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1

equilibrium in macroscopic systems

THERMODYNAMICSELECTROCHEMISTRY

change of concentration as a function of time

(macroscopic) KINETICS(ELECTROCHEMISTRY)

stationary states of particles (atoms, molecules, electrons, nuclei) e.g. during translation, rotation, vibration

STRUCTURE OF MATTERCHEMICAL BOND

•bond breakage and formation

•transitions between quantum states

STRUCTURE OF MATTER(microscopic) KINETICS

CHEMICAL BOND

STATISTICAL THEORY OF MATTER

PHYSICAL CHEMISTRY: An Introduction

macroscopicphenomena

microscopicphenomena

static phenomena dynamic phenomena

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This Physical Chemistry lecture uses graphs from the following textbooks:

P.W. Atkins, Physical Chemistry, 6th ed., Oxford University Press, Oxford 1998

P. Atkins and J. de Paula, Atkins’ Physical Chemistry, 7th ed., Oxford University Press, Oxford 2002

G. Wedler, Lehrbuch der Physikalischen Chemie, 4th ed., Wiley-VCH, Weinheim 1997

Copies of the transparencies are intended only for personal use and in connection with the lectures “Chemistry” and “Physikalische Chemie” at the University of Magdeburg. They may not be distributed to others. No other use is allowed without the prior permission (in writing) of the Institute of Chemistry at the Otto-von-Guericke-UniversitätMagdeburg.

© Helmut Weiss, 2015

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Matter: Substance, intensive and extensive properties, molarity and molality

Substance A substance is a distinct, pure form of matter. The amount of a substance, n, in a sample is reported in terms of the unit called a mole (mol). In 1 mol

are NA=6.0221023 objects (atoms, molecules, ions, or other specified entities). NA is the Avogadro constant.

Extensive and intensive properties An extensive property is a property that depends on the amount of substance in the sample.

Examples: mass, volume… An intensive property is a property that is independent on the amount of substance in the sample.

Examples: temperature, pressure, mass density… A molar property Xm is the value of an extensive property X divided by the amount of substance, n:

Xm=X/n. A molar property is intensive. It is usually denoted by the index m, or by the use of small letters. The one exemption of this notation is the molar mass, which is denoted simply M.

A specific property Xs is the value of an extensive property X divided by the mass m of the substance: Xs=X/m. A specific property is intensive, and usually denoted by the index s.

Measures of concentration: molarity, molality and mole fraction The molar concentration (‘molarity’) of a solute in a solution refers to the amount of substance of the

solute divided by the volume of the solution. Molar concentration is usually expressed in moles per litre (mol L-1 or mol dm-3). A molar concentration of x mol L-1 is widely called ‘x molar’ and denoted x M.

The term molality refers to the amount of substance of the solute divided by the mass of the solvent used to prepare the solution. Its units are typically moles of solute per kilogram of solvent (mol kg-1).

The mole fraction, xJ, is the amount of J expressed as a fraction of the total amount of moles, n, in the sample: xJ = nJ/n = nJ/(nA+nB+…)

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Some fundamental terms:

System and surroundings:For the purposes of Physical Chemistry, the universe is divided into two parts, the system and its surroundings. The system is the part of the world, in which we have special interest. The surroundings is where we make our measurements.

The type of system depends on the characteristics of the boundary which divides it from the surroundings:

(a) An open system can exchange matter and energy with its surroundings.(b) A closed system can exchange energy with its surroundings, but it

cannot exchange matter. (c) An isolated system can exchange neither energy nor matter with its

surroundings.

Except for the open system, which has no walls at all, the walls in the two other have certain characteristics, and are given special names:

A diathermic (closed) system is one that allows energy to escape as heat through its boundary if there is a difference in temperature between the system and its surroundings. It has diathermic walls.

An adiabatic (isolated) system is one that does not permit the passage of energy as heat through its boundary even if there is a temperature difference between the system and its surroundings. It has adiabatic walls.

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5

Homogeneous system:The macroscopic properties are identical in all parts of the system.

Heterogeneous system:The macroscopic properties jump at the phase boundaries.

Phase:Homogeneous part of a (possibly) heterogeneous system.

Equilibrium condition: The macroscopic properties do not change without external influence. The system returns to equilibrium after a transient perturbation. In general exists only a single true equilibrium state.

Equilibria in Mechanics:

unstable (labile) metastable

Equilibria in Thermodynamics:

stable

H2O (water)25°C1 bar

stable

H2 + ½ O2

25°C1 bar

metastable

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The concept of “Temperature”: Temperature is a thermodynamic quantity, and not known in mechanics. The concept of temperature springs from the observation that a change in physical state (for example, a

change of volume) may occur when two objects are in contact with one another (as when a red-hot metal is plunged into water):

If, upon contact of A and B, a change in any physical property of these systems is found, we know that they have not been in thermal equilibrium.

A B A B+

The Zeroth Law of thermodynamics:If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, than C is also in thermal equilibrium with A. All these systems have a common property: the same temperature.

Energy flows as heat from a region at a highertemperature to one at a lower temperature if thetwo are in contact through a diathermic wall, asin (a) and (c). However, if the two regions haveidentical temperatures, there is no net transfer ofenergy as heat even though the two regions areseparated by a diathermic wall (b). The lattercondition corresponds to the two regions beingat thermal equilibrium.

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7

The thermodynamic temperature scale:

In the early days of thermometry (and still in laboratory practice today), temperatures were related to the length of a column of liquid (e.g. Mercury, Hg), and the difference in lengths shown when the thermometer was first in contact with melting ice and then with boiling water was divided into 100 steps called ‘degrees’, the lower point being labelled 0. This procedure led to the Celsius scale of temperature with the two reference points at 0 °C and 100 °C, respectively.

Assumption:Linear relation between the Celsius temperature and an observable quantity x, like the length of a Hg column, the pressurep of a gas at constant volume V, or the volume V of the gas for constant pressure p:

0 C

100 C 0 C

(x) a x b

x x(x) 100 C

x x

For the volume V (left graph), this transforms to:

Observation:

For all (ideal) gases one finds

Introduction of the thermodynamic temperature scale (in ‘Kelvin’):

and

Left: The variation of the volume of a fixed amount of gas with the temperature constant. Note that in each case they extrapolate to zero volume at -273.15 C.Right: The pressure also varies linearly with the temperature, and extrapolates to zero at T= 0 K (-273.15 C).

0 C

100 C 0 C 100 C 0 C

VV100

C V V V V

0 C

100 C 0 C

V100 273.15 0.01

V V

T

K =

θ

°C + 273.15

T

K = 100∙

V

V100°C-V0°C

When the energy of a system is changed as a consequence of a temperature difference between it and the surroundings, the energy has been transferred as heat. When, for example, a heater is immersed in a beaker with water (the system), the capacity of the water to do work increases because hot water can be used to do more work than cold water.Heat transfer requires diathermic walls.

The fundamental physical property in thermodynamics is work: work is done when an object is moved against an opposing force. (Examples: change of the height of a weight, expansion of a gas that pushes a piston and raises the weight, or a chemical reaction which e.g. drives an electrical current)

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Work, heat, and energy:

The energy of a system is its capacity to do work. When work is done on an otherwise isolated system (e.g. by compressing a gas or winding a spring), its energy is increased. When a system does work (e.g. by moving a piston or unwinding the spring), its energy is reduced.

A process that releases energy as heat is called exothermic, a process that absorbs energy as heat endothermic.

(a) When an endothermic process occurs in an adiabatic system, the temperature falls; (b) if the process is exothermic, then the temperature rises. (c) When an endothermic process occurs in a diathermic container, energy enters as heat from the surroundings, and the system remains at the same temperature; (d) if the process is exothermic, then energy leaves as heat, and the process is isothermal.

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Work, heat, and energy (continued):

Molecular interpretation In molecular terms, heat is the transfer of energy that makes use of chaotic molecular motion (thermal

motion). In contrast, work is the transfer of energy that makes use of organized motion. The distinction between work and heat is made in the surroundings.

When a system does work, it stimulates orderly motion in the surroundings. For instance, the atoms shown here may be part of a weight that is being raised. The ordered motion of the atoms in a falling weight does work on the system.

When energy is transferred to the surroundings as heat, the transfer stimulates disordered motion of the atoms in the surroundings. Transfer of energy from the surroundings to the system makes use of disordered motion (thermal motion) in the surroundings.

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State functions and state variables

STATEMENT

If only two intensive properties of a phase of a pure substance are known, all intensive properties of this phase of the substance are known, or

If three properties of a phase of a pure substance are known, all properties of this phase of the substance are known.

example: - p and T as independent variables means: Vm (=v) = f(p,T),i.e. the resulting molar volume is pinned down, or

- p, T, n as independent variables means: V = f(p,T,n)

The resulting function is termed a state function. The variables which describe the system state, are termed

- state variables, and are related to each other via the - state functions.

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The thermal equation of state and the perfect gas equation

Thermal equation of state: The thermal equation of state combines

volume V, temperature T, pressure p, and the amount of substance n:

V = f(p,T,n) or Vm = v = f(p,T)

The “perfect gas” (or “ideal gas”): mass points without expansion no interactions between the particles a real gas, an actual gas, behaves more and

more like a perfect gas the lower the pressure, and the higher the temperature

Some empirical gas laws:V = f(T) for p=const.:

“isobars”

V = const. ( + 273.15°C)= const. T

p = f(T) for V=const.: “isochors”

p = const. ( + 273.15°C)= const. T

p = f(V) for T=const.: “isotherms”

p V = const.

