Envr 210, Chapter 3, Intermolecular forces and partitioning Free energies and equilibrium partitioning chemical potential fugacity activity coef.

Transcription

1 Envr 20, Chapter 3, Intermolecular forces and parttonng Free energes and equlbrum parttonng chemcal potental fugacty actvty coef. phase transfer- actvty coef and fugactes more on free energes and equlbrum constants

2 Much of ths class deals wth the parttonng of an organc compound between two phases A+B C K eq = [C]/{[A][B]} K eq = [ phase ]/[ phase 2 ] When we deal wth ar lqud parttonng K al = C a /C L Octanol-water K ow = C o /C w Sold-water K d = C s /C w We wll fnd that often for classes of compounds log K d = a log K ow + b Why??? 2

3 For a compound to move between one phase and another, the ntermolecular forces that hold a molecule n one phase need to be broken and others reformed n the other phase Smply ths can be represented as: :: + 2:2 : + 2::2 (absorpton) f the phase change s from molecule n phase to the nterface or surface between and 2, then :: + :2 : + ::2 (adsorpton) What s the nature of the bonds that are beng broken or formed??. Nonspecfc nteractons (van der Walls nteractons) 3

4 a. related to a compound s polarzablty (a) or the extent to whch an uneven electron dstrbuton results n response to an mposed electronc feld on tmescales of0-5 sec; the ntermolecular attracton energy s related to the product of the a s of the nteractng set of atoms London dspersve energes b. dpole-nduced nteractons (Debye energes) resultng from electron dstrbuton dfferences n one molecule (carbon and oxygen bond) nducng a charge dstrbuton n the adjacent molecule. The strength of the nteracton should be a functon of the dpole moment, m= qr,n the dpole molecule, tmes the polarzablty of the charge nduced molecule. 4

5 c. dpole-dpole nteractons: strength of attracton proportonal to m x m 2 2. Specfc nteractons: ntermolecular attractons between electron rch and electron poor stes of correspondng molecules hydrogen bondng between the electron poor hydrogen of a carbon hydrogen bond and the unpared oxygen electrons n an adjacent molecule electron donor or acceptor nteractons In the absence of electron donor or accepter nteractons, London dspersve energes can be used to characterzed the attractons of many molecules to ther surroundngs wth respect to equlbrum parttonng 5

6 Consder a molecule movng from the gas to a lqud phase, (g) + :(L) :: (L) when dssolves n solvent, the dspersve attracton energy per nteracton, D dsp g s gven as (Israelachvle, 992) as a functon of polarzablty,a, and the st onzaton energes, I, of compounds and solvent ; I= I + I /( I I ) D dsp g = -(3/2) I /s 6 a a /(4p e 0 ) 2 Vsble lght has frequences (and ts changng electrc felds) on the order of 0-5 cycles /sec. A materal s ablty to respond to lght s related to ts ndex of refracton, n D, and n D s related to that materal s polarzablty va the Lorenz- Lorenz relatonshp 6

7 a /(4p e 0 )= [n 2 D -]/ [n 2 D +2]x(3M /4prN a ) Assumng sphercal molecules, and the nduced temporary dpoles dstances are dameters of the molecules (see page 64 of text) 2 D D D dsp g per I x n 2 nteracton = -3 / nd + 2 nd for a mole of nteractons we need to consder the total surface area (TSA) of the solvated molecule and the contact area (CA) t has wth solvent molecules 2 D D D dsp G N A TSA CA I x n 2 for one mole = - / 3 / nd + 2 nd Snce N A, CA, 3 and I are relatvely constant, n n

8 If the equlbrum s domnated by dspersve forces, ths free energy, D dsp G can be related to the equlbrum of ths process by D dsp G = - RT ln K eq where K eq = [ phase ]/[ phase 2 ] For an organc gas n equlbrum wth a pure lqud the equlbrum s: K al = C a sat /C L = M p * L/[r L RT] D D D dsp G const xtsa n 2 x n 2 for one mole = nd + 2 nd

9 D dsp G = - RT ln K eq we should be able to plot calculated ln K al = M p * L/[r L RT] vs. TSA n n 2 D 2 D x n n 2 D 2 D for a pure solvent nteractng wth the gas phase, =

10 Table 3. page 65 0

11 Fgure 3.4page 69 new text Parttonng n ar-pure solvent vs. ndex of refracton term Parttonng n Hexane? Parttonng n water?

