2 –component solutions General: A (solvent) and B (solute) arbitrary if the solutions are completely miscible properties that define solution: P, T,  A,  B, (V) Examples: benzene & toluene ethanol & chloroform CH 3 -CH 2 -OH CHCl 3 CH 3 | Ideal solution?

Add water to a 10 ml graduated cylinder until it reaches the 5 ml mark. Add denatured ethanol to a 2 nd 10 ml graduated cylinder until it reaches the 5 ml mark. Measure the temperature of the water and ethanol. Add the ethanol solution into the graduated cylinder with the water. Mix the solution with the temperature probe. Measure the volume and temperature of the resultant solution. TableVolumeT (before)T (after) LF CF RF LB CB RB Average

TableVolumeT (before)T (after)moles water LF CF  (H2O) 0.76 RF moles ethanol = LB CB  (eth) 0.24 RB Average Stdev =  V mix = -0.4  T mix = 5.5  S mix = 0.20J (if ideal)  G mix = -59J (if ideal) we could calculate actual  G if we knew Cp for water/ethanol mixture Raoult’s Law P i =  i P i * Applies to ideal solutions

is the same as.... ideal solution = solvent (A) = solute (B) (or i) B & A have same size & shape B--A interactions same as B--B  V mix =  H mix =  U mix = 0  S mix = - n A R ln  A – n B R ln  B  G mix = RT (n A ln  A + n B ln  B ) CH 3 |

Raoult’s Law  i,v (P) =  i,v  + RT ln(P/P  ) ideal gas P i =  i P i * Vapor pressure solution mole fraction pure liquid VP From (dG/dP) T = V we can know how the chemical pressure of an IG varies with P i. If the vapor is ideal that expression also applies to gas mixtures.  i,sln =  i * + RT ln  i ideal solution For a liquid mixture the chemical potential of one component is reduced by mixing relative to the pure liquid form of that component. Follows from  G mix equation.  i,sln =  i,v equilibrium condition equate 1 & 2 When a solution comes to equilibrium with the vapor phase, the chemical potential of each component is the same in the solution and in the vapor. Here is what we already know from previous chapters …..

Raoult’s Law  i,sln =  i * (P i ) + RT ln  i ideal solution  i,v (P i ) =  i,v  + RT ln(P i /P  ) ideal gas  i,sln =  i,v equilibrium condition equate 1 & 2  i *(P i ) + RT ln  i =  i,v  + RT ln(P i /P  )  i *(P i ) -  i *(P i *) + RT ln  i = RT ln(P i /P i *) (d  G/dP) T =  V (the red part is ~ 0)  i * (P i * ) =  i,v * (P i * ) =  i,v ° + RT ln(P i * /P°) pure liquid at T, note P i ≠ P i * subtract from above RT ln  i = RT ln(P i /P i *)  RT then e x (inverse ln)  i = P i /P i * P i =  i P i * =  i,v P from Dalton’s Law of Partial Pressures

P  B 1 PB*PB* PA*PA* PBPB P P =  B (P B * - P A * ) + P A * P = P A + P B =  A P A * +  B P B * PAPA P A =  A P A * P B =  B P B *

P  B 1 Bubble point line: P =  B (P B * - P A * ) + P A * liquid f = c – p + 2 Vapor begins to form At some lower P all of the liquid will be converted into vapor. This is called the dew point. PB*PB* PA*PA*

all liquid P > ‘bubble point’ 1 st vapor at bubble point 2 phases dew point < P P < bubble point Last drop at dew point

TableVolumeT (before)T (after)moles water LF CF  (H2O) 0.76 RF moles ethanol = LB CB  (eth) 0.24 RB Average Stdev =  V mix = -0.4  T mix = 5.5  S mix = 0.20J (if ideal)  G mix = -59J (if ideal) we could calculate actual  G if we knew Cp for water/ethanol mixture Raoult’s Law P i =  i P i * Applies to ideal solutions

P  B 1 PB*PB* PA*PA* Bubble point line: P =  B (P B * - P A * ) + P A * liquid P = P B * P A *  B,v (P A * - P B * ) + P B * Dew Point line vapor both f = c – p + 2 The original vapor is enriched in the more volatile component The last drop of liquid is enriched in the less volatile component At a P where f = 2 (both phases present), a phase diagram allows you to determine the composition of each phase. tie line

