CHEMICAL EQUILIBRIUM

Transcription

1 14 CHAPTER CHEMICAL EQUILIBRIUM 14.1 The Nature f Chemical Equilibrium 14. The Empirical Law f Mass Actin 14.3 Thermdynamic Descriptin f the Equilibrium State 14.4 The Law f Mass Actin fr Related and Simultaneus Equilibria 14.5 Equilibrium Calculatins fr Gas-Phase and Hetergeneus Reactins 613

2 THE NATURE OF CHEMICAL EQUILIBRIUM [C(H O) 6 ] Cl - [CCl 4 ] H O A B C D [C(H O) 6 ] + [CCl 4 ] - Add HCl t (a): Add water t (b): Sme C(II) Sme C(II) [CCl 4 ] - [C(H O) 6 ] + Lavender clr f (c) & (d): [CCl 4 ] - + [C(H O) 6 ] Fig Time dependence f reactants and prducts in the spntaneus reactin: [C(H O) 6 ] Cl - [CCl 4 ] H O. (a) Partial cnversin f [C(H O) 6 ] + int [CCl 4 ] -. (b) Partial cnversin f [CCl 4 ] - int [C(H O) 6 ] +.

3 617 Characteristics f the Equilibrium State HO() l HO( g) Frward reactin: Evapratin f liquid water t water vapr Backward reactin: Cndensatin f water vapr t liquid water At equilibrium, the frward and backward rates becmes equal. Fundamental Characteristics f equilibrium states 1. N macrscpic evidence f change. Reached by spntaneus prcesses 3. Dynamic balance f frward and reverse prcesses 4. Same regardless f directin f apprach 14. THE EMPIRICAL LAW OF MASS ACTION 618 Expressins f Equilibrium Cnstant: Law f Mass Actin Reactins in slutin (C.M. Guldberg & P. Waage, 1864) a A+b B c C+d D K C c [C] eq[d] = [A] a [B] eq d eq b eq ~ dimensins f (cnc) c+d-a-b Reactins in the gas phase K P c ( P ) ( P ) = ( ) a P ( P ) d C eq D eq b A eq B eq ~ dimensins f (press) c+d-a-b

4 Law f Mass Actin fr Gas-Phase Reactins 619 Thermdynamic Equilibrium Cnstant, K ~ dimensinless c ( PC / Pref ) ( PD / Pref ) a ( P / P ) ( P / P ) A ref B ref d b K c PP C P P d D ( c+ d-a- b) KPref K a b P A B Fr P ref = 1 atm, K = K P numerically. Mass actin law fr a general reactin invlving ideal gases c ( PC) ( PD) a ( P ) ( P ) A B d b K Law f Mass Actin fr Reactins in Slutin c d ([C]/ cref ) ([D]/ cref ) c d c 1 M K ref [C] [D] a b K ([A]/ cref ) ([B]/ cref ) a b [A] [B] 619 Law f Mass Actin fr Gas-Phase Reactins 1. Gases appear in K as partial pressures, measured in atm.. Disslved species enter as cncentratins, in mles per liter. 3. Pure slids, pure liquids, slvent in chemical reactin d nt appear in K. 4. Partial pressures and cncentratins f prducts appear in the numeratr, and thse f reactants in the denminatr; each is raised t a pwer equal t its cefficient in the balanced chemical equatin fr the reactin.

5 A388 Hmgeneus equilibria reactants and prducts all in the same phase Hetergeneus equilibria reactants and prducts with different phases Activities f pure slids and liquids = 1 Ca(OH) (s) Ca + (aq) + OH - (aq) - When the slutin is very dilute, the slvent is treated as a pure substance and ignred when writing K. frm the net inic equatin THERMODYNAMIC DESCRIPTION OF THE EQUILIBRIUM STATE Dependence f Gibbs Free Energy f a Gas n Pressure At cnstant T, P 1 P (ideal gas) G = (H TS) = H T S = TS (H = 0 at cnstant T fr an ideal gas) V P P 1 P S nrln nrln nrln G nrt ln V1 P P1 P1 G f taking the gas frm the reference state (P ref = 1 atm) t any P: P G nrt ln nrt ln P P ref

