hapter 14: hemical Equilibrium 46 hapter 14: hemical Equilibrium Sectio 14.1: Itroductio to hemical Equilibrium hemical equilibrium is the state where the cocetratios of all reactats ad products remai costat with time. osider a sample of N O 4 gas. Whe this gas is placed i a closed evacuated cotaier at 100, NO a reddish brow gas is formed by decompositio of N O 4 (colorless. This is represeted by: N O 4 (g NO (g Forward Reactio Iitially, this is the oly reactio occurrig. However, as more ad more NO molecules form, they ca react with each other to produce N O 4. This is represeted by the reverse reactio: NO (g N O 4 (g Reverse Reactio As the cocetratio of N O 4 decreases, the forward reactio slows dow. oversely, as the cocetratio of NO icreases, the rate of reverse reactio icreases. At some poit, the rates of the forward ad reverse reactios become equal. The, we say that the reactio system has reached equilibrium. A chemical reactio at equilibrium is represeted by forward ad reverse arrows: N O 4 (g NO (g Graphically, the approach to equilibrium is depicted as: After the reactio has ru for a while, the cocetratios of NO ad N O 4 remai costat with respect to time.
hapter 14: hemical Equilibrium 47 hemical equilibrium is also defied as the state where the forward ad reverse reactios occur at the same rate. We say that the reactio has reached a state of chemical equilibrium. Sectios 14. - 14.3: Equilibrium ad Equilibrium ostat I 1864, Guldberg ad Waage proposed the Law of Mass Actio as a geeral descriptio of the equilibrium coditio. osider a geeral reactio: j A + k B l + D Where A, B, ad D represet chemical species, ad j, k, l ad are the coefficiets of the balaced chemical equatio. The Law of Mass Actio states that a certai ratio of reactat ad product activities reaches a costat value whe the reactio is at equilibrium. The ratio of these activities at equilibrium is called the equilibrium costat, K, ad is expressed as: (a (a K (a (a l D A j B k K is called the equilibrium costat. The thermodyamic defiitio of K ivolves activities a A, a B, a ad a D of the chemical species ivolved i the reactio. The activity of a compoet of a ideal gas mixture is the ratio of its cocetratio, [ ], or partial pressure,, to the stadard cocetratio of 1 M or to the stadard pressure of 1 atm. Thus, the activity of the reactio compoets ca be expressed as cocetratios or partial pressures. If the activities are expressed as cocetratios, the, l [] [D] K j k [A] [B] This is the equilibrium expressio where the activities are expressed i terms of cocetratio. The umerical value of K c at a give temperature ca be calculated if the equilibrium cocetratios of the reactio compoets are give. The equilibrium costat, K c, is expressed without uits. For ay pure liquid or pure solid, the activity is take as 1. osider a geeral reactio: j A (s + k B (aq l (aq + D (l
hapter 14: hemical Equilibrium 48 Recall the defiitio of activity. For a ideal solutio, the activity of a compoet is the ratio of its cocetratio to the stadard cocetratio of 1 M. For a pure liquid or a pure solid, this ratio is 1. Hece, whe writig equilibrium expressios ivolvig cocetratios, pure solids ad liquids are NOT icluded. j A (s + k B (aq l (aq + D (l The equilibrium costat for this reactio would be expressed as: K [] [B] l k Example: Write the expressio for K for the followig reactio: 4 (s + 5 O (g 4 O 10 (s Remember: For ay pure liquid or pure solid, the activity is take as 1. Thus, 1 K [ O ] 5 osider a geeral reactio: j A (g + k B (g l (g + D (g If the activities of the reactio compoets are expressed as partial pressures, the K l ( j (A (D k (B I this expressio, A, B, ad D are the partial pressures of the reactio compoets A, B, ad D, expressed i the same uit. Remember: For ay pure liquid or pure solid, the activity is take as 1. Now, cosider a geeral reactio: j A (s + k B (g l (g + D (l
