What is Thermo, Stat Mech, etc.? - Wake Forest Universityusers.wfu.edu/macoskjc/Courses/ALL2.pdf · What is Thermo, Stat Mech, etc.? ... The cubic form of the equation predicts 0

What is Thermo, Stat Mech, etc.?

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Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you

go through it, you think you understand it, except for one or two ll i h hi d i h h i ksmall points. The third time you go through it, you know you

don't understand it, but by then you are so used to it, it doesn't bother you any more. -- Arnold Sommerfield

Ludwig Boltzmann who spent much of his life studyingLudwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul

Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. Perhaps it will be wise to

approach the subject cautiously. -- David L. Goodsteinapproach the subject cautiously. David L. Goodstein

Our approach will be to focus on the macroscopic, thermodynamic picture with occasional insight from the microscopic picture via statistical mechanics

© Paul Percival 9/10/2008Modified by Jed Macosko

the microscopic picture via statistical mechanics.

Energy, Work and Heat

Energy is the capacity to do work.

Its classification into:kinetic potential(by motion) (by position)

e.g. thermal chemical, electricalIs purely arbitrary!

Heat and work are not “types” of energy, but are processes???????

Heat and work are not types of energy, but are processes involving transfer of energy. They appear and disappear at the system boundary. They are path variables.

Heat is the transfer of energy from one body to another of lower temperaturelower temperature.

Convention: if heat flows into the system, q > 0.

Work is the transfer of energy by some mechanism other than temperature difference.p

Convention: if work is done on the system, w > 0.

Heat stimulates random motion.

Work stimulates organized motion.g

Work “degrades” into heat.qualitative observations by Count Rumford (Ben Thompson)quantitative measurements by James Joule


Terminology 1

A system is a particular sample of matter or region of space.

An isolated system does not interact with its surroundings.

system + surroundings = universesystem + surroundings = universe

A closed system does not allow passage of mass over its boundaries, in contrast to…

An open system.An open system.

An adiabatic system has boundaries which permit no flow of heat. It is insulated.

A system is in a definite state when all its properties have y p pdefinite values.

A system at equilibrium is time independent; it is not affected by the history of the system.

Total Property property(part)= ∑

Extensive properties depend on the amount of substance in the system, e.g. n, V.

Intensive properties are independent of amount, e.g. P, T.

Total Property property of part=


open systemmatter heat

surroundings

closed systemmatter heat

surroundings

isolated systemmatter heatsystem

surroundings

matter heat


Terminology 2

State variables (state functions) uniquely determine the state of a system at equilibrium. Two samples of a substance with the same state variables are in the same state.

The change in a state variable depends only on the initial and finalThe change in a state variable depends only on the initial and final states, independent of path.

Path functions depend on the process and therefore vary with path.

A li i i hi h h i i i l d fi l hA cyclic process is one in which the initial and final states are the same, i.e. no change in the state variables.

In contrast, path functions generally have non-zero values for cyclic processes, dependent on the path.

A reversible process is one that can be reversed by an infinitesimal modification of a variable. The system is in equilibrium with the surroundings at all times. This is an idealized situation, useful as a theoretical limit, but…

All real processes are irreversible. It is possible to restore the system or the surroundings to their original states but not both.

An equation of state is the functional relationship between the properties of a system, e.g, the ideal gas law.


Ideal Gases -- Review

Ideal gases obey the ideal gas law:PV nRT=

Pressure Pa ≡ 1 N m-2 ≡ 1 J m-3 ≡ 1 kg m-1 s-2

Volume m3 ≡ 103 dm3 ≡ 103 litres

Temperature K

number of moles

gas constant

236.022 10w NnM L

L= = ×=

-1 -18.3145 J K molR =

no. of atoms in 12 g 12C

but if P is in atm (≡ 1.01325 × 105 Pa) and V in litres,3 -1 -10.08206 dm atm K molR =

P V1

V2

V3

V P1

P2

P3

PT3

T

3

T

3

V

T1

isotherms isobars isochors


isotherms isobars isochors

Mixtures of Ideal Gases

If the ideal gas law applies to each component, i i iPV n RT=

n RT n Ppartial pressure tottot

tot

i ii i

n RT n PP PV n

= = = χ

totPP n P= =∑ ∑Dalton’s Law oftot

toti i

i iP n P

n= =∑ ∑Dalton s Law of

Partial Pressures

e.g. for two components:

pressure

A BP P P= +

A AP P= χ

χA0 1χB 01

B BP P= χ

χB 01

composition

Real gases are ideal only at the low density limit Why?


Real gases are ideal only at the low density limit. Why?

Real Gases

• have non-zero volume at low T and high P• have repulsive and attractive forces between molecules

short range,important at high P

longer range,important at moderate P

At low pressure, molecular volume and intermolecular forces can often be neglected, i.e. properties → ideal.

1 B CPV RT ⎡ ⎤+ + + VV V

Virial Equations

21PV RTV V

⎡ ⎤= + + +⎢ ⎥⎣ ⎦K

21PV RT B P C P′ ′⎡ ⎤= + + +⎣ ⎦K

mV Vn

= =

B is the second virial coefficient.C is the third virial coefficients.They are temperature dependent.

Van der Waals Equation

( )2aP V b RT

V⎛ ⎞+ − =⎜ ⎟⎝ ⎠


Compressibility Factoralso known as compression factor

ideal

PV VZRT V

= =

2.0H2

C2H4 CH42.02.0

H2

C2H4 CH4

Z

1.0

NH3Z

1.01.0

NH3

P0

P00

PP

ZT1

T2

TZ

T1

T2

T

The curve for each gas becomes more ideal as T →

P

1.0T3

P

1.0T3ideal as T → ∞


PP

The van der Waals Equation 1

( )2aP V b RT

V⎛ ⎞+ − =⎜ ⎟⎝ ⎠

Intermolecular attraction “molecular volume”Intermolecular attraction= “internal pressure”

molecular volume ≈ excluded volume

( )3 34 23 32 / 2rπ = πσ

RT aP 2PV b V

PV V aZRT V b RTV

= −−

= = −−

( )2

311 2a a ab P b P

RT RT RTRT⎛ ⎞ ⎛ ⎞= + − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

K

1Z a∂⎛ ⎞ ⎛ ⎞

(boring algebra)

The initial slope depends on a, b and T:

1 ...T

Z abP RT RT

∂⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠⇒

/b RT l l i d i titi f /b a RT> molecular size dominant

forces dominant

Boyle Temperature

• positive for

• negative for

• zero at

/b a RT<

/T a Rb=~ ideal behaviour over wide range of P


~ ideal behaviour over wide range of P

Condensation of Gases

Real gases condense… don’t they?

supercritical fluid

P

P1

Pc

P2TcT2

gasliquid

1

VVc

T1

T2

gasliquid

Tc, Pc and Vc are the critical constants of the gas.

Above the critical temperature the gas and liquid phases areAbove the critical temperature the gas and liquid phases are continuous, i.e. there is no interface.


The van der Waals Equation 2The van der Waals Equation is not exact, only a model. a and b are empirical constant.

3 2 0RT a abV b V VP P P

⎛ ⎞− + + − =⎜ ⎟⎝ ⎠⎝ ⎠

P

The cubic form of the equation predicts

Vb0

3 solutions??????? How to solve?

( )2 32 0

T

P RT aV VV b

∂⎛ ⎞ = − + =⎜ ⎟∂⎝ ⎠ −

There is a point of inflection at the critical point, so…

slope:( )

( )

2

2 3 42 6 0

T

T

V VV b

P RT aV VV b

∂⎝ ⎠

⎛ ⎞∂= − =⎜ ⎟∂ −⎝ ⎠

curvature:

c c c2

c c

832727

3 27

a aP V b TRbb

PV aZ T T

= = =

= = = =

⇒


c B cc 8 8

Z T TRT Rb

= = = =

The Principle of Corresponding States

Reduced variables are dimensionless variables expressed as fractions of the critical constants:

P V TP V T= = =r r rc c c

P V TP V T

= = =

Real gases in the same state of reduced volume and reduced temperature exert approximately the same reduced pressure.

They are in corresponding states.

If the van der Waals Equation is written in reduced variables,

3⎛ ⎞( )r r r2r

3 3 1 8P V TV

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠

Since this is independent of a and b, all gases follow the ( i t l )same curve (approximately).

1.0Tr = 1.5

T = 1 2Z

Tr 1.2

Tr = 1.0


Pr

Properties of Real Gases as P → 0

Real gases have interactions between molecules. These change when the gas is compressed, but … …they need not go to zero as P → 0.

e.g. consider T

ZP

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

For an ideal gas: ( )1 0T

Z PVP P RT P

∂ ∂ ∂⎛ ⎞⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

For a real gas: 2 1

2

PVZ B P C PRT

Z B C P

′ ′= = + + +

∂⎛ ⎞ ′ ′= + +⎜ ⎟⎝ ⎠

K

K

and in the limit:0

2

lim 0

T

P T

B C PP

Z BP→

+ +⎜ ⎟∂⎝ ⎠

∂⎛ ⎞ ′= ≠⎜ ⎟∂⎝ ⎠

K

Not all properties of real gases tend to ideal values as P → 0.


van der Waals

Johannes Diderik van der Waals, 1837 - 1923

Nobel Prize in Physics 1910

htt // h / t ht l


http://www.s-ohe.com/stamp.html

Carbon Dioxide

Image copied from:

http://

Original source:

http://science.nasa.gov/headlines/y2003/20aug_supercriticalco2.htm

http://www.chem.leeds.ac.uk/People/CMR/criticalpics.html


Partial Differentiationfor functions of more than one variable: f=f(x, y, …)

Take area as an example A xy=

For an increase y constant1in , x x A y xΔ Δ = Δ

For a simultaneous increase

For an increase x constant2 in , y y A x yΔ Δ = Δ

( )( )A x x y y xyΔ = + Δ + Δ −2AΔ

xΔyΔ

x( )( )

1 2

y y yy x x y x y

A Ax y x yx y

= Δ + Δ + Δ ΔΔ Δ

= Δ + Δ + Δ ΔΔ Δ

A xy=

2AΔ

AΔ

yΔ

yy

In the limits 0, 0x yΔ → Δ →

A AA dA dx dy∂ ∂⎛ ⎞⎛ ⎞Δ → = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1AΔ y

y x

yx y⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

partial differentialtotal differential

( ) ( )0

, ,lim

f x x y f x yfx xδ →

+ δ −⎧ ⎫∂⎛ ⎞ = ⎨ ⎬⎜ ⎟∂ δ⎝ ⎠ ⎩ ⎭

for a real single-value function f of two independent variables,

p


0xyx xδ →∂ δ⎝ ⎠ ⎩ ⎭

Partial Derivative Relations

y x

z zdz dx dyx y

∂ ∂⎛ ⎞⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

( , , ) 0 ( ,, )f x y z z z x y= =Consider so

• Partial derivatives can be taken in any order.2 2z z

x y y x∂ ∂

=∂ ∂ ∂ ∂ yx y x

z zx y y x

⎡ ⎤⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠⎣ ⎦ ⎣ ⎦x y y x∂ ∂ ∂ ∂ yx y xy y⎝ ⎠⎣ ⎦ ⎣ ⎦

• Taking the inverse:1

y y

z xx z

−⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

• To find the third partial derivative:

0yx

z zdz dy dxy x

∂ ∂⎛ ⎞ ⎛ ⎞= = −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⇒

⎝ ⎠( )( )

//

x

yz xy

z yx z xy z x y z

∂ ∂∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞• Chain Rule:

and

1x yz

x y zy z x

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

1y x z∂ ∂ ∂⎛ ⎞⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟


1z y xx z y⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Partial Derivatives in ThermodynamicsFrom the generalized equation of state for a closed system,

( ), , 0f P V T =

six partial derivatives can be written:

P V T P V T

V T P T P VT P V V T P

∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1V T

−∂ ∂⎡ ⎤⎛ ⎞ ⎛ ⎞

1V T P∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

but given the three inverses, e.g

and the chain rule

P P

V TT V

∂ ∂⎡ ⎤⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

P V TT P V⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

there are only two independent “basic properties of matter”. By convention these are chosen to be:

the coefficient of thermal expansion(isobaric), and

1P

VV T

∂⎛ ⎞α = ⎜ ⎟∂⎝ ⎠

1 V∂⎛ ⎞⎜ ⎟

the coefficient of isothermal compressibility.

The third derivative is simplyTV P

⎛ ⎞κ = − ⎜ ⎟∂⎝ ⎠

( )/ PV TP ∂ ∂∂ α⎛ ⎞ = − =⎜ ⎟

p y


( )/V TT V P⎜ ⎟∂ ∂ ∂ κ⎝ ⎠

The Euler Relation

Suppose ( ) ( ), ,z A x y dx B x y dyδ = +

Is δz an exact differential, i.e. dz?

dz is exact providedyx

A By x

∂ ∂⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠cross-differentiation

2z A z∂ ∂ ∂⎛ ⎞⎛ ⎞⎜ ⎟because then

2

y x

z A zAx y y x

z B zBy x x y

∂ ∂ ∂⎛ ⎞⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

∂ ∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ yxy x x y∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

The corollary also holds.

State functions have exact differentials.

Path functions do not.

New thermodynamic relations may be derived from the Euler relation.

S V

dU TdS PdVT PV S

= −

∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

e.g. given that

it follows that


S V⎝ ⎠ ⎝ ⎠

Energy, Work and Heat

Energy is the capacity to do work.

