CHAPTER 14 Chemical Equilibrium: Equal but Opposite Reaction Rates

Transcription

1 CHATER 14 Chemial Equilibrium: Equal but Opposite Reation Rates Collet and Organize For two reversible reations, we are given the reation profiles (Figure 14.1). The profile for the onversion of A to B shows that reatant A has a lower free energy than produt B. The profile for the onversion of C to D shows that C has a higher free energy than D. From the profiles, we are to determine whih reation has the larger k f, the smaller k r, and the larger value of. The magnitude of the rate onstant is inversely related to the magnitude of the ativation energy, E a. The reation with the larger k f, smaller k r, and the larger is that reation with the lowest E a for the forward reation and where k f > k r. The reation C D has the larger k f, the smaller k r, and the larger (beause it has the lowest ativation energy for the forward diretion). Remember that a large k (rate onstant) means that the reation is fast and therefore the reation has a low ativation energy Collet and Organize From Figure 14., depiting the molar onentrations of A (red spheres) and B (blue spheres) over time, we are to determine whether the reation reahes equilibrium, in what diretion that equilibrium is ahieved (A B or B A), and what is the value of the equilibrium onstant. (a) At equilibrium, the relative onentrations of A and B do not hange. (b) From Figure 14., we see that more A is present at the beginning of the reation, but more B is present at later times. () The equilibrium onstant an be alulated from the onentrations of A and B at equilibrium: [ produt] reatant [ ] (a) No. Beause the last two boxes in Figure 14. do not ontain the same number of A and B, we annot tell whether this reation reahes equilibrium. (b) A B () If we assume that the last box in Figure 14. represents equilibrium, then [ B] 4 M 1.5 A 16 M [ ] Equilibrium ould be approahed in this system starting with B reating to give A. If we write that reation as B A, would be [ ] A B M 4 M [ ] Collet and Organize From Figure 16.3, showing 6 blue spheres (produt B) and 13 red spheres (reatant A), we are to write the hemial equation of the equilibrium reation and alulate the value of. 11

2 Chemial Equilibrium 113 (a) In this reation, A is transformed into B. We represent a system at equilibrium by using double-headed reation arrows between the reatants and produts. We will assume that one moleule of A produes one moleule of B in the reation. (b) The value of is the ratio of the onentration of the produts (number of B spheres) and the onentration of the reatants (number of A spheres) raised to their respetive stoihiometri oeffiients from the balaned hemial equation. (a) A B 6 (b).0 13 If the hemial equation were written as A B, then would be [B] [A] (13) Collet and Organize We are to determine whether the reation mixture depited in Figure 14.4 is at equilibrium by omparing the reation quotient to the given equilibrium onstant. If the reation mixture is not at equilibrium, we are to predit the diretion in whih it shifts in order to ahieve equilibrium. Figure 14.4 shows 18 red spheres (A), 4 blue spheres (B), and 10 blue-red pairs (AB). If the reation quotient [ AB] Q [ A][ B] is equal to 3.0, then the reation is at equilibrium. If Q is less than 3.0, the reation shifts to the right. If Q is greater than 3.0, the reation shifts to the left. 10 Q This reation is not at equilibrium, as Q, and beause Q <, the reation shifts to the right to attain equilibrium. When equilibrium is attained, the number of A, B, and AB moleules remains onstant even though, at the moleular level, the onversions of A and B to AB and AB to A and B ontinue (but at equal rates) Collet and Organize By omparing the relative distributions of reatants A and B with produt AB at two different temperatures, as shown in Figure 14.5, we an determine whether the reation is endothermi or exothermi. From the equation Δ G RT ln ΔH T Δ S we see that as temperature rises for an exothermi reation, ΔG beomes less negative and therefore dereases. If, however, the temperature is raised on an endothermi reation, ΔG beomes more negative and inreases. Therefore, if produts inrease upon raising the temperature, the reation is endothermi; if produts derease, the reation is exothermi.

3 114 Chapter 14 At 300 the equilibrium mixture is 6 A, 10 B, and 5 AB. This gives an equilibrium onstant of At 400, the equilibrium mixture is 3 A, 7 B, and 8 AB. This gives an equilibrium onstant of As temperature inreases for this reation, inreases, indiating that more produts form at higher temperatures, so this reation is endothermi. We assume in this problem that the differene in entropy for the two temperatures at whih this reation is run is minimal, so only H ontributes to the differene in G at the two temperatures Collet and Organize From the Arrhenius plot of ln versus 1/T (Figure 14.6), we are to determine whether the reation is endothermi or exothermi. The Arrhenius equation is ln ΔH " 1 % $ '+ ΔS R # T & R where y ln, x 1/T, m ΔH R, and b ΔS R. The slope (m) of the graph is positive, so ΔH R is negative. The reation is exothermi. If the y-axis in Figure 14.6 showed values of ln, we would be able to alulate the slope of the line and therefore estimate the value of ΔH for the reation Collet and Organize We are asked to determine from Figure 14.7 whether the reation depited reahes equilibrium. Equilibrium is a state where the omposition of the reation is not hanging. A reation is at equilibrium when the rate of the forward reation equals the rate of the reverse reation. As a result, the onentration of neither the produts nor the reatants will hange at equilibrium. In Figure 14.7, we see that both the onentrations of the reatants and the produts level off after 50 miroseonds. This is the point of equilibrium. At 0 miroseonds the onentrations of the reatants and the produts are still hanging, so at this point, the reation is not at equilibrium. Chemial equilibrium is a dynami proess. At the moleular level the forward and reverse reations are still ourring. Beause they our at the same rate, however, we observe no marosopi hanges in the onentrations of the reatants and produts in the mixture.

