chemistry 232. physical chemistry ii · 2020-01-06 · thomas engel and philip reid available with...

Instructor: Derek Leaist ([email protected])

Office PSC 3072, Lab PSC 3020

Lectures: Monday 8:15 am, Tuesday 10:15 am, Thursday 9:15 am

(MULH 3030)

Lab/Tutorial: Friday 2:15 pm (Lab PSC 3037 / Tutorial PSC 1072)

Course Notes: https://people.stfx.ca/dleaist/Chem232/

(tutorial problems and answers, problem assignments and tests from

2016-2019, equations sheets, and reading material also posted)

Textbook: Thermodynamics, Statistical Thermodynamics and

(optional) Kinetics, 3rd Edition, Thomas Engel and Philip Reid

Chemistry 232. Physical Chemistry II

Chapter 9. Ideal and Real Solutions

Chapter 10. Electrolyte Solutions

Chapter 11. Electrochemical Cells, Batteries and Fuel Cells

Chapter 16. Kinetic Theory of Gases

Chapter 17. Transport Processes (Diffusion, Heat Flow, Viscosity

Electrical Conductivity)

Chapter 18. Elementary Chemical Kinetics

Chapter 19. Complex Reactions

Course Outline

(about five lectures per Chapter)

Assignments (about ten) 10 %

Labs (five, bi-weekly) 15 %

Term tests (two, 15 % each) 30 %

Final Exam 45 %

100 %

Term tests: Friday 07 February

Friday 20 March

Final exam: date TBA (April exam period)

Marking Scheme

Friday Lab (PSC 3037) or Tutorial (PSC 1072)

Jan. 10 no lab or tutorial

Jan. 17 Dissociation Constant of a Weak Acid Lab

Jan. 24 Tutorial #1

Jan. 31 Dissociation of an Iron (II) Complex Lab

Feb. 07 Tutorial #2 and Term Test 1

Feb. 14, 21 no lab or tutorial

Feb. 28 Primary Kinetic Salt Effect Lab

March 06 Tutorial #3

March 13 Phase Diagrams from Cooling Curves Lab

March 20 Tutorial #4 and Term Test 2

March 27 Partial Molar Volumes Lab

April 03 Tutorial #5

Chem 232 Lab and Tutorial Schedule (2020)

Textbook (optional)

(for Chem 231 and Chem 232)

Thermodynamics, Statistical

Thermodynamics and Kinetics

3rd Edition ($126 Amazon.ca)

Thomas Engel and Philip Reid

available with “free” online

Mastering Chemistry resource

material

or Physical Chemistry, 3rd Edition,


includes chapters on quantum mechanics and

spectroscopy for Chemistry 331 and 332

(but not required for these courses)

Same textbook used previously.

Used copies may be available.

Also available online:https://www.academia.edu/14903550/Thermodynamics_Statistical_Thermodynamics_and_Kinetics_THIRD_EDITION

https://www.academia.edu/14903550/Thermodynamics_Statistical_Thermodynamics_and_Kinetics_THIRD_EDITION

https://www.academia.edu/14903550/Thermodynamics_Statistical_Thermodynamics_and_Kinetics_THIRD_EDITION

Student Solutions Manual

(for Chem 231 and Chem 232)

Thermodynamics, Statistical

Thermodynamics and Kinetics

3rd Edition ($34 Amazon.ca)


worked solutions to

end-of-chapter problems

Schaum’s Outline of Physical Chemistry

2nd edition ($25 Amazon.ca)

Clyde A. Metz

Thermodynamics, electrochemistry,

kinetics, and transport properties

for Chem 231 and Chem 232.

concise summaries, worked problems.

Also covers: quantum mechanics

spectroscopy

crystallography

polymers

Chem 232: Course Material

https://people.stfx.ca/dleaist/Chem232/

lab and tutorial schedule

course notes

tutorial problems with answers

problem sets with answers from 2016, 2017, 2018, 2019

term tests with answers from 2016, 2017, 2018, 2019

equation sheets

pdf copy of the textbook: T. Engel, P. Reid, Thermodynamics, Statistical

Thermodynamics and Kinetics, 3rd Ed., Pearson, Boston, 2013.

Chapter 9. Ideal and Real Solutions

Chem 231 Last term, thermodynamics of pure substances and ideal gas mixtures. But what about liquid mixtures? They are “everywhere”, very important, and often strongly nonideal.

Solutions

mixtures of two or more different chemical components

form a single phase

uniform chemical and physical properties on the microscopic scale

a solution can be a: gas (e.g., air – N2, O2, H2O, Ar, CO2, …)

liquid (e.g., NaCl dissolved in water)

solid (e.g., brass – a copper/zinc alloy)

Ideal Gas Solutions

simplest of all solutions, but very important

no molecular interactions

from Chem 231:

ideal gases always mix

0lnln BBAAmix xRnxRnS

0lnln BBAAmix xRTnxRTnG

Liquid Solutions

even more important than gas solutions

rarely ideal

example: oil and water do not mix. Why?

in addition to p, V, T, composition variables are required,

such as mole fraction (xi), molality (mi), molarity (ci)

k

k

ii

n

nx

solvent kilograms

ii

nm

solution of liters

ii

nc

Chapter 9 – Liquid Solutions

Why study solution thermodynamics? To help understand:

vapor pressures of solutions (and fractional distillation)

solubilities (and purification by re-crystallization)

freezing point depression (why salt melts ice)

osmotic pressure (how desalination works)

solid-liquid-vapor phase diagrams

properties of nonideal solutions

multicomponent phase rule F = C + 2 P

chemical reaction equilibrium in liquid solutions

Ideal Solution of Gases A and B

no interactions between molecules A and B

a poor approximation for liquid solutions of A and B

A and B molecules attract each other to form liquid solutions

Section 9.1 Defining Ideal Solutions

Ideal Solution of Liquids A and B

equal interactions between molecules A and B

a reasonable approximation “similar” A and B molecules

examples: benzene + toluene or C6H6 + C6H5D

Vapor Pressures of Ideal Liquid Solutions: Raoult’s Law

Liquid A + B Mixtures

If A and B molecules have similar:

sizes

A-A, A-B and B-B interactions

expect the vapor pressures of

A and B to be proportional to the

mole fractions of A and B. (Why?)