Its classification into:kinetic potential(by motion) (by position)

e.g. thermal chemical, electricalIs purely arbitrary!

Heat and work are not “types” of energy, but are processesHeat and work are not types of energy, but are processes involving transfer of energy. They appear and disappear at the system boundary. They are path variables.

Heat is the transfer of energy from one body to another of lower temperaturelower temperature.

Convention: if heat flows into the system, q > 0.

Work is the transfer of energy by some mechanism other than temperature difference.p

Convention: if work is done on the system, w > 0.

Heat stimulates random motion.

Work stimulates organized motion.g

Work “degrades” into heat.qualitative observations by Count Rumford (Ben Thompson)quantitative measurements by James Joule


The First Law of Thermodynamics0 f h t fl i t th t

For finite changes of state:

F i fi it i l h

U q wΔ = +

dU δ δ

q > 0 for heat flow into the systemw > 0 for work done on the system

For infinitesimal changes:

U is the internal energy of the system.

When a system changes from one state to another along an adiabatic path the amount of work done is the

dU q w= δ + δ

along an adiabatic path, the amount of work done is the same, whatever the means employed.

For

Forad final initial0,q w U U U= = − = Δ

0q q U w w w≠ = Δ − = −For

The energy of an isolated system is constant.

ForN t l ti hi !

ad0,q q U w w w≠ = Δ − = −

0, 0 0q w U= = Δ =⇒ No perpetual motion machines!

In any cyclic transformation the work done by a system on its surroundings is equal to the heat withdrawn from the surroundings.g

The energy of the universe is constant.

0w q dU− δ = δ ⇔ =∫ ∫ ∫

U UΔ = Δ


system surroundingsU UΔ = −Δ

Temperature

Two systems in thermal equilibrium are at the same temperature.

If system A is in thermal equilibrium with system B, and A is in thermal equilibrium with C,

then B must be in thermal equilibrium with C.

This is a statement of the zeroth law of thermodynamics.

PV⎛ ⎞Ideal gas temperature:0

limP

PVTnR→

⎛ ⎞= ⎜ ⎟⎝ ⎠

Unit of temperature:( )triple point of w

Keat

lviner

1 273 16

T=p

273.16

The freezing point of water at 1 atm is 273.15 K.The boiling point of water at 1 atm is 373.12 K.g p

The Celsius scale is defined as / C / K 273.15t T° = −

It is possible to define an absolute temperature scale(Kelvin scale) by considering the work done in an isothermal reversible expansion/compression.

1

2

ln Vw nRTV

=


2V

Pressure–Volume Work 1

1 /P mg A>

Expansion

T1,P1,V1

m T2,P2,V2

m

h A = area of piston

2 ex /P P mg A= =

( )w mg h= −

1 1 1

P11 equation of state

(need not be isotherm)( )

( )ex

ex 2 1

P A hP V V

= −

= − −P2

2

( )

V1 V2

Expansion into a vacuum0 0P w= ⇒ =

P22

ex 0 0P w= ⇒ =

Compression

P P<

P11

( )ex 2 1w P V V= − −

1 exP P<

2 exP P=


V2 V1

( )ex 2 1w V V


Multi-stage Expansion

1( )w P V V=

P

53

2

4

( )( )

( )( )

2 2 1

3 3 2

4 4 3

w P V VP V V

P V V

− = −

+ −

+ −

V( )5 5 4P V V+ −

Reversible Expansion

P

1Reversible Expansion

Make steps so small that0, 0dP dV

w P dV→ →

−δ =Then

V

2( )ex

int

int

w P dVP dP dVP dV

δ =

= +

→

Then

2

∫ ∫V2

intpath 1w dw P dV= = −∫ ∫

For ideal gases int /P nRT V=

and at fixed temperature 2 2ln lnV Pw nRT nRT⎛ ⎞ ⎛ ⎞

= =⎜ ⎟ ⎜ ⎟


and at fixed temperature rev1 1

ln lnw nRT nRTV P

= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


4compression

P

1

2expansion

P

3

Consider the cyclic path 1 → 2 → 3 → 4 → 1

( ) ( )3 3 1 1 1 30 0w P V V P V V− = + − + + −

V

Consider the cyclic reversible path 1 → 3 → 1

( ) ( )( )( )

3 3 1 1 1 3

3 1 3 1

0 0w P V V P V V

P P V V

+ + +

= − −

3 1

1 3

3 3 0

V V

V V

V V

w PdV PdV

PdV PdV

− = +

= − =

∫ ∫

∫ ∫Even for a cyclic process w depends on path

0dU w q= ⇔ − δ = δ∫ ∫ ∫

1 1V V∫ ∫


q∫ ∫ ∫

Energy vs. Enthalpy

For a change in state at constant volume, no expansion work is done, so

However, for a change in state at constant pressure,

,V VU q dU qΔ = = δ

,P PU q w dU q PdVΔ = + = δ −

( )

2

1

2 2

1 1

V

P VU dU q PdVΔ = = δ −∫ ∫ ∫

( )2 1 2 1PU U q P V V− = − −

( ) ( )( ) ( )

2 2 1 1

2 2 1 21 12 1

P

P

U PV U PV qU PV P P PU PV q

+ − + =

+ − = =+ =

P constant

( ) ( )2 2 1 21 12 1 Pq

H U PV= +Enthalpy

,P PH q dH qΔ = = δ

H, being a function of state variables only, is also a state variable.

For a general change of state (P and V may both change),For a general change of state (P and V may both change),

( )H U PVU P V V P P V

dH dU PdV VdP

Δ = Δ + Δ

= Δ + Δ + Δ + Δ Δ


dH dU PdV VdP= + +

Heat Capacity

D fi at constant volume no workC dTδ

Transfer of heat to a system may result in a rise in T.

path function, so C depends on conditions.q CdTδ =

Define: at constant volume, no work

at constant pressure, only PV workV V

P P

q C dTq C dT

δ =

δ =

From 1st Law, exdU q w q P dV= δ + δ = δ − assume no th k

VV

UCT

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

f r 0o VdU q dV= δ = other work

Similarly( )

for 0P

dH dU PdV VdPq PdV PdV VdP

q dP

= + −= δ − + −= δ = for 0Pq dPδ

PP

HCT

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

For ideal gases ( )P V

dH dU d PV dU nRdTC dT C dT nRdT

= + = +

= +

P VC C R= +


P VC C R+

The Relation Between CP and CV

∂ ∂⎛ ⎞ ⎛ ⎞P V

P V

H UC CT TU V UPT T T

∂ ∂⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

H U PV= +

P P VT T T⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

V T

U UdU dT dVT V

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

But since

P V T P

U U U VT T V T

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

U V VC PC ∂ ∂⎛ ⎞ ⎛ ⎞ ∂⎜ ⎟ ⎜ ⎟= ⎛ ⎞⇒ ⎜ ⎟+V

PP

T PV TC PC

T⎜ ⎟ ⎜ ⎟∂ ∂− =⇒ ⎜ ⎟∂⎝⎝

+⎠ ⎠⎝ ⎠

work needed to overcome

intermolecular forces

expansion per degree

∂⎛ ⎞T

UV

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠“internal pressure”

C C R⇒0U V nR∂ ∂⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

For idealP VC C nR⇒ − =0,

T PV T P⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

For ideal gases

V∂⎛ ⎞⎜ ⎟⎝ ⎠

C C≈For liquids is so small that


PT⎜ ⎟∂⎝ ⎠ P VC C≈and solids is so small that

Adiabatic Expansion 10 0δ

adiabatic = insulated:0, 0q q

U w dU w= δ =

Δ = = δ

U UdU dT dV∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1

2

1for ideal gas(0 )es

if is independent fo

T

VTw dU C dT

C T C T

= = +

Δ

∫ ∫ K

V TdU dT dV

T V+⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

if is independent f oV VC T C T= Δ

For adiabatic expansion 2 10, 0w U T T⇒Δ„ „ „

Free expansion: 0 0 0P w T= = Δ⇒ ==Free expansion: ex 0 0, 0P w T= = Δ⇒ =

Fixed pressure: ex

V

w P Vw U C T

= − Δ

= Δ = Δ

ex

V

P VTC

⇒Δ

Δ = −

Reversible expansion: exP P=

( )2

1

,V

Vw P V T dV= −∫

Substitute appropriate equation of state.N t f l if T h


Not useful if T changes.

Adiabatic Expansion 2R ibl di b ti i f id lReversible adiabatic expansion of ideal gases:

V

dU wnRTC dT PdV dVV

= δ

= − = −

0qδ =

ideal gases only

( ) ( )

2 2

1 1

2 1 2 1

1 1

ln / ln /

V

V

V

V

C dT R dVT V

C T T R V V

= −

= −

∫ ∫( ) ( )

( ) ( )( ) ( ) ( )

2 1 2 1

2 1

2 1 2 1

ln / ln /

ln /

ln / 1 ln /

V

P V

C T T R V V

C C V V

T T V V

= − −

= − γ − P

V

CC

γ =

12 1

1 2

T VT V

γ−⎛ ⎞

= ⎜ ⎟⎝ ⎠

PV T2 2 2

1 1 1

PV TPV T

=

2 1 orP V PV PVγ

γ γ⎛ ⎞= =⎜ ⎟

Also, since

P 1PV

∝

2 11 1 2 2

1 2

or PV PVP V

γ γ= =⎜ ⎟⎝ ⎠

( 1) /2 2

1 1

T PT P

γ− γ⎛ ⎞

= ⎜ ⎟⎝ ⎠

V2V1

1PV γ∝


1 1T P⎝ ⎠

Types of Expansion Work ― Summary

exw P dVδ = −For all types of expansion

( )ex ,P P P V T= =For reversible changes in general

ex rev, ( ) ( )P P w w< − < −In an irreversible expansion

For isothermal expansions T is held constant by leakage of heat into the system from the surroundings.

0q =In an adiabatic expansion

0, 0,T

U U q wV

∂⎛ ⎞ = Δ = =⎜ ⎟∂⎝ ⎠

y g

For an ideal gas

w U= Δ0q

so the internal energy must provide for the work:p

For reversible expansion of an ideal gas through a given f

isothermal adiabatic( ) ( )w w− > −volume change from the same initial state,

since U is continuously replenished by heat intake to keep th t t t tthe temperature constant.


The Joule Expansion Experiment

V T

U UdU dT dVT V

U

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂⎛ ⎞

For a closed system, the state function U is determined by Tand V alone:

VT

UC dT dVV

∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

Joule tried to measure this partial derivative.

T

A B

1. Gas in A, vacuum in B.

2. Open valve.

3. Any change in T?

He found no change in temperature when the gasHe found no change in temperature when the gas expanded from VA to VA+VB, i.e. q = 0. Also, no work was done (free expansion), so w = 0. Conclusion: ΔU = 0 and hence

0U∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠( )U U T=or only

TV⎜ ⎟∂⎝ ⎠( )or only

Strictly only true for ideal gases.

Not true for liquids and solids, but since and ΔV is very small the effect of ΔV on U is usually ignored

( )/ TU U V VΔ ≈ ∂ ∂ Δ


and ΔV is very small, the effect of ΔV on U is usually ignored.

Pressure Dependence of EnthalpyFor a closed system, the state function H is determined by Tand P alone:

P T

H HdH dT dPT P

H

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂⎛ ⎞

PT

HC dT dPP

∂⎛ ⎞= + ⎜ ⎟∂⎝ ⎠

For ideal gases H U PV U nRT= + = +H U∂ ∂⎛ ⎞ ⎛ ⎞ 0

T T

H UP P

∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Not true for real gases, liquids and solids!

P VT T

H UC dT dP C dT dV PdV VdPP V

∂ ∂⎛ ⎞ ⎛ ⎞+ = + + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

dH dU PdV VdP= + +

At fixed temperature, dT = 0

T T T T

H U V VP VP V P P

V∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎝ ⎠≈

⎠

small for liquids and solids

Investigate real gases at constant enthalpy, i.e. dH = 0

PH TC∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⇒


PT HP P⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

The Joule-Thompson Effect (not in text)T∂⎛ ⎞

JTH

TP

∂⎛ ⎞μ = ⎜ ⎟∂⎝ ⎠Joule-Thompson Coefficient

Pump gas through throttle (hole or porous plug) P1 > P2

Keep pressures constant

P1

by moving pistons.

P

Work done by system

2 2 1 1w PV PV− = −Si 0 P2Since q = 0

2 1 1 1 2 2U U U w PV PVΔ = − = = −

2 2 2 1 1 1U PV U PVH H

+⇒ + =

t t th l2 1H H= constant enthalpy

Measure change in T of gas as it moves from side 1 to side 2.

JTTΔ⎛ ⎞μ = ⎜ ⎟

⎝ ⎠JT0PP Δ →

μ ⎜ ⎟Δ⎝ ⎠

A modern, more direct experiment uses similar apparatus but with a heater to offset the temperature drop 0

limP

H qP PΔ →

∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ Δ⎝ ⎠ ⎝ ⎠


heater to offset the temperature drop. 0PTP PΔ →∂ Δ⎝ ⎠ ⎝ ⎠

The Linde RefrigeratorFor most gases at room temperature so sudden (adiabatic) expansion results in a drop in T.