4 Chemial Equilibrium Collet and Organize For the reation rates shown in Figure 14.7, we are asked how the forward and reverse rates at 30 µs ompare for the forward and the reverse reations. We see on the plot that the onentration of the produt B is higher than the onentration of the reatant A at 30 µs, so the reation is well on its way to formation of produt B from reatant A. The reation, however, has not yet ome to equilibrium, where the rate of the forward reation is equal to the rate of the reverse reation. Beause the reation is still making produt B and losing reatant A, the rate of the forward reation must still be larger than the rate of the reverse reation. When the reation reahes equilibrium, the onentrations of the reatants and produts are unhanging. For this reation, equilibrium ours after 50 µs Collet and Organize We are to desribe an everyday experiene of dynami equilibrium. In a dynami equilibrium, the forward and reverse proesses our at the same rate so that we observe no hanges in the onentrations of reatants and produts. Any proess in whih something is removed and replaed immediately would be an example. A ommon example is that of an unopened soda bottle in whih the dissolved arbon dioxide is ontinuously entering the gas phase at the same rate at whih undissolved arbon dioxide enters the liquid phase in a dynami equilibrium proess. One the soda bottle is opened, however, the rate of CO esaping the liquid phase is faster than the rate of CO entering the liquid phase, and the soda eventually goes flat Collet and Organize We onsider whether the sum of the onentrations of reatants equals the sum of the onentrations of the produts at equilibrium. Equilibrium is defined by an unhanging omposition of a reation mixture brought about by the rate of the forward reation equaling the rate of the reverse reation. No. The reation may heavily favor either produts or reatants. The definition of equilibrium does not speify the distribution of the reatants and produts. At any given temperature for a reation at equilibrium, the value of does not hange. The ratio [produts] x /[reatants] y (where x and y are stoihiometri oeffiients from the balaned hemial equation) does not hange Collet and Organize For a reation where k f > k r, we are to determine whether is greater than, less than, or equal to 1.

5 116 Chapter 14 The equilibrium onstant defined in terms of the rate onstants for a reation is kf k When k f > k r, is greater than 1. When > 1, more produts are present at equilibrium than reatants Collet and Organize Using the relationship between and k f and k r, we are to explain why may be large even though k f and k r are small. The equilibrium onstant defined in terms of the rate onstants for a reation is kf k The ratio k f to k r determines the magnitude of. For example, if k f 1 10 and k r , then Likewise, if k f and k r are both large, the value of might be small, as in the following example: Collet and Organize For the deomposition of N O to N and O, we are to identify the speies present after 1 day from the given molar masses. The initial reation mixture ontains 15 N O, N, and O. The only isotope in the reation for oxygen is 16 O, but for nitrogen both 15 N and 14 N are present at the beginning of the reation. After 1 day, the 14 N will be inorporated into N O and 15 N will be inorporated into N. r r Molar Mass Compound How resent 8 14 N Originally present 9 15 N 14 N From deomposition of 15 N 14 NO N From deomposition of 15 N O 3 O Originally present N O From ombination of 14 N and O N 14 NO From ombination of 15 N 14 N and O N O Originally present The redistribution of 15 N from N O to N and of 14 N from N to N O demonstrates that both forward and reverse reations our in a dynami equilibrium proess.

6 Chemial Equilibrium Collet and Organize From the equation CO + O CO and starting the reation with 13 CO, 1 CO, and O we are to identify the ompounds present after 1 day and explain their appearane. The masses of the atoms ombine to give the molar masses of the ompounds seen in the reation mixture: 1 C 1 g/mol 13 C 13 g/mol 16 O 16 g/mol (the most abundant O isotope) The ompounds mathed with their molar masses are as follows: 8 g/mol 1 CO 9 g/mol 13 CO 3 g/mol 16 O 44 g/mol 1 CO 45 g/mol 13 CO The reation mixture initially ontains 13 CO, 1 CO, and 16 O. The 1 CO is produed in the reation mixture through the reverse reation 1 CO O + 1 CO and the 13 CO is produed through the forward reation 13 CO + O 13 CO A reation, even at equilibrium, is dynami and isotopially labeled elements will, to some extent, be srambled throughout the reatants and produts Collet and Organize From the rate laws and rate onstants for the forward and reverse reations for A B we are to alulate the value of the equilibrium onstant. At equilibrium, the rate of the forward reation equals the rate of the reverse reation. The equilibrium onstant is the ratio of the onentrations of the produts raised to their oeffiients to the onentrations of the reatants raised to their respetive oeffiients. For the reation A B we are given that the forward reation is first order in A: Rate f k 1 [A], where k s 1 and that the reverse reation is first order in B: Rate r k 1 [B], where k s 1 At equilibrium, rate f rate r, so k 1 [A] k 1 [B] Rearranging this to give the equilibrium onstant expression allows us to solve for : 1 [ B] k s A k s [ ] 1

7 118 Chapter 14 For the reverse equilibrium the equilibrium onstant is 1/0.333, or B A Collet and Organize Given the equilibrium onstant ( ) and the rate onstant for the reverse reation (k r 0.54 M 1 s 1 ) for NO + O NO we are to alulate the rate onstant for the forward reation. At equilibrium, the rate of the forward reation is equal to the rate of the reverse reation, or k f [NO] [O ] k r [NO ] Rearranging gives the equilibrium onstant expression: [ NO ] kf [ NO] [ O ] kr Therefore, k f k r. k f M 1 s M 1 s 1 Beause the equilibrium onstant is large, it is onsistent that k f > k r Collet and Organize From the equation relating and p, we an determine under what onditions p. The equation relating and p is ( ) n p RT Δ p equals when Δn 0. This is true when the number of moles of gaseous produts equals the number of moles of gaseous reatants in the balaned hemial equation. The value of p may also be less than (for Δn < 0) or greater than (for Δn > 0) Collet and Organize We are asked to determine at whih temperature the values of and p are equal for the reation of nitrogen with oxygen to give gaseous NO. The equation relating and p is ( ) n p RT Δ For this reation Δn 0, so the value of (RT) Δn 1. This means that p (RT ) Δn (RT ) 0 1 p Therefore, and p are equal at all temperatures.