(Charles, 1798; Gay-Lussac, 1802) (Boyle, 1664; Mariotte, 1672)

1 2 3

(Amontons, lived 1663 - 1705)

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Combination of 1 and 3 for:

1 mol gas at

p0 = 1.013 bar

T0 = 273.15 K

v0 = 22.42 l

Step 1: Isobaric change

00

00

0T vvv

,T

,p T

Step 2: Isothermal change

0 0pT vv p,p , v

00

0

vp v p T

T}

= const. !

p v = R T

p V = n R T‘perfect gas equation’ R : ‘gas constant’

(= 8.31434 J K-1 mol-1)

Sections through the surface shown in the figure at constant temperature give the isotherms shown for the Boyle-Mariotte law and the isobars shown for the Gay-Lussac law.

A region of the p,V,T surface of a fixed amount of perfect gas. The points forming the surface represent the only states of the gas that can exist.

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13

Swap the changes: a combination of 3 and 1 for

1 mol gas at

p0 = 1.013 bar

T0 = 273.15 K

v0 = 22.42 l

00

v ''vT ,p,v '' T

T

Step 2: Isobaric change

Step 1: Isothermal change

0 00 0 0 p vT ,p ,v '' p v

00

0

vp v p T

T}

= const. !

same result !!!

The change of a state variable is independent of the path, on which the change of the state has been made, as long as initial and final state are identical.

Some mathematical consequences:(i) The change can be described as an ‘exact differential’,

i.e. the variables can be varied independently; e.g. for z=f(x,y):

(ii) The mixed derivatives are identical (Schwarz’s theorem):

(iii) Upon variation of x, y for z=const (Euler’s theorem):

y yx x

z z z zdz dx dy ; , : partial differentials

x y x y

2 2z z

x y y x

y

y zx

x

zz z y x

dz 0 dx dyx y x z

y

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Mathematical consequences: The exact differential

assume: V x, T y, U z:

T V

U UdU dV dT

V T

+

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A more general approach to thermal expansion and compression:

V = f(T) for p=const.: V = V0 (1 + )

(Charles)

: (thermal) expansion coefficient

p = f(T) for V=const.: p = p0 (1 + )

(Amontons)

p = f(V) for T=const.: pV = const. and d(pV) = pdV + Vdp = 0

(Boyle)

: (isothermal) compressibility

generally valid!

Due to generally valid!

p p0 0

1 V 1 V

V V T

V V0 0

1 p 1 p

p p T

T

1 V

V p

Exact differential of V=f(p,T):

p T

V VdV dT dp

T p

V dT V dp

pT V

V V T 1(Euler)

p T p p

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Mixtures of gases: Partial pressure and mole fractions

Dalton’s law:

The pressure exerted by a mixture of perfect gases is the sum of the partial pressures of the gases.

The partial pressure of a gas is the pressure that it would exert if it occupied the container alone. If the partial pressure of a gas A is pA, that of a perfect gas B is pB, and so on, then the partial pressure when all the gases occupy the same container at the same temperature is

where, for each substance J,

A Bp p p ...

JJ

n R Tp

V

The partial pressures pA and pB of a binary mixture of (real or perfect) gases of total pressure p as the composition changes from pure A to pure B. The sum of the partial pressures is equal to the total pressure. If the gases are perfect, then the partial pressure is also the pressure that each gas would exert if it were present alone in the container.

The mole fraction, xJ, is the amount of J expressed as a fraction of the total amount of moles, n, in the sample:

When no J molecules are present, xJ=0; when only J molecules are present, xJ=1. Thus the partial pressure can be defined as:

JJ A B

nx n n n ...

n

J Jp x p

A B

A B

p p ...

(x x ...) p p

and

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Real gases: An introduction

The variation of the potential energy of two molecules on their separation. High positive potential energy (at very small separations) indicates that the interactions between them are strongly repulsive at these distances. At intermediate separations, where the potential energy is negative, the attractive interactions dominate. At large separations (on the right) the potential energy is zero and there is no interaction between the molecules.

Molecular interactions

Real gases show deviations from the perfect gas law because molecules (and atoms) interact with each other: Repulsive forces (short-range interactions) assist expansion, attractive forces(operative at intermediate distances) assist compression.

Compression factor

For a perfect gas, Z=1 under all conditions. Deviation of Z from 1 is a measure of departure from perfect behaviour.

p vZ

R T

The variation of the compression factor Z = pv/RT with pressure for several gases at 0C. A perfect gas has Z = 1 at all pressures. Notice that, although the curves approach 1 as p 0, they do so with different slopes.

At very low pressures, all the gases have Z1 and behave nearly perfect. At high pressure, all gases have Z>1, signifying that they are more difficult to compress than a perfect gas, and repulsion is dominant. At intermediate pressure, most gases have Z<1, indicating that the attractive forces are dominant and favor compression.

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Real gases: The virial equation of state

Experimental isotherms of carbon dioxide at several temperatures. The `critical isotherm', the isotherm at the critical temperature, is at 31.04 C. The critical point is marked with a star.

Below, some experimental isotherms of carbon dioxide are shown. At large molar volumes v and high temperatures the real isotherms do not differ greatly from ideal isotherms. The small differences suggest an expansion in a series of powers either of p or v, the so-called virial equations of state:

2

2

p v R T (1 B p C p ...)

B Cp v R T 1 ...

v v

The second virial coefficient, B, is usually more important than the third one, C, in the sense that at typical molar volumes C/v2<<B/v. In simple models, and for p 0, higher terms than B are therefore often neglected.

The virial equation can be used to demonstrate the point that, although the equation of state of a real gas may coincide with the perfect gas law as p 0, not all of its properties necessarily coincide. For example, for a perfect gas dZ/dp = 0 (because Z=1 for all pressures), but for a real gas

as p 0, and

as v , corresponding to p 0.

dZB 2pC ... B

dp

dZB

d(1/ v)

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Real gases: The Boyle temperature

Because the virial coefficients depend on the temperature (see table above), there may be a temperature at which Z1 with zero slope at low pressure p or high molar volume v. At this temperature, which is called the Boyle temperature, TB, the properties of a real gas coincide with those of a perfect gas as p 0, and B=0. It then follows that pvRTB over a more extended range of pressures than at other temperatures.

The compression factor approaches 1 at low pressures, but does so with different slopes. For a perfect gas, the slope is zero, but real gases may have either positive or negative slopes, and the slope may vary with temperature. At the Boyle temperature, the slope is zero and the gas behaves perfectly over a wider range of conditions than at other temperatures.

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Real gases: Condensation and critical pointReconsider the experimental isotherms of carbon dioxide. What happens, when gas initially in the state A is compressed at constant temperature (by pushing a piston)?

• Near A, the pressure rises in approximate agreement with Boyle’s law.

• Serious deviations from the law begin to appear when the volume has been reduced to B.

• At C (about 60 bar for CO2), the piston suddenly slides in without any further rise in pressure. Just to the left of C a liquid appears, and there are two phases separated by a sharply defined surface.

• As the volume is decreased from C through D to E, the amount of liquid increases. There is no additional resistance to the piston because the gas can respond by condensation. The corresponding pressure is the vapour pressure of the liquid at this temperature.

• At E, the sample is entirely liquid and the piston rests on its surface. Further reduction of volume requires the exertion of a considerable amount of pressure, as indicated by the sharply rising line from E to F. This is due to the low compressibility of condensed phases.

The isotherm at the temperature Tc plays a special role :

• Isotherms below Tc behave as described above.

• If the compression takes place at Tc itself, a surface separating two phases does not appear, and the volumes at each end of the horizontal part of the isotherm have merged to a single point, the critical point of the gas. The corresponding parameters are the critical temperature, Tc, critical pressure, pc, and critical molar volume, vc, of the substance.

• The liquid phase of a substance does not form above Tc.

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Real gases: Critical constants, compression factors, Boyle temperatures, and the supercritical phase

A gas can not be liquefied if the temperature is above its critical temperature. To liquefy it - to obtain a fluid phase which does not occupy the entire volume - the temperature must first be lowered to below Tc, and then the gas compressed isothermally.

The single phase that fills the entire volume at T> Tc may be much denser then is normally considered typical of gases. It is often called the supercritical phase, or a supercritical fluid.

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The van der Waals equation of gases: A model

Starting point: The perfect gas law pv = RT

Correction 1: Attractive forces lower the pressure

replace p by (p+), where is the ‘internal pressure’. More detailed analysis shows that =a/v2.

Correction 2: Repulsive forces are taken into account by supposing that the molecules (atoms) behave as small but impenetrable spheres

replace v by (v-b), where b is the ‘exclusion volume’. More detailed analysis shows that b is approximately the volume of one mole of the particles.

a, b: van der Waals coefficients

van der Waals equation

2

2

2

or

a(p ) (v b) R T

v

a n(p ) (V n b) n R T

V

The surface of possible states allowed by the van der Waals equation.