12 Fgure 3.6 page 7 ar-hexane, top, ar-water, bottom 2

13 Chapter 3, then uses thermodynamcs to quantfy molecular energes and equlbrum parttonng Secton 3.3 starts wth: µ = N ( G = n ) T, P, n j and G ( = µ n p,t,n n 2,n...n N ) N = How do we get to these equatons and what do they mean?? 3

14 Chapter 3 The Frst Law U 2 - U = q - w change n nternal energy heat of an object work reservor object b U = q -w U = q 2 -w 2 a 4

15 For example one gram of H 2 O at 25 o C s evaporated and condensed; the condensed gram of water at 25 o C wll have the same nternal energy as t dd prevously. If only pv work s done and the pressure of the system s constant w rev = pdv What s the work of a reversble expanson of a mole of an deal gas at 0 o C from 2.24 to 22.4 lters? W rev = V 2 V pdv pv = nrt V2 V2 Wrev = nrt dv / V = nrtln ( ) V V W rev = mole x.987 cal K - mole - x 273 K x log (22.4/2.24) W rev =.25 Kcal mole - 5

16 Internal energy, heat and work when one mole of water s vaporzed at 00 o C the work s w = p D V = RT =.987 cal K - mole - x 373.5K w= 74.4 cal mole - The energy or heat requred to vaporze water at 00 o C requres energy to separate the lqud molecules; that s cal g - q = 8.02 g mole - x 539 cal g - = 9725 cal mole - ; For a mole of water, the nternal energy U = q - w U = 9725 cal mole cal mole - U = 8984 cal mole - 6

17 Enthalpy U = q - D(pV) at constant pressure q= (U 2 + pv 2 ) - (U + pv ) We defne U + pv as the enthalpy, H q = H 2 -H = DH or the heat adsorbed n a process at constant pressure There are usually two types of calormetrc experments used to determne heat, one at const volume (no PV work, so U=q) and one at constant pressure. The heat of combuston of CO n a constant vol calormeter s kcal mol -. Calculate the enthalpy of combuston n a const pressure. CO(g) + ½ O2 CO2(g) Work done s (n2-n)rt, n2 s moles of products, n s moles of reactants. DH =DU + (n-n2)rt = (0.5 mol) kcal K - mol - ) 298K DH = kcal mol - 7

18 Standard Heats of formaton kcal/mole kcal/mole C graphte 0 H2O(g) CO(g) H2O(l) CO2(g) ethene Benzene ethane H 2 C=CH 2 + H 2 ---> H 3 C-CH 3 DH total = DH f (H 3 C-CH 3 ) -DH f (H2) - DH f (H 2 C=CH 2 ) DH= (+2.5) +0 = kcal 8

19 Heat Capacty Heat Capacty, C = rato of heat absorbed/mole to the temperature change = Dq/DT At constant pressure so Dq = DU+pDV = DH C p = dh/dt.e. the calores of heat adsorbed/mole by a substance/ o C DH= C p (T 2 -T ) At constant volume DU = Dq - pdv DU = Dq C v = du/dt What s the relatonshp between C p and C v? 9

20 The Second Law Lord Kelvn ( ): It s mpossble by a cyclc process to take heat from a reservor and convert t nto work wthout at the same tme transferrng heat from a hot to a cold reservor Clausus: It s mpossble to transfer heat from a cold to a hot reservor wthout at the same tme convertng a certan amount of the work to heat.e. work can only be obtaned from a system when t s not at equlbrum It can be shown (see any p-chem book) that the max. effcency of a sequence of sothermal and adabatc process s eff = (T H -T L )/T H = (q H + q L )/q H rearrangng q T H H ql + = 0 T L 20

21 q T H H ql + = 0 T L q = T 0 = dqrev T 0 defne ds = dq/t and S2 S = S = 2 dqrev T at absolute zero the entropy s assumed to be zero 2