8 Chemial Equilibrium 119 As we saw in roblem 14.17, equals p only when Δn Collet and Organize Given three reations involving nitrogen oxides, we are to write and p expressions for eah reation. The expression uses onentration (molarity) units for the reatants and the produts, whereas the p expression uses partial pressure. The expression for a balaned hemial reation takes the general form (a) [N O ] 4 [N ][O ] and NO4 p (b) () [NO ][N O] and wa + xb yc + zd y z [C] [D] and [A] [B] ( ) ( N )( O ) ( NO )( ) NO 3 p 3 [NO] ( NO ) ( N ) ( ) O and p NO [N ] [O ] [N O] ( ) y ( C) ( D) ( ) ( ) w x p w x A B uses onentration units (usually in moles per liter), whereas p uses partial pressure units (usually in atmospheres) Collet and Organize For three balaned reations, we are to write the equilibrium onstant expressions. The expression uses onentration (molarity) units for the reatants and the produts, whereas the p expression uses partial pressure. The expression for a balaned hemial reation takes the general form [ ClO][ O ] (a) and Cl O [ ][ 3 ] [ Cl][ O] (b) [ ClO] 3 [ O ] () and [ O3 ] and p ( ClO )( O ) ( Cl )( O ) 3 ( Cl )( ) O ( ) p ClO 3 ( O ) p O 3 ( ) wa + xb yc + zd y z [C] [D] and [A] [B] y ( C) ( D) ( ) ( ) w x p w x A B z z

9 10 Chapter 14 uses onentration units (usually moles per liter), whereas p uses partial pressure units (usually atmospheres) Collet and Organize Given a plot of the onentration versus time for the deomposition of N O to N and O (Figure 14.1), we are to estimate the value of. The amounts of N O, N, and O are given in onentration units and the form of the expression is 1/ [ N][ O] [ NO ] The onentrations of eah speies at equilibrium an be read from the graph as those onentrations that are no longer hanging with time. This ours after 5 s and gives [N ] M, [O ] M, and [N O] M. [ ][ ] [ ] 1/ 1/ N O ( M)( M) 0.50 NO ( M ) 6 Beause this equilibrium onstant is neither high nor low, at equilibrium the distribution of the reatants and produts is expeted to be roughly equal. A general rule is that if < 0.01, the reatants are favored in the equilibrium, and if > 100, the produts are favored in the equilibrium Collet and Organize From the onentrations of O, NO, and NO at equilibrium shown in Figure 14., we are to estimate for NO + O NO. Equilibrium is ahieved when the onentrations of the reatants and produts in the reation mixture no longer hange. For this reation this is at [O ] M, [NO ] M, and [NO] M.

10 Chemial Equilibrium 11 The equilibrium onstant expression for this reation is NO NO ( M ) M [ ] [ ] [ O ] ( ) ( M ) 300 Beause is greater than 1, the produt is favored in this reation Collet and Organize Given the balaned hemial equation and the equilibrium partial pressures of all speies in the deomposition reation of H S to give H and S, we are to alulate the value of p for the reation. The form of the p expression for this reation is p H ( ) S ( H S ) ( ) p H ( ) S ( H S ) ( ) (0.045 atm)(0.030 atm) (0.00 atm) Remember that equilibrium onstants,, do not have any units Collet and Organize Given the partial pressures for H O, H, and O for the equilibrium reation H O H + O we are to alulate the value of p. The p expression for this reation is ( H ) ( ) O ( HO) p

11 1 Chapter 14 ( ) ( ) ( HO) ( atm) ( atm) (0.040 atm) H O 5 p Even though the partial pressure for eah gas is in atmospheres, the equilibrium onstant itself has no units beause eah pressure in the expression is atually a ratio (whih gives the ativity ) in whih the units anel Collet and Organize Given the equilibrium molar onentrations of N, O, and NO, we are to alulate for N + O NO The equilibrium onstant expression for this reation is [ NO] N O [ ][ ] 3 ( M ) ( M)( M) Be areful to aount for the oeffiients in the mass ation equation when alulating. In this problem, we must be sure to square the equilibrium onentration of NO Collet and Organize Given the equilibrium molar onentrations of N O 4 and NO, we are to alulate for the reation NO N O 4 The equilibrium onstant expression for this reation is [ NO 4] [ NO ] 3 (.9 10 M ) 3 ( M ) 160 Be areful to aount for the oeffiients in the mass ation equation when alulating. In this problem, we must be sure to square the equilibrium onentration of NO Collet and Organize Given the initial moles of H O and CO and the moles of CO present at equilibrium, we are to determine for H O + CO H + CO

12 Chemial Equilibrium 13 Beause the ratios of reatants and produts are 1:1:1:1, we do not need the volume of the vessel beause that volume would anel in the final expression. We an therefore simply use the mole amounts. Furthermore, we an alulate the amounts of reatants and produts present at equilibrium from the initial amounts of the reatants, the given moles of the produt CO formed ( mol), and the stoihiometry of the reation. Reation H O + CO H + CO H O CO H CO Initial (mol) Change (mol) x x +x +x Equilibrium (mol) x x x x At equilibrium, we know that [CO ] x mol. This gives H O CO x mol H CO x mol 3 3 [ H][ CO] ( mol)( mol) H O CO mol mol [ ][ ] ( )( ) We must have a balaned hemial reation so that we an stoihiometrially relate the quantity of produts to the reatants and orretly alulate the value of the equilibrium onstant Collet and Organize Given the initial moles of NO and H ontained in a 100 ml vessel along with the onentration of NO at equilibrium, we are to alulate the value of for H + NO H O + N The initial onentrations of the reatants are mol NO 0.60 M NO L mol H M H L Beause we are given the [NO] at equilibrium, we an alulate the amount of NO reated. From that we an alulate by stoihiometry the equilibrium onentrations of H, H O, and N and alulate the value of for the reation. Reation H + NO H O + N [H ] [NO] [H O] [N ] Initial Change x x +x +x Equilibrium x 0.60 x x x At equilibrium, we know that [NO] M, so [NO] M 0.60 x x