Get:

pA = xA pA*

pB = xB pB*

Section 9.2 Chemical Potentials of Ideal Solutions

The chemical potentials of solution components are useful for

analyzing physical and chemical equilibrium processes.

Raoult’s law gives the chemical potentials for ideal liquid solutions:

Chemical Potential of Component A in an Ideal Gas Mixture

(from Chem 231)

Pure Liquid A (vapor pressure pA*) in Equilibrium with Vapor

A (pure liquid, xA = 1) A (gas, pA*)

A(l) *(xA=1) = A(g)(pA*)

A(l) *(xA=1) = A(g)o(po) + RT ln(pA*/po)

A(g)(pA) = A(g)o(po) + RT ln(pA/po)

A in Ideal Liquid Solution in Equilibrium with Vapor

A (in liquid solution, xA < 1) A (gas, pA)

at equilibrium:

A(l)(xA) = A(g)(pA)

A(l)(xA) = A(g)o(po) + RT ln(pA/po)

use Raoult’s (pA = xApA*) for ideal solutions to get:

A(l)(xA) = A(g)o(T, po) + RT ln(xApA*/po)

= A(g)o(T, po) + RT ln(pA*/po) + RT ln xA

A(l)(xA) = A(l) *(xA=1) + RT ln xA

Significance: the thermodynamics of processes involving

gases and liquid solutions can now be analyzed.

Ideal Liquid Solutions

Example nA moles of pure liquid A and nB moles of pure

liquid B are mixed at temperature T to form an

ideal liquid solution. Exercise Show:

0lnln BBAAmix xRnxRnS

0lnln BBAAmix xRTnxRTnG

Identical results for mixing ideal gases at fixed T and p.

Why is it unnecessary to specify the pressure for mixing liquids?

Section 9.3 Ideal Binary (Two-Component) Solutions

Applications: liquid-vapor equilibrium

liquid-vapor phase diagrams

purification by fractional distillation

Example: Benzene(1) + Toluene(2) Mixtures

at liquid-vapor equilibrium, total vapor pressure:

p = p1 + p2

= x1 p1* + x2 p2*

= x1 p1* + (1 x1) p2*

p = p2* + x1 (p1* p2*)

Benzene(1) + Toluene(2) Solutions (Fixed Temperature)

p = p2* + x1 (p1* p2*)

= x1p1* + x2p2*

all

linear in x1 = xbenzene

(benzene mole fraction in the liquid)

allVapor

reduce the pressure on a liquid

benzene(1) + toluene(2)

solution of composition x1.

First vapor appears at pressure

“bubble point” line25 oC

Benzene(1) + Toluene(2) Solutions

notation convention

x1 is the liquid-phase benzene mole fraction

y1 is the gas-phase benzene mole fraction

21

1

21

1

21

11

)gas()gas(

)gas(

pp

p

RT

Vp

RT

VpRT

Vp

nn

ny

**

*

2211

111

pxpx

pxy

*)/1(*)/1(

*)/1(

2211

111

pypy

pyx

weighted average of p1* and p2*

(tendency to vaporize)

weighted average of 1/p1* and 1/p2*

(tendency to condense)

Also useful:

**

*

21

21

pp

ppx

*

1

*

1

*

11

21

21

pp

ppy

Benzene(1) + Toluene(2) Solutions (Fixed Temperature)

all

Liquid

nonlinear in y1 = ybenzene

(benzene mole fraction in the vapor)

all

increase the pressure on a


vapor of composition y1.

First liquid appears at pressure

“dew point” line

1

2

2

1

1

1

21

1

2

**

*

1

*

1

*

1

p

y

p

y

ppy

pp

25 oC

Liquid-Vapor Phase Diagram: Pressure vs Composition

(combine bubble-point and dew-point curves on one plot)

25 oC

Zbenzene = Z1


gas

2

liq

2

gas

1

liq

1

gas

1

liq

1

21

1

nnnn

nn

nn

n

convenient

visual

presentation

of liquid-

vapor

equilibrium

all

all

average benzene mole fraction Z1

y1(gas)x1(liq)

What happens when

benzene + toluene vapor

is compressed at 25 oC?

initially all vapor at pressure pinitial

pfinal (all liquid)

last vapor at pbubble

first liquid at pressure pdew

at pressure pb

liquid (l) vapor(v)

equilibrium

Horizontal

tie line l - b - v

connects

liquid l and vapor v

in equilibrium

Lever Rule

Relative Amounts of Liquid and Vapor Phases

liquid

x1

vapor

y1

overall

Z1

(horizontal tie line)

liquid

total

liquid

1

total

total1

11n

n

n

nxZ

total

total1

vapor

total

vapor

111

n

n

n

nZy

total

total1

vapor

total

vapor

1vapor

totalliquid

total

liquid

1

total

total1liquid

total11

vapor

total11

liquid

total )()(n

n

n

nn

n

n

n

nnZynxZn

0)( vapor

1

liquid

1

total

total1vapor

total

liquid

total

total

total1vapor

total

vapor

1

liquid

1

total

total1liquid

total nnn

nnn

n

nnnn

n

nn

Lever RuleRelative Amounts of Liquid and Vapor Phases

nliquid

x1

nvapor

y1

ntotal

Z1

(horizontal tie line)

nliquid(Z1 x1) = nvapor(y1 Z1)

Z1 – x1 y1 – Z1

fulcrum

Example 0.350 moles of benzene and 0.15 moles of

toluene are loaded into a container at 25 oC.

The total pressure measured with a gauge is

60.0 Torr.

Describe the phases present. Liquid?

Vapor? Or both phases? Compositions?

average benzene mole fraction

700.0150.0350.0

350.0

total2 total1

total11

nn

nZ

A liquid benzene(1) + toluene(2) solution at 25 oC and

benzene mole fraction 0.700 starts to vaporize at pressure

p = (0.700)(96.4 Torr) + (0.300)(28.9 Torr) = 76.2 Torr

At the lower 60.0 Torr pressure in this example, some vapor exists.

A benzene(1) + toluene(2) vapor at 25 oC and benzene

mole fraction 0.700 starts to condense at pressure

p = [(0.700)(96.4 Torr)1 + (0.300)(28.9 Torr) 1]1 = 56.7 Torr

At the higher 60.0 Torr pressure in this example, some liquid exists.