6 Equilibrium Expressin fr Reactins in the Gas Phase 65 Ex. 3 NO(g) N O(g) + NO (g) Fig A three-step prcess (red arrws) t calculate G f a reactin (blue arrw) fr which reactants and prducts are nt in their standard states f 1 atm. P P ref G1 3RT RT PNO PNO ref Step 1: ln ln Step : G G 3 65 PNO P P P NO NO NO Pref Pref Pref Pref Step 3: G RT ln RT ln RT ln 3 G = G 1 + G + G 3 G RT ln PNO/ P P / P ref NO ref 3 PNO/ Pref At equilibrium, G = 0 (cnst T & P). PNO/ P P / P ref NO ref 3 PNO/ Pref G RT ln RT ln K( T )

7 67 Fr the general reactin, aa+bb cc +dd At equilibrium, G c PC / Pref PD / Pref a P / P P / P ln d RT b A ref B ref RT ln K( T ) Reactins in Ideal Slutin Fr the general reactin, aa+bb At equilibrium, cc +dd G c [C] / cref [D] / cref a [A] / c [B] / c d RT ln RT b ref ref ln K( T ) 68 Activity, a G nrt ln P/P nrt ln P (ideal gas) ref nrt ln c/c nrt ln c (ideal slutin) ref G nrtln a (nnideal system) Activity cefficient, I ( i = 1 fr the reference state) a i = i P i /P ref (gas) = i c i /c ref (slutin) General expressin fr the equilibrium cnstant a a c C a A a a d D b B K

8 14.4 THE LAW OF MASS ACTION FOR RELATED AND SIMULTANEOUS EQUILIBRIA Relatinships amng Equilibrium Expressins Reversed reactin Inversed K H (g) + O (g) H O(g), K 1 = P(H O) / P(H ) P(O ) H O(g) H (g) + O (g), K = P(H ) P(O ) / P(H O) = K Multiplied by a cnstant K raised t a pwer equal t the cnstant H (g) + (1/) O (g) H O(g), K 3 = P(H O) / P(H )P(O ) 1/ = K 1 1/ 631 Additin (r Subtractin) f reactins Multiplicatin (r Divisin) f K s BrCl(g) Cl (g) + Br (g), K 1 = P(Cl )P(Br ) / P(BrCl) Br (g) + I (g) IBr(g), K = P(IBr) / P(Br )P(I ) BrCl(g) + I (g) IBr(g) + Cl (g), K 3 =? K 3 = K 1 K = P(IBr) P(Cl ) / P(BrCl) P(I )

9 63 EXAMPLE 14.7 At 5 C, + 1 () = () = () =? = = EQUILIBRIUM CALCULATIONS FOR GAS- PHASE AND HETEROGENEOUS REACTIONS 63 Step 1 Step Step 3 Balanced chemical equatin Partial pressures; (a) initial (b) changes (c) equilibrium Apprximatin schemes f neglecting a very small quantity

10 Evaluating Equilibrium Cnstants frm Reactin Data CO(g) + Cl (g) COCl (g) phsgene At 600 C, P 0 (CO) = 0.60 atm, P 0 (Cl ) = 1.10 atm, initially. P(COCl ) = 0.10 atm at equilibrium. K =? EXAMPLE 14.8 CO(g) + Cl (g) COCl (g) Initial Change Equilibrium K P COCl PCO PCl (0.10) 0.0 (0.50)(1.00) 633 Calculating Equilibrium Cmpsitins When K is knwn 635 EXAMPLE H (g) + I (g) HI(g) At 400 K, P 0 (H ) = 1.30 atm, P 0 (I ) = atm, in a sealed tube. At 600 K, K = 9.6. P(H ), P(I ), and P(HI)? At 600 K, frm the ideal gas law at cnst V, P 0 (H ) = 1.30 atm x (600 K / 400 K) = atm P 0 (I ) = atm x (600 K / 400 K) = atm