JT 0μ >

This is the basis of operation of the Linde refrigeratorand gas liquefaction.

cooling fins cold gas

compressorexpansion nozzle

In general, µJT depends on T and can even change sign.

i i t t

T

lines of0T∂⎛ ⎞ >⎜ ⎟

inversion temperature

lines of constant enthalpy

0HP

>⎜ ⎟∂⎝ ⎠

0H

TP

∂⎛ ⎞ <⎜ ⎟∂⎝ ⎠


P

The Molecular Interpretation of UWhat is U? ΔU can be related to thermochemical observables:

2

1

,T

V VTV

UC U C dTT

∂⎛ ⎞= Δ =⎜ ⎟∂⎝ ⎠ ∫( ) ( )

T

∫

for ideal gases

( ) ( )0

0 VU T U C dT− = ∫Can U be calculated from molecular properties?

l lU u N u= = < >∑ ignoring intermolecular forces

molecules

Classically, <u> is given by the Equipartition Law:“The average energy of each different mode of motion of a molecule is ½kT.”

k = Boltzmann constant = R/L

Implicit in this “law” is the concept that there is no coupling between molecular modes of motion.

k = Boltzmann constant = R/L=1.381 × 10-23 J K-1

tr ro v lib et ( )u = ε + ε + +εε2 2 2 21 1 1 1

tr 2 2 2 2mv mx my mzε = = + +& & &2 2 21 1 1

rot 2 2 2x x y y z zI I Iε = ω + ω + ω

Total:

Translation:

Rotation:2 21 1

vib 2 2 per modeT V mx kxε = + = +&

y y

Vibration:

for non-linear moleculesfor linear

3 6 3 5 moleculesnn

−⎧= ⎨ −⎩

no. of vib. modes


for linear 3 5 molecu les n −⎩

The Molecular Interpretation of U (cont.)

When quantum effects can be ignored, the average energy of every quadratic term in the energy expression has the same value, ½kT.

3 1 5Uu kT C R∂⎛ ⎞< > = = =⎜ ⎟Monoatomic gases 2 , 1.5VV

u kT C RT

< > = = =⎜ ⎟∂⎝ ⎠( )3

2 1 1 , 3.5Vu kT C R< > = + + =

( )32 1 3 5n kTu< > = + + −

Monoatomic gases

Diatomic gases

Linear polyatomics

( )3 32 2

f

3

3 6

6

0

uCkT

Rn< > = + + −

( )2

e.g. for 3, 6.5Vn C R==

Non-linear molecules

e.g. for 3, 6.0Vn C R= =

Experimental observations do not agree with these predictions, except for monoatomic gases. In particular:

The predicted values are too highThe predicted values are too high.Experimental values are temperature-dependent.

Accurate calculations are provided by statistical mechanics, where it can be shown that the equipartition principle onlywhere it can be shown that the equipartition principle only holds in the limit of high temperature, specifically for the condition kT >> ΔE, where ΔE is the appropriate spacing of energy levels.

E i titi f t f f b t d


Equipartition ≡ free transfer of energy between modes

Temperature Dependence of CV

( )0 tr rot

tr rot vib el

, whereVV

V V V V

UC U U NT

C C C C

∂⎛ ⎞= − = < ε > + < ε > +⎜ ⎟∂⎝ ⎠

= + + +

K

{rotrot

linear non-line f 1. o5 ar r

V

RC kTR= Δε––

tr 1.5V

C R= for all T

rot 0V

C = at very low T, where rot 0< ε >→

vibV

C R= for each vibrational mode, at high Tel 0C l t l i l t i it tiel 0V

C = almost always, since electronic excitation takes great energy

t te.g. for a diatomic molecule:

→ two atomsVC

R 725 t + r

2tt + r + v

5232

0

t

t + r


T0

0

Temperature Dependence of HInteractions between molecules also contribute to the heat capacity of real systems. In particular, first-order phase changes often involve large energy changes. These are usually measured at constant pressure and expressed as y p penthalpies.

gas

H

solid 2liquid

vHΔ

solid 1

T0

0

solid 2fHΔ

s-sHΔ

T0

2H a bT cTH

= + + +

∂⎛ ⎞

KIn general, for each phase

2PP

HC b cTT

∂⎛ ⎞= = + +⎜ ⎟∂⎝ ⎠K


ThermochemistryThe study of energy changes that occur during chemical reactions:

at constant volume ΔU = qV no work

at constant pressure ΔH = qP only PV work

For practical reasons most measurements are made at constant P, so thermochemistry mostly deals with ΔH.

reactionproducts reactants

H H HΔ = −∑ ∑

If ΔH > 0 the reaction is endothermic.

If ΔH < 0 the reaction is exothermic.

For comparison purposes we need to refer ΔH to the same T and P To define a standard reaction enthalpysame T and P. To define a standard reaction enthalpy each component of the reaction must be in its standard state – the most stable form at 1 bar pressure and (usually) 25°C.

1 bar = 105 Pa 1 atm = 1.01325 bar


Reaction Enthalpy 1

Hess’s Law

The standard enthalpy change in any reaction can be expressed as the sum of the standard enthalpy changes, at the same temperature, of a series of reactions into which the overall reaction can be formally divided.

Combine chemical equations as if mathematical equations,

A + B CC + D E + F

F B + G

→→→

1

2

3

HHH

ΔΔΔ

q qe.g.

F B GA + D E + G

→→

3HH

ΔΔ

1 2 3H H H HΔ = Δ + Δ + Δ

Standard Reaction Enthalpy

oHΔ reaction enthalpy at 1 baro298o500

HH

ΔΔ

… and at standard T

or some other T


Reaction Enthalpy 2Standard (molar) enthalpy of formation o o

f fH HΔ ≡ ΔHeat of formation of a substance from its elements, all substances being in their standard state.

B d fi iti f ll l t oBy definition, for all elements of 0HΔ =

o oc cH HΔ ≡ ΔEnthalpy of combustion

ΔH° for total oxidation of a substance

e.g. C6H12O6 + 6O2 → 6CO2 + 6 H2O ΔcH° = -2808 kJ mol-1

Enthalpy of hydrogenationΔH° when an unsaturated organic compound becomes fully g p ysaturated

e.g. C6H6 + 3H2 → C6H12 ΔH° = -246 kJ mol-1

Enthalpy of atomization ≡ Bond dissociation enthalpyEnthalpy of atomization ≡ Bond dissociation enthalpyΔH° for the dissociation of a molecule into its constituent gaseous atoms

e.g. C2H6 (g) → 2C(g) + 6H(g) ΔH° = 2883 kJ mol-12 6

Bond strength ≡ single bond enthalpyAn average value taken from a series of compounds and often combined for a rough estimate


ge.g. ΔH°(C2H6) = ΔH°(C-C) + 6 ΔH°(C-H)

Temperature Dependence of ΔH°

2( ) ( )T

H T H T C dT∫

The temperature dependence of reaction enthalpies can be expressed in terms of the T dependence of the enthalpies of the reaction components:

1

2

1

2 1

2 1

( ) ( )

( ) ( )

h

pT

T

pT

H T H T C dT

H T H T C dT

C C C

= +

∴ Δ = Δ + Δ

Δ

∫

∫∑ ∑

products reactants wher e p p pC C CΔ = −∑ ∑

This general phenomenon is known as Kirchoff’s Law.

( )

( )

1

2

A B C D

A B C D

H T

H T

+ → +

+ →

Δ °

Δ↓

+↓

°↑ ↑

e.g.

( )( )

( )( )

2 1 2

1

( ) (A) (B)

( )

(C) (D)

p pH T C C T T

H T

C C T T

Δ ° = + −

+Δ °

+ +( )( )

( )2 1

1products reactants

(C) (D)

( )p p

p p

C C T T

H T C C T

+ + −

= Δ ° + − Δ∑ ∑

ass ming that the C al es are T independent


assuming that the Cp values are T independent.

Reactions at Constant Volume

( ) ( )r r products reactantsH U PV PVΔ = Δ + −

For solids and liquids ( ) , so 0PV H UΔ ≈ Δ ≈ Δ

For ideal gases ( ) gas

gas

,

so

PV n RT

H U n RT

Δ = Δ

Δ ≈ Δ + Δ

93 6 2 2 22 C H (g) O (g) 3CO (g) 3H O(l)e.g. + → +3 6 2 2 22(g) (g) (g) ( )g

( )5r r 2-H U RTΔ = Δ +

The relationship between ΔH and ΔU is particularly important when relating thermochemical enthalpies to molecular properties,

e.g. for a single bond energy ΔU = ΔH – RT

as seen in the case of O2(g) → 2O(g).

In practice, RT is usually so much smaller than ΔH that it is often ignored.


Enthalpies of Ions in SolutionEnthalpy of solution ΔH° for solution of a substance in a stated amount of solventEnthalpy of dilution ΔH° for dilution of a solution to a lower concentrationEnthalpy of solution to infinite dilution for an infinite amount of solvent

osolnHΔ

The enthalpy of formation for a species in solution can be found by combining with the of the gaseousoHΔ oHΔfound by combining with the of the gaseous species:

osolnHΔ o

fHΔ

o -1lHCl(g) H 75.Cl(aq) 14 kJ molHΔ = −→

o -11 12 22 2 fH (g) Cl (g) HCl(g 92.31 kJ m) olHΔ = −+ →

solnHCl(g) H 75.Cl(aq) 14 kJ molHΔ→

1 12 22 2H (g) Cl (g) HCl(aq)+ → o o o

f f soln-1

(ion)

167.45 kJ mol

H H HΔ = Δ + Δ

= −

for individual ions in solution can only be found if one is arbitrarily fixed. By convention this is H+(aq).

ofHΔ

( )of a

+122 qH (g) H (aq) He 0H +→ + Δ =–

( ) ( ) ( ) ( )o o o o+( ) ( ) ( ) ( )o o o of aq f aq f aq f aqCl HCl H HClH H H H+Δ = Δ − Δ = Δ–

The standard state for a substance in solution (not just ions) is a concentration of 1 mole solute in 1 kg solution (1 molal).


g ( )

Enthalpy of Formation of an Ionic SolidConsider individual steps in the formation of NaCl.

1. Na(s) → Na(g) ( )osubl NaHΔ

2. Na(g) → Na+(g) + e– ( )o NaH I RTΔ = +( )3. ½Cl2(g) → Cl(g) ( )o1

2 Cl-ClHΔ

4. Cl(g) + e– → Cl–(g) ( )oA ClH E RTΔ = − −

5. Na+(g) + Cl–(g) → Na+(aq) + Cl–(aq)≡ NaCl(aq) ( ) ( )o o

sol solNa ClH H+ −Δ + Δ

Na(s) + ½Cl2(g) → NaCl(aq) ( )of aqNaClHΔ

( ) ( ) ( )

( ) ( )

o o o1f aq subl 2

o oA sol sol

NaCl Na (Na) Cl-Cl

(Cl) Na Cl

H H I H

E H H+ −

Δ = Δ + + Δ

− + Δ + Δ

leading us to the enthalpy of formation of solid NaCl:

5'. Na+(g) + Cl–(g) → NaCl (s) ( )olattice NaClHΔ

Step 5 could be creation of solid NaCl instead of solution

g py

( ) ( ) ( )( )

o o o1f s subl 2

o

NaCl Na (Na) Cl-Cl

(Cl) NaCl

H H I H

E H

Δ = Δ + + Δ

+ Δ

Na(s) + ½Cl2(g) → NaCl(s)


( )A lattice(Cl) NaClE H− + Δ

A Born-Haber Cycle for NaCl

Na+(g) + e– + Cl(g)

( )A ClE RT+( )NI RT

Enthalpy changes can also be expressed in a diagram, e.g.

Na+(g) + Cl–(g)

Na(g) + Cl(g)

( )A

( )o NaHΔ

( )NaI RT+

Na(s) + Cl(g)

oHΔ

osolHΔ

( )subl NaHΔ

( )o12 Cl-ClHΔ

Na(s) + ½Cl2(g)

NaCl(aq)

olatticeHΔ

( )of aqNaClHΔ

( q)

NaCl(s)

( )of sNaClHΔ

Since H is a state variable, the sum of enthalpy changes around the cycle must be zero. Consequently, if all but one of the enthalpy changes is known, it can be readily calculated.

Thi i i l t t i H ’ L t ti t


This is equivalent to using Hess’s Law to sum reaction steps.

Spontaneous Change(So why do we need entropy anyway?)(So, why do we need entropy, anyway?)

direction of spontaneous

change

The direction of spontaneous change is that whichThe direction of spontaneous change is that which

leads to chaotic dispersal of the total energy

moves from a state of low intrinsic probability towards one of greater probability. g p y

Work is needed to reverse a spontaneous process.

We need a quantity – entropy – to describe energy dispersal, i.e. the probability of a state.p p y

Spontaneous processes are irreversible.They “generate” entropy

Reversible processes do not generate entropy – but they


may transfer it from one part of the universe to another.

Entropy 1Entropy is a state variable (property) which determines if a state is accessible from another by a spontaneous change.

Entropy is a measure of chaotic dispersal of energy.

The natural tendency of spontaneous change is towards states of higher entropy.

There are both thermodynamic (how much heat is produced?) and statistical definitions (how probable is aproduced?) and statistical definitions (how probable is a state?). They both become equivalent when statistics is applied to a large number of molecules.

Consider a falling weight which Co s de a a g e g t cdrives a generator and thus results in heat q being added to the reservoir (the surroundings).