13 14 Chapter 14 Therefore, the equilibrium amounts of all speies are [H ] x M [NO] M [H O] x M [N ] x M [ ] [ ] [ H ] [ NO] ( M) ( M) ( M) ( M) HO N We must have a balaned hemial reation so that we an stoihiometrially relate the quantity of produts to the reatants and to orretly alulate the value of the equilibrium onstant Collet and Organize We are given p 3 for the following reation at 98 : A + B AB We are to alulate the value of, whih we an do using the relationship RT Δn p ( ) The hange in the number of moles of gas for this reation (Δn) is 1 mol AB (1 mol A + 1 mol B) 1 L atm mol When Δn is positive, p is greater than, but when Δn is negative, p is less than. When Δn is zero, p equals Collet and Organize We are given for the following reation at 500 : CD C + D We are to alulate the value of p, whih we an do using the relationship RT Δn p ( ) The hange in the number of moles of gas for this reation (Δn) is (1 mol C + 1 mol D) (1 mol CD) 1 p p L atm mol When Δn is positive, p is greater than, but when Δn is negative, p is less than. When Δn is zero, p equals.

14 Chemial Equilibrium Collet and Organize We are given p N + 3 H NH 3 We are to alulate the value of. Rearranging the equation relating p and to solve for gives p ( RT) Δn The hange in the number of moles is ( mol NH 3 ) (1 mol N + 3 mol H ) The temperature in kelvins is L atm mol Here, p < beause Δn is negative. When Δn 0, p Collet and Organize Given for the following reation at 98, CO + O CO we are to alulate the value of p, whih we an do using the relationship RT Δn p ( ) The hange in the number of moles of gas for this reation (Δn) is ( mol CO ) ( mol CO + 1 mol O ) 1 p p L atm mol The lower value of p is onsistent with Δn being negative Collet and Organize We are asked to determine in whih of the three reations the values of and p are equal. p and are equal when Δn 0 in the relationship p ( RT ) Δn where Δn is the hange in the number of moles of gas in the reation: Δn (moles of gaseous produts) (moles of gaseous reatants)

15 16 Chapter 14 (a) Δn 3 1 (b) Δn () Δn 0 Reations (b) and () have p. Only reation (a) has p. Remember that Δn is the differene in the moles of gaseous produts and gaseous reatants. In reation (b), Fe(s) and FeO(s) are not onsidered Collet and Organize For three reations we are to determine for whih the values of and p are different. p and are different when Δn 0 in the relationship p ( RT ) Δn where Δn is the hange in the number of moles of gas in the reation: Δn (moles of gaseous produts) (moles of gaseous reatants) (a) Δn 1 1 (b) Δn 3 1 () Δn 3 1 For all three reations (a ), p. For reations (a) and (), p >, whereas for reation (b), p < Collet and Organize Given Cl + CO COCl we are to alulate the value of p RT Δn p ( ) For this reation Δn (1 mol COCl ) (1 mol Cl + 1 mol CO) 1 p L atm mol 1 Beause Δn is negative, the value of p is less than that of Collet and Organize Given p 3.45 for the following reation at 98 : SO + NO NO + SO 3

16 Chemial Equilibrium 17 we are to alulate the value of for the reverse reation. We an use the relationships p ( ) RT Δn reverse 1 forward For this reation Δn (1 mol NO + 1 mol SO 3 ) (1 mol SO + 1 mol NO ) 0 Beause Δn 0, p (RT) 0 1 and the value of p, so 3.45 for the forward reation. For the reverse reation, 1 1 reverse forward Only when Δn 0 would and p be different in value Collet and Organize We onsider how hanges when the oeffiients of a balaned reation are saled up or down. To answer the question, let s onsider the following reations: (1) A + B C () 4 A + B C The equilibrium onstant expressions for the reations are [C] [C] 1 and [A] [B] [A] [B] These are related by ( 1 ) 4 When we sale the oeffiients of a reation up or down, the new value of the equilibrium onstant is the first raised to the power of the saling onstant. Saling reation 1 by 1 / 3 gives 3 A B 1 3 C [C] 1 3 [A] 3 [B], or ( )1/ Collet and Organize We are to explain why the temperature and a written form of the equilibrium equation are needed when reporting the value of. As we saw in roblem 14.37, saling a hemial equation up or down hanged the value of the equilibrium onstant. Temperature and the value of are related through Gibbs free energy: Δ G ln RT

17 18 Chapter 14 The temperature and form of the equilibrium reation must be given with a value of beause both the stoihiometri saling of the reation and the temperature hange the value of. From the Gibbs free energy equation, we see that the value of inreases with an inrease in temperature for a spontaneous reation (negative ΔG) and dereases with an inrease in temperature for a nonspontaneous reation (positive ΔG) Collet and Organize Given the equilibrium onstant for the reation of 1 mol I with 1 mol Br to give mol IBr, we are to alulate the value of the equilibrium onstant for the reation of ½ mol of I and Br to give 1 mol of IBr. The expressions for these reations are IBr 1 and I Br for the seond reation is, therefore, related to that of the first by ( ) I IBr 1 Br ( 1 ) 1 (10) When we multiply a hemial reation by a number, the new value of the equilibrium onstant is the first equilibrium onstant raised to that number Collet and Organize Given the equilibrium onstant for the reation of 1 mol N with 3 mol H to give mol NH 3, we are to alulate the value of the equilibrium onstant for the reation of 1 / mol of N with 3 / mol of H to give 1 mol of NH 3. The expressions for these reations are NH 1 3 and 3 N H for the seond reation is, therefore, related to that of the first by ( ) 1 1 N NH H ( 1 ) 1 ( ) When we multiply a hemial reation by a number, the new value of the equilibrium onstant is the first equilibrium onstant raised to that number.