Conclusion: liquid and vapor are present

A liquid benzene(1) + toluene(2) solution at 25 oC and

pressure 60.0 Torr has benzene mole fraction x1.

60.0 Torr = x1p1* + x2p2* = x1(96.4 Torr) + (1x1)(28.9 Torr)

Solve for liquid composition x1 = 0.461

The liquid benzene(1) + toluene(2) solution at benzene mole

fraction 0.700 is in equilibrium with vapor at benzene mole fraction

Solve for vapor composition y1 = 0.740

Torr0.60

)Torr4.96)(461.0(

**

*

2211

11

21

11

pxpx

px

pp

py

Example (cont.) 0.350 moles benzene and 0.15 moles toluene are

in a container at 25 oC total pressure 60.0 Torr

overall benzene mole fraction Z1 = 0.700

two phase are in equilibrium:

liquid with benzene mole fraction x1 = 0.461

vapor with benzene mole fraction y1 = 0.740

167.0461.0700.0

700.0740.0

vapormoles

liquid moles

11

11

xZ

Zy

Lever Rule:

moles liquid + moles vapor = 0.500

Solve for nliquid = 0.072 mol and nvapor = 0.428 mol

Plan B Use the Liquid-Vapor Phase Diagram

Rapid, convenient graphical method

25 oC


all

all

average benzene mole fraction Z1

y1(gas)

x1(liq)

given overall

composition:

Z1 = 0.700

at pressure

p = 60.0 Torr

(point c)

read from

the graph:

x1 = 0.461

y1 = 0.740

Phase Rule

One-Component Systems C = 1 (e.g., pure water)

P = number of phases

F = degrees of freedom (number of independent variables)

(from C231)

Application: Determines how many state variables must be

specified to describe a pure substance.

Avoids hidden variables.

What about multicomponent systems?

F = 3 P

Phase Rule

Multicomponent Systems

C = number of independent chemical components

Example

Benzene + toluene (C = 2) in the liquid-vapor region (P = 2):

F = 2 + C P = 2 + 2 2 = 2

Two degrees of freedom, such as T and p.

F = 2 + C P

Used by material scientists metallurgists

chemists geologists

chemical engineers physicists

industrial chemists crystallographers

Phase Rule: Benzene + Toluene at 25 oC

F = 2 + C P = 4 P

at 25 oC (fixed T, so one less variable):

F = 3 P

F = 2

F = 2

F = 1 (e.g., p, Zbenzene)

(e.g., p, Zbenzene)

(e.g., p)

(P = 1)

(P = 1)

(P = 2)

Phase Rule

F = 2 + C P

Where does this rule come from?

The number of degrees of freedom F is the number of variables

required to describe the state of each phase minus the number

of equilibrium constraints on the variables

Total Number of Variables

The state of each phase is described by C + 1 variables:

T, p, x1, x2, x3, … xC1

for a total of P(C + 1) variables. C 1 mole fractions are adequate

to describe the composition because xC = 1 – x1 – x2 … xC1.

Phase Rule

Number of Constraints

Thermal Equilibrium [ P 1 constraints ]

Tphase 1 = Tphase 2 = Tphase 3 = … = Tphase P

Mechanical Equilibrium [ P 1 constraints ]

pphase 1 = pphase 2 = pphase 3 = … = pphase P

Chemical Equilibrium [ C(P – 1) constraints ]

1 phase 1 = 1 phase 2 = 1 phase 3 = … = 1 phase P

2 phase 1 = 2 phase 2 = 2 phase 3 = … = 2 phase P

C phase 1 = C phase 2 = C phase 3 = … = C phase P

Phase Rule

F = number of variables number of constraints

= P(C + 1) – (P – 1) – (P – 1) – C(P – 1)

F = 2 + C P

Example A liquid benzene + toluene mixture (P = 1, C = 2)

F = 2 + C P = 2 + 2 1 = 3

Three degrees of freedom, such as T, p, xbenzene

Section 9.4 Temperature-Composition Diagrams

Benzene + Toluene at 1 atm ( ideal liquid mixtures)

all

allliquid

+ vapor

pure benzene boils

pure toluene boils

F = 2

F = 1F = 2

p = 1 atmC = 2

(two components)

F = C + 2 – P

= 4 – P

But the pressure

is fixed, so:

F = 3 – P

Application: Purification by Fractional Distillation

liquid a vapor b

vapor b liquid c

liquid c vapor d

vapor d liquid e

etc.

a

bc

de

fg

hi

1 atm

mostly

toluene

mostly

benzene

heat

k j

Laboratory fractional distillation:

small-scale

batch

cooling

water

Industrial fractional

distillation:

large-scale

continuous flow

Using Raoult’s law, an ideal liquid solution of

benzene(1) + toluene(2) with benzene mole fraction x1

is in equilibrium with vapor of benzene mole fraction

21

1

21

1

21

11

)gas()gas(

)gas(

pp

p

RT

Vp

RT

VpRT

Vp

nn

ny

*)/*(**

*

1221

1

2211

111

ppxx

x

pxpx

pxy

Why does fractional distillation work?

Because x1 + x2 = 1: y1 > x1 if p2* < p1*

y1 < x1 if p2* > p1*

Conclusion: For ideal liquid solutions, the vapor is

always richer in the more volatile component.

To compare liquid and vapor compositions, can also show:

*

**1

1

1

122

1

1

p

ppx

x

y

*

**1

1

122

1

1

p

ppx

y

x

Real solutions can be strongly ideal, in some cases forming

azeotropes (equal liquid and vapor compositions).

!!! Warning !!!

Maximum Boiling Azeotrope

At the azeotrope (point a):

xA(liquid) = yA(vapor) = xaz

The azeotrope temperature Ta

is higher than the boiling points

of the pure liquids.

vapor richer in

component A

vapor richer in

component B

Ta

xaz

Impossible to purify

component A starting with

liquid of composition xA < xaz

Real solutions can be strongly ideal, in some cases forming

azeotropes (equal liquid and vapor compositions).

!!! Warning !!!

Minimum Boiling Azeotrope

At the azeotrope (point a):

xA(liquid) = yA(vapor) = xaz

The azeotrope temperature Ta

is lower than the boiling points

of the pure liquids.

vapor richer in

component B

vapor richer in

component A

Ta

xaz

Impossible to purify

component A starting with

liquid of composition xA < xaz

distillation

stops here

Would you expect chloroform and acetone to form

a maximum or minimum boiling azeotrope?