Define a system variable S

reservoir

Define a system variable S

Use stored energy to restore the weight to its original height The reservoir gives up δq to

generator(surr) /dS q T= −δ

original height. The reservoir gives up δqrev to the system, and there is no overall change in the universe.

rev(sys) (surr) qdS dST

δ= − =


( y ) ( )T

Entropy 2In general, Equality for reversible

processes only(sys) (surr) 0(sys) (surr)

dS dSdS dS

+−

……

or for the system Clausius inequalityqdS δ

or, for the system, Clausius inequalitydST

…

For an isolated system, q = 0 hence 0SΔ …

Isothermal Processes /S q TΔ ==Isothermal Processes

e.g.

Trouton’s Rule:

rev /S q TΔ ==

( ) ( ) vapfus

m b

fusion vapHHS S

T TΔΔ

Δ = Δ == =

( ) -1 -1vap 85 J K molSΔ ≈=Trouton s Rule:

Can be used to estimate ΔHvap if Tb is known. Not good for associated liquids.

( )vap 85 J K molSΔ

Temperature Variation revq CdTδ =revq

( ) 2

1

T VV T

CS dTT

Δ = ∫( ) 2T P

P T

CS dTT

Δ = ∫S

liquid

gas

vHT

Δand1

P T T∫Absolute Entropy

0( ) (0)

T PCS T S dTT

= + ∫ T0

0

solid

qbT


0( ) ( )

T∫ T0

Entropy 3E t d d P b bilitEntropy depends on Probability.Consider the number of ways Ω of arranging n molecules between two sides (A and B) of a container. The probability PA that all molecules are on side A depends Aon the ratio of ΩΑ to the total number of arrangements.

A B

1A tot A 21 2Ω = Ω = =PA tot A 2

1A tot A 41 4Ω = Ω = =P

1A tot A 16

A tot A

1 16

1 2 2n n−

Ω = Ω = =

Ω = Ω = =

P

P

State A becomes less and less probable as n increases. Conversely, the probability of the less ordered, roughly evenly distributed states, increases.

A tot A

Since entropy is a measure of disorder, it follows that Sdepends on Ω.

Boltzmann equation lnS k= Ω


( ) x y x y x ySince x y , ln ln ln+= = +P P P P P PAND

The Second Law of Thermodynamics“An isothermal cyclic process in which there is a net conversion of heat into work is impossible.”

“No process is possible in which the sole result is the absorption of heat from a reservoir and its conversion intoabsorption of heat from a reservoir and its conversion into work.” It is possible to convert all work into heat!

“It is impossible for heat to be transformed from a body at a lower temperature to one at a higher temperature unless work is done.”

“The entropy of an isolated system increases during any natural process.” The universe is an isolated system.

ΔS( ) < 0 i ll d id d ΔS( ) + ΔS( ) > 0ΔS(sys) < 0 is allowed provided ΔS(sys) + ΔS(surr) > 0

“All reversible Carnot cycles operating between the same two temperatures have the same thermodynamic efficiency ”efficiency.“There is a state function called entropy S that can be calculated from dS = δqrev/T. The change in entropy in any process is given by dS ≥ δq/T, where the inequality refers to a spontaneous (irreversible) process ”

The 1st Law uses U to identify permissible changes of state.

The 2nd Law uses S to identify natural changes among the permissible ones

refers to a spontaneous (irreversible) process.


permissible ones.

The Third Law of Thermodynamics“If the entropy of every element in its stable state at T = 0is taken as zero, every substance has a positive entropy which at T = 0 may become zero, and does become zero for all perfect crystalline substances, including compounds.”for all perfect crystalline substances, including compounds.

Nernst Heat Theorem “The entropy change accompanying transformation between condensed phases in equilibrium, including chemical reactions, approaches zero as T → 0.including chemical reactions, approaches zero as T → 0.

Practical consequence: Set S(0) = 0 for elements by

0lim 0T

S→

Δ =

convention. Apply Nernst to determine S(0) for all else.

“It is impossible to reach absolute zero in a finite number of steps.”

The 1st Law says U cannot be created or destroyed.

The 2nd Law says S cannot decrease.The 2 Law says S cannot decrease.

The 3rd Law says zero Kelvin cannot be reached.


The Clausius Inequalityδ

Substitute into the 1st Law:Fundamental Equation

revw PdVδ = −revqdST

δ==Given and

All exact differentials, so path independent.

Fundamental Equation of ThermodynamicsdU TdS PdV= −=

But In general, exdU q w q P dV= δ + δ = δ −i.e. any path

( )

ex

ex

ex

q qTdS PdV q P dV

P PqdS dVT T

− = δ −

−δ= +

=

⇒

ex( ) 0P P dV−∴ …ex exIf , ; if , 0 0P P dV P P dV> > < <

Equal for

Even more generally,

Equal for reversible change

qdSTδ…Clausius Inequality

Conditions for: thermal equilibrium

mechanical equilibrium

( )surrsurruniv surr

P PT TdS dS dVT T

−−⎛ ⎞= +⎜ ⎟⎝ ⎠


equilibrium equilibrium

The Fundamental Equation of Thermodynamicsqδ

Reversible change but true for all paths since dV exact

Combine with revqdST

δ==dU q PdV= δ −

1dU TdS PdV

PdS dU dV

= −

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

=⇒

dV exactor dS dU dVT T

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

This fundamental equation generates many more relationships.U U∂ ∂⎛ ⎞ ⎛ ⎞Example 1: Comparison with

V S

U UdU dS dVS V

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠and

V S

U UT PS V

∂ ∂⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠⇒

Example 2: Consider that dU is exact and cross differentiate.

a Maxwell relationS V

T PV S

∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Example 3: cyclic rule

and again!

T S V

S T SV V T

P S P

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ and again!V V V

P TP T

P S PS T T

V P V VT PT V

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ∂ ∂∂ ∂ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠


P TP T ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠= α κ

How Entropy Depends on T and V1 P⎛ ⎞ ⎛ ⎞1 PdS dU dVT T

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

VV T T

U U UdU dT dV C dT dVT V V

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⇒1V

T

C UdS dT P dVT T V

∂⎡ ⎤⎛ ⎞= + + ⎜ ⎟⎢ ⎥∂⎝ ⎠⎣ ⎦S S∂ ∂⎛ ⎞ ⎛ ⎞

V T TT V V∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Compare withV T

S SdS dT dVT V

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

VCS∂⎛ ⎞ =⎜ ⎟1S UP⎡ ⎤∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎢ ⎥

VT T=⎜ ⎟∂⎝ ⎠

VV

CS dTT

Δ = ∫

T T

PV T V

= +⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

1P∫ ∫

0 for ideal gases T

( )2 1

1

ln /

TPS dV nR dVT V

nR V V

Δ = =

=

∫ ∫

For any substance, VCdS dT dVT

α= +

κ

For ideal gases,assuming CV is T independent

2 2ln lnVT VS C nRΔ = +


g ,

(eqs 3.7.4, 6.1.6) T independent

1 1V T V

Entropy Depends on T and P (not in text) 1 P⎛ ⎞ ⎛ ⎞

First problem: replace dU ; second problem: replace dV.

1 PdS dU dVT T

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

dU dH PdV VdPU d b th l d!dU dH PdV VdP= − −Use and both are solved!

1 VdS dH dPT T

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

But PP T T

H H HdH dT dP C dT dPT P P

∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1PC HdS dT V dP∂⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⇒ P

T

dS dT V dPT T P

⎛ ⎞= + −⎜ ⎟⎢ ⎥∂⎝ ⎠⎣ ⎦

Compare withP T

S SdS dT dPT P

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠unnatural variables

P T⎝ ⎠ ⎝ ⎠

P

P

CST T

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠1

T T

S H VP T P

∂ ∂⎡ ⎤⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

PP

CS dTT

Δ = ∫ 1T

VS dP nR dPT P

Δ = − = −∫ ∫( ) ( )l / l /S R P P R V VΔ

0 for ideal gases


( ) ( )2 1 2 1ln / ln /TS nR P P nR V VΔ = − =

Entropy Changes in Spontaneous ProcessesE t i t t f tiEntropy is a state function, so

ΔS(sys) = S2 – S1 independent of path

This can be used to calculate ΔS for an irreversible process.

Consider isothermal expansion of a gas from V1 to V2:

For the reversible case

( )2 1(sys) ln /S nR V VΔ = reversible and irreversible cases

(surr) (sys)S SΔ = −Δ( ) ( )S SΔ ΔFor the irreversible case

e.g. for free expansion, w = 0

(surr) (sys)S SΔ > −Δ

(surr) q wST T

Δ = − =

g p ,

Consider freezing of supercooled water at T < 273 K

⇒ (surr) 0, (univ) (sys) 0S S SΔ = Δ = Δ >

t 0°C ice 0°Creversiblewater, 0°C ice, 0°C

water, T ice, Tirreversiblerev. rev.

,ofus( )273(sys) (l) ln (s) ln

273 273P PH TS C C

T−Δ⎛ ⎞ ⎛ ⎞Δ = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

[ ]{ }fus ( ) 1( ) (273) (l) ( ) ( 273)H TS H C C TΔΔ Δ


[ ]{ }fusfus

( )(surr) (273) (l) (s) ( 273)P PS H C C TT T

Δ = = Δ + − −

Entropy of MixingC id th i i f t id lConsider the mixing of two ideal gases :

P, T, V1+V2

n1+n2

P, T, V1

n1

P, T, V2

n2

1 11 1 1 1 1

1 2 1 2

2 2

ln ln ln

l l l

V nS n R n R n RV V n n

V nS R R R

Δ = − = − = − χ+ +

Δ 2 22 2 2 2 2

1 2 1 2

ln ln lnS n R n R n RV V n n

Δ = − = − = − χ+ +

( ) ( )mix 1 1 2 2 1 2 1 1 2 2ln ln ln lnS n R n R n n RΔ = − χ − χ = − + χ χ + χ χ

lnS n RΔ χ χ∑In general mix tot lni ii

S n RΔ = − χ χ∑In general

This expression applies to the arrangement of objects (molecules) just as well as fluids (gases and liquids).For example, arrange N identical atoms in N sites in a crystal:

Compare with the arrangement of two types of atom, A and B.!/ ! 1 ln 0N N S kΩ = = = Ω =

!N ( )A BA B

! ln ! ln ! ln !! !N S k N N N

N NΩ = Δ = − −

Application of Stirling’s approximation ( )ln ! lnz z z z= −

( )l lk


leads to ( )config A A B Bln lnS kNΔ = − χ χ + χ χ

Using Entropy to Achieve Low T

( )2 1ln /PP P

CS dT C T TT

Δ = ≈∫ if CP is ≈ constant

To achieve lower temperatures, S must be reduced.

C S S f( )Choose some property X that varies with S, i.e. S = f(X,T).

This could be the pressure of a gas or, for example, the magnetic moment of a paramagnetic salt (whose energy varies with magnetic field).g )

1. Alter X isothermally. Entropy changes.

2. Restore X by a

,TT

SS X q T SX

∂⎛ ⎞Δ ≈ Δ = Δ⎜ ⎟∂⎝ ⎠

X

2. Restore X by a reversible adiabatic process.

3. Repeat cycle.

0, 0q S= Δ =

S

X1

X2curves coincide at0 K a consequence

T0

0 T

0 K, a consequence of the Nernst Heat Theorem


T0 T1

Heat Enginesf fA heat engine is a system capable of transforming heat into

work by some cyclic process.The 2nd Law states that an isothermal cyclic process can not produce net work.The efficiency of a heat engine is defined as the ratio of the work produced to the heat input:

H L L1q q qwq q q

−ε = = = −

high temperature reservoir @ TH

low temperature reservoir @ TL

engine qLqH

H H Hq q q

A heat pump is a heat engine in reverse. Work is needed to transfer heat from a lower to a higher temperature reservoir.

wout = qH - qL

PCarnot Cycle

ideal gasall steps reversible

isotherm @ TH

adiabatic all steps reversible expansion

isotherm @ TL

adiabatic compression


V

The Carnot Cycle

PA

BTH

CD

TL

l ( / ) 0A Bstep w q U

RT V VΔ

→

V

H B A AB

BC H L

L D C CD

ln( / ) 00 ( )

ln(

A B

/ ) 00 ( )

BC DD

C

A

V

nRT V V wU C T TnRT V V wU C T T

− −Δ − −− −Δ

→→→→ DA H L

H L A B cyc

0 ( )( )Total ln( / )

D0

A VU C T TnR T T V V wΔ −

−→

−

t H L A B L( ) ln( / )w nR T T V V T−out H L A B L

in H A B H

( ) ln( / ) 1ln( / )

w nR T T V V Tq nRT V V T

ε = = = −

H L( )T TT−

ε =for best efficiency,

maximize THi i i T


HT minimize TL

Entropy Changes in the Carnot Cycle

P1

2TH cycle

area of plot

w d

P

P V

V

− =

=∫

34

TL

cycleq=

VIsothermal expansion (1 → 2):

0 0 0qU w q SΔ = − = > Δ =⇒ >⇒0 0 0U w q ST

Δ = = > Δ =⇒ >⇒

Isothermal compression (4 ← 3): 0SΔ <

Adiabatic steps (2 → 3 and 1 ← 4): 0 0q S= ⇒ Δ =

S23 ( )

( )tot H 2 1

L 4 3

q T S ST S S

= −

+ −

14

TT

( )( )H L 2 1

cycle

area of plotST T S S

wT

= − −

== −


T THTL

Spontaneous ChangeFor a system in thermal equilibrium with its surroundings,

At constant volume:

Clausius inequalityqdSTδ…

At constant volume:

no work

isolated system( )00

Vdq dUTdS dU

dS

=

− …0d 0

At constant pressure:

isolated system( )( )

,

,

0

0U V

S V

dS

dU

d …0

„

PV work only

( ) ,

00

P

S P

dq dHTdS dH

dH

=− …0

„

For convenience, define:

A U TSG H TS

= −= −

dA dU TdS SdTdG dH TdS SdT

= − −= − −

Then the conditions for spontaneous change become:

G H TS dG dH TdS SdT

( ) , 0T VdA „ ( ) , 0T PdG „


Helmholtz Energy F (or A)H l h lt H l h lt f H l h lt f ti

0dA dU TdS= −

Helmholtz energy; Helmholtz free energy; Helmholtz function; Maximum work function (F for Function or A for Arbeit)

For spontaneous change at constant T and V0dA dU TdS= − „

Note that it is the total function A that tends to a minimum; this is not the same as minimizing U and maximizing S.

dU q w= δ + δ TdS qδ…

dU TdS w+ δ„

Maximum WorkCombine and

⇒ equality for ibl h

( )w TdS dU−δ −„

max rev rev rev( )w TdS q w w−δ = − δ − δ = −δ

Word done bythe system

reversible change

rev rev( )TdA dU TdS dU q w= − = − δ = δBut

A system does maximum work when it is operating reversibly.

