Physical Chemistry I for Biochemists Chem340. Lecture 32 (4/4/11)

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1 Physical Cheistry I for Biocheists Che340 Lecture 32 (4/4/11) Yoshitaka Ishii Ch If you have a note 33, skip printing p he Gibbs-Duhe Equation In Ch 6, we learned dg = -Sd + VdP + i dn i For a binary syste at constant and P. dg = 1 dn dn 2 (1) On the other hand, G = 1 n n 2, dg = d 1 n 1 + d 2 n dn dn 2 (2) (2) (1) yields 0 = (d 1 )n 1 + (d 2 )n 2 d 2 = -(n 1 /n 2 )d 1 = -(x 1 /x 2 )d 1 Changes in 1 and 2 are NO independent d 2 = -(x 1 /x 2 )d( 1* + Rlnx 1 ) = -(x 1 /x 2 )R(dx 1 /x 1 ) 1

2 Relation of olar fractions in gas phase (y) and solution phase (x) P total = P 1 + P 2 = x 1 P 1* + (1-x 1 )P 2 * =P * 2 + x 1 (P [Q1] 1 *-P 2 *) y 1 = P 1 /P total = x 1 P 1* /{P 2* +x 1 (P 1 *-P 2 *)} [Q2] x 1 = y[q3] 1 P 2* /{P 1* +y 1 (P 2 *-P 1 *)} P total = P 2* + x 1 (P 1 *-P 2 *) = P[Q4A] 1* P 2* /{P 1* +y 1 (P[Q4B] 2 *-P 1 *)} y 1 = (P 1* P total P 1* P 2* )/{P total (P 1 * - P 2 *)} P8.4) A and B for an ideal solution. At a total pressure of bar, y A = and x A = Using this inforation, calculate the vapor pressure of pure A and of pure B. y B = 1 y A & x B = 1 -x A So there are two unknown. P A * and P B * We need two uations: (1) P total = x A P A * + x[q1] B P A * (2) y A = x[q2] A P A * /P total 2

3 P8.5) A and B for an ideal solution at 298 K, with x A = 0.600, and P * * A 105 orr, P B 63.5 orr. a. Calculate the partial pressures of A and B in the gas phase. P = * P =(1-x * A [Q1] x A P A B [Q1] A )P B b. A portion of the gas phase is reoved and condensed in a separate container. Calculate the partial pressures of A and B in uilibriu with this liquid saple at 298 K. Assue that x A (new solution) = y A (old gas) and x B (new solution) = y B (old gas). P A (new) = x A (new)p A * = y A (old)p A * y A (old) = x[q2] 1 (old)p 1* /{P 2* +x 1 (old)(p 1 *-P 2 *)} How pressure P total depends on olar fractions in gas phase (y) and solution phase (x)? X benzene [Q1 ] [Q2 y Y ] benzene P total = P 1* P 2* /{P 1* +y 1 (P 2 *-P 1 *)} P total =P 2* + x 1 (P 1 *-P 2 *) Data fro Benzene + oluene 3

4 P-Z (average coposition) diagra P-X Z 1 = (n 1 liquid + n 1 gas )/(n 1 liquid + n 1 gas + n 2 liquid + n 2 gas ) P ressure P-Y Q. How is the phase changed along the line fro a to d? Q2. How uch is X benezen and Y benzene at the point b? Q3. Is n liquid > n gas correct at the point b? lb = Z B X B = n B total /n total -n B liq /n B+ liq b = Y B Z B = n B gas /n B+ gas -n B total /n total Gas Liq ie Line Lever Rule X B Z B Y B Q. What is the olar ratio between liquid and vapor phases (Z B -X B ) / (Y B -Z B ) = n Gas total / n Liq (n gas /n liq ) at point b (B & all ixed)? total 4

5 Physical Cheistry Fundaentals: Figure 5.33 Q. How is the phase changed along the line fro a to e? p b p c p d X 1b Ab X Ac X Ad b d y Ac c y Ab e Q2. How uch is n gas /n liquid at the point c? 2009 W.H. Freean Physical Cheistry Fundaentals: Figure W.H. Freean 5

6 8.7 Freezing Point Depression & Boiling Point Elevation () = *( 0 ) + S liq (- 0 ) + Rln(x ) Q1. How uch is Rln(x ) for pure liquid? 0 Q2. For a ixture, what is the sign of frl Rln(x )? Negative xsolv 1 Freezing Point Depression (Derivation) At a freezing point, liq () = solid *() * () + Rln(x ) = solid *() ln(x solv ) = - { solv* () - solid *()}/R = -(G fusion, )/R Let s find d vs d(lnx) d(ln x d(ln x 1 ( Gfusion, / ) ( Hfusion, ) ) d d R R solv 2 solv 1 ( H ) R 1 ( H R f fusion, ) 1 fusion, ) d 2 f, pure R ln x solv fusion fusion, pure Hfusion, Gibbs-Helholtz. 1 fusion 1 fusion, pure 6

7 1 1 R ln x solv f f, pure Hfusion, 1 1 Using ~ Q. Does this 2 f f, pure ff, pure f, pure depend and lnx = ln(1-x ) ~ -x on? x f R H f, pure x fusion, n n ~ n n n f R f, pure H M fusion, n M w Molality of Solute (ol/kg of ) K f M Boiling Point Elevation At a freezing point, liq () = gas *() * () + Rln(x ) = gas *() ln(x solv ) = { gas* () - *()}/R = (G vap, )/R b R b, pure H M vap, K b It does NO depend on! f f, purem R H fusion, K f 7