P  ben P at which vapor forms bubble point P =  ben (P ben * - P tol * ) + P tol * 74.7 torr 22.3 torr Benzene/Toluene (ideal solution) (T ~ 18ºC)

P  ben Benzene/Toluene (ideal solution) (T ~ 18ºC) Composition of initial vapor P = P ben * P tol *  ben,v (P tol * - P ben * ) + P ben * 74.7 torr 22.3 torr

P  ben Benzene/Toluene (ideal solution) 3 theoretical plates Pressure Distillation 74.7 torr 22.3 torr

P  ben Benzene/Toluene 18°C (ideal solution) Lever Rule n l l l = n v l v vapor liquid 74.7 torr 22.3 torr

Pb*Pb* Pt*Pt* bb real ideal Bubble point line Dew point line Benzene-Toluene 120ºC P i = a i P i * real solutions – a i = activity P i =  i P i * applies to ideal solutions only a i =  i  i  i = activity coefficient 2.98 atm 1.34 atm Activity, a i The mole fraction a component would have if it behaved ideally.

Pb*Pb* Pt*Pt* bb real ideal Bubble point line Dew point line Benzene-Toluene 120ºC P i = a i P i * real solutions – a i = activity P i =  i P i * applies to ideal solutions only a i =  i  i  i = activity coefficient 2.98 atm 1.34 atm

 chl P (torr)P eth P chl P eth (id)P chl (id) P i = a i P i * a i =  i  i Ethanol – chloroform mixture P (torr)  chl

P i = a i P i * a i =  i  i Ethanol – chloroform mixture P (torr)  chl What is the activity of ethanol when  chl = 0.8? a) 0.2 b) 0.58 c) 0.42 c) 0.90

 chl P (torr)P eth P chl P eth (id)P chl (id) P i = a i P i * a i =  i  i Calculate a eth and  eth when  chl = 0.8 is?

acetone - chloroform aa P ideal real Azeotrope = Constant boiling mixture liquid and vapor phases have same concentration Vapor becomes enriched in Chl Vapor becomes enriched in acetone

Benzene - Toluene: theoretical  ben T liquid vapor  A,l = {P - P B * (T)}/{P A * (T) - P B * (T)}  A,v = {P A * (T)/P}  A

HCl - water azeotrope T  HCl

7.12 & 7.16 A solution contains a 1:1 molar ratio of hexane and cyclohexane. P hex * = torr and P cyc * = 97.6 torr. What is P? What is  v,hex and  v,cyc ? 7.14 A solution of ethanol and methanol has P = torr at 50°C. P meth * = torr and P eth * = torr. What is the composition of the solution? Assume ideal solution Set up expression for P meth and P eth using Raoult’s law Dalton’s Law of partial pressures: P = P meth + P eth — This was used to derive bubble point line Reduce one variable by letting  eth = 1-  meth. Solve for  meth Find P hex and P cyc using Raoult’s law Use Dalton’s Law to find P Since P  n, use P ratios to find  v,hex and  v,cyc

P i = a i P i * a i =  i  i Ethanol – chloroform mixture P (torr)  chl Use the graph to estimate a chl and  chl when  chl = 0.8 a chl ~ 0.89 and  chl = 1.11

 chl P (torr)P eth P chl P eth (id)P chl (id) P i = a i P i * a i =  i  i Ethanol – chloroform mixture Use the data to calculate a chl and  chl when  chl = 0.8 compare your results a chl = 384/433.5 = 0.89 and  chl = 1.11 What is a eth and  eth when  chl = 0.8?

Ideally dilute solution ideal solution = solvent (A) = solute (B) (or i) B & A have same size & shape B--A interactions same as B--B B and A are very different Standard state ≠ pure B but rather ideally dilute solution extrapolated to fictitious ‘pure’ B state where each B behaves as if it were surrounded by A molecules. Raoult’s Law: P i =  i P i * Real soln: P i = a i P i * Henry’s Law: P i =  i K i Applied to extremely non-ideal solutions particularly when solute is not miscible in solvent. Particularly useful for gases dissolved in water. Raoult’s law is still used (and very accurate) for the solvent.