hapter 14: hemical Equilibrium 49 For gases i a ideal mixture, the activity of each compoet is take to be the ratio of its partial pressure to the stadard pressure of 1 atm. For a pure liquid or pure solid, this ratio is 1. Hece, K l ( k (B Example: Write the expressio for K for the followig reactio, 4 (s + 5 O (g 4 O 10 (s K 1 5 (O Note: I order to cotiue further with equilibrium reactios, it is very importat for you to be able to write the correct equilibrium expressio. I Sectio 14.3, practice the Iteractive roblems. Sectios 14.4-14.5: The Equilibrium ostat Expressio hagig the hemical Equatio The equilibrium costat, K, is meaigless uless accompaied by a chemical equatio. K depeds o the form of the chemical equatio used to describe the equilibrium system. osider the chemical reactio: N (g + 3 H (g NH 3 (g The equilibrium costat for this reactio is expressed i terms of cocetratios as: Now, look at the reverse reactio: NH K [ 3 ] [ N ] [ H ] 3 NH 3 (g N (g + 3 H (g The equilibrium costat for this reactio is expressed i terms of cocetratios as:
hapter 14: hemical Equilibrium 50 K ' [ N] [ H] 3 [NH 3 ] Thus, K 1 K ' osider agai the chemical reactio: N (g + 3 H (g NH 3 (g NH K [ 3 ] [ N ] [ H ] Now, multiply this equatio by a factor. 3 Therefore, N (g + 3 H (g NH 3 (g The equilibrium costat for this reactio is expressed i terms of cocetratios as: NH K'' [ 3 ] [ N ] [ H ] Thus, the relatioship betwee K ad K is: K'' K 3 Thus, if the coefficiets of a balaced chemical equatio are multiplied by a factor, the equilibrium costat is raised to the th power. The same cocept applies to the equilibrium expressio i terms of partial pressure,, (i.e. K. Example: NH 3 is maufactured by the Haber process accordig to the reactio: N (g + 3 H (g NH 3 (g Take the equilibrium cocetratios at 130 to be: [NH 3 ] 6. x 10 - M [N ] 1.70 x 10 - M [H ] 6. x 10-3 M a alculate K at 130. NH K [ N H 3 ] [ 6. x 10 ] [ ] [ ] [. 170 x 10 ][ 6. x 10 ] 3 3 3 9.5 x 10 5
hapter 14: hemical Equilibrium 51 b alculate K at 130 for the reactio: NH 3 (g N (g + 3 H (g Recall: The equilibrium costats for forward ad reverse reactios are reciprocals of each other. We kow, K 9.5 x 10 5 ad K ' 1 1 11. x 10 K 95. x 105 c alculate K at 130 for the reactio: ¼ N (g + ¾ H (g ½ NH 3 (g ompare this equatio to the origial equatio. The origial equatio is multiplied by a factor,. ¼ Recall: If the coefficiets i a balaced equatio are multiplied by a factor, the equilibrium costat is raised to the th power. Thus, K '' ( K ( 9. 5 x 105 1/ 4 31 6 I Sectio 14.5, practice the Iteractive roblems. Sectios 14.6-14.7: The Equilibrium ostat Expressio Addig hemical Equatios If the overall chemical reactio is expressed as the sum of two or more reactios (called the elemetary steps, the K or K for the overall reactio is the product of the equilibrium costats for the idividual reactios (elemetary steps. This is called the rule of multiple equilibria. Example: osider the followig reactios: Ol (g O (g + l (g K p 8.8 x 10-1 (s + O (g O (g K p.4 x 10-8 alculate K for the followig reactio: (s + O (g + l (g Ol (g Look at the overall reactio: ad O are reactats ad they appear as reactats i the secod reactio. However, l appears as a reactat i the
hapter 14: hemical Equilibrium 5 overall reactio, ad as a product i the first reactio. Also, Ol appears as a product i the overall reactio, ad as a reactat i the first reactio. Thus, you must reverse the first reactio ad multiply it by a factor of sice there are Ol ad l i the overall reactio. The first reactio is reversed: O (g + l (g Ol (g K K' 1 1 11. x 10 K 88. x 10 1 Now, multiply this reactio by a factor of. The reactio multiplied by a factor of is: 11 O (g + l (g Ol (g K K (K Here,. K (1.1 x 10 11 K 1. x 10 These are the reactios: O (g + l (g Ol (g K 1. x 10 (s + O (g O (g K p.4 x 10-8 Addig these reactios gives the overall reactio: (s + O (g + l (g Ol (g Accordig to the rule of multiple equilibria: K K x K (overall reactio (first reactio (secod reactio K 1. x 10 x.4 x 10-8 K.9 x 10 14 Note: The same cocepts ca be applied to calculate K c for the overall reactio. I Sectio 14.7, practice the Iteractive roblems.