18 Chemial Equilibrium Collet and Organize For the reation of NO with NO 3, we are to explain how for the reverse reation relates to of the forward reation by writing their equilibrium onstant expressions. The form of the equilibrium onstant expression for the forward reation is the ratio of the produt of the onentration of the produts raised to their stoihiometri oeffiients to the produt of the onentration of the produts raised to their stoihiometri oeffiients: [produts]y [reatants] x The form of the equilibrium onstant expression for the forward reation is,forward [NO ] [NO][NO 3 ] For the reverse reation, the equilibrium onstant expression is [NO][NO 3],reverse [NO ] Examining these two expressions, we see that 1,reverse,forward Another way to think about this relationship is,forward,reverse Collet and Organize Given p for the reation p for at the same temperature. N + 3 H NH 3 NH 3 N + 3 H The p expressions for these reations are p is, therefore, related to p1 by p1 ( NH 3 ) ( ( N )( H ) and N )( H ) 3 3 p ( NH 3 ) p 1 p1 p When we reverse a reation, the new equilibrium onstant is the inverse of for the forward reation.

19 130 Chapter Collet and Organize For two reations of different stoihiometry for the reation of SO with O to produe SO 3, we are to write the equilibrium onstant expressions and explain how they are related. Equilibrium onstant expressions take the form wa + xb yc + zd y z [C] [D] w x [A] [B] The equilibrium expressions for the reations are [SO 3] [SO 3] and 1 [SO][O] [SO][O] These expressions are related by ( ) If the seond reation were then 1 SO + 1 O 1 SO 4 3 [SO!! 3 ] 1 [SO ] 1 [O ] Collet and Organize Given 0.11 for the reation NO + H N + H O at a given temperature, we are to alulate for NO + H 1 N + H O at the same temperature. The expressions for these reations are 1 N H O and N 1 H O NO H NO H for the seond reation is, therefore, related to that of the first by ( ) 1 1 ( 0.11) When we multiply a hemial reation by a number, the new value of the equilibrium onstant is the first equilibrium onstant raised to the number Collet and Organize Given the value of for the reation SO + O SO 3 as , we are to alulate for three other forms of this reation at the same temperature.

20 Chemial Equilibrium 131 When a reation is multiplied by a fator x, the new equilibrium onstant is ( ) x. When a reation is reversed, the new equilibrium onstant is 1/. (a) This reation is the original equation multiplied by 1. The new is ( ) (b) This reation equation is the reverse of the original equation. The new is ( ) () This reation equation is the reverse of the original equation multiplied by 1. The new is 3 ( ) Beause we an relate the equilibrium onstants for different forms of a reation, we need to tabulate only one of the equilibrium onstants. All others an be alulated from that value Collet and Organize Given the value of for the reation NO + O NO as , we are to alulate for three other forms of this reation at the same temperature. When a reation is multiplied by a fator x, the new equilibrium onstant is ( ) x. When a reation is reversed, the new equilibrium onstant is 1/. (a) This reation is the original equation multiplied by 1 /. The new is ( ) (b) This reation equation is the reverse of the original equation. The new is 1 ( ) () This reation equation is the reverse of the original equation multiplied by 1 /. The new is ( ) 1 Beause we an relate the equilibrium onstants for different forms of a reation, we need to tabulate only one of the equilibrium onstants. All others an be alulated from that value Collet and Organize We are asked to alulate for a reation given values for two other reations. We an add the two equations and reverse the resulting equation and then alulate the equilibrium onstant for this new hemial equation.

21 13 Chapter 14 The expressions for the reations and the overall reation are C A + B C A B C D D C A + B D 3 D A B where 3 1. The equilibrium onstant for the reation D A + B is the inverse of When we add reations, the new equilibrium onstant is the produt of the equilibrium onstants of the reations that were ombined Collet and Organize We are asked to alulate for a reation given values for two other reations. We an add the two equations and reverse the resulting equation and then alulate the equilibrium onstant for this new hemial equation. The expressions for the reations and the overall reation are G H 1 H G H E + F E F.8 10 H G E + F 3 E F G where 3 1. The equilibrium onstant for the reation E + F G 1. will be ( ) ( ) Remember that when we reverse an equation, the equilibrium onstant is the inverse of the forward reation equilibrium onstant.

22 Chemial Equilibrium Collet and Organize / We are to define the term reation quotient. The reation quotient is the ratio of the onentrations of the produts raised to their stoihiometri oeffiients to the onentrations of reatants raised to their stoihiometri oeffiients. The reation quotient has the same form as the equilibrium onstant expression, but the reation onentrations (or partial pressures) are not neessarily at their equilibrium values. If the reation quotient, Q, is greater than, the reation mixture must redue its onentration of produts to attain equilibrium. When Q is less than, the reation mixture must inrease its onentration of produts to attain equilibrium Collet and Organize We are asked how, in most ases, the equilibrium onstant and the reation quotient Q differ for a given equilibrium system. Both and Q take the form of the ratio of the onentrations of produts raised to their stoihiometri oeffiients to the onentrations of reatants raised to their stoihiometri oeffiients. The reation quotient Q is for a reation mixture not neessarily at equilibrium. Only if the system is at equilibrium is Q. If the reation quotient, Q, is greater than, the reation mixture must redue its onentration of produts to attain equilibrium. When Q is less than, the reation mixture must inrease its onentration of produts to attain equilibrium Collet and Organize We are asked what it means when Q. Both and Q take the form of the ratio of the onentrations of produts raised to their stoihiometri oeffiients to the onentrations of reatants raised to their stoihiometri oeffiients. When Q the system is at equilibrium. Whenever Q, the reation is not at equilibrium and it adjusts its relative onentrations of reatants and produts so that Q Collet and Organize We are to explain how omparing numerial values of and Q allows us to determine whether a reation is at equilibrium and whih diretion a reation will shift to attain equilibrium. Both and Q take the form of the ratio of the onentrations of produts raised to their stoihiometri oeffiients to the onentrations of reatants raised to their stoihiometri oeffiients. When Q <, the reation shifts to the right, forming more produts. When Q >, the reation shifts to the left, forming more reatants. When Q, the reation is at equilibrium.

