Lecture 36. The hase Rule number f phases C number f cmpnents (chemically independent cnstituents) F number f degrees f freedm C, the mle fractin f cmpnent C in phase The variables used t describe a system in equilibrium: 11, 1, 31,..., C 1,1 phase 1 1,, 3,..., C 1, phase 1, 3,..., C 1, T,, phase Ttal number f variables (C-1) Cnstraints n the system: m11 m1 m13 m1, m1 m m3 m, mc,1 mc, mc,3 mc, - 1 relatins - 1 relatins - 1 relatins 1
Ttal number f cnstraints C( - 1) Degrees f freedm variables - cnstraints F(C-1) - C( - 1) FC- Single Cmpnent Systems: F 3 - In single phase regins, F. Bth T and may vary. t the equilibrium between tw phases, F 1. Changing T requires a change in, and vice versa. t the triple pint, F 0. T t and t are unique.
Fur phases cannt be in equilibrium (fr a single cmpnent.) Tw Cmpnent Systems: F 4 - The pssible phases are the vapr, tw immiscible (r partially miscible) liquid phases, and tw slid phases. (Of curse, they dn t have t all eist. The liquids might turn ut t be miscible fr all cmpsitins.) 3
Liquid-Vapr Equilibrium ssible degrees f freedm: T,, mle fractin f mle fractin f in the liquid y mle fractin f in the vapr z verall mle fractin f (fr the entire system) We can plt either T vs z hlding cnstant, r vs z hlding T cnstant. Let be the mre vlatile substance: * > * B and T b, <T b,b ressure-cmpsitin diagrams Fi the temperature at sme value, T. ssume Rault s Law: * B * B * B * (1 - ) B * ( * - B * ) 4
Cmpsitin f the vapr y > * * B * B * ( ) y < * B B * * B B B B * B ( ) int a: One phase, F 3 (T,, ) int b: Liquid starts t vaprize, F (T,; nt free.) z b, y y b Vapr is rich in. int c: Liquid has lst s much that its cmpsitin is c. The vapr is nw prer in, y y c Rati f mles in the tw phases is given by the lever rule: nliq cc n cc vap 5
intd:liquidisalmstallgne, d, y y d z Fr pints belw d, nly the vapr is present (F3). Numerical eample: benzene-tluene at 0 C Eercise 8.4b. Let B tluene and benzene. Given: * 74 Trr, B * Trr, z 0.5 Q. t what pressure des the miture begin t bil?. t pint b, 0.5 0.574 48 Trr. Q. What is the cmpsitin f the vapr at this pint?. B 0.5 11, 0.574 37, y 37/48 0.77 Q. What is the cmpsitin and vapr pressure f the liquid when the last few drps f liquid bil?. int d. y z 0.5 74 74 (1 ) 0.9, B 0.771 0.974 0.771 33.9 Trr 6
Lecture 37. Cnstructin f the Temperature- Cmpsitin Diagram Fi the ttal pressure at sme chsen value,. Biling des nt ccur at a unique temperature, but rather ver a range f temperatures. (Cntrast with a pure substance.) The vapr pressure curve is determined by the liquid cmpsitin: (T) B (T) (T) is the vapr pressure f pure at temperature T. 7
8 ssume Rault s Law: ) ( ) ( * * T T B B ) ( ) (1 ) ( * * T T B )) ( ) ( ( ) ( * * * T T T B B The vapr pressure pure liquid at any temperature T is given by the Clapeyrn-Clausius equatin: 0, 0 * * 1 1 ) ( ) ( ln T T R H T T vap ) ( 0 * T 1atm,T 0 nrmal biling pint f. The same reasning applies t substance B. Once we have ) ( * T and ) ( * T B, Rault s Law gives us as a unique functin f T and. This allws us t cnstruct the liquid curve.