Therefore, for macroscopic changes

maxw A T S U− = −Δ = Δ − Δ constant T

(-w) can be more or less than ΔU according to the sign of ΔS.F S 0 h t fl i t th t t f l th t k


For ΔS > 0, heat flows into the system to fuel the extra work.

Gibbs Energy GGibbs free energy Gibbs functionGibbs free energy, Gibbs function

Very important in chemistry since it tells whether a particular reaction can proceed at a given T and P.

For spontaneous change,

( ) , 0T PdG „ products reactants 0G G GΔ = − „

( ) ( ) ( )G T H T T S TΔ = Δ − Δo o o

for reactions can be calculated from tabulated data,T PGΔ

If ΔH is and ΔS is then ΔG the reaction proceeds-ve +ve < 0 at all temperatures+ve -ve > 0 at no temperatures-ve -ve … if T < ΔH / ΔS+ve +e … if T > ΔH / ΔS

,

( )( )

T

T P

dG dH TdSdG dU PdV TdS

PdV TdS

= −= + −

δ δ

Maximum Work:

rev rev

rev

max (non- )

q w PdV TdSw PdVw PV

= δ + δ + −

= δ +

= δ


max (non- )w PV GΔ = Δ constant T, P

Basic Thermodynamic Relations 1L dU PdVδ (1)

Definitions H U PV= + (3)K

Laws

rev /dU q PdVdS q T

= δ −= δ

(1)(2)

K

K

Fundamental Equations

A U TSG H TS

= −= −

(4)(5)

K

K

dU TdS PdVdH TdS VdPdA SdT PdV

= −= +

= − −

(6)(7)(8)

K

K

K

U UT P∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

Partial Differentials

dG SdT VdP= − + (9)K

V S

P S

T PS VH HT VS P

= = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

V T

A AS PT VG GS V

∂ ∂⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟


P T

S VT P⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Basic Thermodynamic Relations 2

(10)K

Maxwell Relations

From (6)S V

T PV S

∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(11)

(12)

KFrom (7)

From (8)

S P

T VP SS P

∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ (12)

(13)

K

K

From (8)

From (9)

T V

T P

V TS VP T

⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Thermodynamic Equation of State

(6)dU TdS PdVU S

= −

∂ ∂⎛ ⎞ ⎛ ⎞

K

T T

U ST PV V

∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

U PT P∂ ∂⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟Substituting (12)

T VT P

V T= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

U T PV

∂ α⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ κ⎝ ⎠ ⎝ ⎠

Substituting (12)


TV∂ κ⎝ ⎠ ⎝ ⎠

How Free Energy Depends on T

G H TSG H T S

dG VdP SdT

= −Δ = Δ − Δ

= −

(1)(2)(3)

K

K

K

definition

constant T

fundamental eqn.

P

G G HST TG G

∂ −⎛ ⎞ = − =⎜ ⎟∂⎝ ⎠∂⎛ ⎞

from (1)

2( / ) 1

P

G G HT T TG T G G

T T T T

∂⎛ ⎞ − = −⎜ ⎟∂⎝ ⎠∂ ∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(4)

(5)

K

KBut

∴

2P PT T T T⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

2( / )

P

G T HT T

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠Gibbs-Helmholtz Equation

substitute (4) in (5)

( / )(1/ ) P

G T HT

∂⎛ ⎞=⎜ ⎟∂⎝ ⎠

alternative form:

2( / )G T H∂ Δ Δ⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

By applying the Gibbs-Helmholtz equation to both reactants and products of a chemical reaction,

⇒


2PT T⎜ ⎟∂⎝ ⎠

How Free Energy Depends on PG∂⎛ ⎞

2

2 1 1

T

G VP

G G G VdP

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

Δ = − = ∫For solids and liquids V does not change much with P, so

F f t

2 1 2 1( ) ( ) ( )G T G T V P P≈ + −

nRTVFor a perfect gas VP

=

2 1ln( / )TG nRT P PΔ =

where G° is the standard free energy defined at P°=1 bar

( ) ln( / )G G T nRT P P= +o o

The chemical potential for a pure substance is the molar Gibbs energy:

GGn

μ = =

∴ For a perfect gas

ln( / )RT P Pμ = μ +o o like eq 5.3.5


The Chemical PotentialSo far, our thermodynamic relations apply to closed systems. However, G, U, H, etc. are extensive properties.

Partial molar quantities are used to describe quantities which depend on compositiondepend on composition.

, , j i

ii T P n

i i

JJn

J n J≠

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

= ∑

, , , , ,J U H S A G V=

i ii

J n J∑

1 2 1 2, , ... , , ...T n n P n n

G GdG dP dTP T

∂ ∂⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞

In general

2 1 3

1 21 2, , ... , , , ...T P n T P n n

G Gdn dnn n

⎛ ⎞ ⎛ ⎞∂ ∂+ + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

K

G⎛ ⎞∂Chemical Potential

, , j i

ii P T n

Gn

≠

⎛ ⎞∂μ = ⎜ ⎟∂⎝ ⎠

Fundamental Equations i idU TdS PdV dn= − + μ∑i idH TdS VdP dn= + + μ∑

U H A⎛ ⎞ ⎛ ⎞∂ ∂ ∂⎛ ⎞μ = = = ⎜ ⎟⎜ ⎟ ⎜ ⎟

etc.

⇒


,, ,i

i V Ti iV S P Sn n nμ = = = ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

⇒

Free Energy of Mixing 1Consider the mixing of two ideal gases A and B at constant Tand P.

Before:

h1 A A B B(pure) (pure)

( ) l ( / )

G n n

RT P P

= μ + μo

{ }{ }

A A A A 0

A A 0 A

ln( / )

ln( / ) ln

G n RT p P

n RT P P RT

= μ +

= μ + + χ

o

o

where A A 0(pure) ln( / )RT P Pμ = μ +o

After mixing:

{ }A A 0 A

A A A A

2 A B 1

ln( / ) ln

(pure) ln

n RT P P RT

n n RTG G G G

μ + + χ

= μ + χ= + < negative

Change:

( )

mix 2 1

A A B B

A A B B

ln lnln ln

G G Gn RT n RTnRT

Δ = −= χ + χ

= χ χ + χ χ

( )mix A A B Bln lnG nRTΔ = χ χ + χ χ

mix 0GΔ < Mixing is spontaneous.

Entropy of Mixing ( )mix mix ,/ lni

i iP nS G T nRΔ = − ∂Δ ∂ = − χ χ∑Enthalpy of Mixing mix mix mix 0H G T SΔ = Δ + Δ =

Si il l 0U VΔ Δ f id l d l ti


Similarly, mix mix 0U VΔ = Δ = for ideal gases and solutions

Free Energy of Mixing 2F bi t ΔG i t it i i

A B 0.5χ = χ =For a binary system, ΔGmix is at its minimumand ΔSmix at its maximum when

0

ΔSmix ΔGmix

0 0.2 0.4 0.6 0.8 1χΑ

0

χΒ 01

0 0.2 0.4 0.6 0.8 1χΑχΒ 01

( ) ( )A B mix(pure) (pure)

l lG G G G= + + Δ

A BConsider an equilibrium between two ideal gases

Equilibrium and Mixing:

( ) ( )A A B B A A B Bln lnn nRT= χ μ + χ μ + χ χ + χ χ

0

G G The equilibrium

Gmix

G Gpure

Gtotal

qcomposition is determined by the minimum in Gtotal.


composition BA

Reaction Equilibrium 1A BC id A BConsider

Suppose an amount dx of A turns into B.Then ( ) A A B B A B,T PdG dn dn dx dx= μ + μ = −μ + μ

G∂⎛ ⎞ lik h t iB A

,T P

Gx

∂⎛ ⎞ = μ − μ⎜ ⎟∂⎝ ⎠⇒

µA and µB depend on composition, and therefore change during the reaction.

like what is done in 2.5.10

If µA > µB the reaction proceeds from A → B.If µA < µB the reaction proceeds from B → A.

G slope Gx

¶=¶

A Bμ = μ

At equilibrium

x BA

0 1eq

( ) ( )( ) ( )( )

eq eqA A B B

eq eqA B B A rxn

ln / ln /

ln /

RT p P RT p P

RT p p G

μ + = μ +

= μ − μ = Δ

o o o o

o o o

( )eqB exp /p G RTΔ o see 5 3 6


( )Brxneq

A

exp /G RTp

= −Δ see 5.3.6

Reaction Equilibrium 2A reaction such as 2A + 3B → C + 2D can be written as

0 = –2A –3B +C +2D

or completely generally as 0 R 0i i= ν =∑p y g y i ii

∑

Then,

i iiT P

G∂⎛ ⎞ = ν μ⎜ ⎟∂ξ⎝ ⎠∑

and at equilibrium, 0i ii

ν μ =∑eqiln 0pRT

⎛ ⎞ν μ + ν =⎜ ⎟∑ o

rxneq

ln 0

ln 0

i i ii

ii

i

RTP

pG RTP

ν

ν μ + ν⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞Δ + =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

∑

∏

o

o

oq⎣ ⎦

iν⎛ ⎞

rxn lnG RT KΔ = −o

eq

ii

i

pKP

ν⎛ ⎞= ⎜ ⎟⎝ ⎠

∏ o


Equilibrium Constants

( ) ( )Y Zeq eq/ /

y zi

ip P p PpK

ν⎛ ⎞⎜ ⎟∏

o o

o

aA bB yY zZ y z a b+ + Δν = + − −Consider

Thermodynamic Equilibrium Constant

( ) ( )eq eq

eq A Beq eq/ /

ia b

i

pKP p P p P

⎛ ⎞= =⎜ ⎟⎝ ⎠

∏o

o o o

( ) ( )Y Zy zp p

Some books still refer to the pressure equilibrium constant

which in general has( ) ( )( ) ( )

Y Zeq eq

A Beq eq

p a b

p pK

p p= which in general has

pressure units

The equilibrium constant can also be expressed

( ) ( )( ) ( )

C tot D toteq eq tot

A tot B toteq eq

/ /

/ /

y z

xa b

P P P P PK KPP P P P

Δνχ χ ⎛ ⎞= = ⎜ ⎟⎝ ⎠χ χ

o o

o

oo o

in mole fractions:

eq eq

/i i ip n RT V c RT= =

i ii ip c RT c RTK K

ν ν ∑ν⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∏ ∏

oo

and in concentrations:

???? confusing notation

eq eqc

i iK K

P P P⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∏ ∏o o o

( ) ( )( ) ( )

eq eq[ ]/ [ ]/

[ ]/ [ ]/

y z

c a b

C c D cK

A B=

o o

o owhere for this example

notation


( ) ( )eq eq

[ ]/ [ ]/A c B c

Equilibrium Calculations

2 NO + Cl2 → 2NOCl

2 1 2

2 1 0

− − +stoichiometry

initial moles

( )

( )

2 1 0

2 1 1 2

2 1 1 2

a a a

a a a

− −

− −

initial moles

during reaction

mole fraction ( )

( )3 3 3

2 1 1 23 3 3

a a a

a P a P a Pa a aP P P

− − −

− −− − −o o o

mole fraction

partial pressure

( ) ( )

2

22

3 3aa a a

K

⎧ ⎫⎨ ⎬− −⎩ ⎭( )

( )( )

( )( )

( )2 312 1 13 3

xKaa a

a a

⎩ ⎭= =−⎧ ⎫− −⎧ ⎫

⎨ ⎬ ⎨ ⎬− −⎩ ⎭⎩ ⎭

P⎛ ⎞ox

PK KP

⎛ ⎞= ⎜ ⎟

⎝ ⎠

oo

The value of a at equilibrium (and thus the equilibrium composition of the reaction mixture) depends on pressure


composition of the reaction mixture) depends on pressure.