23 134 Chapter 14 When Q, the reation shifts until Q. At that point no more hanges in the relative onentrations of the reatants and produts our as long as reation onditions remain onstant Collet and Organize We are asked whether the reation A(aq) B(aq) where [A(aq)] 0.10 M, [B(aq)].0 M, and is at equilibrium. If the reation is not at equilibrium, we are to state in whih diretion the reation will proeed to reah equilibrium. The reation quotient (Q) is [ B] [ A] Q where [B] and [A] are the onentrations of A and B in the reation mixture. If Q >, the reation will proeed to the left to reah equilibrium. If Q <, the reation will proeed to the right. If Q, the reation is at equilibrium..0 M Q M No, the reation is not at equilibrium. Q <, so this reation proeeds to the right to reah equilibrium. In this reation, more B forms as the reation proeeds to equilibrium Collet and Organize We an ompare Q versus to determine in whih diretion the reation will shift to attain equilibrium. We are given The form of the reation quotient for this reation is Q [ D][ E] [ C] If Q >, the reation will proeed to the left to reah equilibrium. If Q <, the reation will proeed to the right. If Q, the reation is at equilibrium.

24 Chemial Equilibrium (5 10 M)(5 10 M) 4 Q 1 (5 10 M ) Beause, Q > ( ), this reation proeeds to the left to reah equilibrium, forming more of reatant C. To alulate the amount of eah reatant and produt at equilibrium, we solve for x in the following equation: 4 3 (5 10 M x) (5 10 M + x) Collet and Organize Given two sets of reatant and produt onentrations for the reation of A and B to form C (where p 1.00), we are to determine whether either reation mixture is at equilibrium. The temperature is 300. These systems are at equilibrium when Q p For the reation where A, B, and C are expressed in terms of molarity, we must onvert p to by using p ( RT ) Δn where for this reation Δn atm (a) Q p 1.0 ( 1.0 atm)( 1.0 atm) This reation mixture is at equilibrium. (b) Rearranging the equation to solve for gives p n 1 ( RT ) Δ L atm mol 1.0 M Q 1.0 ( 1.0 M)( 1.0 M) Beause Q <, this reation mixture is not at equilibrium and shifts to the right to attain equilibrium. Both p and are lose to 1, and so these reation mixtures have roughly equal proportions of reatants and produts when they reah equilibrium.

25 136 Chapter Collet and Organize For the reation A + B C for whih p 1.00 at 300, we are to alulate Q for different partial pressures or onentrations of A, B, and C to determine in whih diretion eah reation mixture will proeed to reah equilibrium. If Q >, the reation will proeed to the left to reah equilibrium; if Q <, the reation will proeed to the right; if Q, the reation is at equilibrium. Beause we are given p for the reation, we have to alulate for the onditions of part b from the relationship RT Δn (a) Q p ( C ) 1.0 atm ( )( ) ( 1.0 atm)( 0.50 atm).0 A B p ( ) For this reation mixture, Q >, so the reation proeeds to the left. (b) p n RT Δ ( ) where R (L atm)/(mol ), T 300, and n L atm ( 300 ) mol 1.0 M Q M 1.0 M For this reation mixture, Q <, so the reation proeeds to the right. Beause and p are both greater than 1, the produts are favored for this reation Collet and Organize By omparing Q versus for the reation of N with O to form NO, we an determine in whih diretion the reation will proeed to attain equilibrium. The form of the reation quotient for this reation is Beause n 0 for this reation, p Q p ( ) NO ( N )( ) O 3 ( ) ( )( ) Q 1.00 p 3 3 Beause Q >, the system is not at equilibrium and proeeds to the left. For this problem p. This is not always the ase, so be areful to notie whih value is provided for and in what units the amounts of the reatants and produts are expressed.

26 Chemial Equilibrium Collet and Organize Given p at 650 for the reation of N with H to produe NH 3, we an use the amounts of the reatants and produts present to alulate Q and then ompare that value to to determine whether more ammonia will form in the reation (reation shifts to the right). Beause the amounts of N, H, and NH 3 are expressed in terms of molarity, we need to find the value of from p by using the equation RT Δn p For this reation, Δn and is 4 p n ( RT ) Δ L atm mol The reation quotient is ( M ) Q (0.010 M)(0.030 M) For this reation mixture, Q <, so the reation proeeds to the right. Yes, more ammonia will form. If we had not onverted p to, we would have reahed the opposite onlusion that less ammonia would be present one the reation attained equilibrium Collet and Organize Given initial onentrations of the reatants X and Y and produt Z and the value of the equilibrium onstant ( 1.00 at 350 ), we an use the reation quotient to determine in whih diretion the reation will shift to reah equilibrium. The reation quotient for this reation has the form Q ( ) [Z] [X][Y] [Z] 0. M Q 5 [X][Y] 0. M 0. M Beause Q >, the reation proeeds to the left (a), produing more X and Y. To alulate the equilibrium onentration of eah reatant and produt, we solve for x in the equation 0. x x ( ) Collet and Organize From the result of roblem 14.59, we are to determine how the onentration of reatant X hanges from the initial onditions to equilibrium. We determined in roblem that the equilibrium shifts to the left.