11 635 H (g) + I (g) HI(g) Initial Change x x +x Equilibrium x x x = = 9.6 x = r.35 (unphysical!) P(H ) = atm, P(I ) = 0.06 atm, P(HI) = atm A399 Fr ideal gases,

12 14 CHAPTER CHEMICAL EQUILIBRIUM 14.6 The Directin f Change in Chemical Reactins: Empirical Descriptin 14.7 The Directin f Change in Chemical Reactins: Thermdynamic Explanatin 14.8 Distributin f a Single Species between Immiscible Phases: Extractin and Separatin Prcesses 14.6 THE DIRECTION OF CHANGE IN CHEMICAL REACTIONS: EMPIRICAL DESCRIPTION The Reactin Qutient, Q aa+bb cc +dd 639 c PC PD a Q= P P A B Reactin qutient d b K= eq c eq PC PD eq a eq PA PB Equilibrium cnstant d b N (g) + 3 H (g) NH 3 (g), P(N ) : P(H ) = 1 : 3 PNH P P 3 NH3 NH3 K P P P /3 P P /3 N H H H H P P NH3 H

13 639 P P NH3 H Fig N (g) + 3 H (g) NH 3 (g) (a) Q < K : G < 0, Q must increase, frward reactin, Q > K : G > 0, Q must decrease, reverse reactin (b) Frm initial nnequilibrium cnditins n either side f the parabla, the partial pressures apprach equilibrium alng lines with slpe /3, because three mles f H are cnsumed t prduce tw mles f NH Hemglbin and Oxygen Transprt Myglbin Hemglbin Hemglbin A large prtein (glbin) + 4 irn-cntaining hemes Each heme carries a mlecule f O (near the lung) Myglbin ~ Only ne heme grup (near cells)

14 S -shaped fractinal saturatin curve fr HB 641 A plt f the fractin f the xygen binding sites f Hb and Mb that are ccupied as a functin f the partial pressure f O. External Effects n K: Principle f Le Châtelier 643 Fig Partial pressure versus time fr the equilibrium: H (g) + I (g) HI(g) (1) LHS f the dashed line: Apprach t equilibrium (Ex ) () Abrupt perturbatin by increasing P(H ) t.0 atm. (3) Le Châtelier principle wrks n the RHS f the dashed line: Decrease in P(I ) and increase in P(HI), resulting in the decrease in P(H ) t cunteract the perturbatin. (4) Apprach t a new equilibrium!

15 643 Le Châtelier s principle (1884) A system in equilibrium that is subject t a stress will react in a way that tends t cunteract the stress. Le Châtelier s principle predicts the directin f change f a system under an external perturbatin. Henry Le Châtelier (Fra, ) Effects f Changing the Cncentratin f a Reactant r Prduct 643 EXAMPLE H (g) + I (g) HI(g) An equilibrium mixture at 600 K (Ex ): P(H ) = atm, P(I ) = atm, P(HI) = atm K(600 K) = 9.6 External perturbatin (additin f H ) Abrupt increase f P(H ) t.000 atm New equilibrium reached New equilibrium partial pressures?

16 643 H (g) + I (g) HI(g) Initial Change x x + x Equilibrium.000 x x x = =9.6 x = r.99 (unphysical!) At new equilibrium, P(H ) = 1.86 atm, P(I ) = atm, P(HI) = 3.9 atm Effects f Changing the Vlume P ( g) P ( g) Q P / P Fig An equilibrium mixture f P and P 4 (center) is cmpressed (left) r expanded (right). Cmpressin Equilibrium shifts tward the frward directin. Expansin Equilibrium shifts tward the backward directin.