Comparison to the virial equation of state:

aB b

RT

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Analysis of the van der Waals equation of gases

(1) Perfect gas isotherms are obtained at high enough temperatures and large molar volumes.

Van der Waals isotherms at several values of T/Tc. The van der Waals loops are normally replaced by horizontal straight lines. The critical isotherm is the isotherm for T/Tc = 1.

(3) The critical constants are related to the van der Waals constants.

At the critical point the isotherm has a flat inflexion. An inflexion of this type occurs if both the first and second derivative are zero:

at the critical point. The solution is

and the critical compression factor, Zc, is predicted to be equal to

for all gases.

2 3

2

32 4

dp RT 2a0

dv vv b

d p 2RT 6a0

dv vv b

c c c2

a 8av 3b p T

27b 27Rb

c cc

c

p v 3Z

RT 8

(2) Liquids and gases coexist when cohesive and dispersing effects are in balance. The ‘van der Waals loops’ are unrealistic because they suggest that under some conditions an increase in pressure results in an increase of volume. Therefore they are replaced by horizontal lines drawn so the loops define equal areas above and below the lines (‘Maxwell construction’)

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The principle of corresponding states

Idea

If the critical constants are characteristic properties of gases, than characteristic points, like melting or boiling point, should be unitary defined states. We therefore introduce reduced variables

and obtain the reduced van der Waals equation:

r r rc c c

p v Tp v T

p v T

rr 2

r r

8T 3p

3v 1 v

b) Reduced melting temperature

c) Reduced boiling temperature

d) Trouton’s rule (or: Pictet-Troutons’s rule)

A wide range of liquids gives approximately the same standard entropy of vaporization of

S 85 J K-1 mol-1

But:

approximation!

works best for gases composed of spherical particles

fails, sometimes badly, when the particles are non-spherical or polar

Examples

a) Compression factors

The compression factors of four gases, plotted for three reduced temperatures as a function of reduced pressure. The use of reduced variables organizes the data on to single curves.

Tm

Tc ≅ 0.44 at 1.013 bar

TbTc

≅ 0.64 at 1.013 bar

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The First Law of Thermodynamics

In thermodynamics, the total energy of a system is called its internal energy, U.

The internal energy is the total kinetic and potential energy of the molecules (atoms) composing the system. We denote by U the change in internal energy when a system changes from an initial state i with an internal energy Ui to a final state f of internal energy Uf: U = Uf – Ui.

The internal energy is a state function in the sense that its value depends only on the current state of the system, and is independent of how this state has been prepared. The internal energy is an extensive property.

Please note: In thermodynamics, only changes of state functions are of importance. Their absolute values are usually not known.

The internal energy of an isolated system is constant.

The First Law of Thermodynamics:

The sum of work W and heat Q which is transferred between a system and its surroundings is equal to its resulting change in internal energy:

U = Q + W

or:

or:

No perpetual motion machine (perpetuum mobile) of the first kind, i.e. a machine that does work without consuming fuel or some other source of energy, has ever been build.

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Rationale for the path independency of U, and the path dependencies of Q and W

It is found that the same quantity of work must be done on an adiabatic system to achieve the same change of state even though different means of achieving that work may be used. This path independence implies the existence of a state function, the internal energy. The change in internal energy is like the change in altitude when climbing a mountain: its value is independent of path.

As the volume and temperature of a system are changed, the internal energy changes. An adiabatic and a non-adiabatic path are shown as Path 1 and 2, respectively: they correspond to different values of q and w but to the same value of U.

Please note: Although the internal energy U is a state function, and depends only on the current state and not on how this state has been prepared, the exchange work W and the exchange heat Q depend on the path!

In the language of infinitesimal changes, this is often expressed by the notation dU = Q + W, where dUcharacterizes an exact, i.e. path-independent differential, while the changes Q and W are inexact, or path-dependent, differentials.

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27

Isochoric versus isobaric changes: ‘internal energy’ and ‘enthalpy’

Since U is an extensive quantity, it can be described by two state variables and the amount of substance:

U = f(T, V, n1, n2, … nk)

This equation is often called the ‘caloric equation of state’. Since U is a state function, its change can be described via an ‘exact differential’:

j i

iiV,n T,n i T,V,n

U U UdU dT dV dn

T V n

a) isochoric changes of state (dV = 0): dU = Q (UII – UI) = Q

U: exchanged heat

Closed system (dni = 0):

b) isobaric changes of state (dp = 0): dU = Q – pdV

(UII – UI) = Q – p(VII – VI)

(UII + pVII) - (UI + pVI) = Q

(U + pV): exchanged heat

introduction of the enthalpy H = U + pV

As in case of U, H is an extensive quantity, which is usually described by the two state variables T and p and the amount of substance:

H = f(T, p, n1, n2, … nk)

and:

j i

iip,n iT,n T,p,n

H H HdH dT dp dn

T p n

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The temperature dependence of internal energy and enthalpy: Heat capacities

experimentally, cp is often easier to determine than cv calculate cv from cp

p vT p

2

v p

U vc c p

v T

p v vT T

T dT

2

2 1

1

2

2 1

1

T

T T v

T

T

T T p

T

u u c d T

h h c d T

assume: 1 phase, 1 mol, pure substance

vv

pp

uV const.: du Q dT c dT

T

hp const.: dh Q dT c dT

T

cv: molar heat capacity at constant volume

cp: molar heat capacity at constant pressure

Perfect gas: the internal energy is independent of the volume v

T v

U pT p 0

v T

p vp

vc c p R

T

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29

Reaction Enthalpy and Internal Reaction Energy

so far considered: pure, homogeneous systems U=f(V,T,n) and H=f(p,T,n)

now considered: - closed system, pure substance, two (or more) phases physical changes or- closed system, one phase, several components chemical changes

U=f(V,T,n1,n2, … nk) and H=f(p,T,n1,n2, … nk)

closed system: changes in the amounts of substance, ni, are linked to each other !

e.g. for a reaction: AA + BB CC + DD we find

: ‘extent of reaction’note: stoichiometric coefficients of reactants have negative, of products positive sign

C C!

A A

B B D D

CA B D i

A B C D i

dndn dndndn

dn dn

dn dd

n

it is: dni = ni – nistart = id and I = const.

U=f(V,T,n1start,n2

start, … nkstart,) and H=f(p,T,n1

start,n2start, … nk

start,)

since all nistart = const.:

V, T, T,V

p, T, T,p

U U UdU dT dV d

T V

H H HdH dT dp d

T p

30

Reaction Enthalpy and Internal Reaction Energy (cont’d)

T,V

U T,p

H

What is the meaning of and ? a) isothermal and isochoric processes

b) isothermal and isobaric processes

T,V

T,V

UdU Q d

T,p

T,p

HdH Q d

and correspond to the heat q produced or absorbed by a

chemical reaction or a phase transition for one formula conversion.

The expression / is usually abbreviated by the Greek U, H

T,V

Ud

T,p

Hd

1) U and H are state functions, i.e. path independent.

2) U and H are composed of the internal energies and enthalpies of reactants and products:

U = iui H = ihi

for e.g. ammonia synthesis: 1/2 N2 + 3/2 H2 NH3

H = h(NH3) – 1/2 h(N2) – 3/2 h(H2)

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31

Standard Enthalpy Changes and Transition Enthalpies

Changes in enthalpy are normally reported for processes under a set of standard conditions:

The standard state of a substance at a specified temperature is its pure form at 1 bar.

The standard enthalpy change is denoted by H°.

The different types of enthalpies encountered in ‘thermochemistry’ – the study of the heat produced or required by physical changes or chemical reactions - is listed in the table below.

Since H is a state function, it is e.g.

subH = fusH + vapH

32

The correlation between H and U

A constant-volume bomb calorimeter. The `bomb' is the central vessel, which is massive enough to withstand high pressures. The calorimeter (for which the heat capacity must be known) is the entire assembly shown here. To ensure adiabaticity, the calorimeter is immersed in a water bath with a temperature continuously readjusted to that of the calorimeter at each stage of the combustion.

for dT = dp = 0 holds:

T,

UH U p V

V

U is markedly different from H only in case of noticeable changes V of the volume.

for the perfect gas we find because of (U/V)T=0 :

H = U + pV

= U + pivi

= U + iRT

e.g. ammonia synthesis (1/2 N2 + 3/2 H2 NH3):

I = 1 – 1/2 – 3/2 = -1

H = U - RT

The most common device for measuring U is the adiabatic bomb calorimeter, where the change in temperature, T, of the calorimeter is proportional to the heat that a reaction releases or absorbs. The above correlation allows to determine also H.

17

33

The temperature dependence of reaction enthalpies H:Kirchhoff’s law

An illustration of the content of Kirchhoff's law. When the temperature is increased, the enthalpies of the products

and the reactants both increase, but may do so to different extents. In each case, the change in enthalpy depends on

the heat capacities of the substances. The change in reaction enthalpy reflects the difference in the changes of

the enthalpies.

Standard reaction enthalpies at different temperatures may be estimated from heat capacities and the reaction enthalpy at some other temperature.