Cmpsitin f the vapr: y * ( T ) H RT 1 1 T Tb vap * y ( T ), This result is equivalent t e µ * *, vap RT ln y µ, liq RT ln This last result is the starting pint fr the biling pint elevatin frmula, ecept that there we assume y 1. 9
Recipe fr cnstructin f the phase diagram: 1. Calculate T b, and T b,b at the pressure f the diagram.. Chse a temperature T b, <T<T b,b 3. Calculate * ( T ) and * ( T ). * B ( T ) 4. * * ( T ) ( T ) 5. y * (T)/. B B 10
Cling f a vapr miture int a. ure vapr, y z int b. Vapr begins t cndense at z b,y z z b int c. Cmparable amunts f the tw phases are nvapr cc present. Lever rule: n cc liquid int d. Vapr is almst all gne; z d z,y z d int e. Only liquid is present. 11
Distillatin int a. Miture starts t bil, with z,y z b ints b-c. Vapr is cndensed t frm a liquid with z b z c int c. The liquid that was cllected in the previus step is biled t frm a vapr with z d Cndensatin f the last bit f vapr prduces a liquid very rich in either (if T b, <T b,b n the left) r B (if T b, >T b,b n the right). 1
Nn-ideal slutins Left: Impssible phase diagram, because at ma,where the liquid f this cmpsitin just starts t bil, there is n crrespnding pint n the vapr curve. (There is n tie line.) The vapr shuld always lie belw the liquid in a pressure-cmpsitin diagram. t an etremum they must tuch. Center: Vapr pressure reaches a maimum because f repulsin between and B. This is an azetrpe, where the liquid and vapr have the same cmpsitin. (Nte an errr in the drawing: The vapr curve des nt have a cusp, but rather is tangent t the liquid curve.) Right: Vapr pressure reaches a minimum because f attractin between and B. This is als an azetrpe. 13
Distillatin f nn-ideal slutins Left diagram: lw biling azetrpe. slutin f cmpsitin z a first bils at T a. The vapr cmes ff with cmpsitin z b. It is cndensed and then bils at T c The final vapr t cme ff has the azetrpic cmpsitin, and the remaining liquid is enriched in B (r, depending n which side f the azetrpe the prcess started). Right diagram: high biling azetrpe. bils at T a s befre, z a first The final vapr t cme ff is enriched in B (r ), and the remaining liquid has the azetrpic cmpsitin. 14
Equilibrium between immiscible liquids Upper Critical Temperature Lwer Critical Temperature Duble Critical Temperature The tw phase regin is always described by a tie line. It is a n man s land. 15
Biling f immiscible liquids Melting f immiscible slids Miture f nn-reactive slids. Slids and B react t frm cmpund B. 16
Lecture 38. Equilibrium Cnstants Cnsider the fllwing reactin: H Cl Ø HCl Des it ccur spntaneusly? Let s calculate DG fr ne mle f reactin. 0 G ( ) f H 0 0 ( Cl ) 0 G f 0 G f ( HCl) 95.30kJ / ml 0 G r 190.6 It seems that the reactin ccurs spntaneusly. But what if we start with pure HCl. Shuldn t sme f it react t frm H and Cl? 17
T determine when the reactin ccurs spntaneusly, we must take the partial pressures int accunt: G G f f ( Cl ) ( HCl) 0 RT ln 95.30 RT ln Cl Cl HCl HCl H HCl r 190.6 RT ln RT ln RT ln 0 t equilibrium, DG r 0. Thatis, G Cl 190.6 G r RT ln ( H HCl ) Cl RT ln H HCl Cl RT ln G 0 r ln 76.93 at 98.5710 33 18
S what happens if we start with pure HCl? Suppse that initially 1barand H 0 HCl Cl t equilibrium, HCl 1 z and z H Cl ( 1 z) 33 z.5710 1.5710 z 33 z1.97 10-17 This reactin ges nearly t cmpletin. But this wn t be the case at higher T. 19
nther eample: N O 4 F NO G 0 f 97.89 and 51.31 kj/ml 0 G r 51.31 97.89 47.30 ln -4.730/RT 1.909 NO 0.148 NO 4 NO N O 4 What is the cmpsitin f bar f this material? Let be the mle fractin f NO. ( ) ( ) 1 Nting that 1bar, 0
1 - - 0 3 4 ± Fr bar, 0.38. (Only the sign is physically meaningful.) Fr0.01bar,0.940 What is the cmpsitin f a very lw pressure gas? / 4 1 ( ) 1 / / 1 (Here we used the series (1e) 1/ 1e/ -e /8 ) Fr 0.01 bar bar, 966 0.