hapter 14: hemical Equilibrium 53 Sectio 14.8: hemical Equilibrium alculatios (Iteractive I this topic, calculatios ivolvig chemical equilibrium are discussed. Example: osider the followig reactio at 350. O (g + l (g Ol (g K c.4 x 10 3 Assume the equilibrium cocetratio of l ad Ol are the same at 350. alculate the equilibrium cocetratio of O. First, write the expressio for the equilibrium costat. Ol K c [ ] [ O] [ l ] 4. x 10 3 Sice the equilibrium cocetratio of Ol ad l are the same, they cacel out from the expressio. 1 4. 10 3 [ O ] x [ ] 1 O 4. x 10 3 [O] 4. x 10-4 M I this sectio, practice the Iteractive roblems. Sectio 14.9: More hemical Equilibrium alculatios I this sectio, calculatios ivolvig chemical equilibrium are discussed. Example: A 5.00 L flask cotais 7.00 mol HI, 8.0 mol H ad 0.60 mol I i equilibrium at a particular temperature. alculate K c at this temperature for the reactio. H (g + I (g HI (g First, write the expressio for the equilibrium costat. HI K c [ ] [ H ][ I ] Remember, [ ] are used to express Molarity, M, ad M mol/l. The volume of the flask is 5.00 L.
hapter 14: hemical Equilibrium 54 7.00 8.0 Thus, [HI] [H ] [I ] 5.00 5.00 Substitutig these values ito the above equatio gives: K 7.00 5.00 8.0 0.60 5.00 5.00 c 9.96 0.60 5.00 Note: I this problem, you could have used oly moles to calculate K c sice the volume of the flask cacels out. However, it is a good practice to express [ ] as Molarity (M. I this Sectio practice the Iteractive roblems. Sectios 14.10-14.11: The Relatioship betwee K ad K c osider the reactio at a give temperature, T: The expressio for K c is: ad the expressio for K is: H (g + F (g HF (g HF K c [ ] [ H ] [ F ] K HF ( ( ( H F Recall, for a ideal gas, V RT or Apply this relatioship to the K expressio: RT V Molarity (M i.e. the cocetratio V Let (the cocetratio, [ ] V The, R T
hapter 14: hemical Equilibrium 55 K ( HFRT ( RT( RT H F ( ( RT HF Rearrages to: K ( ( ( RT H F Thus, ( HF K ( ( H F However, HF, H, ad F are writte as [HF], [H ], ad [F ], respectively. This meas [ HF] K K [ H ] [ F ] c For this reactio, K K c For the geeral reactio: j A + k B l + m D K K (RT Memorize this! sum of the coefficiets sum of the coefficiets - of the gaseous products of the gaseous reactats Therefore, for the geeral reactio: (l + m (j + k R gas costat 0.081 L.atm.mol -1.K -1 T absolute temperature Example: For the reactio, NO (g + l (g NOl (g at temperature, T K K c (RT ( ( + 1-1 Thus, K K c (RT -1 This equatio ca also be writte as: K c K p or K c K RT p (RT
hapter 14: hemical Equilibrium 56 Example: osider the followig equilibrium at 9. N O 4 (g NO (g K 1.3 alculate K c for this equilibrium reactio. K K c (RT 1 1 Thus, for this equilibrium reactio K K c R T R 0.081 L.atm.mol -1.K -1 T 9 + 73 365 K Thus, K c 13. 0. 0440 ( 0. 081 x ( 365 I Sectio 14.11, practice the Iteractive roblems. Sectios 14.1-14.13: The Reactio Quotiet (Q osider the geeral gas phase reactio: j A (g + k B (g l (g + m D (g K [] [A] l j [D] [B] m k ad K l ( j (A m (D k (B For a give system at a particular temperature, the values of K c ad K are fixed. Whe the reactats ad products for a give reactio are mixed, it is useful to kow whether the reactio system is at equilibrium. If the system is ot at equilibrium, the it is useful to kow the directio i which the reactio must proceed for the system to reach equilibrium. The reactio will occur i the directio favorig either the reactats or the products. To determie the directio i which a reactio proceeds, oe eeds to determie the reactio quotiet, Q. Q ca be expressed as Q c or Q i the same way as we expressed K as K c or K. For the geeral gas phase reactio:
