial 1mol HS Latm 1.06g HS 1670 nrt.08g HS mol {H Sg }= = 8.5atm V L initial L atm mol S 1670 nrt mol equil{s g }= = atm P V L Equation: Hg + Sg HSg Initial : 16 atm 0 atm 8.5 atm Changes : atm atm atm Equil : 16 atm atm 8.5 atm P{HS g } 8.5 P = = =1.79 {H g } P{S (g)}

15 19. (M) PCl5 (a) = L PCl O = NM QP L Cl O NM QP 0. (M) F HG b g b g gpcl 1 mol PCl 50. L 08. g 0. 0 gpcl 1 mol PCl 50. L 17. g (b) = RT n = = I F J HG 1. gcl 1mol Cl 50. L g 1 I J 6. 1 mol ICl 0.68 g ICl 16.6 g ICl - [ICl] initial = = M 0.65 L Reation: ICl(g) I (g) + Cl (g) Initial: M 0 M 0 M Change -x +x +x Equilibrium M -x x x 1 mol I 0.08 g I g I - [I ] equil = =.1 10 M 0.65 L = x xx (.110 ) (6.710 x) (6.710 (.110 )) 6 1. (E) Fe Fe H Fe M. (E) NH (aq) P P NH (g) 9 NH (g) NH (g) P

16 Equilibrium Relationshis. (M) SO SO SO L =81= = = = mol SO SO O SO. (M) 0.7 mol I I V V 1L I 1.00 mol I V V 5. (M) (a) A ossible equation for the oxidation of NH g 7 NH g + O g NO g + H O g b g to NO b gg follows. (b) We obtain for the reation in art (a) by aroriately ombining the values of given in the roblem NH g + O g NO g + H O g = NO g + O g NO g = (D) (a) (b) () net: NHg + Og NOg + HOg = = We first determine [H ] and [CH ] and then [C H ]. [CH ] = [H ] = CH 0.10 H CH H CH = C H = = = 1.5 M 0.10 mol 1.0 L 19 = 0.10 M In a 1.00 L ontainer, eah onentration numerially equals the molar quantities of the substane. 1.5 mol CH {CH } = = mol C H mol CH mol H The onversion of CH (g) to C H (g) is favored at low ressures, sine the onversion reation has a larger sum of the stoihiometri oeffiients of gaseous roduts () than of reatants (). Initially, all onentrations are halved when the mixture is transferred to a flask that is twie as large. To re-establish equilibrium, the system reats to the right, forming more moles of gas (to omensate for the dro in ressure). We base our solution on the balaned hemial equation, in the manner we have used before. 681

17 Equation: CH (g) CH (g) + H Initial: 0.10 mol 1.5 mol 0.10 mol.00l.00l.00l = M = 0.75 M = M Changes: x M + x M + x M Equil: M M M x x x CH H x x CH x = = = 0.15 We an solve this th -order equation by suessive aroximations. First guess: x = M x = Q = = = x = 0.00 Q = = = x = Q = = = x = Q = = = This is the maximum number of signifiant figures our system ermits. We have x = M. CH = 0.08 M; CH = M; H = M Beause the ontainer volume is.00 L, the molar amounts are double the values of molarities mol CH 0.08 mol CH.00 L = 1.51 mol CH.00 L = mol CH 1 L 1 L mol H.00 L 1 L =0.1mol H Thus, the inrease in volume results in the rodution of some additional C H. 68

18 7. (M) (a) CO nh O n COHO V V COH n{co } nh V V Sine V is resent in both the denominator and the numerator, it an be striken from the exression. This haens here beause n g = 0. Therefore, is indeendent of V. (b) Note that = for this reation, sine n gas =0. 0. mol CO 0. mol H O = = = mol CO 0.76 mol H 8. (M) For the reation CO(g) + H O(g) CO (g) + H (g) the value of =. [CO ][H ] The exression for Q is. Consider eah of the rovided situations [CO][HO] (a) P CO = P HO = P H = P CO ; Q = 1 Not an equilibrium osition PH P CO (b) = = x ; Q = x If x =., this is an equilibrium osition. PHO P CO () PCO P H O = PCO P H ; Q = 1 Not an equilibrium osition PH P CO (d) = = x ; Q = x If x =., this is an equilibrium osition. P P CO HO Diretion and Extent of Chemial Change 9. (M) We omute the value of Q for the given amounts of rodut and reatants. 1.8mol SO SO 7.L Q = SO O.6mol SO.mol O 7.L 7.L The mixture desribed annot be maintained indefinitely. In fat, beause Q, the reation will roeed to the right, that is, toward roduts, until equilibrium is established. We do not know how long it will take to reah equilibrium. 68