Suppse yu smehw arranged t start with pure N O 4. What is the cmpsitin f an equilibrium miture? N O 4 NO Initial 0 0 Final 0 -z z (z) z 0 Fr 0.148 and 0 1bar, z0.175 1 0.175 0.85 N NO O 4 0.175 tt 1.1744 0.349 /1.1744 NO 0.349 0.97
The bk defines the fractin f dissciatin, a. Initial n. mles Final n. mles Final mle fractin N O 4 NO N 0 (1-a)n 1α 1 1α α 1 α α an α 1 α NO N O 4 NO N O 4 4α 1α But (1 a) 0 Substituting this result gives the same equatin as befre. This methd is useful if the ttal pressure is specified. Hw des change if is increased? Hw des the cmpsitin change if is increased? 3
Mre cmple stchimetry: H SCH 4 F 4H CS Set 1bar,sthat 4 H H S CS CH 4 4 H H S CS CH 4 rblem: Given that T 700 C and 76 Trr Initial Final H S 11.0 mml CH 4 5.48 H 0 CS 0 0.711 Find and DG r 4
5 Slutin: H SCH 4 F 4H CS 0 4 4 ) ( 4 4 CH S H CS H CH S H CS H Set 1bar,sthat rblem: Given that T 700 C and 76 Trr 1.01 bar Initial Final H S 11.0 mml 9.598 0.536 CH 4 5.48 4.769 0.66 H 0.844 0.159 CS 0 0.711 0.040 Find and DG r 3.3410-4 3.43 10-4 DG r 0 -RT ln 94.5 kj/ml
Lecture 39. Equilibrium Cnstants II Derivatin f fr H Cl F HCl G µ µ m m H nh µ Cl n Cl H dnh µ Cl dn Cl dg µ µ HCl n HCl HCl dn HCl Let indicate the etent f reactin. It isa device fr handling the stchimetry f the reactin. The chemical ptentials are dn H dξ dn Cl dξ dn HCl dξ H µ H G f m H RT, ( ) ln Cl µ Cl G f m Cl RT, ( ) ln µ HCl G f, m( HCl) RT ln HCl 6
7 utting all the pieces tgether: dξ RT HCl G RT Cl G RT H G dg HCl m f Cl m f H m f m } ln ) ( ln ) ( ln ) ( {,,, ln Cl H HCl r m RT G d dg ξ t equilibrium, 0 dξ dg m G r RT ln Cl H HCl
8 Nte that is independent f pressure (fr ideal gases). This des nt mean that the equilibrium ratis f mle fractins r cncentratins are independent f. Let s use H SCH 4 as an eample. CS CS etc. 4 4 4 CH S H CS H In general, ν
Fr eample, fr N O 4 F N O G, NO N O 4 e r m / RT lim NO 4 0 0 lim 0 NO Increasing the pressure reduces the ttal number f mles, in accrd with le Chatlier s principle. 9
30 Temperature dependence f RT G m r, ln dt T G d R dt d m r ) / ( 1 ln, Gibbs-Helmhltz equatin:, 1 ln T H R dt d m r, ln T dt R H d m r Van t Hff equatin: 1, 1 1 1 ) ( ) ( ln T T R H T T m r
Suppse that the reactin is ethermic, s that H r < 0. Fr T >T 1, ln ( T ( T 1 ) ) < 0 In ther wrds, increasing T shifts the reactin t the left. Otherwise, the reactin wuld run away, in accrd with le Chatlier. 31
Eample f H Cl F HCl H, 184.6 r m kj/ml t 5000 (assuming that DH iscnstant), 184,60 1 1 ln (5000) 76.93 8.31451 5000 98 6.86 95 Nwifwestartffwith1barfpureHCl,the equilibrium miture cntains 0.03 bar f H and 0.03 bar f Cl. 3