It is:

and:

Integration yields Kirchhoff’s law for the standard reaction enthalpy change from rH°(T1) to

where rCp° is the difference in the molar heat capacities of products and reactants weighted by the stoichiometric coefficients that appear in the chemical equation.

e.g. ammonia synthesis (1/2 N2 + 3/2 H2 NH3):

rCp° = cp(NH3) – 1/2 cp(N2) – 3/2 cp(H2)

p T

H Hd H dT dp

T p

2 2p

p i p,ip pp p

CH H HC c

T T T

2

1

To

r 2 r 1 r p

T

H (T ) H (T ) C dT

A similar expression is found for rU°:

2

1

To

r 2 r 1 r v

T

U (T ) U (T ) C dT

34

The pressure dependence of reaction enthalpies H

It is: 2

1

p

r 2 r 1pp

VH (p ) H (p ) V T dp

T

for perfect gases H is independent of pressure, since T(V/T)p = V

for condensed phases V is very small, and H almost independent of pressure

Hess’s law

Standard enthalpies of individual reactions can be combined to obtain the enthalpy of another reaction. This application of the First Law of Thermodynamics is called Hess’s law (1840):

The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.

The thermodynamic basis is the path-independence of rH. The individual steps need not be realizable in practice: they may be hypothetical reactions, the only requirement being that their chemical equations should balance.

The importance of Hess’s law is that information about a reaction of interest, which may be difficult to determine directly, can be assembled from information on other reactions.

Example: formation of carbon monoxide

1. C + ½ O2 CO H1, immeasurable2. CO + ½ O2 CO2 H2 = -283.1 kJ mol-1

3. C + O2 CO2 H3 = -393.7 kJ mol-1

H1 = H3 - H2 = -110.6 kJ mol-1

IC+O2

IIICO2

IICO+ ½O2

3

1? 2

18

35

Standard enthalpies of formation

The standard enthalpy of formation, fH°, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference state.

The reference state of an element is its most stable state at the specified temperature and 1 bar.

Examples: at 298 K the reference state of nitrogen is a gas of N2 molecules, that of mercury liquid Hg, that of carbon is graphite, and that of tin the white (metallic) form.

Only exception: The reference state of phosphorous is the white form since this allotrope is the most reproducible form of this element.

The standard enthalpies of formation of elements in their reference states are zero at all temperatures.

The reaction enthalpy in terms of enthalpies of formation:

Conceptually, a reaction can be regarded as proceeding by decomposing the reactants into their elements, and then forming those elements into the products. The value for fH° is the sum of these ‘unforming’ and ‘forming’ enthalpies.

Illustration: The standard reaction enthalpy of

2HN3(l) + 2NO(g) H2O2(l) + 4N2(g)

is calculated as follows:

rH° = {fH°(H2O2,l) + 4fH°(N2,g)}

- {2fH°(HN3,l) + 2fH°(NO,g)}

= {-187.78 + 4(0)} – {2(264.0) + 2(90.25)}

= -892.3 kJ mol-1

36

Conversion of heat and expansion work

General expression for work: dw = -F dz

where dw is the work required to move an object a distance dz against an opposing force F.

When a piston of area A moves out through a distance dz, it sweeps out a volume dV = A dz. The external pressure, pex, is equivalent to a weight pressing on the piston, and the force opposing expansion is F = pex A.

Expansion work:

Assume a system with massless, frictionless, rigid, perfectly fitting piston of area A und an external pressure pex.

f

i

ex

V

ex

V

dw p dV

w p dV

i: initial statef: final state

All processes considered here: perfect gas in cylinder with perfect piston work done by the gas is stored in a (virtual) ‘work storage’ modus operandi either isothermal

(in a thermostat; left) or adiabatically (right)

Starting point:V T

U UdU Q pdV dT dV

T V

=0 for a perfect gas

19

37

- Free expansion

i.e. expansion against zero opposing force, or pex = 0 w = 0

The work done by a gas when it expands against a constant external pressure, pex, is equal to the shaded area in this example of an indicator diagram.

The work done by a perfect gas when it expands reversibly and isothermally is equal to the area under the isotherm p = nRT/V. The work done during the irreversible expansion against the same final pressure is equal to the rectangular area shown slightly darker. Note that the reversible work is greater than the irreversible work.

- Expansion against constant pressure

Piston e.g. pressed on by the atmosphere, which exerts the same pressure throughout the expansion constant external pressure pex, which can be taken outside the integral:

This type of expansion is irreversible.

f

i

V

ex ex f i

V

ex

w p dV p (V V)

w p V

- Isothermal reversible expansion

A reversible change is a change that can be reversed by an infinitesimal modification of a variable. The system is in equilibrium with its surroundings, and the pressure p=pex at each stage p=nRT/V:

The maximum work available from a system operating between specified initial and final states and passing along a specified path is obtained when the change takes place reversibly.

f

i

V

f

iV

dV Vw nRT nRT ln

V V

38

Adiabatic changes

When a perfect gas expands adiabatically, a change in temperature is to be expected:

• Because work is done, the internal energy falls, and therefore the temperature of the working gas also falls.

• In molecular terms, the kinetic energy of the molecules falls, so their average speed decreases, and hence the temperature falls.

Starting point: no exchange with surroundings, i.e. q = 0

V

UdU w pdV dT

T

i.e.: expansion work at the expense of U temperature change of the perfect gas

compression heatingexpansion cooling

it is:V

V

V

pdV nc dT

RTdV c dT

VRdlnV c dlnT

for a change of state from Vi, pi, Ti Vf, pf, Tf:

v

Rc

f f i fV

i i f i

V T V TR ln c ln

V T V T

Since R = cp – cv:

p

v

c1

ci f

f i

V T

V T

With T = pV/nR and the heat capacity ratio = cp/cv we find:

1 1f f i i

1 1f f i

f i i

i

fp V p V

T V T V const.

p T

cons

p T cons

t.

t

The variation of temperature as a perfect gas is expanded reversibly and adiabatically. The curves are labelled with different values of c = cV/R. Note that the temperature falls most steeply for gases with low molar heat capacity.

‘Poisson’s law’

20

39

The work of (reversible) adiabatic changes

To achieve a change of state from one temperature and volume to another temperature and volume, we may consider the overall change as composed of two steps. In the first step, the system expands at constant temperature; there is no change in internal energy if the system consists of a perfect gas. In the second step, the temperature of the system is increased at constant volume. The overall change in internal energy is the sum of the changes for the two steps.

An adiabat depicts the variation of pressure with volume when a gas expands reversibly and adiabatically. (a) An adiabat for a perfect gas. (b) Note that the pressure declines more steeply for an adiabat than it does for an isotherm because the temperature decreases in the former.

VW pdV dU nc dT

f

i

V

V f i

V

W pdV U nc (T T)

Starting point:

Provided the heat capacity is independent of temperature, the adiabatic expansion work is

40

The Carnot Cycle

Starting point: - ideal gas

- alternating isothermal and adiabatic expansion and compression

1) Isothermal expansion from V1 to V2

dT = 0, dU = 0

2) Adiabatic expansion from V2 to V3

Q = 0

3) Isothermal compression from V3 to V4

dT = 0, dU = 0

4) Adiabatic compression from V4 to V1

Q = 0

11

c2 1 2

1 h 4 3

TT V V

T T V V

Determination of T2 (Tc):

21

41

The Carnot Cycle (cont’d)

Step GasHot

SourceCold sink Work storage

1Isothermal, reversible expansion

+ Q(T1)-nRT1ln(V2/V1)

- Q(T1) -+nRT1ln(V2/V1)

2Adiabatic, reversible expansion

-ncv(T1 – T2) - -+ncv(T1 – T2)

3Isothermal, reversible compression

- Q(T2)+nRT2ln(V3/V4)

- + Q(T2) -nRT2ln(V3/V4)

4Adiabatic, reversible compression

+ncv(T1 – T2) - --ncv(T1 – T2)

0 - Q(T1) + Q(T2) +nR(T1-T2)ln(V2/V1)

The Carnot cycle transports heat from the reservoir 1 into the colder reservoir 2, and delivers work

21 2

1

VW nR(T T )ln

V

Suppose an energy qh (for example, 20 kJ) is supplied to the engine and qc

is lost from the engine (for example, qc

= -15 kJ) and discarded into the cold reservoir. The work done by the engine is equal to qh + qc (for example, 20 kJ + (-15 kJ) = 5 kJ). The efficiency is the work done divided by the heat supplied from the hot source.

Reversal: heat pump

42

The Carnot Cycle (cont’d)

Efficiency of the Carnot-Cycle:work output

supply of heat from the hot reservoir

21 2

1 1 2 2

21 1 11

1

VnR(T T )ln

W V T T T1

VQ(T ) T TnRT lnV

becomes larger with increasing T1 and decreasing T2

T2/T1 is the fraction of Q(T1) transferred as heat to the cold reservoir

The direction of spontaneous change

Experience 1

All spontaneous processes in nature proceed only in one direction.

A B A B+TC

with TA < TC < TB

TA TB

with TA < TB

Experience 2

All spontaneous processes in nature are irreversible. They cause a loss of useable work and lead to a gain in heat.

not required by the First Law of Thermodynamics!

22

43

The Second Law of Thermodynamics

This means:

A gain of work is possible only with simultaneous transport of heat (efficiency )

It is impossible to build an engine with a higher efficiency than the Carnot engine. This implies that a perpetual motion machine (perpetuummobile) of the second kind is not feasible.

