Chemical Equilibrium. Problem Set: Chapter 16 questions 25, 27, 33, 35, 43, 71

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Chemil Equilibrium roblem Set: Chpter 16 questions 5, 7, 33, 35, 43, 71

Exmples of Equilibrium Wter n exists simultneously in the gs nd liquid phse. The vpor pressure of H O t given temperture is property ssoited with n equilibrium ondition. H O (l) H O (g) I originlly dissolved in wter (left) will prtition between the CCl 4 nd H O liquids suh tht [I ] CCl4 / [I ] HO 86. The distribution oeffiient of solute between two immisible liquids is n equilibrium property. N O 4 (g) (olourless) deomposes to give moleules of NO (g) (brown gs). At low T, we hve mostly N O 4. At high tempertures we hve mostly NO. Under given set of onditions, the onentrtion of eh is in equilibrium.

Dynmi Equilibrium Equilibrium is dynmi ondition. Even though the system does not pper to be hnging, on mirosopi level it is onstntly hnging. In eh se shown previously, the forwrd nd reverse proesses re ourring t equl rtes, suh tht the mrosopi system ppers to be stti. We symbolize this equilibrium by the double rrow, reminding us tht the forwrd nd reverse retions re proeeding t equl rtes, resulting in blned hemil system.

roof of Dynmi Equilibrium Non-rdiotive AgI rdiotive Ag 131 I AgI/Ag 131 I Mixture AgI (s) Ag 131 I (s) Ag + (q) + I - (q) Ag + (q) + 131 I - (q) Add sturted solution mde from rdiotive Ag 131 I to sturted nonrdiotive AgI solution is rdiotive, should solid be? grdul pperne of rdiotivity in the solid AgI is proof tht 131 Iis inorported into solid. Tht is the equilibrium is dynmi

Approh to Equilibrium Exmple: Methnol prodution using the synthesis retion CO (g) + H (g) CH 3 OH (g) regrdless of our initil strting onditions, hnge in the system slows down nd ppers to ome to equilibrium. there does not pper to be onsistent finl ondition, i.e. reltionship between onentrtions.but there is!!

Finl stte for Synthesis retion [CH3OH] [CO][H ] 14.5 14.4 14.5 Equilibrium onstnt K

A Generl Expression for K A + bb + C For the generl retion gg + hh + ii The equilibrium onstnt expression hs the form [G] [H] [I] K [A] [B] [C] g h i eq eq eq b eq eq eq denotes tht onentrtions re expressed s molrities. the equilibrium onstnt is temperture dependent Equilibrium onentrtions do not onfuse the generl expression for equilibrium with the generl expression for the rte lw.here the onentrtions ARE rised to the stoihiometri oeffiients. Remember this ws not true in generl for the kineti rte lw.

Thermodynmi Equilibrium Constnt, K eq The thermodynmi equilibrium onstnt is slightly different thn wht we just desribed For the generl retion A + bb + C gg + hh + ii K eq ( ( G A ) ) g ( ( H B ) ) h b ( ( I C ) ) i A unitless tiviy of A B unitless tiviy of B, et.

Ativities nd K eq tivities re unitless (by definition γ on./stnd. refer. on.) stnd. refer. on. is typilly hosen s 1 M tivities re defined suh tht tivities of dilute solutes in solution re generlly numerilly equivlent to their onentrtions expressed s molrities (γ 1) tivities of gses re pproximtely equl to their prtil pressure expressed in tmospheres. tivities of pure solids nd pure liquids re defined to be unity (i.e. pure liquid pure solid 1.00) thermodynmi equilibrium onstnts re unitless.

Working with Equilibrium Constnts The equilibrium onstnt for reverse retion is given by the inverse of the forwrd retion equilibrium onstnt ( ) 1 1 K, reverse K K exmple N + 3H NH 3 K K [NH ] 3.6 10 M 3 3 [N ][H ] 8 NH 3 N + 3H 3 [N ][H ] 1 1,reverse 8 [NH 3] K 3.6 10.8 10 M 9

ont d The equilibrium onstnt for retion tht hs been multiplied through by onstnt, x, is given by : K ' ( ) x K Exmple multiply eq from previous exmple N + 3H NH 3 by ½ to produe only 1 mole of mmoni 1 N ( ) 3 + H NH3 [NH] 3 [NH] 3 1/ K ' 1/ 3/ 3 ( K) [N ] [H ] [N ][H ] 8 1/ 4 1 3.6 10 1.9 10 M 1/