Important Application of Distillation: Petroleum Refining

Section 9.5 Gibbs-Duhem Equation:

The Chemical Potentials of Solution Components are Related

Differential of the Gibbs energy of a binary solution

2211 ddddd nnpVTSG

at fixed temperature and pressure (dT = 0, dp = 0) is

2211 ddd nnG

2211 ddd nnG

Integrating at constant composition (1 and 2 fixed)

2211 nnG

Differentiating this expression for the Gibbs energy gives

222211112211 dddd)(d)(dd nnnnnnG

Eq. (I)

Eq. (II)

Gibbs-Duhem Equation:

The Chemical Potentials of Solution Components are Related

Two equations for dG. Subtracting Eq. (I) from Eq. (II) gives

Equation Duhem-Gibbs dd0 2211 nn

Significance The chemical potentials of the solution components

cannot be varied independently.

Example: changing the chemical potential of component 1 by

d1 produces the change d2 = (n1/n2)d1 in the

chemical potential of component 2.

Gibbs-Duhem Equation: Application

Suppose component 1 of a binary mixture is volatile and

obeys Raoult’s law, p1 = x1 p1*, which gives the chemical

potential of component 1 in the liquid phase

11liquid1

1

o

1

oo

vapor1

o

11

oo

vapor1

o

1

oo

vapor1vapor1liquid1

ln)1(*

ln)/*ln()(

)/*ln()(

)/ln()(

xRTx

xRTppRTp

ppxRTp

ppRTp

Component 2, however is nonvolatile (e.g., sucrose). What is its

chemical potential? (Can’t be determined from vapor pressure

measurements)


The Gibbs-Duhem equation 0 = n1d1 + n2d2 provides

a useful expression for the differential of the chemical potential

of nonvolatile component 2.

)ln*d(dd 1liquid1

2

1liquid1

2

1liquid2 xRT

x

x

n

n

x1 + x2 = 1 gives dx1 + dx2 = 0 and dx1 = dx2

2

1

1

1

2

11

2

1 ddlnd

x

xRT

x

x

x

xRTx

x

xRT

2

2

2liquid2 lnd

dd xRT

x

xRT

2

2

2

2

ln

0ln

2

1

liquid2 lndd

x

x

x

x

xRT 22liquid2liquid2 ln)1(* xRTx


Integrating d2 for the nonvolatile component in terms of

d1 from the measured vapor pressure of the volatile component:

22liquid2liquid2 ln)1(* xRTx

Salts, sugars, proteins, polymers, and many other solutes

are not volatile.

The chemical potentials of these important components are

determined using the Gibbs-Duhem equation and the solvent

chemical potential measured by vapor pressure techniques

and other methods (Sections 9.6 to 9.8).

Several important properties of dilute solutions

collectively (leading to the terminology “colligative”)

depend only on the number of solvent and solute molecules,

not on their chemical composition or molecular structure.

vapor pressure lowering

freezing point depression

boiling point elevation

osmotic pressure

Why?

Section 9.6 Colligative Properties

Example The vapor pressure of pure water at 25 oC is

23.76 Torr.

Estimate the vapor pressure of water over a

solution prepared by dissolving 1.00 mol sucrose

(or 1.00 mol of any other nonelectrolyte solute)

in 1.00 kg of water at 25 oC.

The mole fraction of water in the water(1) + sucrose(2) solution is

Vapor Pressure Lowering

982.000.1)mol g015.18/g1000(

)mol g015.18/g1000(1

1

21

11

nn

nx

Assuming an ideal solution (no other information), Raoult’s law gives

p1 = x1p1* = (0.982) (23.76 Torr) = 23.34 Torr

Measure 23.30 Torr. Why different?

Freezing Point Depression

Salt (NaCl) spread on winter roads melts ice. Why?

But if it gets really cold (< 15 oC), no melting. Why?

Can other compounds melt ice?

Thermodynamics of automotive antifreeze products?

F.p. depression is used to measure molecular weights. How?

Solid-liquid phase diagrams introduced

How does purification by re-crystallization work?

Section 9.7 Freezing Point Depression

and Boiling Point Elevation


First, consider pure liquid water in equilibrium with ice

(pure solid water) at a given pressure (often 1 atm).

H2O(liquid) H2O(solid)

)solid pure*,(*)liquid pure*,(* s11 TTl

F = 1 + C – P

= 1 + 1 2

= 0

invariant point

(zero degrees of freedom)

Example

T* = 273.15 K (0 oC)

for water at 1 atm

pure

solid (1)

pure

liquid (1)

x1 = 1

(asterisk designates “pure”)

T = T*

1 atm


Next, consider a liquid water(1) + solute(2) solution and ice

at T* = 273.15 K.

pure

solid (1)

water(1) + solute(2) solution

x1 < 1

T = T*

solute lowers the

chemical potential

of water in the solution

the system is not at

equilibrium at T*

ice spontaneously

melts, transferring

water to the lower

chemical potential

recall 1l*(T*, pure liquid) = 1s*(T*, pure solid)

so 1l(T*, x1) = 1l*(T*, pure liquid) + RT lnx1 < 1s*(T*, pure solid)

1l (T*,in solution) < 1s*(T*,pure solid)

Freezing Point Depression T

Consider a liquid water(1) + solute(2) solution and ice

at equilibrium:

pure

solid (1)

water(1) + solute(2) solution

x1 < 1

T < T*

spontaneously

melting ice

absorbs heat

temperature drops

below T*.

equilibrium reached

at T = T* T

freezing point

depression T = T* T

1l*(T, pure liquid) + RT lnx1 = 1s*(T, pure solid)

1l(T,in solution) = 1s*(T,pure solid)


Example A water (1) + sucrose (2) solution in contact

with ice at temperature T.

At equilibrium, the chemical potential of water in the solution

and in the ice are identical.