Temperature Dependence of KP

2( / )

P

G T HT T

∂ Δ Δ⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

rxn lnG RT KΔ = −o

Gibbs-Helmholtz

substitute rxn

r2

(ln )P

HKT RT

Δ∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

o

van’t Hoff Equation

r(ln )(1/ ) P

HKT R

Δ∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠

o

ln K

slope HR

Δ= −

o

or

1/T1/T

Exothermic reactions: ΔH < 0 so K falls with increasing T.Endothermic reactions: ΔH > 0 so K rises with increasing T.

2

1 2 1

( ) 1 1ln( )

K T HK T R T T

⎧ ⎫ ⎡ ⎤Δ= − −⎨ ⎬ ⎢ ⎥

⎩ ⎭ ⎣ ⎦

o

Integrating,

H SΔ Δo o

lG RT KΔ o


ln ( ) H SK TRT R

Δ Δ= − +or from rxn lnG RT KΔ = −o

Pressure Dependence of Equilibrium

( )/K K P P−Δν

= o oBut

( )exp /K G RT= −Δo o 0T

KP

⎛ ⎞∂=⎜ ⎟∂⎝ ⎠

o

Since , ???????

( )( )

/

ln ln ln /

ln ln ln

x

x

K K P P

K K P P

K P P

= − Δν

= − Δν + Δν

o o

o o

But,

lnln

x

T

KP

∂⎛ ⎞ = −Δν⎜ ⎟∂⎝ ⎠

ln xK V∂ Δν Δ⎛ ⎞⎜ ⎟ ideal gas

???????

x

TP P RT⎛ ⎞ = − = −⎜ ⎟∂⎝ ⎠

ideal gas

Although the thermodynamic equilibrium constant does not depend on pressure the K for mole fraction does if 0Δν ≠depend on pressure, the K for mole fraction does if 0Δν ≠

The equilibrium composition depends on pressure if 0Δν ≠

Le Chatelier’s PrincipleA system at equilibrium, when subjected to a perturbation, responds in a way that tends to minimize the effect.


Equilibria Involving Condensed Matter

3 2CaCO (s) CaO(s) CO (g)+e.g.

( ) ( ) ( )3 2CaCO CaO COT P

G∂⎛ ⎞ = −μ + μ + μ⎜ ⎟∂ξ⎝ ⎠ ,T P⎝ ⎠

( ) ( ) ( )( ) ( )

22 2 CO

3 3

CO CO ln /

CaCO CaCO

RT p Pμ = μ +

μ ≈ μ

o o

o

where

but

( ) ( )CaO CaOμ ≈ μo

,

0T P

G∂⎛ ⎞ =⎜ ⎟∂ξ⎝ ⎠At equilibrium,

,

( )CO2

eqln / 0G RT p PΔ + =o o

( )CO2

eqK p P=o o

⇒

K depends only on the partial pressures of the gaseous reaction components.

A special case is the evaporation of a liquid: L(l) G(g)

( )eqGK p P=o o

vap2

ln ln HK PT T RT

Δ∂ ∂= =

∂ ∂

oo


2T T RT∂ ∂

Phase Equilibria

Consider a closed system of a single component. The chemical potential determines which phase is stable at a particular T and P. µ tends to a minimum.At the melting point T µ(s) = µ(l)At the melting point Tm, µ(s) = µ(l)At the boiling point Tb. µ(l) = µ(g)These points depend on temperature and pressure.

dG VdP SdTdG VdP SdT= −

P

ST

∂μ⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

µ

liquidsolid

T

VP

∂μ⎛ ⎞ =⎜ ⎟∂⎝ ⎠

Tm

gas

liquid

TbmT

b

µphase 1

higher pressure

T

phase 2

T '


Tx Tx' phase transition at higher T

The Clapeyron Equation

Consider two phases α and β in equilibrium:

If small changes in T and P are made such that α and β are still in equilibrium:

( ) ( ), , , ,T P T Pμ α = μ β

still in equilibrium:( ) ( )

[ ] [ ]

, , , ,( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

d T P d T PS dT V dP S dT V dP

V V dP S S dT

μ α = μ β

− α + α = − β + β

β β[ ] [ ]( ) ( ) ( ) ( )V V dP S S dTα − β = α − β

dP S HdT V T V

Δ Δ= =

Δ Δ

Melting

( )H T P⎧ ⎫Δ

m

m m

HdPdT T V

Δ=

Δ

Integrating, m m 22 1

m m 1

m

( )ln( )

H T PP PV T PH TV T

⎧ ⎫Δ− = ⎨ ⎬Δ ⎩ ⎭

Δ Δ≈

Δ

⇒ Tm increases with pressure not for water!

m mV TΔ

m 0HΔ > m 0VΔ >and usually


The Clausius-Clapeyron Equation

vap vap

vap ( )H HdP

dT T V T V gΔ Δ

= ≈Δ

Vaporization

( ) /V g RT P=Assuming the vapour is an ideal gas,

vap2

ln Hd PdT RT

Δ=

1 1HP Δ⎛ ⎞ ⎡ ⎤vap2

1 2 1

1 1lnHP

P R T TΔ⎛ ⎞ ⎡ ⎤

= − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

Integrating,

The normal boiling point is the temperature at which the vapour pressure becomes standard, i.e. 1 bar.

Sublimation solid ↔ gasThe liquid is not stable at any temperature.q y p

Triple Point: solid, liquid and gas are all in equilibriumThis happens at the pressure where the sublimation t t d th b ili t t i idtemperature and the boiling temperature coincide.At the triple point,

vapour pressure of liquid = vapour pressure of solidTtriple and Ptriple are fixed.


Ttriple and Ptriple are fixed.

The Phase RuleH i t i i bl d d t d ib f llHow many intensive variables are needed to describe fully a system of C components and P phases?

Two for temperature and pressure.How many for the composition of each phase?

but since the phases are in equilibrium,

How many for the composition of each phase?Take mole fractions of each component in each phase

( )1P C⇒ × − C-1 because for each phase 1iχ =∑

( ) ( )phase 1 phase 2μ = μ =Kp q ,

(P – 1)C variables are redundant

∴ Number of independent ( ) ( )1 1P C P C C P∴ Number of independent concentration variables

∴ Total number of variables (degrees of freedom)

( ) ( )1 1P C P C C P= − − − = −

2F C P= − +(degrees of freedom)

Phase: A state of matter that is uniform throughout, in both chemical composition and physical state.

Component: The number of components is theComponent: The number of components is the minimum number of independent species necessary to define the composition of all phases in the system.Reactions and phase equilibria must be taken into account.


Phase Diagrams of Pure Materials2 h 1 3witF C P C F P⇒= − + = = −

For single phase regions there are 2 degrees of freedom.

For phase boundaries there is 1 degree of freedom.

At the triple point there is no freedom.

Pressuree.g. CO2

Pressure

supercritical

Critical pointliquid

pfluid

Pc

P3 Triple pointsolid

gas

q

Temperature

g

T3 Tc


p

The Phase Diagram of Water

Pressure / atm

C iti l

FLUID

218

1 0 Triple

Critical point

ICELIQUID

1.0 Triple point

VAPOURSTEAM

Temperature /°C0 100 374

There are other solid phases at much higher pressures.


Kinetics and Mechanismto determine / describe / predict the course of reactions

qualitatively: identify reactants, products, intermediatesidentify elementary reaction steps

quantitatively: measure reaction rates ⇒ rate constantsexplore effects of T, P, [H+], …relate rate constants to molecular propertiesdevelop theories ⇒ make predictionsdevelop theories ⇒ make predictions

Rate data (growth of products, decay of reactants) are the basic experimental input to reaction kinetics.The shapes of kinetic data plots depend on rate constantsThe shapes of kinetic data plots depend on rate constants and concentrations. Their functional form ⇒ reaction order

conc.A + B → C


time

Empirical Chemical Kineticsf

[A

e.

]rat 1 [B] 1 [Y] [Z

g. f A + 2B 3Y + o Zr

]e d d d d= − = − = =

→

time dependence of reactant and product concentrations

rat2 3

edt dt dt dt

= = = =

Th t t f ti

In general, a chemical equation is written 0 Xi ii

= ν∑( ) ( )0tThe extent of a reaction

(the advancement)( ) ( )0i i

i

n t n−ξ =

ν

For an infinitesimal advancement dξ the concentration f h t t/ d t h b [ ]d dξof each reactant/product changes by [ ]Xi id d= ν ξ

[ ]X1 i

i

dddt dtξ

= =ν

rateBy definition,

order in B

t t l d

Reaction rates usually depend on reactant concentrations,

[ ] [ ] A Bx yk=rate

rate constant total order = x + y

In elementary reaction steps the orders are always integral, but they may be fractional in multi-step reactions.


The molecularity is the number of molecules in a reaction step.

Measurement of Reaction Rates

Sampling of reaction mixture, followed by chemical analysis, chromatography, spectroscopy

Quenching of whole reaction or sample, followed by …

Matrix isolation is a particular quenching technique used to study otherwise short-lived species.

FlowFlowA B

mixing chamberThe time between mixing mixing chamber

detector

and measurement is varied by changing the distance or the flow rate.

Real time analysis involves measurement of a quantity that varies throughout the reaction. This could be:

a general property of the systema general property of the systempressure, volume, conductance, optical rotation, …a molecular property: usually by spectroscopy

Accurate initiation is necessary for real time experiments.


Simple Rate Laws

Zero order: 0

0 0

[A]

[A] -[A]t

d kdt

k t

− =

=

[A]

12

0

0

[A]2

tk

= t0

[A]d!st order:

1

1

0

[A] [A]

[A] [A] e k tt

d kdt

−

− =

=

[A] ln[A]

12

1

ln 2tk

=t0 t0

[A]d[A]

half-life

2nd order: 22

0

[A] 2 [A]

[A][A]1 2 [A]t

d kdt

k t

− =

=+

t0

1

2 0

-1 -10 2

1 2 [A]

[A] [A] 2

1t

k t

k t

t

+

= +

=

[A]-1


12

2 02 [A]t

k t0

Second-order Kinetics: [A] ≠ [B]

A B products + k⎯⎯→

[ ], rat ]e A [Bda kab a bdt

= − = = =

( )( )0 0

dt

k a x b x= − −

0But sinc ,e dx daa a x= − = −

( )( )0 0 dx k a x b xdt

∴ = − −

0 But sinc ,e a a xdt dt

( )( )0 00

1 1 1

x

x

dxkta x b x

⎧ ⎫= ⎨ ⎬− −⎩ ⎭

⎧ ⎫

∫

( ) ( ) ( )

( )

0 0 0 00

0

1 1 1 d

1 ln

xa b a x b x

a b

⎧ ⎫−= −⎨ ⎬− − −⎩ ⎭

⎧ ⎫= ⎨ ⎬

∫

( )0 0 0a b a b⎨ ⎬− ⎩ ⎭

( )00 0

0

ln lna a a b ktb b

⎛ ⎞⎛ ⎞ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠or more usefully,


0b b⎝ ⎠ ⎝ ⎠

Determination of Order 1I t l M th dIntegral MethodsTest the data against an appropriate integral rate law.e.g. ln[A] vs t or 1/[A] vs t.

Differential Methodse e a e odsPlot the rate directly:

log log log log

x yka bk x a y b

ρ =ρ = + +

Determine initial slopes or tangents from a single curvep g gfor different initial concs.

a(t) a(t)

t t

and plot versus log ρconcentration:

slope = x


log a0

Determination of Order 2H lf lif M th dHalf-life MethodZero-order: the reactant is used up in two half-lives.

1st-order: the half-life is constant in time:

2 d d th h lf lif i ith ti

12 1ln 2 /t k=

2nd order: the half-life increases with time

12

1

10

2 1( 1)

n

ntn ka

−

−

−=

−In general,

a(t) a(t)

2nd order1st-order

t½ 2t½ t½ 2t½

Isolation Method together with one of the other methodsAll except one reactant is added in large excess, so that their concentrations do not vary significantly. Then the other reactant has pseudo-first order kinetics:

2[A][B] [B for] [A] [B]kρ = = λ


2[ ][ ] [ ] [ ] [ ]ρ

Data Analysis“Classical” methods of data analysis are often useful to explore the order of reactions, or to display the results (e.g. a semi-log plot to demonstrate exponential decay).

However these methods should be avoided forHowever, these methods should be avoided for quantitative data analysis, since errors (and thus weighting) can be distorted.

e.g.

[A] [A]-1

t0 t0

Modern data analysis uses computer methods for direct curve fitting, e.g. by chi-square minimization.

22 expt theory

tlχ

-æ ö= ç ÷è øådata points exptl. errorçè ø

A good fit has χ2/(no. of degs. freedom) = 1.


Temperature Dependence

a /e E RTk A −= aln ln ,Ek ART

= -

The Arrhenius “law” is an empirical description of the Tdependence of the rate constant:

⇒RT

The pre-exponential factor is often interpreted as a collision rate. Collision theory predicts

ln kCollision theory predictsT½ dependence for A.

1/T0

The exponential factor describes the fraction of collisions with sufficient energy for reaction, as

n(E)

predicted by the Boltzmann distribution:

EEa

The activation energy is defined by 2a

lnd kE RTdT

=

Curvature in the Arrhenius plot is often attributed to tunneling, but other reasons include complex reactions and T dependence of the


other reasons include complex reactions, and T dependence of the pre-exponential factor, which arises naturally in most theories.