19 0. (M) We omute the value of Q for the given amounts of rodut and reatants mol NO NO 5.5 L Q NO 0.750mol NO 5.5L The mixture desribed annot be maintained indefinitely. In fat, beause Q <, the reation will roeed to the right, that is, toward roduts, until equilibrium is established. If E a is large, however, it may take some time to reah equilibrium. 1. (M) (a) (b) We determine the onentration of eah seies in the gaseous mixture, use these onentrations to determine the value of the reation quotient, and omare this value of Q with the value of mol SO 0.18 mol O SO = = 0.9 M O = = M 1.90 L 1.90 L mol SO SO 0.99 SO = = 0.99 M Q = = = L SO O Sine Q = =, this mixture is not at equilibrium. Sine the value of Q is smaller than that of, the reation will roeed to the right, forming rodut and onsuming reatants to reah equilibrium. b g of the substane divided by its molar mass and further divided by the volume of the ontainer.. (M) We omute the value of Q. Eah onentration equals the mass m 1 mol CO 1 mol H m m.0 g CO.0 g H 1 COH = V V.0.0 Q = (value of ) COH O 1 mol CO 1 mol HO m m 8.0 g CO 18.0 g H O V V (In evaluating the exression above, we anelled the equal values of V, along with, the equal values of m.) Beause the value of Q is smaller than the value of, (a) the reation is not at equilibrium and (b) the reation will roeed to the right (formation of roduts) to reah a state of equilibrium. 68

20 . (M) The information for the alulation is organized around the hemial equation. Let x = mol H (or I ) that reats. Then use stoihiometry to determine the amount of HI formed, in terms of x, and finally solve for x. x Equation: H bgg Ibgg HIbg g HI.5 L Initial: mol mol mol HI x x Changes: x mol x mol +x mol.5l.5 L Equil: x x x x Then take the square root of both sides: x x = x x = = mol, amount HI = x = mol = 0. mol HI 9.09 H = amount I = x mol = mol = 0.0 mol H (or I ) amount. (M) We use the balaned hemial equation as a basis to organize the information Equation: SbCl5g SbClg + Clg Initial: 0.00 mol 0.80 mol mol.50 L.50 L.50 L Initial: M 0.11 M M Changes: + x M x M x M Equil: xm b0.11 xg M b0.060 xg M SbClCl 0.11 x0.060 x x + x = 0.05= = = SbCl x x = = 0 x x x x x b b a x = = = 0.06 or a The seond of the two values for x gives a negative value of Cl = M, and thus is hysially meaningless in our universe. Thus, onentrations and amounts follow. SbCl = x = 0.06 M amount SbCl =.50 L 0.06 M = mol SbCl SbCl = 0.11 x = M amount SbCl =.50 L M = 0.17 mol SbCl Cl = x = M amount Cl =.50 L M = 0.00 mol Cl 685

21 5. (M) We use the hemial equation as a basis to organize the information rovided about the reation, and then determine the final number of moles of Cl g resent. Equation: COg + Cl g COCl g Initial: mol mol mol Changes: + x mol + x mol x mol Equil.: x x x mol mol mol COCl COCl 1.10 x.050 (0.500 x) x ( x) mol.050 L ( x)mol x mol.050l.050 L Assume x This rodues the following exression x = x = =.010 mol Cl We use the first value we obtained, x = =.0 10 mol Cl (= ), to arrive at a seond value. Beause the value did not hange on the seond iteration, we have arrived at a solution. 6. (M) Comute the initial onentration of eah seies resent. Then determine the equilibrium onentrations of all seies. Finally, omute the mass of CO resent at equilibrium g 1 mol CO CO int = = 0.05 M 1.1 L 8.01 g CO 1.00 g 1 mol H H = = 0.5 M int 1.1 L.016 g H 1.00 g 1 mol H O HO = = 0.09 M int 1.1 L 18.0 g H O Equation : CO g + HO g CO g + H g Initial : 0.05 M 0.09 M M 0.5 M Changes : x M x M + x M + x M Equil : 0.05 M 0.09 M M M x x x x CO H x x 0.5 x + x = =. = = CO HO 0.05 x 0.09 x x+ x x x x x x x = = 0 686