hapter 14: hemical Equilibrium 57 Q [] [A] l 0 j 0 [D] [B] m 0 k 0 ad Q ( ( A l 0 j 0 ( D ( B m 0 k 0 Recall that [ ] 0 is defied as the iitial cocetratio, ad ( 0 is defied as the iitial partial pressure. By comparig the umerical value of Q (Q c or Q with K (K c or K, it is possible to decide the directio i which the reactio will proceed i order to achieve equilibrium. ase 1: If Q c < K c or Q < K If Q < K, there is more reactat ad less product i the iitial coditios tha at equilibrium. Thus, the reactio will favor the formatio of products (i.e. proceed to the right so that Q icreases ad evetually becomes equal to K. ase : If Q c > K c or Q > K If Q > K, there is less reactat ad more product i the iitial coditios tha at equilibrium. Thus, the reactio will favor the formatio of reactats (i.e. proceed to the left so that Q decreases ad evetually becomes equal to K. Note: If Q K, the reactio system is already at equilibrium uder iitial coditios ad there is o chage. Example: For the sythesis of ammoia at 300, N (g + 3 H (g NH 3 (g K c 5. x 10 - redict the directio i which the reactio will proceed to reach equilibrium whe the iitial cocetratios of NH 3, N ad H are 0.001 M, 0.0001 M, ad 0.00 M, respectively. First, calculate Q c. [NH 3 ] 0 0.001 M [N ] 0 0.0001 M [H ] 0 0.00 M NH Q c [ 3 ] 0 [ N ] [ H ] 0 0 3 ( 0. 001 Q c 15. x 106 K c 5. x 10 - ( 0. 0001 ( 0. 00 3 Thus, Q c > K c
hapter 14: hemical Equilibrium 58 Q is greater tha K. Hece, there is more product ad less reactat i the iitial coditios tha at equilibrium. Thus, the reactio will proceed to the reactat side (i.e. to the left. I Sectio 14.13, practice the Iteractive roblems. Sectio 14.14: alculatig Equilibrium ocetratios Example: Gaseous l 5 decomposes accordig to the reactio l 5 (g l 3 (g + l (g At 300 K, a.00 L flask iitially cotaied 0.00435 mol l 3 ad 0.98 mol l 5. After the reactio system reached equilibrium, 0.0000 mol l is foud i the flask. a alculate the equilibrium cocetratios of all species. Iitial ocetratios: 098. mol [ l 5 ] 0 0. 149 00. L 0. 00435 mol M [ l 3 ] 0 0. 0018 M 00. L [l ] 0 0 M (iitially, there is o l At equilibrium, the cocetratio of l is: 0. 0000 mol [ l ] 0. 00100 M 00. L l 5 (g l 3 (g + l (g Iitial: 0.149 M 0.0018 M 0 M At equilibrium: 0.00100 M The iitial ad equilibrium cocetratio of l are 0 M ad 0.00100 M, respectively. [l ] produced is 0.00100 M 0 M 0.00100 M. Now, look at the stoichiometry. Oe mole of l 5 gives oe mole l 3, ad oe mole l. Sice 0.00100 M l is produced, the 0.00100 M l 3 is also produced ad 0.00100 M l 5 is cosumed. [l 3 ] 0.0018 M + 0.00100 M 0.00318 M [l 5 ] 0.149 M 0.00100 M 0.148 M