22 Consder mole of H 2 0 (l)---> H 2 0 g at 00 o C S H20 = dq/t = /T dq = /T n= n = 0 Hdn = dh vap /T = 9,720 cal/373k= + 26cal/degK mole s surroundngs = a negatve 26cal/degK S total = zero When spontaneous processes occur there s an ncrease n entropy When the net change n entropy s zero the system s at equlbrum If the calculated entropy s negatve the process wll go spontaneously n reverse. 22

23 Entropy cont. S = dq/t. What s t about a gas that makes t have more entropy when t s expanded, then when t s compressed or n the lqud state? Let s say that n the reacton of A ---> B B has more entropy than A 2. What s t about B that gves t more entropy? st consder a box wth a 4 pennes; f we place them wth heads up and then shake the box, we get: # combnatons 4 heads, 0 tals 3 heads, tal 4 2 heads, 2 tals 6 head, 3 tals 4 0 heads, 4 tals we mght consder ths to be the normal state, or equlbrum state, because there are more combnatons to go to 23

24 At a molecular level, A ---> B can be understood f A and B have states wth equal energes, and f B has more energy levels to go to wthn these states. Movng to the B energy states brngs the system to hgher entropy. the average translatonal energy of a gas n one drecton s gven by /2 RT If the energy level has the form e n = n 2 h 2 /(8ma 2 ) there wll be a certan # of e n levels for the gas as gven by the quantum #s n The sum energes n each e n level wll be /2 RT If the gas s expanded n the a drecton, we decrease the spacng between the energy levels, whch permts more energy states 24

25 Fgure 6.7, Physcal Chemstry, Barrow, McGraw Hll, New York, 963, page 47 25

26 The concept of free energy comes from the need to smultaneously deal wth the enthalpy energy and entropy of a system G = H -TS G = U+PV - TS dg= du + PdV + VdP -TdS -SdT dh = du +pdv at const temp and pressure G= H -T S What s the free energy for the process of convertng mole of water at 00 o C and one atm. to steam at one atm. H= H vap S vap = /T dq = H vap /T T S = H vap G= H vap - T S G= H vap - H vap = 0 26

27 Equlbrum Constants dg= du +VdP + pdv -TdS -SdT for a reversble process TdS = dq du -dq+dw = 0 so dg= +VdP -SdT at const temp ( G/ P) T = V; and f const. temp s stated all the tme dg/dp= V dg =nrt dp/p G2 -G = nrt ln(p 2 /P ) At standard state G = G o + nrt ln(p) 27

28 G = G o + nrt ln(p) for a reacton for A aa + bb--> cc + dd we have G A =G o A +art lnp A t s the free energy of the products mnus the reactants that s of nterest G =G prod - G react for reactants A and B G AB = G o A + G o B+ RT lnp A a + RT lnp B b For aa + bb--> cc + dd o (P G = Σ G + RT ln (P f the reacton goes to completon G = zero G P P = RTln ( ) ( ) ( P ) ( P ) o C c D d A a B b C A ) ) c a (P (P D B ) ) d b 28

29 G O = -RT lnk eq 29

30 Equlbrum Constants and Temperature 30

31 We are now ready for the st two equatons of Chapter 3, Secton 3.3 µ = N ( G = n ) T, P, n j G ( = µ n P,T,n n 2...n...n N ) N = 3

32 Closed Systems From the frst law du = dq - pdv from the defnton of entropy ds = dq/t du = TdS - pdv If we dfferentate by parts,.e. separately hold dv and ds constant ( U ) T S V = ; and U du S ds U = V + V ( ) ( ) U ( ) S = p V S dv Envronmental systems are often open systems,.e. materal s beng added or removed, and/or materal s reacted If a homogeneous system contans a number of dfferent substances ts nternal energy may be consdered to be a functon of the entropy, the volume and the change n the # moles 32

33 U du S ds U = V + V du ( ) ( ) S dv k U n = U U = V, n ds + S, n dv + S V ( ) ( ) ( ) S, V, n j dn k U s call the Chemcal Potental S V n n= j the term ( ) n,,, µ. du = TdS pdv + k ( U n = ) S, V, n j dn k µ U = n n= ( ),, S V n j du = TdS pdv + k µ dn = 33