(a) The demonstration of the equivalence of the efficiencies of all reversible engines working between the same thermal reservoirs is based on the flow of energy represented in this diagram. (b) The net effect of the processes is the conversion of heat into work without there being a need for a cold sink: this is contrary to the Kelvin statement of the Second Law.

Question:

Which state function lets us assess the direction of spontaneous change?

Answer:

The Entropy S !

Kelvin statement (Second Law of Thermodynamics):

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.

44

The Entropy S

Derivation via the Carnot process:

Q = 0 for adiabatic processes:

Extension to general cycles:

21

1 1 1

22 22

1

VnRT ln

Q(T ) V TVQ(T ) TnRT lnV

1 2

1 2

Q(T ) Q(T )

T T

“reduced heat”;state function!

1 2

1 2

1 2

1 2

Q(T ) Q(T )0 for reversiblechanges

T T

Q(T ) Q(T )< 0 incaseof irreversiblecontributions

T T

A general cycle can be divided into small Carnot cycles. The match is exact in the limit of infinitesimally small cycles. Paths cancel in the interior of the collection, and only the perimeter, an increasingly good approximation to the true cycle as the number of cycles increases, survives. Because the entropy change around every individual cycle is zero, the integral of the entropy around the perimeter is zero too.

Each cyclic process can be regarded as a sequence of two lines of the Carnot cycle:

dQrev

T = 0 

dQirrev

T < 0

23

45

The Entropy S (cont’d)

Introduction of the Entropy S as new state function, with (Clausius)

Consider e.g. an isothermal cyclic process, carried out irreversible from 1 to 2, and reversible from 2 to 1:

Isolated system: dQ = 0 S2 – S1 0 for irreversible processes

S2 – S1 = 0 for reversible processes

2irrev

2 1

1

dQ< S S

T

Other formulation of the Second Law of Thermodynamics:

The entropy of an isolated system increases in the course of a spontaneous change:

Stot, irrev > 0

It remains constant in the course of a reversible change:

Stot, rev = 0

dQdS

T

or, more general: Clausius inequality

dQ

T =

dQirrev

T

2

1

+ dQrev

T

1

2

= dQirrev

T +

2

1

dS < 0

1

2

revdQdS

T

46

Entropy changes accompanying specific processes

General expression:

Perfect gases:

Reactions:

Phase transitions:

CAUTION: Chemical reactions are (in general) irreversible, and can not be described via

rev

V T

rev

p T

dQS S dU pdVdS dT dV

T V T T

dQS S dH VdpdS dT dp

T p T T

v

p

cdS dT dV

T

cdS dT V dp

T

v pdS nc dlnT nRdlnV nc dlnT nRdlnp

2 2 2 22 1 v p

1 1 1 1

T V T pS S nc ln nRln nc ln nRln

T V T p

i ip, T, T,p T,p

S S S SdS dT dp d S s

T p

for cv, cp = const.

trstrs

trs

HS

T

for p, T = const.

revdQdS

T

24

47

The variation of entropy with temperature: a more general approach

usually cp is not independent of temperature

This leads to the

f f

i i

T Tprev

f i i

T T

C dTdQS(T ) S(T) S(T)

T T

The determination of entropy from heat capacity data. (a) The variation of Cp/T with the temperature for a sample. (b) The entropy, which is equal to the area beneath the upper curve up to the corresponding temperature, plus the entropy of each phase transition passed.

Nernst heat theorem:

The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero: S 0 as T 0 provided all the substances involved are perfectly ordered.

Third Law of Thermodynamics:

The entropy of all perfect crystalline substances is zero at T = 0.

48

Entropy: the statistical approach (or: Entropy as a measure for order)

Assume at total of N particles in k different states, with W being the “weight of a configuration”. An example: N = 5, k = 2

General expression:

(approximation via Stirling formula: ln W N·ln N - Ʃ ni·ln ni

for very large N)

Boltzmann derived the entropy to

S = k · ln W

with the Boltzmann constant

k = R/NA = 1.38066·10-23 J K-1

W =N!

n0!n1!n2!…

e.g. N = 10 and equal distribution {2,2,2,2,2} between 5 states: W = 362,880 different configurations! This leads to incredibly huge numbers for chemically relevant particle numbers!

n0 = 5, n1=0: W = 1

3 5 3 4 2 5 2 4 2 3 1 5 1 4 1 3 1 2 1 2 3 4 5 1 2 4 1 2 5 1 3 4 1 3 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5

n0 = 4, n1=1: W = 5

5 4 3 2 1 2 3 1 5 1 4 1 3 1 2 1 2 3 4 1 2 3 5 1 2 4 5 1 3 4 5 2 3 4 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5

n0 = 3, n1=2: W = 10

4 5 3 5 3 4 2 5 2 4 2 3 1 5 1 4 1 3 1 2 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5

n0 = 2, n1=3: W = 10

3 4 5 2 4 5 2 3 5 2 3 4 1 4 5 1 3 5 1 3 4 1 2 5 1 2 4 1 2 3 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5

n0 = 1, n1=4: W = 5

2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4 2 3 1 5 1 4 1 3 1 21 2 3 4 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5

n0 = 0, n1=5: W = 1

1 2 3 4 5 3 5 3 4 2 5 2 4 2 3 1 5 1 4 1 3 1 2 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5

Zentralfriedhof (Central Cemetery), Vienna

25

49

Concentrating on the system: The Helmholtz and Gibbs energies

• Entropy is the basic concept for the direction of spontaneous processes, but requires the simultaneous consideration of system and surroundings

• In the surroundings, any change is reversible.• Idea: focus the attention on the system, and simplify the evaluation of spontaneity.

Approach: System Surroundings

SurroundingsSystem

dS dS dS

dQdS 0

T

2nd law of thermodynamics

Surroundings SystemdQ dQ

1st law of thermodynamics

+

SystemSystem

dQdS 0

T Clausius inequality

(i) Isothermal and isobaric heat transfer

Qsystem = dH TdS – dH 0

from G H - T·S Gibbs Energy

follows (dG)p,T 0

(ii) Isothermal and isochoric heat transfer

Qsystem = dU TdS – dU 0

from A U - T·S Helmholtz Energy

follows (dA)V,T 0

50

Concentrating on the system: The Helmholtz and Gibbs energies (cont’d)

• The change in the Helmholtz function is equal to the maximum work accompanying a process: dWmax = dA

• The change in the Gibbs function is equal to the maximum non-expansion work accompanying a process: dWmax, non-expansion = dG

Top: In a system not isolated from its surroundings, the work done may be different from the change in internal energy. Moreover, the process is spontaneous if, overall, the entropy of the global, isolated system increases. In the process depicted here, the entropy of the system decreases, so that of the surroundings must increase in order for the process to be spontaneous, which means that energy must pass from the system to the surroundings as heat. Therefore, less work than U can be obtained.

Bottom: In this process, the entropy of the system increases; hence we can afford to lose some entropy of the surroundings. That is, some of their energy may be lost as heat to the system. This energy can be returned to them as work. Hence the work done can exceed U.

Summary:

dG,dA < 0: irreversible, spontaneous process

dG,dA = 0: reversible process, equilibrium

dG,dA > 0: thermodynamically impossible process

26

51

The “characteristic functions” and the “Maxwell relations”

• The ‘characteristic functions’ relate energies and enthalpies to other state variables.

• The ‘Maxwell relations’ yield additional interrelations between state functions and state variables

dU = Q + W = TdS – pdV

dH = dU + d(p·V) = TdS + Vdp

dA = dU – d(S·T) = -SdT – pdV

dG = dH – d(S·T) = -SdT + Vdp

S V

pS

V T

p T

T p

V S

T V

p S

p S

T V

V S

T p

S U V

+ H A -

p G T

The “Guggenheim scheme”

52

The ‘chemical potential’ and the ‘fundamental equations of thermodynamics’: Treatment of simple mixtures

In mixtures consisting of k components in one phase, which may react with each other, as well as in multi-phase systems the Gibbs and the Helmholtz energy may be written as

G = G(p,T,n1,n2,…nk)

A = A(V,T,n1,n2,…nk)

The partial derivative of G and A with respect to the amount of substance ni of the component i is called the chemical potential i:

j i j i

ii ip,T,n V,T,n

G A

n n

That is, the chemical potential is the slope of a plot of e.g. Gibbs energy against the amount of substance of the component i, with the pressure and temperature (and the amounts of all other substances) held constant. The chemical potential of a substance in a mixture is the contribution of that substance to the total Gibbs energy of the mixture, and plays the central role in chemical thermodynamics.

The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials are positive.

The Gibbs energy depends on the composition, pressure, and temperature of the system. Thus, G may change when these parameters change, and

dG = -SdT + Vdp + AdnA + BdnB + …

This expression is the fundamental equation of chemical thermodynamics.

A similar expression describes the change of the Helmholtz energy:

dA = -SdT - pdV + AdnA + BdnB + …

27

53

The ‘chemical potential’ (cont’d)

Change of G and A in chemical reactions:

It holds:

The dependence of µ on temperature T, pressure p and composition x

i i i ii ip,T V,T

G AG A

• In equilibrium (i.e. dG=0) the chemical potential i of substance imust be the same in all parts of the system under study.