Equilibri with Gses Mixtures of gses re in mny wys nlogous to mixtures in solution. Conentrtions of gses n be expressed s molrities OR by prtil pressure due to the idel behviour of gses. [G] ng G (rell tht V V RT nrt) Equilibrium onstnts for retions involving gses re often expressed s K p, where the subsript p denotes the use of pressure units. A (g) + bb (g) + C (g) For the generl retion K p g G A h H b B gg (g) + hh (g) + ii (g) i I C

Reltionship between K C nd K p... RT RT RT RT RT RT [C] [B] [A] [I] [H] [G] K C b B A i I h H g G b i h g n p b i h g C b B A i I h H g G RT 1 K RT 1 + + ( ) n p RT K K where: n is the differene in stoihiometri oeffiients of produts nd retnts ( n g+h+i b ) R is the gs onstnt (0.0806 L tm mol -1 K -1 )

Equilibri with ure Liquids or Solids Equilibrium onstnt expressions do NOT ontin terms for solids or liquid phses of single omponent (not mixture). Why? 1) onentrtions of single-omponent solids nd liquids do not hnge. ) tivities of pure liquids nd solids re defined to be 1 Exmple: C(s) + H O CO(g) + H (g) ol vpour [CO][H] [CO][H] KC [C][H O] [H O] ure solid Ativity of pure solid eq CO C H H O CO K 1 H H O CO H H O

Exmples of Equilibrium Constnts A retion is often sid to go to ompletion if the numeril vlue of the equilibrium onstnt is very lrge (i.e. K > 10 10 ) Note: tht lrge K is onsistent with lrge vlue in the numertor (produts) nd/or smll vlue in the denomintor (retnts). H (g) + O (g) H O (l) (written with single rrow)

Exmple A few liters of liquid wter re dded to the lrgest experimentl ontiner tht n be found in the hemistry deprtment t York (CAC smog hmber; 9.0 m 3 ) t 98K. If we wited lrge period of time suh tht the system omes to equilibrium (or dd hypothetil tlyst to speed up the retion), wht would be the equilibrium onentrtion of H nd O? How mny moleules of H nd O would exist? H 0 (l) H (g) + O (g) Let O x; H x H O 83 1 K p (1.4 10 ) 1 84 3 7.1 10 tm 3 84 8 K p (x) x 4x 7.1 10 x 1.1 10 tm 8 V (1.1 10 tm)(9000 L) 6 n 4.46 10 mol 1 1 RT (0.0806 Ltm mol K )(98K) # O n x N A (4.46x10-6 mol)(6.0x10 3 mol -1 ) 0.07 moleules Conlusion t equilibrium, only frtion of moleule of O will exist. Sine this is physilly impossible, we onlude the retion lies ompletely to the right nd does not proeed to the left to ny meningful extent. H 0 (l) H (g) + O (g)

The Retion Quotient (kwo:shent) Often we would like to predit the diretion of hemil hnge for system tht is not t equilibrium. This predition is trivil for the ses of systems in whih we hve only retnts or only produts. But wht bout n intermedite se in whih we strt with mixture of retnts nd produts? To mke this predition, we use the retion quotient, Q or Q p or Q eq. For the generl retion : [G] Q [A] g init init [H] [B] h init b init [I] [C] i init init where init implies initil onentrtions for the nonequilibrium system Note: we n lso write nlogous equtions for Q p using prtil pressures nd Q eq using tivities

rediting the Diretion of Chnge We ompre the retion CALCULATED quotient to the KNOWN equilibrium onstnt to predit the diretion of hemil hnge in retion. If Q < K the retion will proeed to the right If Q > K the retion will proeed to the left If Q K the retion is t equilibrium?????

Le Chtelier s riniple When n equilibrium system is subjet to some hnge in onditions, the system will respond by minimizing the effet of tht hnge. etrui sttes it s: When n equilibrium system is subjeted to hnge in temperture, pressure, or onentrtion of reting speies, the system responds by ttining new equilibrium tht prtilly offsets the impt of the hnge.

Effet of Chnge in Amount If the mount of one speies in n equilibrium system is inresed, the system will try to derese it ording to the Le Chtelier s riniple Indeed the equilibrium will shift in the diretion tht fvours the removl of tht speies. Alterntively, if the mount of the speies is deresed, the equilibrium shifts in the diretion tht inreses the mount of tht speies. Exmple: rell the synthesis retion CO (g) + H (g) CH 3 OH (g) K 14.5 Wht would hppen to the equilibrium system if we dded old finger to the retion vessel, llowing us to ondense out methnol? Let us presume tht we n remove 80% of the gseous methnol s liquid, in eh of severl ondenstion steps. Strt with experiment 1 onditions.