Hw d real gases and cndensed phases behave? µ i µ i RT ln a i Gases: a i f i /, lim 0 f i i Slvent: a g, lim 1 a Slute: a B B / B gb B, lim B 0 a B B We can als write a γ b B B B Nte: B n n B n B b B nb n M The activity f a pure slid r liquid is 1. Why? Because its activity desn t change, and n crrectin term is needed. 33
Eample: Liquid-vapr equilibrium: CCl 4 (liq) F CCl 4 (vapr) at 98 avapr vapr / a 1 liq vapr in bar DG vap,m(98) 4.46 kj/ml ln G RT vap, m 4460 8.31498 1.80 0.165 fl vapr pressure 14 Trr Nte: Frm the van t Hff equatin, 0 Gvap ( T ) 0 1 1 H vap RT T Tb This is the same result that we get frm the Clapeyrn- Clausius equatin. 34
Lecture 40. Equilibrium f Elecytrlytes Slubility rduct Fr a saturated slutin, gcl(s) Ø g (aq) Cl - (aq) S γ b γ b g ( b ) Cl Given that 1.810-10 and g g - 1, b g b b g Cl b g b 5 1.35 10 S, and assuming that ml/g 35
Debye-Híckel limiting law: lg `10 γ ± z z I I inic strength f the slutin (dimensinless) 0.5096 fr water at 5 C. 1 z i charge number f in i I zi ( bi / b ) Here, z 1,z - -1, I b @ 1.3510-5 lg 10 g -0.00187 g 0.9957 i b g 1.3510 5 ml/g 36
Eample: Determine the slubility f Ni 3 (O 4 ),given that S 4.7410-3. ssume first that g 1. One mle f electrlyte prduces 3 mles f Ni and tw mles f O 4. Let b the mlality f the Ni 3 (O 4 ) that gets disslved. The activity f the disslved Ni is 3gb, and the activity f the disslved O - 4 is 3g-b. S 3 b b b 3γ γ 108 b b b 5 b.1310-7 ml/kg Nw calculate the inic strength. z,z - 3I 0.5{3b() b(3) } 3.19610-6 lgγ ± 0.5096()(3) I 0.005463 5 3 γ ± γ γ g 0.9875 S 5 5 b 108γ ± b b.1610-7 ml/kg 37
Inizatin f water Water: H O F H 3 O OH - w a O ( H3O ) a( OH ) [ H 3O ][ OH ] [ H3 [ a( H O)] ] Nte: ph meters are calibrated fr activity. [ 3 H O ] W What is w at 0 0 C? W (5 0 C) 10-13.997 @ 10-14 [H 3 O ]10-7 M ph -lg 10 [H 3 O ]7 We need DG(0 0 C) fr acid dissciatin. Must crrect fr DC. G T H d T dt 38
0 ) ( T ) H ( T ) H ( T C T 0 T T T G( T ) G( T 0 ) H ( T0 ) 1 C T T0 T ln T0 T0 T0 DG(5 0 C) 79.868 kj/ml DH(5 0 C) 56.563 kj/ml DC (5 0 C) 19.7 j/ml/deg DG(0 0 C) 78.17 kj/ml w 1.19810-15 ph 7.47 39
Lecture 4. cid-base Equilibria Strng acid: H H O Ø H 3 O - [H 3 O ][H] 0 >> 10-7 ph - lg[h] 0 Strng base: [OH - ]>>[H 3 O ] [H 3 O ] W /[OH - ] ph p W lg[b] Weak acid: H H O F H 3 O - [ H O ][ [ H] ] [ H O ] [ H] 3 3 a 0 [ H 7 3O ] a[ H] 0 >> 10 ph ½ p a - ½ lg 0 40
Weak base: - H O Ø H 3 O OH - b [ H][ OH [ ] ] [ OH [ ] ] a b w [ ] bb0 B0 OH / w a [ H B 3 O ] w /[ OH ] wa / 0 ph ½ p w ½p a - lg B 0 41