hapter 14: hemical Equilibrium 59 l 5 (g l 3 (g + l (g Iitial: 0.149 M 0.0018 M 0 M At equilibrium: 0.148 M 0.00318 M 0.00100 M b alculate K c. l l K c [ 3 ] [ ] [ l ] 5 Note: We must use the equilibrium cocetratios to calculate the equilibrium costat. ( 0. 00318 ( 0. 00100 K c 0. 000015 ( 0. 148 I this sectio, practice the Iteractive roblems. Sectios 14.15-14.16: alculatig Equilibrium ressures Example: osider the followig equilibrium at 383 K. N O 4 (g NO (g K 1 redict the directio i which the reactio will occur to reach equilibrium if iitially there are 0.10 mol N O 4 ad 0.0 mol NO i a 1.0 L cotaier. Assume the gases obey the Ideal Gas law, V RT. N O 4 N O 4 V RT Here, NO4 N O 4 0.10 mol R 0.081 L.atm.mol -1.K -1 T 383 K V 1.0 L 0. 10 mol x 0. 081 L. atm. mol 1. K 1 x 383 K 31. atm 10. L
hapter 14: hemical Equilibrium 60 Similarly, NO NO V R T Here, 0.0 mol NO R 0.081 L.atm.mol -1.K -1 T 383 K V 1.0 L NO 0. 0 mol x 0. 081 L. atm. mol 1. K 1 x 383 K 63. atm 10. L I order to predict the directio i which the reactio will occur, the reactio quotiet, Q, should be calculated. Q ( NO ( 63. 31. NO4 Q 1.8 ad K 1 1. 8 Thus, Q > K Hece, the reactio proceeds i the reverse directio. I this sectio, practice the Iteractive roblem ivolvig the quadratic equatio. I Sectio 14.16, practice the Iteractive roblems to calculate equilibrium molarities. Sectio 14.17: hages Affectig a Gaseous Equilibrium System Oce a reactio system is at equilibrium, it is possible to chage the cocetratios of products ad reactats by chagig exteral coditios. Three differet exteral coditios ca be chaged: 1. Addig or removig a reactat or a product. Expadig or compressig a reactio system 3. hagig the temperature Always remember: If a chage is imposed o a system at equilibrium, the equilibrium will shift i the directio that couteracts the chage. This statemet is kow as Le hâtelier s riciple.
hapter 14: hemical Equilibrium 61 hagig the volume of the reactio system or the cocetratio of reactats ad/or products at costat temperature will ot chage the value of the equilibrium costat, K. Oly chages i temperature will chage the value of K. Sectio 14.18: Addig or Removig a Reactat or a roduct Recall Le hâtelier s riciple: If a chage is imposed o a system at equilibrium, the equilibrium will shift i the directio that couteracts the chage. osider the followig reactio system at equilibrium: N (g + 3 H (g NH 3 (g From here o, a shift towards the product side will be referred to as a shift to the right ad, a shift towards the reactat side will be referred to as a shift to the left. Read Le hâtelier s riciple ad apply the chage to the reactio system at equilibrium. a Add N (g : By addig N (g, the equilibrium will shift to the right to couteract the chage (by cosumig N (g. b Add H (g : By addig H (g, the equilibrium will shift to the right to couteract the chage (by cosumig H (g. c Add NH 3 (g : By addig NH 3 (g, the equilibrium will shift to the left to couteract the chage (by cosumig NH 3 (g. d Remove N (g : By removig N (g, the equilibrium will shift to the left to couteract the chage (by formig more N (g. e Remove NH 3 (g : By removig NH 3 (g, the equilibrium will shift to the right to couteract the chage (by formig more NH 3 (g. Remember: Merely addig or removig a reactat or product at costat temperature i a system at equilibrium does ot chage the umerical value of the equilibrium costat, K.