27 138 Chapter 14 The onentration of X inreases when the equilibrium shifts to the left. At equilibrium, this reation has a dereased onentration of produt Z Collet and Organize For the dissolution of CuS to give Cu + and S in aqueous solution, we are asked to write the expression. Beause any pure solids or liquids do not hange in onentration during a reation, their onentrations do not appear in the mass ation (equilibrium onstant) expression. [Cu + ][S ] Beause the reatant in the dissolution reation is a pure solid, only the onentrations of the produts influene the position of the equilibrium Collet and Organize For the heterogeneous equilibrium reation Al O 3 (s) + 3 H O( ) Al 3+ (aq) + 6 OH (aq) we are to write the expression. Beause any pure solids or liquids do not hange in onentration during a reation, their onentrations do not appear in the mass ation (equilibrium onstant) expression [Al ] [OH ] Beause both reatants are either pure solids or pure liquids for this reation, only the onentrations of the produts influene the equilibrium Collet and Organize For the deomposition reation of CaCO 3 to CO and CaO, we are to explain why [CaCO 3 ] and [CaO] do not appear in the expression. The expression inludes the onentrations of the reatants and produts that may hange during the reation attaining equilibrium. The onentrations (expressed as densities) of pure solids (CaCO 3 and CaO) do not hange during the reation to attain equilibrium, and so they do not appear in the equilibrium onstant expression. The form of for this reation is simply [CO ].

28 Chemial Equilibrium Collet and Organize For the deomposition of NaHCO 3 to Na CO 3, CO, and H O we are to identify the number of partial pressure terms that are part of the p expression. The deomposition of NaHCO 3 is desribed by the stoihiometri equation NaHCO 3 (s) Na CO 3 (s) + CO + H O In the equilibrium expression, pure liquids and solids do not appear, so the only speies that will appear in the p expression are CO and H O. The reation involves the breaking of bonds to form CO from CO 3, so we expet this reation to be endothermi. If we inrease the temperature, more produts will form in this reation by Le Châtelier s priniple. The p expression for the deomposition reation of sodium biarbonate is ( ) ( H O ) p CO Therefore, the p expression has two partial pressure terms. If the water formed in this reation were in the liquid form, the p expression would simply be ( ) p CO Collet and Organize We are asked whether the value of inreases when more reatants are added to a reation already at equilibrium. The general form of the equilibrium onstant expression for a reation is wa + xb yc + zd [C]y [D] z [A] w [B] x No, the equilibrium onstant is not hanged when the onentration of the reatants is inreased. The relative onentrations of the reatants and produts in that ase adjust until they ahieve the value of. The value of the equilibrium onstant is affeted only by temperature. The value of the reation quotient Q dereases below the value of when reatants are added to a system previously at equilibrium, and the reation shifts to the right Collet and Organize Given that adding reatants shifts a reation to the right, we are to determine how adding a reatant to a reation mixture affets the rates of the forward and reverse reations. At equilibrium, the rate of the forward reation is equal to the rate of the reverse reation. For the reation A B, the forward and reverse rate expressions are Rate f k f [A] Rate r k r [B]

29 140 Chapter 14 When a reatant is first added to a reation mixture at equilibrium, the forward rate will inrease in relation to the reverse rate beause we have inreased the onentration of A in the rate equation, thereby inreasing rate f. As more produt B is formed, the rate r inreases as rate f dereases as A is used up in the reation. Eventually, rate f rate r and the reation again reahes equilibrium. Be sure to remember that it is the overall rate f that is equal to rate r at equilibrium. It is not true that the rate onstants, k f and k r, are equal Collet and Organize Given that the for the binding of CO to hemoglobin is larger than that for the binding of O to hemoglobin, we are to explain how the treatment of CO poisoning by administering pure O to a patient works. By giving a patient pure O to breathe, we inrease the partial pressure of O to whih the hemoglobin in the patient s blood is exposed. This oxygen an displae the CO bound to the hemoglobin through appliation of Le Châtelier s priniple. Combining the two equations, we an write the expression for the displaement of CO bound to hemoglobin by O : Hb(CO) 4 Hb + 4 CO Hb + 4 O Hb(O ) 4 Hb(CO) O Hb(O ) CO As the onentration of O inreases, the reation shifts to the right and the CO on the hemoglobin is displaed. If we were given the values of the equilibrium onstants for the reations, we ould alulate the new equilibrium onstant for the overall reation Collet and Organize We onsider the effet of atmospheri pressure on the value of p for the reation NO N O 4 Even though Los Angeles and Denver have different atmospheri pressures, the reation will not be affeted, beause the onentrations or partial pressures of the reations are assumed to be the same. A lower pressure in Denver than in Los Angeles at the same temperature does not hange the value of p. Only temperature affets the value of the equilibrium onstant. As we will see later in the hapter, the value of p is dependent on the value of the free energy, as shown by the equation G RT ln. The free energy is affeted by temperature, as shown by G H T S, and so truly the only parameter that affets is the temperature Collet and Organize We are to interpret Henry s law through Le Châtelier s priniple.

30 Chemial Equilibrium 141 The dissolution of a gas (let s use oxygen in this example) in a liquid (let s use water) an be written as a hemial equation: O O (aq) Aording to Le Châtelier s priniple, an inrease in the partial pressure (or onentration) of O above the water shifts the equilibrium to the right so that more oxygen beomes dissolved in the water. This is onsistent with Henry s law. The solubilities of different gases in a liquid are different from eah other, but all gases are more soluble in a liquid when present at higher partial pressures Collet and Organize We are to explain why adding an inert gas to a reation mixture does not hange the position of equilibrium for the following reation: CO + O CO The equilibrium onentrations of reatants and produts are expressed by the equilibrium onstant equation: ( CO ) ( CO ) ( O ) p Adding an inert gas to a reation at equilibrium does not hange the relative partial pressures of the reatant gases. The partial pressure of the inert gas does not appear in the equilibrium onstant expression, and so the position of the equilibrium is unaffeted.