Effet of Inrementl removls of CH 3 OH [CO] g mol/ L [H ] g mol/ L [CH 3 OH] g mol/l Q -CH 3 OH moles* Comment Exp 1 Strt.0911.08.0089 14.5 Q K, equilib Condense 80%.0911.08.00178.90-0.0714 Q < K ; must shift right Shift right.0863.076.00659 14.5 Q K, equilib. Condense 80%.0863.076.0013.90-0.057 Q < K ; must shift right gin Shift right.085.0651.00507 14.5 Q K, equilib. Condense 80%.085.0651.00101.90-0.0405 Q < K ; must shift right gin Shift right.0795.0591.0040 14.5 Q K, equilib. Eh time we ondense liquid methnol, we remove gseous methnol, mking Q <K. The equilibrium ondition is re-estblished by shift of the retion to the right, suh tht Q K gin. *Inrementl mount of liquid methnol removed in moles (lulted).

[CO], [H], [CH3OH] (mol/l) 0.10 If we ontinuously remove methnol gs, the equilibrium ontinues to shift to the right. We n use this pplition of LeChteliers priniple to inrese the yield of reovered methnol. 0.08 Now we will plot the resultnt equilibrium 0.06onentrtion of the gses (mol/l), nd the umultive mount of methnol 0.04removed from the system ( moles), fter eh suessive ondenstion 0.0(Equilibrium #) 0 5 10 15 0 5 30 Equilibrium # 0.5 0.4 0.3 0. 0.1 0.0 CH3OH removed (moles) [CO] [H] [CH3OH]

Effet of Chnge in ressure If the totl pressure of hemil system is inresed (or deresed) by hnging the volume, the equilibrium will shift in diretion tht fvours redution (or inrese) in the totl pressure. Exmple: SO (g) + O (g) SO 3 (g) K 80 (1000K) A derese in volume inreses the prtil pressure of ll gses in the system. The equilibrium is shifted in diretion tht redues the totl pressure, in this se to the right, sine there re fewer moles of gses on the right.

Another Explntion SO (g) + O (g) SO 3 (g) K 80 (1000K) [SO3] [0.068] K 8 [SO ] [O ] [.03] [.016] After ompression to 1.0L, nd before the system n respond we hve ll onentrtion inresed 10 fold: [SO 3 ] 0.68 M; [SO ] 0.3 M; [O ] 0.16M [SO [SO ] ] [O [0.68] [.3] [.16] 3 Q ] 8. Sine Q < K, equilibrium will shift to the right. Note tht for every x moles of O onsumed, x moles of SO re onsumed nd x moles of SO 3 re produed. Thus, when equilibrium is rehed we must hve: [SO [SO ] ] [O [0.68 + x] [.3 - x] [.16 - x] 3 K ] 8 We will lter disuss how to solve these problems. For now the solution is: x 0.0751 moles; [SO 3 ] finl 0.830M, [SO ] finl 0.1697M, [O ] finl 0.0849M,

Wht We Lerned So Fr eq eq b eq eq i eq h eq g [C] [B] [A] [I] [H] [G] K init b init init i init h init g init [C] [B] [A] [I] [H] [G] Q n p b i h g C b B A i I h H g G RT 1 K RT 1 + + K

Effet of Inresed Totl pressure using spettor gses If we inrese the totl pressure t onstnt volume by dding n inert gs or nother gs not involved in the equilibrium, the prtil pressure of eh speies in the equilibrium will not hnge therefore the retion quotient will not hnge. The system will not respond to hnges in totl pressure hieved by dding inert gses t onstnt volume. Note: In solution, dilution of mixture by ddition of wter WOULD hnge the totl volume, WOULD hnge the onentrtion of eh speies in n equilibrium, nd therefore the retion quotient WOULD hnge in generl. We should expet shift in equilibrium in this se

Effet of Chnge in Temperture To predit the effet tht temperture hs on n equilibrium system, we n invoke Le Chtelier s priniple. If we inrese the temperture, the equilibrium system responds by trying to lower the temperture (utilize the exess energy). It would do this by shifting in the diretion of the endothermi retion. If the temperture is deresed, the equilibrium shifts in the diretion of the retion tht provides exess het (ie- the exothermi retion) N O 4(g) NO (g) H +57. kj mol -1 Q: Will the mount of NO (g) formed be greter t high or low temperture? A: The forwrd retion s written is endothermi. An inrese in temperture will fvour shift to the right in equilibrium. NO (g) will be higher t higher temperture.