)solid pure,(*)solutionin ,( s11 TTl

)solid pure,(*ln)liquid pure,(* s111 TxRTTl

1s11 ln)solid pure,(*)liquid pure,(* xRTTTl

1mfusion, ln)( xRTTG


0)1ln(*)(mfusion, RTTG

At T = T* pure liquid water (x1 = 1) are in equilibrium:

0*)(**)( mfusion,mfusion, TSTTH

*)(*)()( mfusion,

mfusion,mfusion, TTT

GTGTG

p

)(*

0*)()(0 mfusion,

mfusion, TT

HTTS

At T = T* T

TT

H

*

mfusion,


At T = T* T

TT

HxRTTG

*ln)( mfusion,

1mfusion,

)1ln(*

ln*

2

mfusion,

1

mfusion,

xH

RTTx

H

RTTT

For dilute solutions (x2 << 1 and T < 10 K) the approximations:

ln(1 x2) x2 TT* (T*)2 give

2

mfusion,

2*)(x

H

TRT


solution of

solvent (1) + solute (2)


In terms of solute molality

1

2

11

22

solvent of kilograms

solute of moles

M

x

Mn

nm

2

mfusion,

2

1 *)(m

H

TRMT


solution of


solute mole

fraction 1

2

21

22

moles total

solute of moles

n

n

nn

nx

for dilute

solutions

(n2 << n1)

M1 = solvent molecular weight in units of kg mol1

Use x2 = M1m2 to get


Example Calculate the freezing point depression of an aqueous

solution containing 10 % by weight ethanol (2).

Data: T* = 273.15 K ` M1 = 0.018015 kg mol1

Hfusion,m = 6010 J mol1 M2 = 0.046069 kg mol1

9583.00.04607) / kg (0.100 0.018015) / kg (0.900

0.018015 / kg 0.9001

x

)9583.0ln(6010

)15.273(314.8ln

*)( 2

1

mfusion,

2

xH

TRT

water mole fraction

freezing point depression T = 4.40 K

T = (273.15 – 4.40) K = 268.75 K (4.40 oC)

Boiling Point Elevation

Example A water (1) + nonvolatile solute (2) solution in

contact with water vapor at a fixed pressure.

At equilibrium, the chemical potential of water in the solution

and in the vapor are identical.

) vaporpure,(*)solutionin ,( g11 TTl

) vaporpure,(*ln)liquid pure,(* g111 TxRTTl

1g11 ln) vaporpure,(*)liquid pure,(* xRTTTl

1mon,vaporizati ln)( xRTTG

Boiling Point Elevation

The treatment used for freezing point depression leads to

2

mon,vaporizati

2

1 *)(m

H

TRMT

Pure liquid 1 (the solvent) boils at temperature T* and pressure p.

A solution of solvent (1) + solute (2) boils at T, p.

The boiling point elevation T is

T = T T*

Freezing Point Depression and Boiling Point Elevation

2b2

mon,vaporizati

2

1 *)(mKm

H

TRMT

Kf and Kb are the freezing-point and boiling-point constants

2f2

mfusion,

2

1 *)(mKm

H

TRMT

mfusion,

2

1f

*)(

H

TRMK

mon,vaporizati

2

1b

*)(

H

TRMK

Pure solvent (1) and a solution of solvent (1) + solute (2)

are separated by a semipermeable membrane that allows

solvent molecules to pass, but not solute.

Section 9.8 Osmotic Pressure

pure solvent (1) solvent (1) + solute (2)

1l*(T,p) 1l*(T,p) + RTlnx1

1(left compartment) > 1(right compartment)

solvent flows

spontaneously

into the

solution where

its chemical

potential is

lower

The flow of solvent into the solution increases the hydrostatic pressure

on the solution. At equilibrium, the chemical potential of pure solvent

at pressure p equals the chemical potential of solvent in the solution at

pressure p + p. The osmotic pressure is p.

Osmotic Pressure

pure solvent (1)


1l*(T,p)

1l*(T,p+p) + RTlnx1

1(left compartment) = 1(right compartment)

At equilibrium (pure water at pressure p, solution at p + p):

Osmotic Pressure

111 ln),(*),(* xRTppTpT ll

11

11 lnd*

),(*),(* xRTpp

pTpT

T

pp

p

lll

11 lnd*0 xRTpV

pp

p

l

11 ln*0 xRTpV l

1

1

ln*

xV

RTp

l

recall (G/p)T = V

Interpretation: The increase

in water chemical potential

caused by the osmotic pressure

p cancels the decrease in

water chemical potential RTlnx1

caused by mixing with the solute.

Osmotic Pressure

2

1

2

1

1

1 *)1ln(

*ln

*x

V

RTx

V

RTx

V

RT

lll

for dilute solutions of solvent (1) + solute (2) (x2 << 1):

1

2

121

2

1

2

1 *** n

n

V

RT

nn

n

V

RTx

V

RT

lll

solution volume V = n1V1m + n2V2m n1V1l*

22 RTc

V

nRT

c2 is the solute concentration

in units of moles per unit volume

(analogous to the ideal gas law)

Osmotic Pressure

Example Estimate the osmotic pressure of a 0.55 M aqueous

NaCl solution (the approximate composition of

seawater) at 25 oC.

L

mol1.1)K15.298()molKLbar08314.0( 112

V

nRT

Careful! Each mole of dissolved NaCl produces two moles of

dissolved solute ions (Na+ and Cl).

The solute concentration for calculating osmotic pressure is

therefore c2 = 1.1 M (not 0.55 M)

= 27.3 bar

Reverse Osmosis

Desalination: the production

of pure water from seawater

(or other salt water).

A pressure is applied to

seawater, larger than its

osmotic pressure.

Water spontaneously flows

across a semipermeable

membrane, from the saltwater

side to the pure water side.

Less energy required compared to distillation.

p p + p

applied

pressure

p >

Osmotic Pressure

Osmotic pressure can be used to understand:

why sap rises in trees

why plants wilt if exposed to salt water

why salt or sugar prevent microbial spoilage of food

polymer molecular weight determination from data

why injectable drugs are dissolved in “saline” solutions

(about 9 grams NaCl per liter), not in pure water

Ideal Solutions Mix n1 moles of pure liquid 1 and n2 moles

of pure liquid 2 at fixed temperature and pressure.

If the resulting solution is ideal, then:

p1 = x1 p1* (Raoult’s law for component 1)

p2 = x2 p2* (Raoult’s law for component 2)

Gmix = n1 RT lnx1 + n2 RT lnx2 ( = TSmix )

Smix = n1 R lnx1 n2 R lnx2

Hmix = 0

Vmix = 0

Umix = 0

Expect ideal-solution behavior if 1-1, 1-2 and 2-2 molecular

interactions are identical in strength (e.g., benzene + toluene).