22 b b a x = = = M, M a. The first value of x gives negative onentrations for reatants ([CO] = M and [H O] = M). Thus, x = M = CO. Now we an find the mass of CO mol CO.01 g CO 1.1 L 1 L mixture 1 mol CO = 0.99 g CO 7. (D) We base eah of our solutions on the balaned hemial equation. (a) Equation : PCl g PCl g + Cl g 5 Initial : mol mol 0 mol.50 L.50 L.50 L Changes : x mol + x mol + x mol.50 L.50 L.50 L Equil : x mol x mol x mol.50 L.50 L.50 L [PCl ][Cl ].50L.50L x(0.550 x) [PCl (0.550 )mol 5 ] x.50(0.550 x) x mol x mol.50l x x = x x x 0.05 = 0 b b a x = = = mol, mol a The seond answer gives a negative quantity. of Cl, whih makes no hysial sense. PCl 5 mol PCl 5 n = = 0.78 n = = 0.6 mol PCl n Cl = x = mol Cl Equation : PCl5 g PCl g + Cl g Initial : mol.50 L 0 M 0 M Changes : x mol + x mol + x mol.50 L.50 L.50 L Equil : x mol x mol x mol.50 L.50 L.50 L (b) PCl 687

23 ( xmol) ( x mol) PClCl.50L.50 L = =.810 PCl x mol 5.50 L x x = x x x0.058 = x b b a x = = = 0.0 mol, 0.9 mol a amount PCl = 0.0 mol = amount Cl; amount PCl 5 = = 0.1 mol 8. (D) (a) We use the balaned hemial equation as a basis to organize the information we have about the reatants and roduts. Equation: COF g CO g + CF g 0.15 mol 0.6 mol 0.07 mol Initial: 5.00 L 5.00 L 5.00 L Initial: M 0.05 M M And we now omute a value of Q and omare it to the given value of. Q COCF COF = = = = Beause Q is not equal to, the mixture is not at equilibrium. (b) Beause Q is smaller than, the reation will shift right, that is, roduts will be formed at the exense of COF, to reah a state of equilibrium. () We ontinue the organization of information about reatants and roduts. Equation: COF g CO g + CF g Initial: M 0.05 M M Changes: x M + x M + x M x x x Equil: M M M 688

24 COF x CO CF x x x + x = = =.00= x x x + 8 x = x + x 7x 0.99 x = 0 b b a x = = = 0.00 M,0.09 M a 1 The seond of these values for x (0.09) gives a negative COF = M, learly a nonsensial result. We now omute the onentration of eah seies at equilibrium, and hek to ensure that the equilibrium onstant is satisfied. x COF = = = 0.0 M CO = x = = M CF = x = = M CO CF M M = = =.01 COF 0.0 M The agreement of this value of with the ited value (.00) indiates that this solution is orret. Now we determine the number of moles of eah seies at equilibrium. mol COF = 5.00 L 0.0 M = 0.11 mol COF mol CO = 5.00 L M = 0.79 mol CO mol CF = 5.00 L M = mol CF But suose we had inorretly onluded, in art (b), that reatants would be formed in reahing equilibrium. What result would we obtain? The set-u follows. Equation : COF g CO g + CF g Initial : 0.090M 0.05 M M Changes : + y M y M y M Equil : M 0.05 M M y y y COF y CO CF 0.05 y y y + y = = =.00= y + 8 y = y + y 7 y y = 0 y y b b a y = = = 0.00 M, 0.09 M a 1 689