34 The complete expresson for the dfferental of free energy s dg = du +pdv+vdp -TdS- SdT f du = TdS pdv + k ( U n = ) S, V, n j dn Substtutng for du n the free energy expresson dg = SdT + Vdp + k ( = or we could drectly defne k dg = SdT + Vdp+ ( n G = U n ) ) S, V, n T, Pj, nj j dn dn k U µ = ( ) n S, V, n j = = k ( ),, = k µ G = n = G n T P n j ( ),, = T P n j k ( ),, = H n S P n j 34

35 f we rewrte the equaton for the potental energy n ts ntegrated form U = T S p V + k µ = n t can be shown that for an expanson n whch temperature, pressure and the number of elements are proportonately ncreased and that the relatve proportons of the components are kept constant... G U = TS pv snce H= U+ pv k + H = TS+ µ n and G= H TS n= k µ n = k ( P, T, n n... n ) 2 n= = m n 35

36 G k ( P, T, n n... n ) 2 n= = m What ths says s that the total free energy of a system s drectly related to the sum of the ndvdual chemcal potentals tmes the number f moles of each contrbutng entty n Gong back to k µ G = n = ( ),, T P n j what does ths say about systems that are not at equlbrum 36

37 37

38 Chemcal Potentals and Pressure If we go back to the expresson for potental energy U = TS pv + k µ n = du = SdT+ TdS -Vdp-pdV+ µ dn + n dµ k = = k For a closed system whch only does pressure volume work we sad that du = TdS pdv + k µ dn = subtractng 0 = SdT -Vdp+ n dµ k = At constant temperature, one obtans the Gbbs-Duhem Equaton for gases 38

39 k Vdp = = n dµ g and so for just compound Vdp /n = (dm g ) T substtutng for V = nrt/p and ntegratng from a partal pressure of a compound defned as p 0 to p Du g = RT ln p /p 0 f our boundary condtons or lmts start at standard states m g = m o g + RT ln p /p o your book has elected to defne one bar as the standard state for pressure 39

40 What f the system s not deal? Van der Waal s equaton 2 an ( p + )(V nb ) = 2 V nrt nter molecular attracton occuped molecular volume For a non-deal system we could attempt to substtute for V n the chemcal potental relatonshp Vdp /n = (dµ ) T another way s to defne a parameter related to pressure called fugacty where by analogy f µ = µ o +RT ln f / f o 40

41 f =θ p θ s a fugacty coef. n a mxture of gas phase compounds p = x p * the vapor pressure n bars so what s vapor pressure???? whch your book calls p L * one atmosphere =.03 bars one atmosphere supports a 76 cm column of Hg one atmosphere = 760 mm Hg = 760 torr one atmosphere =.03x0 6 dynes/cm 2 derved from the force of mercury on cm 2 bar = 0 5 pascals 33.3 pascals = torr f g = x g θ g p L * where x s the mole fracton x g = n j g n j 4

42 Fugactys of lquds p = X p * (Raoult s Law) for two dfferent lquds wth the same components p p 2 5% 0% A n B A n B µ 2 = µ +RT ln p 2 /p snce p = x p L * and p 2 = x 2 p L * µ 2 = µ +RT ln x 2 /x (Ideal) smlarly µ 2 = µ +RT ln f 2 /f f pure lqud = g pure lqud p L * 42

43 Where g s called an actvty coeffcent f we dscuss compound n a lqud mxture f L = γ X p * L (pure lqud) the fugacty of compound wth respect to the fugacty of the pure lqud can also be wrtten as f = γ X f * L (pure lqud); for deal behavor of smlar compounds lke benzene and toluene n a mxture, γ = 43

44 If we go back to the chemcal potental wth respect to a pure lqud µ = µ pure lqud +RT ln f /f* pure lqud so f = γ X f* pure lqud µ = µ pure lqud +RT ln γ X where γ X s called the actvty, a, of the compound n a gven state wth respect to some reference state n γ X = a the actvty sometmes s called the apparent concentraton because t s related to the to the mole fracton, X or the real concentraton va g 44

45 Phase Transfer Processes Consder a compound,,whch s dssolved n two lquds whch are mmscble lke water and hexane. at equlbrum µ H2O = µ pure lqud +RT lnγ H2OX H2O µ hx = µ pure lqud +RT lnγ hx X hx at equlbrum µ H2O = µ hx RT lnγ hx X hx = RT lnγ H2OX H2O * substtutng γ X = f /f L (pure lqud) 45