8 8.8 he Osotic Pressure Water and Sugar Water are separated by a ebrane that perits only water go through. H 2 O H 2 O + Sugar Q. Which way is H 2 O likely to ove to? Q2. How can we ake the two sections in uilibriu? d(p, ) = V dp S d Elucidation of the Osotic Pressure =P (P+P, )* + Rln(x ) = * (P,) (P, )* + V P + Rln(x ) = * (P, ) V = - Rln(x ) = - (R/V )ln(x ) ln(x ) =ln(1 - x ) ~ -x ~ -n /n = (R/V )(n /n ) =n R/(V n ) = n R/V (Vant Hoff Equation) or = c R, where c = n /V (olarity). 8

9 Osotic Pressures -Molecular weight deterination =n R/V =c R If we use C = w/v = n R/V = (w /M )(R/V) = {(w /V)/M }R = {C /M }R M = RC / P turgor = cell - ediu At isotonic c cell R = c ediu R Q. A ediu is a ixture. How do you define C ediu? Q2. What is C cell? 9

10 Osotic Pressure for a polydisperse or ixture syste otal = i = c i R = (C i /M i )R, where c i and C i are olar and ass concentration of i, respectively. So c ediu or c cell c i Define as otal =(C /M)R and C = C i M = (C i )/(C i /M i ) = (n i M i /V)/(n i /V) = (n i M i )/(n i ) = (n i M i )/(n otal ) P8.36) Assue sucrose and water for an ideal solution. What is the uilibriu vapor pressure of a solution of 2.0 gras of sucrose (olecular weight 342g ol 1 ) at = 293 K if the vapor pressure of pure water at 293 K is orr. Assue g of water. (Note density 1.00 L/g.) Use P /P *= x What is the osotic pressure of the sucrose solution versus pure water? Assue g of water. Just use = nr/v 10

11 Pressure (orr) or r X chlorofor X chlorofor 8.9 Real Solutions Exhibit Deviation fro Raoult s Law Q.Which is Raoult s Law? (a) Pi = x i Pi* (b) Pi = x i P total (c) Pi = n i R/V Q2. Which is for a real solution? Dotted or solid line? Q3. How does b -X cs2 curve look like for the solid curve? 11

12 8.10 he Ideal Dilute Solution solution i = * i + Rln(P i /P i *), where i* = i* (vapor) = i* (liquid) i0 (pure liquid, ) For a non-ideal solution P i x i P i *. For a dilute binary solution, we define activity a as a = P /P*. In general, a i P i /P* i. Naturally, for an ideal solution, a = x For a general case, a and x are related by activity constant = a /x i solution = i * + Rln(a i ). So use a i for a non-ideal solution in place of x i! Henry s law (useful when x i ~ 0) P acetone = x acetone k acetone H as x acetone 0 (or x CS2 1) In general, P i = x i k i H as x i 0 o find P i when x i ~ 0 look up the Henry s law constant for SOLUE An ideal dilute solution in this text eans a syste where Henry s law is valid for a and Raoult s law is valid for 12

13 Ex. Gas in solution x CO2 = P CO2 /k H CO2 In a soda ~3 g of CO2 (0.05 ol) in 450 L (H2O ~ 20 ol ) x CO2 << Activities Are Defined with Respect to Standard States In 8.10 a and were defined for a dilute solution by relations Activity a a i P i /P* i. i = a i /x i In a standard state for dilute solution (x 1) (Raoult s law standard state) a x ~ 1 solution = * + Rln(a ) a i = x i In Raoult s law standard state, the standard cheical potential for (x 1) is solution * 13

14 Activity in Henry s law Standard State Standard µ for a dilute P i = x i k i H solution i = * i + Rln(k Hi x i /P i *). = * i i + Rln(k Hi /P i *) + Rln(x i ) = i *H + Rln(x i ) he above uation is correct as x i 0. (In general, use a i in place of x i ). he standard-state (x i ~1; i.e. hypothetical pure ) cheical potential for a i is given by *H * i i Solute i + Rln(k Hi /P i *). In general, P i = a i k i H i = *H Solute i + Rln(a i ). In Henry s law standard state, (x i ~ 1) a i = P i /k ih (~ x i )& i = a i /x i (~1) Fro Atkins Physical Cheistry 9 th Ed 14

15 8.12 Henry s Law & Solubility of Gas in a Solvent N 2 (aq, c N2 ) N 2 (gas, P N2 ) In this case (x N2 ~0), we use µ following Henry s law µ solution N2 = µ *H N2 (vapor) + Rln(a N2 ) he ole fraction in solution is x N2 aq = n N2 aq /(n N2 aq + n H2O ) ~ n N2 aq /n H2O. he aount of dissolved gas is given by n N2 = n H2O x N2 ac ~ n H2O P N2 /k N2 H Cheical Equilibriu in Solution he concept of activity can be used to express the therodynaics uilibriu constant using activities. When a reaction in solution A+2B 3C + D is in uilibriu 3µ C + µ D -µ A -2µ B = 0 For a general uilibrated reaction in a solution, 0 = i µ i (solution) = i µ *H i (solution) +R ln(a i ) i c D ac ad K A a B A ab = ( i c i /c 0 ) i G 0 reaction = - R ln(a i ) i = - R lnk K = (a i ) i = ( i x i ) i For a dilute solution, i ~ 1. K ~ (c i /c 0 ) i By the way, for a dilute solution C 0 ~ C (~constant). 15