25 b g gives a negative The seond of these values for x 0.09 COF = M, learly a nonsensial result. We now omute the onentration of eah seies at equilibrium, and hek to ensure that the equilibrium onstant is satisfied. y COF = = = 0.0 M CO = 0.05 y = = M CF = y = = M These are the same equilibrium onentrations that we obtained by making the orret deision regarding the diretion that the reation would take. Thus, you an be assured that, if you erform the algebra orretly, it will guide you even if you make the inorret deision about the diretion of the reation. 9. (D) (a) We alulate the initial amount of eah substane. 1 mol C H OH n {C H OH = 17. g C H OH = 0.7 mol C H OH g CH5OH 1 mol CH CO H n{ch CO H =.8 g CH CO H = 0.96 mol CH CO H g CHCOH n{ch CO C H } = 8.6 g CH CO C H 5 5 n{chcoch}=0.55 mol CHCOCH mol CHCOCH g CH CO C H 1 mol H O n H O = 71. g H O =.95 mol H O 18.0 g HO 5 Sine we would divide eah amount by the total volume, and sine there are the same numbers of rodut and reatant stoihiometri oeffiients, we an use moles rather than onentrations in the Q exression. Q n n{ch CO C H H O 0.55 mol.95 mol 5 = = = 1.8 =.0 n{c H5OH n CHCOH 0.7 mol 0.96 mol Sine Q the reation will shift to the left, forming reatants, as it attains equilibrium. Equation: C H OH + CH CO H CH CO C H + H O Initial 0.7 mol 0.96 mol 0.55 mol.95 mol Changes +x mol +x mol x mol x mol Equil (0.7+x) mol (0.96+x) mol (0.55 x) mol (.95 x) mol (b)

26 (0.55 x)(.95 x).18.50x x.0 (0.7 x)(0.96 x) x x x.50 x+.18 = x +.08 x x x 1.59 = 0 b b a x = = = 0.19 moles,.7 moles a 6 Negative amounts do not make hysial sense. We omute the equilibrium amount of eah substane with x = 0.19 moles. n {C H5OH = 0.7 mol mol = 0.56 mol CH5OH 6.07g CH5OH mass CH5OH = 0.56mol CH5OH 6g CH5OH 1 mol C H OH n {CH CO H 5 = 0.96 mol mol = 0.59 mol CH CO H 60.05g CH CO H mass CH CO H = 0.59mol CH CO H 5g CH CO H 1 mol CHCOH n{ch CO C H = 0.55 mol 0.19 mol = 0.6 CH CO C H g CH CO C H mass CH CO C H = 0.6mol CH CO C H g CH CO C H mol CHCOCH5 n{h O =.95 mol 0.19 mol =.76 mol H O 18.0g H O mass H O =.76mol H O 68g H O 1 mol HO 5 n n n CH CO C H H O 0.6 mol.76 mol To hek = = =.1 n C H OH CH CO H 0.56 mol 0.59 mol 5 0. (M) The final volume of the mixture is L +.5 L =.00 L. Then use the balaned hemial equation to organize the data we have onerning the reation. The reation should shift to the right, that is, form roduts in reahing a new equilibrium, sine the volume is greater. Equation: NO g NO g Initial: mol mol.00 L.00 L Initial: 0.M 0.019M Changes: xm xm Equil : (0. x)m (0.019 x)m 691

27 x x x NO = = = =.6110 N O 0. x 0. x x + x = x x x = 0 b b a x = = = M, M a 8 NO = = M amount NO = M.00 L = 0.11 mol NO N O = = 0.16 M amount N O = 0.16 M.00 L = 0.9 mol N O 1. (M) mol HCONH = = M init.16 L Equation: HCONH g NH g + CO g Initial : M 0 M 0 M Changes: xm + xm + xm Equil : ( x) M xm xm NH CO x x x x x x = = =.8 = = HCONH x b b a x = = = 0.08 M,.9 M a The negative onentration obviously is hysially meaningless. We determine the total onentration of all seies, and then the total ressure with x = total = NH + CO + HCONH = x+ x x= = M P tot = mol L L atm mol 00. = 5.58 atm. (E) Comare Q to. We assume that the added solids are of negligible volume so that the initial artial ressures of CO (g) and H O(g) do not signifiantly hange. 1 atm P{H O = 715 mmhg = 0.91 atm HO 760 mmhg Q = P CO P HO =.10 atm CO 0.91 atm HO = = Beause Q is larger than, the reation will roeed left toward reatants to reah equilibrium. Thus, the artial ressures of the two gases will derease. 69