Sections 9.9 to 9.12 Real Solutions

Big Problem: very few solutions are ideal

Even benzene + toluene solutions deviate slightly

(and measurably) from ideal behavior.

Real Solutions

Negative Deviations from Raoult’s Law

pi < xi pi* (e.g., chloroform + acetone)

Why?

Positive Deviations from Raoult’s Law

pi > xi pi* (e.g., CS2 + acetone)

Why?

Vapor Pressures of CS2 + Acetone Solutions at 20 oC

Raoult’s law

for CS2

pCS2 = xCS2pCS2*

Raoult’s law

for acetone

pAc = xAcpAc*

Illustrating positive deviations from Raoult’s law

CS2 mole fraction in the liquid: xCS2

pCS2*

pAc*

Thermodynamics of Real Solutions – Activity Coefficients

Include an activity coefficient “correction factor” i in Raoult’s law

for each component to allow for nonideal solution behavior.

Vapor pressure of nonideal solution component i

pi = i xi pi*

Chemical potential of nonideal solution component i

i = i* + RT ln(i xi )

Interpretation of Raoult’s law activity coefficient i :

i

i

px

p

ii

ii

component of pressure law sRaoult' ideal

component of pressure actual

*

Thermodynamics of Real Solutions – Activities

The chemical potential of nonideal solution component i

i = i* + RT ln(i xi )

is often written as

i = i* + RT ln(ai )

Interpretation of Raoult’s law activity ai = i xi of component i

i

i

p

pa

i

ii

component pure of pressure

component of pressure actual

*

Activity ai is a measure of the “escaping tendency” from the solution

of component i relative to the pure liquid.

Example At 90 oC a liquid water (1) + n-propanol (2) solution

with x1 = 0.741 has a total vapor pressure of 820.3 Torr

and water mole fraction y1 = 0.603.

Data: p1* = 527.76 Torr for pure water

p2* = 577.50 Torr for pure n-propanol

water vapor pressure p1 = y1p = 0.603 (820.3 Torr) = 494.6 Torr

ideal water vapor pressure = x1p1* = 0.741 (527.76 Torr)

= 391.1 Torr

p1 > x1p1* means positive deviation from Raoult’s law

water activity coefficient

water activity a1 = 1x1 = 1.265 (0.741) = 0.937

265.1Torr)(527.76(0.741)

Torr494.6

*11

11

px

p

Henry’s Law

Problem with Raoult’s law: Many important solution components

do not exist as pure liquids.

Example: H2O (1) + O2 (2) solutions at 25 oC

The critical temperature of O2 is 119 oC, so liquid O2

does not exist at 25 oC and p2* in Raoult’s law is not defined.

In cases where Raoult’s law is inconvenient, use Henry’s law

(not x2 1 for pure 2)

k2 is Henry’s law constant for component 2 (measureable).

p2 = k2 x2 in the limit x2 0

Example Calculate the solubility of O2 in water in

equilibrium with air at 1.00 bar and 298 K.

Data: kO2 = 49,500 bar

Assume air is 20 mol % O2.

pO2 = yO2 p = (0.20) (1.00 bar) = 0.20 bar

pO2 = kO2 xO2

Molecular oxygen (nonpolar) is not very soluble in water (polar),

which can be checked later, so Henry’s law should be accurate.

0.20 bar = (49,500 bar) xO2

xO2 = 4.0 106

Only about 4.0 106 mol O2

per mole (18. 0 mL) of water.

cO2 0.00022 mol L1

Why is this important

if you’re a trout?

pB

pB*

kB*

pB*kB

Henry’s Law

pB kBxB

as xB 0

Raoult’s Law

pB pB*xB

as xB 1

B escaping

from A

B escaping

from B

xB 0 xB 1

Comparison of

the Two Laws

pCS2 pCS2*

= 512.3 Torr

kCS2*

= 1750 Torr

pCS2*kCS2

CS2

+

Acetone

at 35 oC

mole fraction CS2, xCS2

Positive or negative

deviations from

Raoult’s law?

Why?

Raoult’s Law Activities

and Activity Coefficients

pCS2 = CS2(RL) xCS2 pCS2*

pressure CS law sRaoult'

pressure CS actual

*

2

2

CS2CS2

CS2S2(RL)

px

pC

CS2 Partial Pressure

CS2 Activity

*CS2

CS2S2(RL)

p

paC

CS2 Activity Coefficient

CS2(RL) 1

as xCS2 1

Henry’s Law Activities

and Activity Coefficients

pCS2 = CS2(HL) xCS2 kCS2

CS2S2(HL)

CS2 CS2

2

2

actual CS pressure

Henry's law CS pressure

C

p

x k

CS2 Partial Pressure

CS2 Activity

CS2S2(HL)

CS2

C

pa

k

CS2 Activity Coefficient

CS2(HL) 1

as xCS2 0

!!! Warning !!!

Instead of using solute mole fractions, Henry’s law is also written

in terms of

solute molarities (moles of solute per liter of solution)

pi = ki(c)ci

solute molalities (moles of solute per kilogram of solvent)

pi = ki(m)mi

Different concentrations (mole fraction, molarity, or molality)

give different Henry’s law activities and different activity coefficients.

Confusing if the concentration scale is not specified.

The reaction

aA + bB + … eE + fF + …

is at equilibrium when

G(reactants) = G(products)

aA + bB + … = eE + fF + …

Using stoichiometric coefficients vA = a, vB = b, … for the reactants

and vE = e, vF = f, … for the products gives the abbreviation:

Section 9.13 Chemical Reaction Equilibrium in Solutions

0i

iiv

liquid solutions are rarely ideal

deviations from ideal solution behavior can be large

accurate treatments of liquid-phase reaction equilibria use

activities (ai) to include nonideal solution behavior

Chemical Reaction Equilibrium in Solutions

iii aRT lno

0ln)ln( oo i

ii

i

ii

i

iii

i

ii avRTvaRTvv

i

vi

i

i

ii

i

ii aavRT

Gv

RTlnln

1 oo

321

321

o )/exp()(vvv

aaaRTGTK

Example Equilibrium Constants Using Raoult’s Law


321 )()()()( 3)RL(32)RL(21)RL(1

vvvxxxTK

io = i* = chemical potential of pure liquid i (vapor pressure pi*)

ai = i (RL) xi

i (RL) = pi / (xi pi*)

equilibrium constant:

Example Equilibrium Constants Using Henry’s Law


321 )()()()( 3)HL(32)HL(21)HL(1

vvvxxxTK

io = chemical potential of hypothetical pure liquid i with

vapor pressure ki (Henry’s law constant for component i).

ai = i (HL) xi

i (HL) = pi / (xi ki)

equilibrium constant:

Example Equilibrium Constant for Water Dissociation


Wait ! Shouldn’t this equilibrium constant be:

No !