Opposing Reactions – Relaxation

da dzk k

1

1

A Zkk−

At equilibrium

1 1 0da dzk a k zdt dt−− = − = =

1 eq 1 eqk a k z−= eq 1

eq 1

[Z][A]

kKk−

= =⇒

If equilibrium is disturbed by an amount x, so that

eq eq( ) ( ), ( ) ( ), dz dx daa t a x t z t z x tdt dt dt

= − = + = = −

( ) ( )

( )

1 eq 1 eq

1 1

da k a x k z xdt

dxk k xd

−

−

− = − − +

= − + =conc

( )

( ){ }1 1

0 1 1expdt

x x k k t

−

−= − +

( ) ( ) ( )1 1eq 0 eq e k k ta a a a −− +− = −

aeq

zeq

( ) ( )q q

( ) ( ) ( )1 1eq eq 0 e k k tz z z z −− +− = − time

This exponential relaxation of concentrations is the basis for several jump methods of studying fast reaction kinetics


for several jump methods of studying fast reaction kinetics.

Parallel Reactions – Competitionk

Consider a molecule that can react by two different routes: A

C

Bkb

kc

Define a = [A] b = [B] c =[C]

The overall decay of A depends on both reactions:

( ) ( )b cb c b c 0e

k k tda k a k a k k a a adt

− +− = + = + =⇒

Define a [A], b [B], c [C].

The rate of formation of each product depends on both rate constants:

( )b ce k k tdb k a k a k d− + ⎫= = ⎪⎪ ∫( )b c

b b 0b b

ccc c 0

e//e k k t

k a k a k adt kb db dtdtdc c k dc dtk adtk a k adt

− +

= = ⎪⎪ = = =⎬⎪= = ⎪⎭

⇒ ∫∫

b B yield of B[ ]kb

c

B yield of BC yie

[ ]ld of C[ ]

kk

= =

This is the basis for competition kinetics, whereby an unknown rate constant is determined from a known rateunknown rate constant is determined from a known rate constant and the ratio of competitive products.

The above treatment assumes kinetic control. In contrast, at equilibrium,

eq eq eq b b -c[B] [C] [B] K k kK K= = = =


q q q b b cb c

eq eq eq c c -b

, ,[A] [A] [C]

K KK k k

= = = =

Consecutive ReactionsSimplest case - two first-order stepsSimplest case - two first-order steps

1 2BA Ck k⎯⎯→ ⎯⎯→

1da k adt

= − 10 e k ta a −=

1 2

dtdb k a k bdtdc k b

= −1 2

1 2

10

2 1

2 1

e e

1

k t k t

k t k t

kb ak k

k k

− −

− −

⎡ ⎤= −⎣ ⎦−

⎡ ⎤+⎢ ⎥2k b

dt= 1 22 1

02 1 2 1

1 e ek t k tc ak k k k

= − +⎢ ⎥− −⎣ ⎦

0da db dcdt dt dt

+ + = 0a b c a+ + =

For k1 >> k2 the kinetics can be considered as two steps:1. At short times b increases as a falls. 2. At longer times (k1t >> 0), c increases as b falls.g ( 1 ),

20 e k tb a −→( )2

0 1 e k tc a −→ −

conc.b

ca


t

The Steady-State Approximationk k

( )2 11 1 21 2 1 0

2 1 2 1

k k tk tdb k kk a k b k a e edt k k k k

− −− −⎡ ⎤= − = +⎢ ⎥− −⎣ ⎦

1 22 1B rC o A fk k k k⎯⎯→ ⎯⎯→

2 1 2 1dt k k k k⎣ ⎦

conc.ca

b

After the induction period, i.e. for k2t >> 0,

t

da dc db k⎡ ⎤1 1 1 1

1 0 1 0 1 02 1

, ,k t k t k tda dc db kk a e k a e k a edt dt dt k k

− − − ⎡ ⎤= − ≈ ≈ − ⎢ ⎥−⎣ ⎦

,db da dcdt dt dt

Although b is not constant, it changes at a much smaller rate than a or c.This is the essence of the steady-state approximation

dt dt dt


This is the essence of the steady state approximation.

An Example of a Complex MechanismConsider the overall reaction

2NO + O2 → 2NO2

It is found experimentally to be third order overall, second order in NO first order in O It is much too fast to be aorder in NO, first order in O2. It is much too fast to be a termolecular process

12 2

-1

2NO N O kk

Test the mechanism:

22 2 2 2N O O 2NOk+ ⎯⎯→

Apply the steady-state approximation to [N2O2]

[ ] [ ] [ ] [ ][ ]

[ ] [ ]

22 2 1 1 2 2 2 2 2 2

21

2 2

N O NO N O N O O 0

NON O

d k k kdt

k

−= − − =

=[ ] [ ]

[ ] [ ][ ] [ ] [ ][ ]

2 21 2 2

21 2 2

2 2 2 2 21 2 2

N OO

NO O1 = NO N O O2

rateO

k k

k kd kdt k k

−

−

=+

= =+ [ ]

In the limit of k-1 >> k2 [O2], [ ] [ ]21 22

1

r t Oa Oe Nk kk−

=

This is an example of a pre equilibrium mechanism


This is an example of a pre-equilibrium mechanism.

Atom/Radical Combination Reactions

UIn low pressure gases, atoms seem to react slower than expected because the combination product falls

r(A-B)

De

combination product falls apart in the period of a molecular vibration (~ 10-14 s).

A + B → (A-B)* → A + B r(A-B)( )

If A and B are polyatomic radicals, (A-B)* may live longer (e.g. 10-9 s), by distribution of De among different vibrational modes.

For efficient reaction a third body is needed:

keff

For efficient reaction a third body is needed:

1.2.

AB*AB

A BM B* A M

++ → +

2

1

[AB] [M][ ]

[A][B][M]

AB*d kdt

kk

ρ = =

= ⋅

Pressure2

1 2

1 2

1 2

[M][M]

[M] [A][B][M]

kk k

k kk k

−

−

= ⋅+

= ⋅+ The effective rate constant

depends on pressure ([M])


depends on pressure ([M]).

Enzyme Kinetics 1Another example of a pre-equilibrium mechanism is one used to model the kinetics of enzyme action:

1 2E + S ES E + P k kk

⎯⎯→1k−

enzyme substrate bound enzyme productenzyme regenerated

Applying the steady-state approximation to the bound state,pp y g y pp

( )

1 1 2

1 1

[ES] [E][S] [ES] [ES] 0d k k kdt

k k

−= − − =

( )1 10

1 2 1 2

[ES] [E][S] [E] [ES] [S]k kk k k k− −

= = −+ +

1 0[E] [S][ES][S]

kk k k

=+ +

Rearranging:1 2 1[S]k k k− + +

1 2 02

1 2 1

[E] [S][P] [ES][S]

k kd kdt k k k−

ρ = = =+ +

Rate:

2 0

M

[E] [S][S]

kK

ρ =+

Michaelis-Menten Michaelis constant

1 2M

1

k kKk

− +=with


Michaelis Menten Michaelis constant

Enzyme Kinetics 2

enzyme

substrate

bound enzyme

productproduct

Write the Michaelis-Menten equation in reciprocal form:

M M

2 0 2 0 max max

1 1 1 1[E] [E] [S] [S]

K Kk k

= + = + ⋅ρ ρ ρ

ρ

ρ

ρmax

ρ-1

M

max

slope K=

ρ

[S] [S]-1

-1maxρ


Lineweaver-Burke plot

Diffusion-limited KineticsF f t ti i li id th t d t i i t

{ }D RA B AB products + k k⎯⎯→

For fast reactions in liquids, the rate-determining step can be diffusion of the reactants to form the encounter pair:

Apply the steady-state approximation to {AB}:

{ } [ ][ ] ( ) { }D -D RA B 0AB ABd k k kd

= − + =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

{ }-D

pk

{ } [ ][ ] ( ) { }

{ } [ ][ ]

D D R

D RR

-D R

rate A BAB

dtk kk

k k

⎣ ⎦ ⎣ ⎦

= =⎡ ⎤⎣ ⎦ +

The effective rate constant has two limits:The effective rate constant has two limits:

D Reff D R -D

-D R

ifk kk k k kk k

k k

= →+

Slow diffusion :

Intermediate situations can be described by:

( / )k k k k k k k k

D Reff {AB} R act -D R

-D

ifk kk K k k k kk

→ = =Fast diffusion :

D R D D R -D D acteff

-D R D D R -D D act

( / )( / )

k k k k k k k kkk k k k k k k k

= = =+ + +

Take the inverse: 1 1 1= +


Take the inverse:eff D actk k k

+

Diffusion-limited Rate ConstantsThe diffusion–limited rate constant can be calculated:

Smoluchowski 3 -1

3 -1 -1AB AB Av

D AB-1

AB4 ( )m molecule s

4 ( )000 dm mol s

k

r

r D

D N

= π

πwith encounter distancecoefficient of mutual diffusion

AB A B

AB A B2 1(m s )

r r r

D D D −

= +

= +

DA, DB can be estimated from the

Stokes-Einstein equation B

S

k TDr

=πβη

Sis the viscosity of the solvent, is the hydrodynamic radiusf id l S k diff i ( i di )

6

rηβ for ideal Stokes diffusion (continuous medium)

for the opposite limit (solute radi6 4 us solvent radiu s )

β =β = ≈

Assuming the hydrodynamic radius ≈ reaction radius,

( )B B B AD A B

A B A B

4 1 1 4 2k T k T r rk r rr r r r

⎡ ⎤ ⎡ ⎤= + + = + +⎢ ⎥ ⎢ ⎥βη βη⎣ ⎦ ⎣ ⎦

For reactants of similar size,

10 -1 -1B

S

16 1 10 M s in water at 300 KDk Tk ≈ ≈ ×

βηThe rate constant is determined by solvent properties!The viscosity dominates the temperature dependence


The viscosity dominates the temperature dependence.

T Dependence of Complex ReactionsAssume some complex reaction

for which the overall reaction rate constant can be expressed in terms of the elementary steps:

1 2A + B C + Dk k⎯⎯→ ⎯⎯→L

expressed in terms of the elementary steps:1 2

3

1 2overall

3

n n

nk kk

kK

K=

If each rate constant obeysIf each rate constant obeys the Arrhenius expression, /e iE RT

i ik A -=

( ){ }1 2

3

1 2overall 1 1 2 2 3 3

3exp - /

n n

nA Ak n E n E n E RT

AK

KK

æ ö= + - -ç ÷è ø3 ø

i.e. The Arrhenius parameters are

overallin

iiA A= Õ

i iiE n E= å

The overall “activation energy” may be negative, if niis negative and the corresponding Ei is large enough. Also, for a pre-equilibrium reaction where overall 1 2k K k=

2 2overall 1a 2 1 2

ln lnd k d KE RT RT E H EdT dT

Δ= = + = +

E i iti b t ΔH b ti


E2 is positive but ΔH can be negative.

Chain Reactions

1

1 2 1

A R

R B R P

R C R P

→

+ → +

+ → +

initiation

propagation2 1 2

1 3

R C R P

2R P

+ → +

→ termination

R is often, but not always, a free radical.I iti ti b th l h t h i l di l iInitiation may be thermal, photochemical, radiolysis, …The overall reaction is determined by adding the propagation steps:

B C P P+ → +

Important examples include polymerization, combustion, photochemical smog production and the depletion of stratospheric ozone by CFCs.

1 2B C P P+ → +

2 2 2

3 2

CF Cl CF Cl Cl

Cl O ClO O

hν⎯⎯⎯→ ⋅ + ⋅

⋅ + ⎯⎯→ ⋅ +

p p y

2ClO O Cl O⋅ + ⋅ ⎯⎯→ ⋅ +

The net effect is catalysis by the CFC of the reaction

O O 2O+ →


3 2O O 2O⋅ + →

Rice-Herzfeld Mechanisme g for the thermal decomposition of acetaldeh dee.g. for the thermal decomposition of acetaldehyde

3 3CH CHO CH CHOCHO H CO

→ ⋅ + ⋅

⋅ → ⋅ +

initiation1

2

3 2 3

3 3 4 3

H CH CHO H CH CO

CH CH CHO CH CH CO

CH CO CH CO

⋅ + → +

⋅ + → +

→ +

&

&

&propagation

3

4

5 3 3

3 2 6

CH CO CH CO

2 CH C H

→ ⋅ +

⋅ → termination

Write steady-state equations for [CHO], [H], [CH3], [CH3CO]:

5

6

3 3

[ ] [ ] [ ]1 3 2CHO CH CHO CHO 0d k kdt

= − = [ ] ( )[ ]1 2 3CHO / CH CHOk k=

[ ] [ ] [ ][ ]2 3 3H CHO H CH CHO 0d k k= − = [ ] ( )1 3H /k k=[ ] [ ] [ ][ ]2 3 3H CHO H CH CHO 0k kdt

[ ] ( )1 3H /k k

[ ] [ ][ ] [ ][ ] [ ]3 4 3 3 3 3 5 3CH CO CH CH CHO H CH CHO CH CO 0d k k kdt

= + − =

[ ] [ ] [ ][ ] [ ] [ ]23 1 3 4 3 3 5 3 6 3CH CH CHO CH CH CHO CH CO 2 CH 0d k k k k

dt= − + − =

[ ] ( )[ ]23 1 6 3CH / CH CHOk k=

[ ] [ ][ ] ( ) [ ]3/ 21/ 2CH CH CH CHrate O / CH CHOd k k k k= = =


[ ] [ ][ ] ( ) [ ]4 4 3 3 4 1 6 3CH CH CH CHrate O / CH CHOk k k kdt

= = =

An Explosive Reaction

2 2 22H O 2H O+ →

2H 2H→ initiation

Overall:

2

2

2 2

H O O OH

O H OH H

OH H H O H

+ → +

+ → +

+ → + propagation

branching

2 2

2 2

OH H H O H

H wall

H O M HO M

+ → +

→

+ + → +termination

p p g

2 2 2 2

2 2 2 2

HO H H H O

HO H O OH H O

+ → +

+ → +slow propagation

At 700 K and 0.1 bar O2,

each initial H atom → 1013 H atoms in 0.3 s.