28 . (M) We organize the solution around the balaned hemial equation. Equation: + Cr aq + Cd(s) + Cr aq Cd aq Initial : 1.00 M 0 M 0 M Changes: xm x x Equil: (1.00 x) M x x + + Via suessivearoximations, one obtains Cr Cd ( x) ( x) = = = 0.88 x= 0.57 M Cr 1.00 x Therefore, at equilibrium, [Cd + ] = 0.57 M, [Cr + ] = 0.51 M and [Cr + ] = 0.86 M Minimum mass of Cd(s) = 0.50L 0.57 M 11.1 g/mol = 10.1 g of Cd metal. (M) Again we base the set-u of the roblem around the balaned hemial equation. 10 = Equation: Pb s + Cr aq Pb aq + Cr aq Initial: 0.100M 0M 0M Changes: xm xm xm Equil : (0.100 x)m xm xm x( x) = =. 10 x (0.100) (0.100) x M Assumtion that x Pb + = x = M, [Cr + ] = M and [Cr + ] = M, is valid and thus 5. (M) We are told in this question that the reation SO (g) + Cl (g) SO Cl (g) has =.0 at a ertain temerature T. This means that at the temerature T, [SO Cl ] =.0 [Cl ] [SO ]. Careful srutiny of the three diagrams reveals that sketh (b) is the best reresentation beause it ontains numbers of SO Cl, SO, and Cl moleules that are onsistent with the for the reation. In other words, sketh (b) is the best hoie beause it ontains 1 SO Cl moleules (er unit volume), 1 Cl moleule (er unit volume) and SO moleules (er unit volume), whih is the requisite number of eah tye of moleule needed to generate the exeted value for the reation at temerature T. 6. (M) In this question we are told that the reation NO(g) + Br (g) NOBr(g) has =.0 at a ertain temerature T. This means that at the temerature T, [NOBr] =.0 [Br ][NO]. Sketh () is the most aurate reresentation beause it ontains 18 NOBr moleules (er unit volume), 6 NO moleules (er unit volume), and Br moleules (er unit volume), whih is the requisite number of eah tye of moleule needed to generate the exeted value for the reation at temerature T. 69

29 7. (E) 8. (E) aonitate itrate Q Sine Q =, the reation is at equilibrium, CONADredoxoglut. itratenadox Q Sine Q <, the reation needs to roeed to the right (roduts). Partial Pressure Equilibrium Constant, 9. (M) The Ibg s maintains the resene of I in the flask until it has all vaorized. Thus, if enough HI(g) is rodued to omletely onsume the I bsg, equilibrium will not be ahieved. 1 atm P{H S} = 77.6 mmhg = atm 760 mmhg HS g + I s HI g + S s Initial: atm 0 atm Changes: x atm +x atm Equil: b0.987 xg atm x atm P{HI} x 5 x = = =1.10 = P{H S} x Equation: x = = atm The assumtion that x is valid. Now we verify that suffiient I (s) is resent by omuting the mass of I needed to rodue the redited ressure of HI(g). Initially, 1.85 g I is resent (given). mass I = tot atm 0.75 L 1 mol I 5.8 g I = g I L atm mol mol HI 1 mol I Ptot = P{H S}+ P{HI} = x + x= x= = atm P = 79.0 mmhg 69

30 50. (M) We first determine the initial ressure of NH. 1 1 nrt mol NH L atm mol 98 P{NH g }= = =0.98 atm V.58 L Equation: NH HS(s) NH (g) + H S(g) Initial: 0.98 atm 0 atm Changes: +x atm +x atm x atm x atm Equil: = P{NH } P{H S} = = x x x x x x b b a b g = = x = = = 0.10 atm, 1.05 atm a The negative root makes no hysial sense. The total gas ressure is obtained as follows. Ptot = P{ NH } + P{ H S} = b xg + x= x= = 1.15 atm 51. (M) We substitute the given equilibrium ressure into the equilibrium onstant exression P{O } P{O } and solve for the other equilibrium ressure. = = 8.5 = P{CO } atm CO P{O } P{O } 8.5(0.071 atm) 0.59 atm O Ptotal = P{CO } + P {O } = atm CO atm O = atm total 5. (M) The omosition of dry air is given in volume erent. Division of these erentages by 100 gives the volume fration, whih equals the mole fration and also the artial ressure in atmosheres, if the total ressure is 1.00 atm. Thus, we have P {O } = atm and P {CO } = atm. We substitute these two values into the exression for Q. Q P{O } atm O = = = = P{CO } atm CO The value of Q is muh larger than the value of. Thus this reation should be sontaneous in the reverse diretion until equilibrium is ahieved. It will only be sontaneous in the forward diretion when the ressure of O dros or that of CO rises (as would be the ase in self-ontained breathing devies). 695