Effet of Ctlyst For given set of retion onditions, the equilibrium mounts of retnts nd produts in reversible retion re independent of whether the retion is homogeneous, heterogeneous or otherwise tlysed. A tlyst does not hnge the position of equilibrium The tlyst only ts to derese the mount of time it tkes to get to equilibrium Sine we know tht tlysts n provide lternte retion mehnisms, our onlusion must be tht the equilibrium ondition is independent of the proess by whih the system moves from initil to finl onditions. We will lern more bout this in the setion on Free Energy

Equilibrium Clultions: Exmple 1 rtie Exmple 16.10 B: 0.100 moles SO nd 0.100 moles O re introdued into n evuted 1.5 L flsk t 900K. When equilibrium is hieved, the mount of SO 3 mesured is 0.0916 moles. Use this dt to determine K p for the retion: SO 3 (g) SO (g) + O (g) Solution Approh: we will use n ICE tble to relte the initil nd equilibrium (finl) onditions. It will be onvenient to strt working in units of moles. Our equilibrium onentrtions n then be expressed s moles L -1. We n then lulte K, nd finlly relte it to K p through the known reltionship; K p K (RT) n.

Amounts nd Conentrtions I nitil (mol) C hnge (mol) E quilibrium (mol) Con. mol L -1 Con. mol L -1 Solution: exmple 1 SO 3 SO 0 0.100 +x -x 0 + x 0.1 x x/v (0.1-x)/V.0606.00553 O 0.100 -x 0.1 x (0.1-x)/V.03566 We re given tht the equilibrium mount of SO 3 0.0916 mole x 0.0916 moles x 0.0458 moles Knowing tht V 1.5 L, we n omplete the lst line in the tble.

K [SO ] [O ] Solution 1, ont d [.00553] [.03566] 4 3.00 10 [SO3] [0.0606] n ( + 1) 1 K p K (RT) n K p 3.00 x 10-4 (0.0806 L tm mol -1 K -1 ) (900K) K p 0.0

Equilibrium Clultions: Exmple rtie Exmple 16.1 A: If 0.150 moles H (g) nd 0.00 moles I (g) re introdued into 15.0 L flsk t 445 o C nd llowed to ome to equilibrium, how mny moles of HI will be present? H (g) + I (g) HI (g) K 50. (445 o C) Solution Approh: We will gin use ICE tble to relte the initil nd equilibrium (finl) onditions. In this se though, we will not know the equilibrium onentrtions, but we know K. We use x s the vrible of hnge, write the equilibrium expression in terms of x, nd then solve for x. We must solve qudrti to get x.

Amounts nd Conentrtions I nitil (mol) C hnge (mol) E quilibrium (mol) Con. mol L -1 Exmple solution H 0.150 -x 0.150-x (0.150-x)/V I 0.00 -x 0.00-x (0.00-x)/V x [HI] V ( x) K [H ][I ] 0.150 - x 0.00 - x (0.150 - x)(0.0- x) V V This n be lgebrilly rrnged to give: -46. x + 17.57 x 1.506 0 This is qudrti eqution -46., b17.57, -1.506 HI 0 +x x x/v 50.

Solution, ont d x b ± b 4 17. 57 ± ( 17. 57) 4( 46. )( 1. 506) ( 46. ) x 0.1305 OR 0.498 tking the + nd root respetively. Only one solution is physilly relisti; x 0.1305. For x 0.5 moles we would end up with negtive mounts of H nd I. Thus x 0.1305 moles nd t equilibrium we hve x moles of HI, At equilibrium we hve 0.61 moles of HI

Complex Formtion A + B k 1 K-1 AB The equilibrium onstnt for omplex formtion is often lled the ssoition onstnt: [AB] eq K M -1 [A] [B] eq eq Even more often we operte with the dissoition onstnt: K d [A] eq [B] [AB] eq eq M K d 1 K

Mss Conservtion riniple for Complex Formtion [A] tot [A] + [AB] nd [B] tot [B] + [AB] Reltionship between the Equilibrium Constnt nd the Rte Constnts At equilibrium the rte of forwrd retion is equl to the rte of reverse retion: k 1 [A] eq [B] eq k -1 [AB] eq K d k -1 /k 1 [A] eq [B] eq /[AB] eq K d k k -1 1