46 RT ln f hx /f L * (pure lqud) = RT lnf H2O /f L *(pure lqud) f hx = f H2O Hnt For your homework: Calculate the actvty coef. g of hexane from ts solublty n water. hexane has some low solublty n water n grams/l H2O; st recall we derved RT lnγ hx X hx = RT lnγ H2OX H2O What s the actvty coeffcent and mol fracton of hexane n hexane? Ths gves the mportant result: g H2O=/ X H2O to calculate the g H2O we need to know X H2O C = sat. conc. = X / molar volume mx (why???) molar vol = lqud vol of one mole (L/mol) 46

47 the V H2O = L/ mol V mx = Σ X V ; typcally organcs have a V of ~0. L/mol V mx» 0. X X H2O ========================= Excess Free Energy, Excess Enthalpy and Excess Entropy Gong back to RT lnγ hx X hx = RT lnγ H2OX H2O rearrangng x RTln x H2O hx RTlnK = (RTlnγ x x H2O hx ' H2O / hx = H2O K RTlnγ ' H2O /hx hx = (RT ln γ ) H2O RTln γ hx ) 47

48 we already know that -RTln K eq = DG and we wll call ths DG, D 2 G, RT lnk ', 2 = ( 2 G ) and, D 2 G = G E -G 2 E n our water, hexane example of dssolvng n both ' RT ln KH2 O / hx = ( RT ln γh2 or E E 2 G G2 = (RT ln γ RT ln γ so we could therefore say G E = RT ln γ and we know G E = RTln g = H E -TS E O RT ln ) γ hx ) 48

49 H E s the partcle molar excess enthalpy of soluton and S E s the partal excess entropy In a calormeter, we could measure H E the heat requred to dssolve a compound n say water or hexadecane, whch s a measure of the total bondng forces that have to be broken and fromed (vdw, polar attractons) Ths would then gve -TS E from G E - H E In the gas phase H E g s ts heat of vaporzaton and G E g we can get from ts equlbrum parttonng const. K al = M P l */(RTρ L ) 49

50 Table 3.3, p83 50

51 By calculatng G E -G 2 E from Table 3.4 page 87 between two dfferent phases we can estmate transfer energes and K 2 5

52 Almost done, For a lqud phase a reacton A + B--> C + D µ A = µ o A +RT ln γ A XA µ A = µ o A +RT ln(γ A [A]V ) mx µ B = dg total = µ products dn-µ reactants dn dg/dn= µ C +µ D -(µ A +µ B ) (dg/dn = G, the molar free energy) G= µ ο C+µ ο D-µ ο A-µ ο B +RT ln V [ C] γ V [ D] γ V [ A] γ V ( B) γ mx c mx d mx A mx B 52

53 f the reacton goes to completon G o G = zero V C V D = RTln [ ] γ [ ] γ V [ A] γ V [ B] γ mx c mx d mx A mx B []γ = an actvty (I) C D Keq = ( )( ) ( A)( B) n the gas phase a reacton aa + bb--> cc + dd o (P G = RTln (P C A ) ) c a (P (P D B ) ) d b 53

54 Usng Lnear free energy relatonshps (LFERs) In envronmental systems, often the free energy descrbng two phases, say water and ar, for a compound s assumed to be drectly related to the free energy n two dfferent phases (whch can often be measured or determned) D 2 G = a D 32 G + const ln K 2 = ln K 32 + const and example s the organc semvolatle gas-partcle parttonng coeffcent K ap, or K p,whch can be related to an ar-octanol parttonng coeffcent K ow 54

55 log K p = a log K ao + b K ao, the ar octanol parttonng coef. can be determned from the rato of Henry s law parttonng K aw,or K H between ar and water dvded by the octanol water parttonng coef., K ow. K aw and K aw are usually known or can be estmated K ao = K aw / K ow In addton, there are lnear free energy technques that permt the estmaton of equlbrum constants base on molecular structure Ths s possble f one assumes that the overall free energy of phase transfer s related to the lnear combnatons of the free energes related to the ndvdual parts of the molecule that are nvolved n the transfer. D 2 G = SD 2 G parts of + specal nteracton terms 55

56 LogK 2 = SLogK parts of 2 + specal nteracton terms 56