31 5. (M) (a) We first determine the initial ressure of eah gas. 1 1 nrt 1.00 mol L atm mol 668 P{ CO} = P{ Cl} = = = 1. atm V 1.75 L Then we alulate equilibrium artial ressures, organizing our alulation around the balaned hemial equation. We see that the equilibrium onstant is not very large, meaning that we must solve the olynomial exatly (or by suessive aroximations). Equation CO(g) + Clg COClg =.5 Initial: 1. atm 1. atm 0 atm Changes: x atm x atm +x atm Equil: 1. x atm 1. x atm x atm P{COCl } x x.5 P{CO} P{Cl } (1. x) ( x x ).5( x x ) x x.5x x x x (Solve by using the quadrati equation) b b a x = = a (.5) x = 0.1,.5(too large) 5 ( 109.5) ( 109.5) (.5)(0) P{CO} = P{Cl } = 1. atm 0.1 atm = 1.16 atm P{COCl } = 0.1 atm (b) Ptotal = P{CO}+ P{Cl }+ P {COCl } = 1.16 atm atm atm =.6 atm 5. (M) We first find the value of for the reation. 6 NO g NO g + O g, = at 18 C = 57. For this reation n gas =+1=+1. n g (RT) = ( ) 1 = To obtain the required reation NOg + O g NO g from the initial reation, that initial reation must be reversed and then divided by two. Thus, in order to determine the value of the equilibrium onstant for the final reation, the value of for the initial reation must be inverted, and the square root taken of the result , final 5 696

32 Le Châtelier's Prinile 55. (E) Continuous removal of the rodut, of ourse, has the effet of dereasing the onentration of the roduts below their equilibrium values. Thus, the equilibrium system is disturbed by removing the roduts and the system will attemt (in vain, as it turns out) to re-establish the equilibrium by shifting toward the right, that is, to generate more roduts. 56. (E) We notie that the density of the solid ie is smaller than is that of liquid water. This means that the same mass of liquid water is resent in a smaller volume than an equal mass of ie. Thus, if ressure is laed on ie, attemting to fore it into a smaller volume, the ie will be transformed into the less-sae-ouying water at 0C. Thus, at 0 Cunder ressure, H O(s) will melt to form H O(l). This behavior is not exeted in most ases beause generally a solid is more dense than its liquid hase. 57. (M) (a) This reation is exothermi with H o = 150. kj. Thus, high temeratures favor the reverse reation (endothermi reation). The amount of H g resent at high temeratures will be less than that resent at low temeratures. (b) () (d) 58. (M) (a) (b) () (d) HOg b g is one of the reatants involved. Introduing more will ause the equilibrium osition to shift to the right, favoring roduts. The amount of H g will inrease. Doubling the volume of the ontainer will favor the side of the reation with the largest sum of gaseous stoihiometri oeffiients. The sum of the stoihiometri oeffiients of gaseous seies is the same () on both sides of this reation. Therefore, inreasing the volume of the ontainer will have no effet on the amount of H g resent at equilibrium. A atalyst merely seeds u the rate at whih a reation reahes the equilibrium osition. The addition of a atalyst has no effet on the amount of H g resent at equilibrium. This reation is endothermi, with H o = +9.5 kj. Thus, a higher temerature will favor the forward reation and inrease the amount of HI(g) resent at equilibrium. The introdution of more rodut will favor the reverse reation and derease the amount of HI(g) resent at equilibrium. The sum of the stoihiometri oeffiients of gaseous roduts is larger than that for gaseous reatants. Inreasing the volume of the ontainer will favor the forward reation and inrease the amount of HI(g) resent at equilibrium. A atalyst merely seeds u the rate at whih a reation reahes the equilibrium osition. The addition of a atalyst has no effet on the amount of HI(g) resent at equilibrium. 697