• In case of dG<0 a spontaneous transition of substance from parts with larger chemical potential to those with lower chemical potential will occur.

• A chemical reaction can only take place when (G)p,T = ii < 0.

(i) pure substances:

p,T

Gµ µ g

n

and therefore

p T

µ µs , v

T p

54

The ‘chemical potential’ (cont’d)

thus for the “standard pressure” p° = 1 bar one finds for a pure and

µ(p) p p

µ(p ) p p

dµ vdp RT dln(p)

ideal

pµ µ RTln

p

real

fµ µ RTln

p

(ii) simple mixtures

assume a perfect gas in a container (C) in an ideal gas mixture, and compare this to the pure gas (A, B):

pure gas ip, T

pure gas ipi, T

gas mixturepi, T, pj=p

A B C*iµ (p,T)

*i iµ (p ,T)

connection permeable only for gas i, and equilibrium between B and C !

and

* * ii i i i

pµ (p ,T) µ (p,T) RTln RTlnx

p

*: pure compound

since

also valid for condensed phases!

in case of real mixtures, xi has to be replaced by an effective mole fraction, called the activity ai.

xi and ai are related by the activity coefficienti: ai=ixi.

a) perfect (ideal) gas:

b) real gas: to adapt the above equation to this case, the true pressure, p, has to be replaced by an effective pressure, called the fugacity f. The fugacity can be written as f=p, where is the fugacity coefficient.

μi p,T,xi  

μi p,T,xi - μi* p,T = 0 

μi p,T = μi* p,T +R∙T∙ln xi

μi p,T = μi* p,T + R∙T∙ln ai

28

55

Simple mixtures: Partial molar quantities

example: the volume of a mixture

a) perfect (ideal) mixtures: no interactions

* * * * *1 2 1 1 2 2 i iV V V ... n v n v ... nv

b) real mixtures: interactions

• Imagine a huge volume of water, and add 1 mol H2O. The volume will increase by 18 cm3, because 18 cm3 mol-1 is the molar volume of water.

• Now, add 1 mol H2O to a huge volume of pure ethanol. The volume will increase by 14 cm3 mol-1, which is the partial molar volume of water in pure ethanol. The reason for the different increase is that the volume occupied by a given number of water molecules depends on the identity of the molecules that surround them.

In general, the partial molar volume of a substance A in a mixture is the change in volume per mole of A added to a large volume of the mixture.

The partial molar volumes of the components of the mixture vary with composition because the environment of each type of molecule changes as the composition changes from pure A to pure B.

The partial molar volumes of water and ethanol at 25°C. Note the different scales (water on the left, ethanol on the right).

(analogous: e.g. U, H, cp, …)

56

Simple mixtures: Partial molar quantities (cont’d)

formal definition of the partial molar volume of a substance i at some general composition:

j i

ii p,T,n

Vv

n

e.g. binary mixture:

B A

A B A A B BA Bp,T,n p,T,n

V VdV dn dn v dn v dn

n n

and, after integration with fixed composition:

A A B B i iV n v n v nv The partial molar volume of a substance is the slope of the variation of the total volume of the sample plotted against the composition. In general, partial molar quantities vary with the composition, as shown by the different slopes at the compositions a and b. Note that the partial molar volume at b is negative: the overall volume of the sample decreases as a is added.

a general approach:

j i

i i ii n

i i i i p,i

n nn

: extensive property of the mixture

: partial molar property, e.g. v ,u ,h ,c ...

experimental determination: e.g. via mean molar quantities

i i 1 1 2 2i

x x x for a binary mixturen

1 2 2 22 2

d dx (1 x )

dx dx

29

57

The thermodynamics of mixing

a) The Gibbs energy of mixing

• assume the amounts of two perfect gases in two containers as nA and nB; both are at the same temperature T and pressure p. The Gibbs energy of this initial system, Gi, is then

• after mixing, the partial pressures of the gases are pA and pB, with p=pA+pB. The total Gibbs energy changes to

• the difference Gf – Gi, the Gibbs energy of mixing, mixG, is therefore

The Gibbs energy of mixing of two perfect gases and (also) of two liquids that form an ideal solution. The Gibbs energy of mixing is negative for all compositions and temperatures, so perfect gases mix spontaneously in all proportions.

i A A B B

A A B B

G n µ n µ

p pn µ RTln n µ RTln

p p

A Bf A A B B

p pG n µ RTln n µ RTln

p p

A Bmix A B

A A B B

p pG n RTln n RTln

p p

nRT x lnx x lnx

mix i ii

G nRT x lnx

58

The thermodynamics of mixing (cont’d)

b) The entropy of mixing

• because (G/T)p,n = -S it follows immediately that for a mixture of perfect gases, but also for an ideal solution, the entropy of mixing, mixS, is

• in case of ideal mixtures, from G = H - TS follows mixH=0. This is to be expected for a system in which there are no interactions.

The entropy of mixing of two perfect gases as well as condensed phases that form an ideal solution. The entropy increases for all compositions and temperatures, so perfect gases mix spontaneously in all proportions. Because there is no transfer of heat to the surroundings when perfect gases mix, the entropy of the surroundings is unchanged. Hence, the graph also shows the total entropy of the system plus the surroundings when perfect gases mix.

A B

mixmix

p,n ,n

A A B B

GS

T

nR x lnx x lnx

mix i ii

S nR x >lnx 0

c) Excess functions

• the thermodynamic properties of real solutions are expressed in terms of the excess functions, XE, the difference between the observed thermodynamic function of mixing and the function for an ideal solution. The excess entropy, SE, for example is defined as SE = mixS - mixSideal .

• deviations of the excess functions from zero indicate the extent to which the solutions are nonideal.

30

59

Real gases: The Joule-Thomson effect

• central to the technological problems associated with the liquefaction of gases

• experiment by Joule and William Thomson (later Lord Kelvin):

A diagram of the apparatus used for measuring the Joule-Thomson effect. The gas expands through the porous barrier, which acts as a throttle, and the whole apparatus is thermally insulated. This arrangement corresponds to an isenthalpic expansion (expansion at constant enthalpy). Whether the expansion results in a heating or a cooling of the gas depends on the conditions.

- expansion of a gas through a barrier from one constant pressure to another in an insulated apparatus, i.e. adiabatic process

- observation of an (in general) lower temperature on the low pressure side, the difference in temperature being proportional to the maintained pressure difference

- the expansion occurs without change of enthalpy:

H = U + pV = const.

A diagram representing the thermodynamic basis of Joule-Thomson expansion. The pistons represent the

upstream and downstream gases, which maintain constant pressures on either side of the throttle. The transition from the left diagram to the right diagram,

which represents the passage of a given amount of gas through the throttle, occurs without change of enthalpy.

The process:

60

Real gases: The Joule-Thomson effect (cont’d)

Thermodynamic treatment:

adiabatic changes (q=0) imply: U = Uf - Ui = w = piVi - pfVf

Ui + piVi = Uf+pfVf Hi = Hf

!

p T

H HdH dT dp 0

T p

with

The isenthalpic Joule-Thomson coefficient µ is then defined as:

T

pH

p

HpT

µHp cT

Simplest approach (Virial equation / van der Waals equation, via pv=RT+Bp and B=b-a/RT):

p

2ab

RTµc

Depending on the identity of the gas, the pressure, the relative magnitudes of the attractive and repulsive forces, the sign of the coefficient may be either positive or negative. At the inversion temperature Ti, µ becomes zero (see figure).

i

2aT

Rb

For a perfect gas, µ = 0.

The sign of the Joule-Thomson coefficient, , depends on the conditions. Inside the boundary, the shaded area, it is positive and outside it is negative. The temperature corresponding to the boundary at a given pressure is the `inversion temperature' of the gas at that pressure. For a given pressure, the temperature must be below a certain value if cooling is required but, if it becomes too low, the boundary is crossed again and heating occurs. Reduction of pressure under adiabatic conditions moves the system along one of the isenthalps, or curves of constant enthalpy. The inversion temperature curve runs through the points of the isenthalps where their slopes change from negative to positive.

31

61

Real gases: Application of the Joule-Thomson effect

The ‘Linde refrigerator’ makes use of Joule-Thomson expansion to liquefy gases. The gas at high pressure, which must have a temperature below the upper inversion temperature, is allowed to expand through a throttle; it cools and is circulated past the incoming gas. That gas is cooled, and its expansion cools it still further. There comes a stage when the circulating gas becomes so cold that it condenses to a liquid.

At a given pressure, a gas typically has two inversion temperatures, one at higher and the other at lower temperature.

The inversion temperatures for three real gases: nitrogen, hydrogen, and helium.

The principle of the Linde refrigerator is shown in this diagram. The gas is recirculated and, so long as it is beneath its inversion temperature, it cools on expansion through the throttle. The cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle.

62

Physical transformation of pure substances: Phase diagrams

• A phase of a substance is a form of matter that is uniform throughout the chemical composition and physical state.

• The phase diagram of a substance shows the regions of pressure and temperature at which its various phases are thermodynamically stable.

• The lines separating the regions, which are called phase boundaries, show the values of p and T at which two phases coexist in equilibrium.