H2O(liquid) H+(aq) + OH(aq)

Kw = [OH] [H+] = cOHcH+ = 1.00 1014 at 25 oC

H2O

HOH

2

w]OH[

]H][OH[

c

ccK

??

from first-year chem:




OH

HOHw

2a

aaK

(exact using activities)

H2OH2O

HHOHOHw

x

ccK

Use Henry’s law for OH and H+. (very low concentrations,

about 10-7mol/L).

Use Raoult’s law for H2O. (nearly pure liquid water).




cOH and cH+ are almost zero, so OH H+ 1

H2OH2O

HHOHOHw

x

ccK

Water (the solvent) is almost pure, so xH2O 1 and H2O 1

HOHw ccK

Strongly nonideal solutions can form two separate liquid phases

Section 9.14 Partially Miscible Liquids

Water + Phenol

Above the upper consolute

temperature (UCT)

at 65.85 oC :

completely miscible

Below TUC, two phases:

water-rich liquid L1 and

phenol-rich liquid L2

At point a (lever rule):

1 atm

al

la

1

2

2

1

L moles

L moles

Water + Triethylamine

Below the lower consolute

temperature (LCT)

at 18.5 oC:

completely miscible

Above TLC two phases:

water-rich liquid L1 and

triethylamine-rich liquid L2

1 atm

Why are LCT’s rarer than

UCT’s?

Water + Nicotine

upper consolute temperature

210 oC

lower consolute temperature

61 oC

Also called upper critical

solution temperature and

lower critical solution

temperature.

1 atm

Even rarer: Both an Upper and Lower Consolute Point!

UCT

LCT

Minimum Boiling Azeotrope and Upper Consolute Point

In this case the UCT is

lower than the minimum

boiling azeotrope temperature.

Why are systems with a

maximum boiling

azeotropes unlikely to

have an upper consolute

temperature?

(1 atm)

Minimum Boiling Azeotrope and Upper Consolute Point

Heating liquid initially

at point a forms vapor

of composition b.

Cooling vapor from

point b to 94 oC forms

vapor of composition i liquids

of composition g and h.

Further cooling produces

immiscible liquids of

composition d and e.

1 atmWater(1) + n-butanol(2)

Section 9.15 Eutectic Points

Adding B depresses

the freezing point

TA* of pure A

Adding A depresses

the freezing point

TB* of pure B.

At the eutectic point E

the two freezing point

depression curves meet.

Below the eutectic,

solid A and solid B exist.

TA*TB*

E

s

Gold + Silicon

Eutectic point

at 363 oC and silicon

mole fraction 0.186.

Near the eutectic

composition, liquid

gold + silicon solutions

exist at hundreds of

degrees below the melting

points of pure gold

and pure silicon.

Eutectic is Greek for

“easily melted”.

1 atm

one-phase

SiLiq

+

Au(s)

Solid Au + Si

xA

0.0 0.2 0.4 0.6 0.8 1.0

t /

oC

-20

0

20

40

60

A + B solid-liquid phase diagram at 1 bar

one-phase liquid

liquid + A(s)liquid + B(s)

A(s) + B(s)

F' = 2

F' = 1F' = 1

F' = 1

measure cooling curves for samples of liquid of different composition

plot temperature as a function of time

use the results to determine phase diagrams

(Chem 232 Lab “Phase Diagrams from Cooling Curves”)

Pressure is fixed.

So the reduced

degree of freedom

is F = F 1.

xA

0.0 0.2 0.4 0.6 0.8 1.0

t /

oC

-20

0

20

40

60

B(l)

B(s)

B(l)+B(s)

F = 0

C = 1

F = 0

C = 1

1

3

2 4

5

A(l)

A(s)

A(l)+A(s)

l (mixture) l (mixture) l (mixture)

l + B(s)

l + B(s) + A(s) l + B(s) + A(s) l + A(s) + B(s)

l + A(s)

B(s) + A(s) B(s) + A(s)

A(s)

+

B(s)

b

a

ta

d

e e e

c

tb

td

te tete

tct / oC

time

F = 0

C = 2F = 0

C = 2

F = 0

C = 2

xA

Cooling Curves

eutectic “halts”

(F = 0) at te

“breaks” (change

in slope) at tb, tc

1

23 4

5

F = 2 - P

F = 2 - P

F = 3 - PF = 3 - P

F = 3 - P

Chapter 10. Electrolyte Solutions

Summary

A whole chapter on just one kind of solution ?

What’s so special about electrolyte solutions ?

Enthalpy, entropy, and Gibbs energy of ions in solutions

Ion solvation

Activities of ions in solution

Debye-Huckel theory for dilute electrolyte solutions

Ionic reaction equilibrium

What is an “electrolyte solution” ?

a solution of ionic species of positive and negative

electrical charge dissolved in a solvent

the ions are mobile and conduct electric current in an

applied electric field (ionic conductivity)

usually ( but not always ) liquid solutions

important examples:

all biological and physiological solutions

strong acids and strong bases

buffer solutions for pH control

seawater, brines, and groundwater

battery electrolytes

molten salts

What’s Special About Electrolyte Solutions ?

thermodynamic properties of nonelectrolyte solution

components are functions of T, p, and composition

the properties of ions also depend on the electric potential

applied electric potential changes the chemical potential

of an ion of charge z by

electrical = zF

Example A 1.5 volt applied electric potential (from a common

AA battery) increases the chemical potential of Na+ ions by

electrical = zF = (+1) (96,485 C mol1) (1.5 V)

= 145,000 J mol1 (significant!!!)