33 (e) The addition of an inert gas to the onstant-volume reation mixture will not hange any artial ressures. It will have no effet on the amount of HI(g) resent at equilibrium. 59. (M) (a) The formation of NO(g) from its elements is an endothermi reation ( H o = +181 kj/mol). Sine the equilibrium osition of endothermi reations is shifted toward roduts at higher temeratures, we exet more NO(g) to be formed from the elements at higher temeratures. (b) Reation rates always are enhaned by higher temeratures, sine a larger fration of the ollisions will have an energy that surmounts the ativation energy. This enhanement of rates affets both the forward and the reverse reations. Thus, the osition of equilibrium is reahed more raidly at higher temeratures than at lower temeratures. 60. (M) If the reation is endothermi (H > 0), the forward reation is favored at high temeratures. If the reation is exothermi (H < 0), the forward reation is favored at low temeratures. (a) H o = H o o o f PCl5 g Hf PCl g Hf Clg o H = 7.9 kj 87.0 kj 0.00 kj = 87.9 kj/mol (favored at low temeratures) o o o o o (b) H =Hf [ HO(g )] Hf[S(rhombi )] Hf [SO (g) ] Hf[ HS(g) ] o H = 1.8 kj kj 96.8 kj 0.6 kj o H = 15.5 kj/mol (favored at low temeratures) () H o =H o o f NOCl g + Hf HO g o o o H N g H O g H HClg H H o o = = f f f kj kj 0.00 kj 0.00 kj 9.1 kj +9.5 kj/mol (favored at higher temeratures) 61. (E) If the total ressure of a mixture of gases at equilibrium is doubled by omression, the equilibrium will shift to the side with fewer moles of gas to ounterat the inrease in ressure. Thus, if the ressure of an equilibrium mixture of N (g), H (g), and NH (g) is doubled, the reation involving these three gases, i.e., N (g) + H (g) NH (g), will roeed in the forward diretion to rodue a new equilibrium mixture that ontains additional ammonia and less moleular nitrogen and moleular hydrogen. In other words, P{N (g)} will have dereased when equilibrium is re-established. It is imortant to note, however, that the final equilibrium artial ressure for the N will, nevertheless, be higher than its original artial ressure rior to the doubling of the total ressure. 698

34 6. (M) (a) (b) Beause H o = 0, the osition of the equilibrium for this reation will not be affeted by temerature. Sine the equilibrium osition is exressed by the value of the equilibrium onstant, we exet to be unaffeted by, or to remain onstant with, temerature. From art (a), we know that the value of will not hange when the temerature is hanged. The ressures of the gases, however, will hange with temerature. (Reall the ideal gas law: P= nrt / V.) In fat, all ressures will inrease. The stoihiometri oeffiients in the reation are suh that at higher ressures the formation of more reatant will be favored (the reatant side has fewer moles of gas). Thus, the amount of D(g) will be smaller when equilibrium is reestablished at the higher temerature for the ited reation. 1 A s B s + C g + D g 6. (M) Inreasing the volume of an equilibrium mixture auses that mixture to shift toward the side (reatants or roduts) where the sum of the stoihiometri oeffiients of the gaseous seies is the larger. That is: shifts to the right if n gas 0, shifts to the left if n gas 0, and does not shift if n gas =0. (a) ngas (b) C s + H O g CO g + H g, 0, shift right, toward roduts Ca OH s + CO g CaCO s + HO g, ngas = 0, no shift, no hange in equilibrium osition. NH g + 5 O g NO g + 6 H O g, n 0, shifts right, towards roduts () gas 6. (M) The equilibrium osition for a reation that is exothermi shifts to the left (reatants are favored) when the temerature is raised. For one that is endothermi, it shifts right (roduts are favored) when the temerature is raised. 1 1 o NO g N g + O g H = 90. kj shifts left, % dissoiation (a) (b) 1 () (d) o SO g SO g + O g H = kj shifts right, % dissoiation o N H g N g + H g H = 95. kj shifts left, % dissoiation o COCl g CO g + Cl g H = kj shifts right, % dissoiation 65. (E) (a) Hb:O is redued, beause the reation is exothermi and heat is like a rodut. (b) No effet, beause the equilibrium involves O (aq). Eventually it will redue the Hb:O level beause removing O (g) from the atmoshere also redues O (aq) in the blood. () Hb:O level inreases to use u the extra Hb. 699