rblem 9.16: cid rain. Calculate the ph f rainwater caused by disslved CO,given: The reactin invlved is 4 CO 3.610 atm H 1.510 6 Trr p a 6.37 H CO 3 (aq) H O(liq) F H CO 3 - (aq) H 3 O (aq) a [ H CO [ H 3 ][ H3O CO ] 3 ] [ H [ H 3 O ] CO ] 3 p a ph lg[h CO 3 ] ssume that [H CO 3 ][CO ] in the liquid phase. CO [ ] CO CO CO nco nh O nh O [ H O] n n [ CO ] 1000/18.015 [ CO ] CO / 0.018015 4
We calculate CO using Henry s law. CO H CO CO 4 CO 3.610 760.10 6 1.510 H 7 7.10 [ CO] 1.10 0.018015 5 M 43
44 Eact slutin: 4 equatins with 4 unknwns [H 3 O ], y [OH - ], z [ - ], [H] w y a z/ yz 0 - z Slutin: y- z w (-z) fl z- w / w a w w w a 0 3 0 w a a a w 0 3 ph ½ pa - ½lg(1.10-5 )5.645
Titratin f a Weak cid with a Strng Base Chemical equatins: BOH Ø B OH - OH HØ - H O - H H O F H 3 O - - H O F H OH - Unknwns: H, H 3 O, -,B,OH - lgebraic equatins: a [ H 3 O ][ [ H] ] (nte: a b w ) b [ H][ OH [ ] ] H 3 O B - OH - [H] [H] 0 [ - ] 45
[B ] [BOH] 0 S Initial pint: ph ½ p a - ½lg 0 Befre the equivalence pint: OH - HØ H O - H H O F H 3 O - [ - ]S [H] 0 - S (S is the cncentratin f salt prduced by BOH H Ø H OB - ) a [ H O ][ [ H] ] 3 [ H3O ] S S 0 ph p a - lg(( 0 - S)/S) 46
t the equivalence pint: H has been all cnverted, and the relevant chemical equatin is - H O F H OH - [ H][ OH [ ] ] [ OH S b ] [ OH ] Sb [ S( w / a 1/ )] poh ½ p w - ½p a - ½lgSp w - ph ph ½ p a ½p w ½lgS (Eercise: Wrk ut the crrespnding equatin fr titratin f a weak base by a strng acid.) End pint: [H 3 O ] w /[OH - ] w /(B 0-0 ) ph p w lg(b 0-0 ) 47
Lecture 43. Standard States The chice f standard state is arbitrary. It affects the value f, but nt what yu actually bserve in the lab. Gas: 1 bar Liquid r slid: the pure material Slute, including H : 1 M H in bichemical reactins: 10-7 M (i.e.,ph 7) Suppse yu lived n a planet where the atmspheric pressure is l bar. Yu my chse t use as yur standard a pressure f l bar. Hw des this affect m and? λ µ µ RT ln µ RT lnλ 48
49 Fr the cnditins f yur eperiment, RT RT RT RT RT RT RT ln ln ln ln ln ln ln µ λ λ λ µ λ λ λ µ µ µ Fr the reactin a bb F cc dd, λ ν lnλ ln ) (,,, RT G RT b a d c G G m r m r m r ν λ λ λ λ λ b B a d D c C reactin that is at equilibrium at will nt be at equilibrium at. \ This idea applies als t inic equilibria. Nrmally we chse fr the standard state c 1M. But fr bichemical systems, we chse fr H M c 7 10
µ( H ) µ ( H ) RT ln a 0 RT ln [ H ].30RT ph c [ H ] µ ( H ). RT µ ( H ) RT ln 30 7 10 c Fr a reactin nh Ø G r m Gr m 7ν. 303RT,, 50