Ethanol – chloroform mixture  chl P (torr)P eth P chl P eth (id)P chl (id) Raoult’s Law: P i =  i P i * Real soln: P i = a i P i * Henry’s Law: P i =  i K i  chl P (torr) K chl ~ 850 torr K eth ~ 400 torr What is Peth when  eth = 0.1? P eth ~ ~ 40 torr P * chl = torr P * eth = torr K chl = P chl /  i = 159.8/0.2 ~ 800

What is the concentration of oxygen dissolved in a lake at 25ºC? K O2 = 4.34 x 10 9 Pa What will happen to K O2 and the concentration of oxygen dissolved if global warming causes an increase in T to 30ºC? Discuss this with a partner and then explain your answer.

d(ln P)/d(1/T) = -  H m /R & P(atm) = exp{-  H m /R (1/293 – 1/T nbp )}  H vap T nbp Water40, Methanol39, P* meth = 89 torr P* H2O = 21.1 torr  meth bubbledew P = P meth * P w *  meth,v (P w * - P meth * ) + P meth * P =  meth (P meth * - P w * ) + P w *

d(ln P)/d(1/T) = -  H m /R & P(atm) = exp{-  H m /R (1/293 – 1/T nbp )} P* meth = 89 torr P* H2O = 21.1 torr 1 st vapor:  meth ~ 0.8 &  H2O ~ 0.2last drop:  meth ~ 0.19 &  H2O ~ 0.81 (2 plates) F = c – p + 2 = 2 – = 2l l =.12, l v =.21n l = 0.64, n v = 0.36 Vapor:  meth ~ 0.61 &  H2O ~ 0.39Liquid:  meth ~ 0.28 &  H2O ~ 0.72 n meth ~ 0.22 & n H2O ~ 0.14n meth ~ & n H2O ~ 0.46

Continuing to assume this is an ideal solution, calculate  H mix,  S mix, and  G mix for the initial liquid when  meth = 0.4, T = 298K and n total = 1 mole.  H mix = 0  S mix = -n H2O ln (  H2O ) - n meth ln (  meth ) = -0.6 ln ln 0.4 = 0.67 J  G mix =  H mix – T  S mix = J If the dissolution process is exothermic, comment on the actual value of  G mix relative to the ideal value.  G mix, real < -200 J since the reduction in  H mix makes the process more favorable.

T BB p = 1 miscible p = 2 TcTc partially miscible liquids  B,2 A sat in B  B,1 B sat in A

T BB B & C solution solid B & C solution + C s soln. + B s 2c solid-liquid phase diagram nfp B nfp C

Colligative Properties solute added to solvent changes  A such that  A,soln <  A * This affects the equilibrium between phases so that component A will ‘escape’ from the phase with the higher  A. Vapor Pressure Lowering Boiling Point Elevation Freezing Point Depression Osmotic Pressure

Vapor Pressure Lowering Derive.....  P = -  B P A * assume nonvolatile solute in IDS Define  P as P - P A * Raoult’s Law: P = P A =  A P A * P = P A =  A P A * sub into above....  P = P - P A *  P =  A P A * - P A * factor.....  P = (  A – 1)P A * = {(1 -  B ) – 1} P A *  P = -  B P A * AAABAAABAAABAAA AABAAABAAABAAAB ABAAABAAABAAAB A BAAABAAABAAAB AA A AA A AAAAAAAAAAAAAAA A AAA A A

Boiling Point Elevation  T b = T b - T b *  A,v (T b,P) =  A(sln) (T b,P) =  A * + RT ln(  A ) Concept: vapor pressure lowering means that a solution will exert less pressure than the pure solvent, therefore it requires a higher T before the vapor P = atm P (i.e. boiling point).  T b = k b m B where k b = M A RT b *2 /(  H vap,A ) 7.49

Freezing Point Depression  T f = -k f m B  T f = T f - T f *  A(s) * (T f,P) =  A(sln) (T f,P)  A(sln) =  A * + RT ln(  A ) k f = M A RT f *2 /(  H fus,A ) AA AlAl AsAs Tf*Tf* A sln TfTf

P R = P atm +  P L = P atm  = n B RT/V = c B RT Osmotic Pressure  = c B RT   L >   R not at equilibrium  A,sln =  A * + RT ln  A  A,L =  A,R or  A * (P) =  A,sln (P +  ) 7.55

Osmotic Pressure  = c B RT Freezing Point Depression  T f = -k f m B Boiling Point Elevation  T b = -k b m B Assume 1 L of solution so c B = n B 55Osmotic PMWTmbmb  (Pa)  (atm) g solute TbTb TfTf E E E-05 Assume m B ~ c B = n B = (30/101325)/( ) = For Wednesday 7.49