application: electrochemical synthesis of sodium metal

Section 10.1 Enthalpy, Entropy, and Gibbs Energy

of Ions in Solution

electrolyte solutions are electrically neutral

impossible to study solutions containing only cations or anions

impossible to independently vary cation and anion concentrations

example aqueous MgCl2 solutions

2cMg2+ = cCl

important result: the internal energy, enthalpy, entropy, Gibbs

energy, volume, … of individual ions cannot be measured

But Many Important Processes Involve Ion Formation

Dissolution

CaCO3(s) Ca2+(aq) + CO32(aq)

Dissociation

H2O(l) H+(aq) + OH(aq)

pH Control (Buffers)

H+(aq) + CH3COO(aq) CH3COOH(aq)

Molten Salts

NaCl(s) Na+(l) + Cl(l)

Billion-Dollar Electrochemical Industries

Chlor-Alkali Production (Cl2, H2, and NaOH from saltwater)

Na+(aq) + Cl(aq) + H2O(l) Na+(aq) + OH(aq) + ½H2(g) + ½Cl2(g)

Aluminum Production (from molten cryolite)

2Al3+(l) + 3O2(l) + 3C(s) 2Al(l) + 3CO2(g)

Corrosion

Fe(s) + H2O(l) Fe2+(aq) + H2(g)

Batteries

PbO2(s) + Pb(s) + 2H2SO4(aq) 2PbSO4(s) + H2O(l)

Thermodynamic Convention for Ion Formation

Hfmo, Gfm

o, and Smo values are useful for many ionic reactions.

But these quantities cannot be measured for individual ions.

For convenient tabulation purposes, the standard enthalpy,

Gibbs energy, and entropy of formation of aqueous H+ ions

are defined as zero at all temperatures.

Hf,mo(H+, aq) = 0 Gf,m

o(H+, aq) = 0 Sf,mo(H+, aq) = 0

Why does this work?

Thermodynamic Convention for Ion Formation

Example Hfmo(H+, aq) = 0 Why is this convention valid?

The zero-point for enthalpy can be chosen arbitrarily to calculate H.

H(products)abs

H(reactants)abs

zero Habs

Hmo(H+,aq)abs

Habs shifted down by Hmo(H+,aq)abs

H(products)

H(reactants)

0 0 (shifted)

Absolute Enthalpies Conventional Enthalpies

H

same H

Example Calculate Ho and Go at 25 oC for the reaction

Cl(aq) + H2O(l) OH(aq) + ½H2(g) + ½Cl2(g)

Data: Tables 4.1 and 10.2

Ho = Hfo(products) Hf

o(reactants)

= Hfmo(OH,aq) + ½ Hfm

o(H2,g) + ½ Hfmo(Cl2,g)

Hfmo(Cl,aq) Hfm

o(H2O,l)

= 230.0 + ½ (0) + ½ (0) (167.2) (285.8) = 223.0 kJ

Go = Gfmo(OH,aq) + ½ Gfm

o(H2,g) + ½ Gfmo(Cl2,g)

Gfmo(Cl,aq) Gfm

o(H2O,l)

= 157.2 + ½ (0) + ½ (0) (131.2) (237.1) = 211.1 kJ

(not spontaneous)

Section 10.2 Ion Solvation

Why are electrolyte solutions “special” ?

For years, many scientists believed solutions of ions couldn’t exist

under ambient conditions (T 300 K). Why?

Consider the formation of gas-phase H+ and Cl ions at 25 oC.

½ H2(g) H(g) Go = +203 kJ

½ Cl2(g) Cl(g) Go = +106 kJ

H(g) H+(g) + e Go = +1312 kJ

Cl(g) + e Cl (g) Go = 349 kJ

Overall:

½ H2(g) + ½ Cl2(g) H+(g) + Cl (g) Go = +1272 kJ

extremely unfavorable

Ion Solvation

For comparison, the formation of aqueous H+ and Cl ions at 25 oC:

½ H2(g) H(g) Go = +203 kJ

½ Cl2(g) Cl(g) Go = +106 kJ

H(g) H+(g) + e Go = +1312 kJ

Cl(g) + e Cl (g) Go = 349 kJ

H+(g) + Cl (g) H+(aq) + Cl (aq) Go = 1403 kJ

Overall:

½ H2(g) + ½ Cl2(g) H+(aq) + Cl (aq) Go = 131 kJ

favorable

solvation

Interaction with water molecules stabilizes the aqueous ions. Why?

Hydration of a Sodium Ion

Section 10.3 Activities of Electrolytes

Why are electrolyte solutions “special” ?

ions in solution interact by long-range electrostatic forces

electrolyte solutions can be strongly nonideal, even if dilute

nonelectrolyte (Chap. 9) and electrolyte activity expressions

are very different

aNaCl = aNa+aCl

aLaCl3 = aLa+++(aCl)3

aAl2(SO4)3 = (aAl+++)2(aSO4 )3

Examples Activities of Aqueous NaCl, LaCl3, and Al2(SO4)3

NaCl = Na+ + Cl

= Na+o + RT lnaNa+ + zNa+F

+ Clo + RT lnaCl + zCl F

= Na+o + Cl

o + RT ln(aNa+aCl)

NaCl = NaClo + RT lnaNaCl

aNaCl = aNa+aCl

Chemical Potential (Molar Gibbs Energy) of Aqueous Sodium Chloride

(chemical) (electrical)

(chemical) (electrical)

Why is there no electric

potential in the expression

for the chemical potential of

aqueous NaCl?

Cv+Av = v+C + vA

= v+( Co + RT lnaC + zCF )

+ v( Ao + RT lnaA + zA F )

= v+Co + vA

o + RT ln(aCv+aA

v)

Cv+Av = Cv+Avo + RT lnaCv+Av

aCv+Av = aCv+aA

v

Chemical Potential of Electrolyte Av+Cv

Cv+Av = Co + A

o + RT ln(aCv+aA

v)

a = (aCv+aA

v)1/v

Mean (Average) Activity of Electrolyte Av+Cv

Useful, but cation and anion activities aC and aA can’t be measured.

For practical calculations, the measurable mean ionic activity ais used, a weighted-average of the cation and ion activities

which gives the mean ionic chemical potential

aRTv

vvln

oAC

v = v+ + v

Cv+Av = vo + RT ln(a

v)

chemistry 232. physical chemistry ii · 2020-01-06 · thomas engel and philip reid available with...

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