17 Effects f Changing the Temperature 645 NO (g) N O 4 (g) high T lw T Exthermic H (5 C) = 58.0 kj ml 1 PNO 4 K P NO K(5 C) = P(N O 4 )/P (NO ) = 8.8 Fig Equilibrium between N O 4 and NO depends n temperature. Right: Ice bath at 0 C, Mstly N O 4, Pale clr Left: Water bath at 50 C, Mstly NO, Deep clr A413 The Haber prcess Irn xide catalyst

18 Maximizing the Yield f a Reactin Haber-Bsch prcess: Fixatin f N frm air N (g) + 3 H (g) NH 3 (g), H < 0 (exthermic) 646 Large K at lw T (slw reactin) and at high P 500C, 00 atm, catalyst, cntinuus NH 3 remval lw T NH 3 at 500 C ttal P NH THE DIRECTION OF CHANGE IN CHEMICAL REACTIONS: THERMODYNAMIC EXPLANATION The Magnitude f the Equilibrium Cnstant 646 K ln K G S H RT R RT G S H exp exp exp RT R RT Large value f K Fr S psitive and large and H negative and large Increasing the number f micrstates (S > 0) and decreasing enthalpy (H < 0)

19 Free Energy Changes and the Reactin Qutient 647 aa+bb cc +dd G = G + RT ln Q Q c PC / Pref PD / Pref a P / P P / P A ref B ref At equilibrium, G = 0 and Q K. G = RT ln K d b G = RT ln K + RT ln Q = RT ln (Q/K) Fig The free energy f a reactin system is pltted against its prgress frm pure reactants(left) t pure prducts (right). Temperature Dependence f Equilibrium Cnstants 648 RT ln K G H TS ln / ln K 1 ln K K G RT H RT 1 H RT H RT S R / / S R S R 1 1 ln K H K R T T 1 1 Van t Hff equatin Fig Temperature dependence f the equilibrium cnstant fr the reactin N (g) + 3 H (g) NH 3 (g)

20 Effect f temperature change n K Depends n the sign f H H < 0 (exthermic) K as T H > 0 (endthermic) K as T 1 1 ln K H K R T T Fig Sketch f ln K against 1/T fr an exthermic and fr an endthermic reactin as predicted by thermdynamics. A41 EXAMPLE The equilibrium cnstant K fr the synthesis f ammnia is 6.8x10 5 at 98 K. Predict its value at 400 K. H f0 (NH 3 (g)) = kjml -1

21 649 Temperature Dependence f Vapr Pressure HO( l) HO( g) PH O(g) K K P H 1 1 ln vap ln K1 P1 R T T1 At the nrmal biling pint, T 1 = T b at P 1 = 1 atm. H vap 1 1 lnp R T Tb DISTRIBUTION OF A SINGLE SPECIES BETWEEN IMMISCIBLE PHASES: EXTRACTION AND SEPARATION PROCESSES Hetergeneus equilibrium Partitining a slute species between tw immiscible slvent phases I in H O and CCl 4 I (aq) I (CCl 4 ) Partitin cefficient, K I 4 I CCl K 85 (at 5 C) 1 aq ~ I mre sluble in CCl 4 than in H O Shifting the equilibrium Add I in the water. I (aq) + I (aq) I 3 (aq) Mre I (aq) in the water cnsumed. Le Châtelier s principle causes mre I t mve frm CCl 4 t H O.

22 Extractin Prcesses 651 EXAMPLE [I (aq)] i =.00 x 10 3 M L f this aq slutin is extracted with L f CCl 4 at 5 C. [I (aq)] f =? n(i ) = (.00 x 10 3 ml L 1 )(0.100 L) =.00 x 10 4 ml Let y mles remain in aqueus slutin. CCl 4 I 4 (.0010 y) / K 85 I y / aq y = 4.6 x 10 6 ml r.3% Fig (a) I (aq) n CCl 4 in a separatry funnel. (b) After shaking. Chrmatgraphic Separatins Separatin technique based n partitin equilibria Cntinuus extractin prcess Exchange f slute species between mbile and statinary phases [A] Partitin rati, K [A] statinary mbile 65

23 65 Paper chrmatgraphy Thin layer chrmatgraphy (TLC) 653 Gas-liquid chrmatgraphy Clumn chrmatgraphy

24 Thank yu fr listening t this lecture, and please cntinue t I t knw a variety f Green Wrld!