The general regions of pressure and temperature where solid, liquid, or gas is stable (that is, has lowest chemical potential) are shown on this phase diagram. For example, the solid phase is the most stable phase at low temperatures and high pressures. Based on thermodynamic considerations, the precise boundaries between the regions can be located.

Critical points and boiling points:

• condition of free vaporization throughout a liquid: boiling.

• Temperature at which the vapour pressure of the liquid is equal to the external pressure: boiling temperature. For the special case of an external pressure of 1 bar, the boiling temperature is called the standard boiling point (for 1 atm: normal boiling point).

• Boiling does not occur when a liquid is heated in a closed vessel. At the critical temperature Tc, the densities of liquid and vapour become equal, and the surface between the two phases disappears. At and above Tc, a single uniform phase called a supercritical fluid fills the container.(a) A liquid in equilibrium with its vapour. (b) When a liquid is heated in a sealed container, the density of the vapour phase increases and that of the liquid decreases slightly. There comes a stage, (c), at which the two densities are equal and the interface between the fluids disappears. This disappearance occurs at the critical temperature. The container needs to be strong: the critical temperature of water is 374C and the vapour pressure is then 218 atm.

32

63

Physical transformation of pure substances: Phase diagrams (cont’d)

Melting points and triple points:

• Temperature at which, under a specified pressure, the liquid and solid phases of a substance coexist in equilibrium: melting temperature. For the special case of an external pressure of 1 bar, the freezing temperature is called the standard freezing point (for 1 atm: normal freezing point).

• Under a certain set of conditions three different phases of a substance (typically solid, liquid, and vapour) simultaneously coexist in equilibrium. These conditions are represented by the triple point. Its temperature is denoted as T3.

Typical phase diagrams:

Right: The experimentally determined phase diagram

for water showing the different solid phases. Note the change of vertical scale

at 2 atm.

Left: The experimental phase diagram for carbon dioxide. Note that, as the triple point lies at pressures well above atmospheric, liquid carbon dioxide does not exist under normal conditions (a pressure of at least 5.11 atm must be applied).

Video on critical phase transition of CO2 : see e.g. (https://www.youtube.com/watch?v=GEr3NxsPTOA)

64

Phase diagrams: Gibbs’ phase rule

General relation between the variance, F, the number of components, C, and the number of phases at equilibrium, P, for a system of any composition:

F = C – P + 2 (phase rule)

Justification: total number of variables (p, T, composition x1, x2, ... xC-1) in all P phasesminus (P-1) equilibrium conditions for p, T, and the chemical potentials µ1, µ2, ... µC

e.g. one-component systems

• P = 1 F=2 p, T independently variable

• P = 2 F=1 p or T variable, the other automatically set

• P = 3 F=0 p and T fixed

• P = 4 impossible

The typical regions of a one-component phase diagram. The lines represent conditions under which the two adjoining phases are in equilibrium. A point represents the unique set of conditions under which three phases coexist in equilibrium. Four phases cannot mutually coexist in equilibrium.

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65

Phase stability and phase transitions: The thermodynamic criterion of equilibrium

At equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.

When two or more phases are in equilibrium, the chemical potential of a substance (and, in a mixture, a component) is the same in each phase and is the same at all points in each phase.

The schematic temperature dependence of the chemical potential of the solid, liquid, and gas phases of a substance (in practice, the lines are curved). The phase with the lowest molar Gibbs energy at a specified temperature is the most stable one at that temperature. The transition temperatures, the melting and boiling temperatures, are the temperatures at which the chemical potentials of the two phases are equal.

66

Phase stability and phase transitions: The location of phase boundaries

When pressure is applied to a system in which two phases are in equilibrium (at a), the equilibrium is disturbed. It can be restored by changing the temperature, so moving the state of the system to b. It follows that there is a relation between dp and dT that ensures that the system remains in equilibrium as either variable is changed.

When two phases and are in equilibrium, their chemical potentials must be equal:

µ(p,T) = µ(p,T)

If either p or T is changed infinitesimally, and equilibrium shall be maintained, the changes in the chemical potentials must be equal, too (‘maintained equilibrium’):

dµ(p,T) = dµ(p,T)

With

dµ = -sdT + vdpfollows

-sdT + vdp = -sdT + vdpand

(v - v)dp = (s - s)dT

The Clapeyron equation is an exact expression and applies to any phase equilibrium of any pure substance!

trs trs

coex trs trs

s hp

T v T v

Clapeyron equation

Negative slope of the water solid-liquid boundary due to trsv < 0

34

67

Phase transitions: The liquid-vapour and solid-vapour boundary

Because the molar volume of a gas is so much greater than the molar volume of a liquid or solid (e.g. H2O at 298 K: vgas/vliquid > 1000), the volume of the condensed phase is often neglected, and the gas phase assumed to behave perfect (v = RT/p):

which arranges into the Clausius-Clapeyron equation for the variation of vapour pressure with temperature:

If temperature independence of the enthalpy of vaporization (sublimation) is assumed, this integrates to

trs

coex

hp

T T(RT /p)

trs trs

2coex coex

h hlnp lnp

T RT 1/ T R

trsh 1 1p p * e

R T T *

linear relation ln(p) 1/T

LEFT: A typical liquid-vapour phase boundary. The boundary can be regarded as a plot of the vapour pressure against the temperature. Note that, in some depictions of phase diagrams in which a logarithmic pressure scale is used, the phase boundary has the opposite curvature. That phase boundary terminates at the critical point (not shown).

RIGHT: Near the point where they coincide (at the triple point), the solid--gas boundary has a steeper slope than the liquid-gas boundary because the enthalpy of sublimation is greater than the enthalpy of vaporization and the temperatures that occur in the Clausius-Clapeyron equation for the slope have similar values.

68

Two-component systems: Ideal solutions and Raoult’s law

At equilibrium, the chemical potential of the gaseous form of a substance A is equal to the chemical potential of its condensed phase. The equality is preserved if a solute is also present. Because the chemical potential of A in the vapour depends on its partial vapour pressure, it follows that the chemical potential of liquid A can be related to its partial vapour pressure.

It was shown before, that for an ideal mixture the chemical potential is given by

From the conditions of equilibrium and maintained equilibrium a lowering of the vapour pressures of the components can be deduced, which is linearly dependent on their mole fractions in the liquid state:

This relation is known as Raoult’s law.

*i i iµ (p,T) µ (p,T) RTlnx

* *A A A B B Bp x p , p x p

The total vapour pressure and the two partial vapour pressures of an ideal binary mixture are proportional to the mole fractions of the components in the liquid phase.

A pictorial representation of the molecular basis of Raoult's law. The large spheres represent solvent molecules at the surface of a solution (the uppermost line of spheres), and the small spheres are solute molecules. The latter hinder the escape of solvent molecules into the vapour, but do not hinder their return.

35

69

Two-component systems: Ideal solutions and Raoult’s law (cont’d)

• Raoult’s law is closely obeyed by chemically similar components over the whole composition range, i.e. for both the solute as well as the solvent (ideal solutions).

• In case of dissimilar liquids significant deviations from Raoult’s law are observed. Nevertheless, the law is obeyed increasingly closely for the component in excess (the solvent) as it approaches purity. The law is therefore a good approximation for the properties of the solvent if the solution is dilute.

Two similar liquids, in this case benzene and methylbenzene (toluene), behave almost ideally, and the variation of their vapour pressures with composition resembles that for an ideal solution.

Strong deviations from ideality are shown by dissimilar liquids (in this case carbon disulfide and acetone (propanone)).

70

Two-component systems: Ideal-dilute solutions and Henry’s law

For real solutions at low concentrations, although the vapour pressure of the solute is proportional to its mole fraction in the liquid phase, the constant of proportionality is not the vapour pressure of the pure substance:

This relation is known as Henry’s law. The Henry constant KA (KB) is an empirical constant chosen so that the plot of the vapour pressure of A (B) against its mole fraction is tangent to the experimental curve at xA=0 (xB=0).

Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law are called ideal-dilute solutions.

A A A B B Bp x K , p x K

When a component (the solvent) is nearly pure, it has a vapour pressure that is proportional to the mole fraction with a slope pB

* (Raoult's law). When it is the minor component (the solute), its vapour pressure is still proportional to the mole fraction, but the constant of proportionality is now KB (Henry's law).

In a dilute solution, the solvent molecules (the blue spheres) are in an environment that differs only slightly from that of the pure solvent. The solute particles, however, are in an environment totally unlike that of the pure solute.

The experimental partial vapour pressures of a mixture of chloroform (trichloromethane) and acetone (propanone). The values of K are obtained by extrapolating the dilute solution vapour pressures as explained above.

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71

Two-component systems: Colligative properties

- elevation of the boiling point

- depression of the freezing point

- osmotic pressure

In dilute solutions, these properties depend only on the number of solute particles present, not their identity. For this reason, they are called

colligative properties}

Assumptions: - solute not volatile, so it does not contribute to the vapour

- solute not dissolvable in the solid solvent; that is, the pure solvent separates when the solution is frozen (quite drastic assumption, helps to deduce simple relations)

The chemical potential of a solvent in the presence of a solute. The lowering of the liquid's chemical potential

has a greater effect on the freezing point than on the boiling point because of the angles at which the lines

intersect (which are determined by entropies).