hapter 14: hemical Equilibrium 6 Sectios 14.19-14.0: Expadig or ompressig the Reactio System Recall Le hâtelier s riciple: If a chage is imposed o a system at equilibrium, the equilibrium will shift i the directio that couteracts the chage. Expasio of the reactio system at costat temperature: Whe the reactio system expads, its volume icreases. As a result, the cocetratios of the reactig species decrease. Recall, ocetratio, To reestablish equilibrium, the reactio proceeds i the directio that icreases the total umber of moles of gas. Example: osider the followig reactio system at equilibrium: V N (g + 3 H (g NH 3 (g How does a expasio of the reactio system affect the cocetratios of reactats ad products at equilibrium? As a reactio system expads, ad its volume icreases, the cocetratios of reactats ad products decrease. Now, read the statemet of Le hâtelier s riciple: If a chage is imposed o a system at equilibrium, the equilibrium will shift i the directio that couteracts the chage. The equilibrium will shift i the directio that icreases the total umber of moles of gas. I this reactio, there are 4 moles of reactats ad moles of product. 4 - I this case, as the system expads, the equilibrium shifts to the left.
hapter 14: hemical Equilibrium 63 ompressio of a reactio system at costat temperature: Whe a reactio system is compressed, its volume decreases. As a result, the cocetratios of reactig species icrease. To reestablish equilibrium, the reactio proceeds i the directio that decreases the total umber of moles of gas. Example: N (g + 3 H (g NH 3 (g - How does the compressio of this reactio system disturb the cocetratios of reactats ad products at equilibrium? Recall: Accordig to Le hâtelier s riciple, a equilibrium shifts i the directio that couteracts the disturbace. Here, the disturbace is a compressio (reductio i volume. Thus, the compressio leads to a icrease i the cocetratios of reactat ad products. Therefore, i this case, as the system is compressed, the equilibrium shifts to the right i order to decrease the total umber of moles of gas. Note: Expasio or compressio of a reactio system at costat temperature does ot chage the umerical value of the equilibrium costat, K. I Sectio 14.0, practice the Iteractive roblems. Sectio 14.1: hagig the Temperature Recall Le hâtelier s riciple: If a chage is imposed o a system at equilibrium, the equilibrium will shift i the directio that couteracts the chage. Whe the temperature is icreased, the equilibrium shifts i the directio for which the reactio absorbs heat (i.e. i the directio for which the reactio is edothermic. osider the reactio: N (g + 3 H (g NH 3 (g H - 9 kj Recall: If H is egative, the reactio is exothermic. If H is positive, the reactio is edothermic. The above reactio is exothermic i the forward directio. Thus, as temperature icreases, the equilibrium shifts to the left to absorb heat.
hapter 14: hemical Equilibrium 64 Now, cosider the reactio: N O 4 (g NO (g H 57. kj I this case, the forward reactio is edothermic. Thus, as the temperature is icreased, the equilibrium shifts to the right so as to absorb heat. Sectio 14.: Summary of Le hâtelier s riciple a hage i ocetratio. Reactats roducts Icrease i cocetratio Equilibrium shifts to the right Decrease i cocetratio Equilibrium shifts to the left Icrease i cocetratio Equilibrium shifts to the left Decrease i cocetratio Equilibrium shifts to the right b Expasio or otractio First, calculate. Reactats roducts If 0 If < 0 If > 0 No shift i equilibrium with expasio or cotractio. Upo expasio, equilibrium shifts to the left. Upo compressio, equilibrium shifts to the right. Upo expasio, equilibrium shifts to the right. Upo compressio, equilibrium shifts to the left. c hagig the Temperature Reactats roducts Look for the sig of H give with the reactio. If H is egative, the reactio is exothermic. As temperature icreases, the equilibrium shifts to the left. If H is positive, the reactio is edothermic. As temperature icreases, the equilibrium shifts to the right.