H HOH + O H OH + O OH + OOH + H OH + O

O2

O2

O2

O2

H2

H2

H2

H2

O2

H H

H2

H2

OH + H OH + O OH + H

H2

H2

OH

H HO

H + O OH + O

H H

H OH + H

O2

H2

O2

H2

O

H

H2

HO

H + O

O2


Branching Chain Reactions

1

1 2 3

A, B R

R A R R

B R Ri i

→

+ → +

+ → +K

1 Initiation

2 Branching

3 Propagation

[R] increases

1

2

R P2R P

i i

i

i

→

→

→

Kp g

4 Termination (wall)[R] decreases

(combination)

The stead state appro imation does not applThe steady-state approximation does not apply.

If n is the number of radicals at time t,

( )I b w gdn k n k k ndt

= ρ + − +( )

( )

I b w g

I

I e 1t

dtn

n φ

= ρ + φρ

= −φ

( )b w gk k kφ = − +

explosion limitsφ

The exponential i i f

keff kb

kg+wincrease in n for φ > 0 leads to explosion.

kg

kw


pressure

Explosion Limits

rate log P steady reaction 2

3

1 2 3 1

2

At pressures below 1,wall termination is dominant

pressure temperature

the limit depends on surface composition and areathe limit is altered by the size of the reaction vessel

Between 1 and 2,is the explosion peninsulais the explosion peninsulathe limits change with temperature because branching reactions are T dependent, diffusion less so

Between 2 and 3,gas phase termination reactions are dominant

At pressures above 3,reaction products are importantheat from exothermic reactions → thermal explosion


heat from exothermic reactions → thermal explosion

Autocatalysis

0 0[A] , [P]ar te da dx a x p xdt dt

= − = = − = +

PA + P2k⎯⎯→

( )( )

( )( )

0 0

x

dx k a x p xdt

dxkta x p x

= − +

⎧ ⎫= ⎨ ⎬

− +⎩ ⎭∫ ( )( )

( ) ( ) ( )

( ) ( )

0 00

0 0 0 00

1 1 1

1

x

x

a x p x

dxa p a x p x

− +⎩ ⎭

⎧ ⎫= +⎨ ⎬+ − +⎩ ⎭

⎡ ⎤

∫

( )( ) ( )

( ) ( )( )

0 0 00 0

00

0 0 0 0

1 ln ln

1 nl

xa x p xa p

p xaa p a x p

= − − + +⎡ ⎤⎣ ⎦+

⎧ ⎫+= ⎨ ⎬

+ −⎩ ⎭

( )

( )

0 0 0 0

0

= , /

1 /ln

Substitute a p k p a

x pt

α + β =

+α = /

asymptote at x = a0

( )0

0

ln1 /

e 1/1 e

t

t

tx p

x pα

α

α =− β

−=

+ β

x/p0

exponential increase


αt

Oscillations in Gas Phase Kinetics

Consider the concentration profile of an intermediate in the H2 + O2 reaction.

conc

exponential rise due to autocatalysis

conc.

decay when reactants used up

t

What if more reactant is supplied?

conc.

t

Examples:

Fl i f h h i l l t d fl k Flaring of phosphorus in a loosely stoppered flask (Robert Boyle, 17th century)

Cool flames = limited combustion of hydrocarbons due to “long-lived” intermediates which damp the explosion.

P i i i ( i i i ) d i “k k” i i


Pre-ignition (autoignition) producing “knock” in auto engines.

Cool Flame OscillationsHydrocarbon fuels spontaneously ignite in the presence of O2at T > 400-500 K.

“True” ignition gives CO, CO2, H2O and T increases ~ 1000 K.

“Cool” flames produce ROH RCHO RCOOH and ΔT ~ 100 KCool flames produce ROH, RCHO, RCOOH and ΔT 100 K

essu

re

ignition

n ra

te

pre

slow combustion

cool flame R

eact

ion

temperature temperature

Oscillations occur because of both chemical and thermal feedbackfeedback.

2R + O ROO⋅ ⋅

high T

low Tchain branchingthermal acceleration

chain terminationthermal damping


Oscillating Reactions

The Lotka-Volterra Mechanism

+ 2X XA →

+ X Y

B

2Y

Y

→

→

[A] is held constant (replenished). [X] and [Y] oscillate.

[X] [Y]conc.

t

[Y]

t[X]


Oscillating Reaction ModelsA

B

2 + 3

+

X

+

X X

C

Y

YX

→

→

→

Brusselator

Prigogine et al

DX →

[A] and [B] are held constant. [X] and [Y] settle down toa limit cycle:a limit cycle:

[Y]

[X]

Oregonator

Noyes et al +

+ C

A X

X

Y

Y

→

→

[X]

The B-Z reaction is of this general form, with

+ 2 + Z

2

X X

X

B

YZ

D

→

→

→

g ,X = HBrO2

Y = Br–

Z = 2Ce4+

18 steps, 21 species!


YZ → p , p

Reaction Dynamics

What is the probability that the systemwill move from reactants to products?

transition state

How is thisenergy

distributedamongst the

What is the structureof the transition state?

How is energysupplied to

overcome thisbarrier?

C hg

reactionproducts?

reactants

Can the systemtunnel throughthe potential

barrier?

products

ti di t

products


reaction coordinate

Collision Theory 1In the simple hard sphere model of molecular collisions, the impact parameter (distance of closest approach) is the sum of the radii of the collision pair.

rA rBd = rA+rBA B

Collision cross-section 2dσ = πarea σ

collision volumecollision volume swept by A per sec.

Collision frequency AB A B relZ N N v= σ

collision frequency per A moleculenumber of collisions per unit timeper unit volume = (m-3) (m-3) (m2) (m s-1) ⇒ m-3 s-1


Collision Theory 21/ 2 1/ 2

relA B

8 8 8RT RT RTvM M

⎡ ⎤ ⎛ ⎞= + = ⎜ ⎟⎢ ⎥π π πμ⎣ ⎦ ⎝ ⎠

Maxwell

A BM Mμ =where is the reduced mass.

Activation Energy: Only a fraction of collisions have sufficient kinetic energy to overcome the activation barrier.

A BM Mμ

+

a /ABreaction rate e E RTZ −=

Bimolecular Rate Constant

[A]d Z mol m-3 s-1

molecules m-3 s-1

a

a

/AB

/A Brel

[A] e

e

E RT

E RT

d Zdt L

N N vL

−

−

− =

= σ mol m-3 s-1

L = Avogadro’s no.

mol m 3 s 1

a

a

6 /rel

3 /rel

10 [A][B] e

10 [A][B] e

E RT

E RT

L

L v

L v

−

−

= σ

= σ

mol m-3 s-1

mol dm-3 s-1

a3 /AB rel10 e E RTk L v −= σ mol-1 dm3 s-1

( )a

1/ 23 2 /

AB 10 8 eA B E RTM Mk Ld RT

M M−+⎛ ⎞

= π⎜ ⎟⎝ ⎠

M-1 s-1


A BM M⎜ ⎟⎝ ⎠

Collision Theory 3

1. Orientation DependenceThe reactions of polyatomic molecules typically depend

Basic collision theory has several deficiencies, which can be partially overcome by making more sophisticated models.

The reactions of polyatomic molecules typically depend on their mutual orientation.

Solution: Replace σ with σ*, the reactive cross-section:

* Pσ = σ* Pσ = σsteric factor

2. Intermolecular Interactions Molecules are not incompressible hard spheres!

uU(r)

rij rijrij

0

Hard sphere Square well Lennard-Jones

3. The collision energy depends on the impact parameter, which is angular dependent.

4. There is no way to predict the activation energy.


4. There is no way to predict the activation energy.

Potential Energy Surfaces 1A ti f i l t f th f ti tA reaction surface is a plot of the energy of a reaction system as a function of all the independent variables (bond lengths and bond angles). Even a collinear triatomic reaction such as

A + BC → AB + Cneeds a 3-D plot:

E

rB-C rA-B

A + BCAB + C

rA-B

or a contour plot: symmetric vibration of A-B-C

minimum energy path


rB-C

Potential Energy Surfaces 2

0.60

0.90

1.20

1.50

1.80r (B-C)

0.60

0.90

1.20

2.10

2.40

2.70

3.00

0.60 0.90

1.20 1.50

1.80 2.10

2.40 2.70

3.00

( )

r (A-B)

1.50

1.80

2.10

2.40

2.70

r (B-C)

3.000.60 0.90 1.20 1.50 1.80 2.10 2.40 2.70 3.00

r (A-B)

0.600.901.201.501.802.102.402.703.00

0.60 0.90 1.20 1.50 1.80 2.10 2.40 2.70 3.00 r (B-C)r (A-B)


Transition State Theory1 A th t i th f ti th i1. Assume that in the course of a reaction there is some

dividing surface (point on a one-dimensional reaction path) past which reaction to products is inevitable.

2. Assume that this transition state is in effective equilibrium with the reactants.

Ukp

K‡

reactants

products

p

A B X P‡ p oduc s

reaction coordinate

A B X P+ →‡

[P]d[P] [X ] [A][B]p pd k k Kdt

= =‡ ‡

2 pk k K= ‡

A full discussion of k and K‡ requires quantum chemistryA full discussion of kp and K‡ requires quantum chemistry and statistical mechanics, and leads to the

Eyring Equation 2RTk KhL

⎛ ⎞= κ⎜ ⎟⎝ ⎠

‡

transmission coefficient


transmission coefficient or tunneling factor

Thermodynamic Formulation of TST

A B X+

- lnG RT KΔ =° °For gases,

( )X /p p p° °⎛ ⎞ p nRT RT⎛ ⎞( )( )( )

Xc

A B

// /

p p pK Kp p p p RT

⎛ ⎞° = = ⎜ ⎟° ° ⎝ ⎠p nRT RTcp Vp p

⎛ ⎞= = ⎜ ⎟° ° °⎝ ⎠

where

Substitute into the Eyring Equation:

2 cRTk KhL

RT RT K

⎛ ⎞= κ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= κ⎜ ⎟⎜ ⎟⎝ ⎠

‡

‡

/e G RT

hL p

RT RThL p

−Δ

⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= κ⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠

‡

“hide” κ in the

/ /e eS R H RTRT RThL p

Δ −Δ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠

‡ ‡ΔS factor

Si 2E U RT H RTΔ Δ‡ ‡ (f l )Since act 2E U RT H RT= Δ + = Δ +‡ ‡ (for gases only)

act2 / /2 e e eS R E RTRT RTk

hL pΔ −⎧ ⎫⎛ ⎞⎛ ⎞= ⎨ ⎬⎜ ⎟⎜ ⎟°⎝ ⎠⎝ ⎠⎩ ⎭

‡


hL p⎝ ⎠⎝ ⎠⎩ ⎭

Electron Transfer Reactions in Solution

Marcus Nobel Prize 1992

e.g. Ce4+ + Fe2+ → Ce3+ + Fe3+

102Ru2+ + 106Ru3+ → 102Ru3+ + 106Ru2+

oΔG ≠ 0

ΔG o = 0

Reaction is slow if solvent reorganization is required in the TS.

Greactants and products

reactants products

G have the same solvent configuration here

reaction coordinate = measure of solvent configuration

Marcus Theory

ln ke

ke is a maximum when ΔGo = -λ,the solvent reorganization term.

ΔG ‡ = 0

ΔG0 G

k A Go−+

FHG

IKJ

RS| UV|λ 12

Δ

© Paul Percival 9/10/2008Modified by Jed Macoskoreaction coordinate

k ARTe = +HG KJS|T| V|W|exp

λ41

The Heat Death of the Universe

"There is an even deeper philosophical implication of bioenergetics. The universe as a whole is an isolated system. The entropy of the whole universe must be increasing. It follows that each of us, as a living organism that locally and temporarily decreases entropy, must produce somewhere in the world around us an increase in entropy. As we metabolize food, for py ,example, we give off heat and increase random molecular motion around us. In a sense, we buy our lives through the entropic death of the universe.“

quotation from a biochemistry text


Essential Concepts

temperature, energy, heat, work

path variable, state variable

internal energy, heat capacity, equipartition

ideal gas, real gas, equation of state

isothermal, adiabatic

enthalpy free energy entropyenthalpy, free energy, entropy

spontaneity, reversible, irreversible

equilibrium

h i l t ti lchemical potential

mixing

rate of reaction, rate constant

reaction order, molecularity

intermediate, steady-state approximation

chain reaction, branching chain

collision frequency, activation. transition state


What is Thermo, Stat Mech, etc.? - Wake Forest Universityusers.wfu.edu/macoskjc/Courses/ALL2.pdf · What is Thermo, Stat Mech, etc.? ... The cubic form of the equation predicts 0

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