35 66. (E) (a) CO (g) inreases as the equilibrium is ushed toward the reatant side (b) Inrease CO (aq) levels, whih then ushes the equilibrium to the rodut side () It has no effet, but it hels establish the final equilibrium more quikly (d) CO inreases, as the equilibrium shifts to the reatants 67. (E) The ressure on N O will initially inrease as the rystal melts and then vaorizes, but over time the new onentration dereases as the equilibrium is shifted toward NO. 68. (E) If the equilibrium is shifted to the rodut side by inreasing temerature, that means that heat is a reatant (or being onsumed). Therefore, HI deomosition is endothermi. 69. (E) Sine ΔH is >0, the reation is endothermi. If we inrease the temerature of the reation, we are adding heat to the reation, whih shifts the reation toward the deomosition of alium arbonate. While the amount of alium arbonate will derease, its onentration will remain the same beause it is a solid. 70. (E) The amount of N inreases in the body. As the ressure on the body inreases, the equilibrium shifts from N gas to N (aq). Integrative and Advaned Exerises 71. (E) In a reation of the tye I (g) I(g) the bond between two iodine atoms, the I I bond, must be broken. Sine I (g) is a stable moleule, this bond breaking roess must be endothermi. Hene, the reation ited is an endothermi reation. The equilibrium osition of endothermi reations will be shifted toward roduts when the temerature is raised. 7. (M) (a) In order to determine a value of, we first must find the CO onentration in the gas hase. Note, the total volume for the gas is 1.00 L (moles and molarity are numerially equal) n P 1.00 atm [CO (aq)].910 M [CO ] M 0.80 V RT Latm [CO (g)] M mol (b) It does not matter to whih hase the radioative 1 CO is added. This is an examle of a Le Châtelier s rinile roblem in whih the stress is a hange in onentration of the reatant CO (g). To find the new equilibrium onentrations, we must solve and I.C.E. table. Sine Q <, the reation shifts to the rodut, CO (aq) side. Reation: CO (g) CO (aq) Initial: mol mol Stress mol Changes: x mol x mol Equilibrium: - ( x) mol.910 x mol 700

36 -.910 x mol - [CO (aq)] L x C 0.80 x 7.10 mol CO [CO ( ) mol (g)] x x L Total moles of CO in the aqueous hase ( L)( ) = moles Total moles of CO in the gaseous hase (1.000 L)( ) = moles Total moles of CO = moles moles = moles There is ontinuous mixing of the 1 C and 1 C suh that the isotoi ratios in the two hases is the same. This ratio is given by the mole fration of the two isotoes. For mol CO in either hase its mole fration is % mol Moles of 1 CO in the gaseous hase = moles = moles Moles of 1 CO in the aqueous hase = moles = moles 7. (M) Dilution makes Q larger than. Thus, the reation mixture will shift left in order to regain equilibrium. We organize our alulation around the balaned hemial equation. Equation: Ag (aq) Fe (aq) Fe (aq) Ag(s).98 Equil: 0.1 M 0.1 M 0.19 M Dilution: 0.1 M 0.08 M M Changes: x M x M x M New equil: (0.1 x) M (0.08 x) M (0.076 x) M [Fe ] x.98 [Ag ][Fe ] (0.1 x) (0.08 x).98 (0.1 x) (0.08 x) x x x.98 x.98 x 1.61 x x 0.07, 0.57 Note that the negative root makes no hysial 5.96 sense; it gives [Fe ] M. Thus, the new equilibrium onentrations are [Fe ] M [Ag ] M [Fe ] M We an hek our answer by substitution M = =.9.98 (within reision limits) M 0.15 M 7. (M) The erent dissoiation should inrease as the ressure is lowered, aording to Le Châtelier s rinile. Thus the total ressure in this instane should be more than in Examle 15-1, where the erent dissoiation is 1.5%. The total ressure in Examle 15-1 was omuted starting from the total number of moles at equilibrium. The total amount = ( ) moles N O mol NO = 0.07 mol gas. 701