31 14 Chapter 14 In this equilibrium mixture, adding either CO or O shifts the equilibrium to the right, whereas adding CO shifts the equilibrium to the left Collet and Organize Of four reations, we are to determine whih will shift its equilibrium to form more produts when the mixture is ompressed to half its original volume. Dereasing the volume by half doubles the partial pressures of all the gaseous reatants and produts in the reations. This would ause a shift in the position of the equilibrium toward the side of the reation with the fewer number of moles of gas. (a) This equilibrium shifts to the left, forming more reatants. (b) This equilibrium shifts to the right, forming more produts. () This equilibrium shifts neither to the left nor to the right, as the number of moles of gas is the same on both sides of the equation. (d) This equilibrium shifts to the right, forming more produts. Reations (b) and (d) will form more produts when the volume of the mixture is dereased by half. The opposite shifts our in the position of the equilibrium when the volumes of the reation mixtures are inreased Collet and Organize Of four reations, we are to determine whih will shift its equilibrium to form more produts when the volume of the reation mixture is doubled. Doubling the volume of a reation will derease the pressure, whih auses the equilibrium to shift toward the side of the reation that has the greater number of moles of gas. (a) This equilibrium will shift to the left, forming more reatants. (b) This equilibrium will not hange. () This equilibrium will shift to the right, forming more produts. (d) This equilibrium will shift to the right, forming more produts. Reations () and (d) will form more produts when the volume of the reation mixture at equilibrium is inreased.

32 Chemial Equilibrium 143 The opposite shifts in equilibrium our if the volume is dereased to inrease the pressure, whih favors the side of the reation with the fewer numbers of moles of gas Collet and Organize We are to predit the effet on the position of the equilibrium O 3 3 O with various hanges in onentration and volume. Inreasing the onentration of a reatant shifts the equilibrium to the right, whereas inreasing the onentration of a produt shifts the equilibrium to the left. An inrease in volume dereases the pressure and shifts the equilibrium toward the side of the reation with the greater moles of gas. (a) Inreasing the onentration of the reatant, O 3, shifts the equilibrium to the right, inreasing the onentration of the produt, O. (b) Inreasing the onentration of the produt, O, shifts the equilibrium to the left, inreasing the onentration of the reatant, O 3. () Dereasing the volume of the reation to 1/10 its original volume shifts the equilibrium to the left, inreasing the onentration of the reatant, O 3. Adding O and dereasing the volume ause the same shift in the position of the equilibrium Collet and Organize We are to predit the effet on the position of the equilibrium NO NO + NO 3 with various hanges in onentration and volume. Inreasing the onentration of a reatant shifts the equilibrium to the right, whereas inreasing the onentration of a produt shifts the equilibrium to the left. An inrease in volume dereases the pressure and shifts the equilibrium toward the side of the reation with the greater moles of gas. (a) Inreasing the onentration of the produt, NO, shifts the equilibrium to the left. (b) Inreasing the onentration of the reatant, NO, shifts the equilibrium to the right. () Beause the moles of gaseous produts equal the moles of gaseous reatants, inreasing the volume of the reation has no effet on the position of the equilibrium. Inreasing the onentration of NO 3 has the same effet on the equilibrium as inreasing the onentration of NO Collet and Organize We are to determine how dereasing the partial pressure of O affets the equilibrium SO + O SO 3 Aording to Le Châtelier s priniple, inreasing the partial pressure or onentration of a reatant shifts the equilibrium to the right. Dereasing the partial pressure of a reatant, then, shifts the equilibrium to the left.

33 144 Chapter 14 Dereasing the partial pressure of O in this reation shifts the equilibrium to the left. At equilibrium, then, we have less SO 3 produt when we redue the partial pressure of O in the reation mixture Collet and Organize We are asked to predit how the onentrations of H, Cl, and HCl in the equilibrium H + Cl HCl will be affeted if the equilibrium onstant for the reation of ammonia with hydrohlori aid NH 3 + HCl NH 4 Cl(s) is muh higher than that of the first reation. If the for the reation of NH 3 with HCl is muh larger than that of H with Cl, then adding NH 3 to a reation mixture of H, Cl, and HCl will substantially remove HCl from that equilibrium mixture. When NH 3 reats with HCl to remove HCl from the equilibrium mixture, the reation to form HCl will shift to the right. Beause mol of HCl is produed in the reation of H with Cl, the relative onentrations of H and Cl will derease at half the rate at whih HCl dereases. By removing a produt of a reation through using a large- proess, we an drive even a low- proess to ompletion Collet and Organize For three proesses at equilibrium, we are to determine whih shows an inreased yield of produt C at inreasing temperatures. Endothermi proesses have heat as a reatant. When the temperature is raised for these reations, the equilibrium shifts to the right, inreasing the amount of produt C formed. Reation (a) is the only endothermi proess ( H > 0), so it is the only proess for whih the produt yield inreases with inreasing temperature. Temperature hanges do not affet the amount of produt formed for reation (b), for whih H Collet and Organize For three proesses at equilibrium, we are to determine whih shows a dereased yield of produt Z at inreasing temperatures. Exothermi proesses have heat as a produt. When the temperature is raised for these reations, the equilibrium shifts to the left, dereasing the amount of produts formed. Reation () is the only exothermi proess ( H < 0), so it is the only proess for whih the produt yield dereases as the temperature of the reation is raised. Temperature hanges do not affet the amount of produt formed for reation (b), for whih H 0.