All the colligative properties stem from the reduction of the chemical potential of the liquid solvent as a result of the presence of the solute:

The reduction in chemical potential of the solvent implies that the liquid-vapour equilibrium occurs at a higher temperature (the boiling point is raised) and the solid-liquid equilibrium occurs at a lower temperature (the freezing point is lowered).

Note: If solutes dissociate in the solvent, this has to be taken into account in the mole fraction of the solvent!

* *i i i iµ (p,T) µ (p,T) RTlnx < µ (p,T)

72

Two-component systems: Colligative properties (cont’d)

a) The elevation of boiling point

(e.g. by sugar in water)

Equilibrium established at a temperature for which

µA*(g) = µA*(l) + RT ln xA

increase in normal boiling point from T* to T*+T, where

Kb is the ebullioscopic constant of the solvent, b its molality (mol solute / kg solvent).

2

bB b

vap

R T *T x K b

H

b) The depression of freezing point

Equilibrium established at a temperature for which

µA*(s) = µA*(l) + RT ln xA

decrease in normal freezing point by T’=T*-T, where

Kf is the cryoscopic constant of the solvent, b its molality (mol solute / kg solvent).

2

fB f

fus

R T *T ' x K b

H

Left: The heterogeneous equilibrium involved in the calculation of the elevation of boiling point is between B in the pure vapour and A in the mixture, A being the solvent and B an involatile solute.

Right: The heterogeneous equilibrium involved in the calculation of the lowering of freezing point is between A in the pure solid and A in the mixture, A being the solvent and B a solute that is insoluble in solid A.

The elevation of boiling point and the depression of freezing point may be used to measure the molar mass of a solute. However, the technique is nowadays of little more than historical interest.

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73

Two-component systems: Colligative properties (cont’d)

c) Osmosis and the van’t Hoff equation

- The phenomenon of osmosis is the spontaneous passage of pure solvent into a solution separated from it by a semipermeable membrane, a membrane permeable to the solvent (A) but not to the solute (B).

- The osmotic pressure, , is the pressure that must be applied to the solution to stop the influx of solvent. At equilibrium:

µA*(p) = µA(xA,p+)

- Thermodynamic treatment:

This law is know as the van’t Hoff equation. nB is the amount of substance of the solute, [B] its molar concentration.

- Importance of osmosis: osmometry, the determination of molar masses of e.g. macromolecules by the osmotic pressure (for example, [B] = 1 mol/dm3 =25 bar!)

The equilibrium involved in the calculation of osmotic pressure, , is between pure solvent A at a pressure p on one side of the semipermeable membrane and A as a component of the mixture on the other side of the membrane, where the pressure is p + .

In a simple version of the osmotic pressure experiment, A is at equilibrium on each side of the membrane when enough has passed into the solution to cause a hydrostatic pressure difference.

V = nBRT or = [B]RT

74

Two-component systems: Vapour pressure diagrams

The variation of the total vapour pressure of a binary mixture with the mole fraction of A in the liquid when Raoult's law is obeyed.

The composition of the vapour

- Partial vapour pressures of the components of an ideal solution of two volatile liquids

Raoult’s law pA = xApA* pB = xBpB*

- Total vapour pressure of the mixture:

p = pA + pB = xApA* + xBpB* = pB* + (pA* - pB*)xA

- Mole fractions y in the gas:

Dalton’s law

The gas is richer than the liquid in the more volatile component!

A BA B

p py y

p p

A AA B A

B A B A

x p *y y 1 y

p * (p * p *)x

The dependence of the total vapour pressure of an ideal solution on the mole fraction of A in the entire system. A point between the two lines corresponds to both liquid and vapour being present; outside that region there is only one phase present. The mole fraction of A is denoted zA.

The general scheme of interpretation of a pressure-composition diagram (a vapour pressure diagram).

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75

Two-component systems: Vapour pressure diagrams (cont’d)

The points of the pressure-composition diagram discussed in the text. The vertical line through a is an isopleth, a line of constant composition of the entire system.

Consider the effect of lowering of the pressure on a liquid mixture of overall composition a:

At p=p1 the liquid can exist in equilibrium with its vapour. The composition of the vapour phase is given by a1’. The line joining two points representing phases in equilibrium is called a tie line. At this pressure, there is virtually no vapour present (see lever rule).

Lowering the pressure to p2 takes the system along the isoplethto an overall composition a2’’. This new pressure is below the vapour pressure of the original liquid, so it vaporizes until the pressure of the remaining liquid reaches p2. The composition of such a liquid must be a2. The equilibrium composition of the corresponding vapour is a2’.

Since two phases are in equilibrium F=1 between the lines, i.e. for a given pressure p the vapour and liquid phases have fixed compositions.

Further reduction of pressure to p3 leads to compositions a3 and a3’ for the liquid and vapour phase, respectively. The amount of liquid present is now virtually zero.

Further decrease in pressure takes the system to a4. Only vapour is present and its composition equal to the initial composition.

The lever rule. The distances l and l are used to find the proportions of the amounts of phases (such as vapour) and (for example, liquid) present at equilibrium. The lever rule is so called because a similar rule relates the masses at two ends of a lever to their distances from a pivot(ml = ml for balance).

76

Two-component systems: Temperature-composition diagrams and distillation

The temperature-composition diagram corresponding to an ideal mixture with the component A more volatile than component B. Successive boilings and condensations of a liquid originally of composition a1 lead to a condensate that is pure A. The separation technique is called fractional distillation.

A temperature-composition diagram is a phase diagram in which the boundaries show the composition of the phases that are in equilibrium at various temperatures and a given pressure, typically 1 atm. The liquid phase now lies in the lower part of the diagram.

Temperature-composition diagrams are particularly important for the distillation of mixtures. Consider the heating of a liquid of composition a1:

At the temperature T2, the liquid will start to boil. It has the composition a2 (=a1), the vapour (which is presently only as a trace) has composition a2’. The vapour is richer in the more volatile component A.

In a simple distillation, the vapour is withdrawn and condensed. This technique is used to separate a volatile liquid from a non-volatile solid or liquid. In fractional distillation, the boiling and condensation cycle is repeated successively. In the example, the first condensate boils at T3. It has the composition a3 (=a2’), the

vapour has composition a3’. Its first drop condenses to a liquid of composition a4.

The number of theoretical plates is the number of steps needed to bring about a specified degree of separation of two components in a

mixture. The two systems shown correspond to (a) 3, (b) 5 theoretical plates.

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77

Two-component systems: Temperature-composition diagrams and azeotropes

Many temperature-composition diagrams resemble the ideal version. In case of strong interactions between A and B molecules maxima or minima may occur:

Attractive interactions: The A-B interactions stabilize the liquid and reduce the vapour pressure of the mixture below the ideal value, corresponding to an increase in boiling temperature. Examples of this behaviour include trichloromethane/propanone and nitric acid/water mixtures.

Repulsive interactions: The A-B interactions destabilize the liquid relative to the ideal solution. They increase the vapour pressure of the mixture above the ideal value, corresponding to a decrease in boiling temperature. Examples of this behaviour include dioxane/water and ethanol/water mixtures.

At a maximum or minimum, the composition of liquid and vapour is the same. The mixture is said toform an azeotrope. A mixture of azeotropic composition can not be separated by distillation.

Left: A high-boiling azeotrope. When the liquid of composition a is distilled, the composition of the remaining liquid changes towards b but no further.

Right: A low-boiling azeotrope. When the mixture at a is fractionally distilled, the vapour in equilibrium in the fractionating column moves towards b and then remains unchanged.

78

Liquid-solid phase diagrams

Consider systems where solid and liquid phases may both be present at temperatures below the boiling point. In the example to the left, the following changes occur upon cooling the two-component liquid of composition a1:

Knowledge of the temperature-composition diagrams for solid mixtures guides the design of important industrial process, such as the manufacture of semiconductors and liquid crystal displays.

a1 a2. The system enters the two-phase region labelled ‘Liquid + B’. Pure solid B begins to come out of the solution and the remaining liquid becomes richer in A.

a2 a3. More of the solid forms, and the relative amounts of liquid and solid (which are in equilibrium) are given by the lever rule. At this stage there are roughly equal amounts of each. The liquid phase (composition b3) is richer in A than before because some B has been deposited.

a3 a4. At the end of this step, there is less liquid than at a3, and its composition is given by e. This liquid, which has the eutectic composition, now freezes to give a two-phase system of pure B and pure A.

The temperature-composition phase diagram for two almost immiscible solids and their com-pletely miscible liquids. The isopleth through e corresponds to the eutectic composition, the mixture with lowest melting point.

Eutectics

A solid with the eutectic composition melts without a change in composition at a single temperature, the lowest temperature of any mixture. The corresponding liquid freezes at the same single temperature, without previously depositing solid A or B. Eutectics are often of technological importance (solder 67% tin, 33% lead Tm=183°C; 23% rock salt, 77% water Tm=-21.1°C)

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79

Liquid-solid phase diagrams (cont’d)

The previous discussion was based on the assumption, that the components are completely miscible in the liquid state, and almost immiscible in the solid state.

In reality, often complete or partial miscibility is given in the solid state, too. Selected examples are given below.

The temperature-composition diagram of germanium – silicon. The components are completely miscible in the liquid as well as the solid state, and no eutectic behaviour is observed.

The temperature-composition diagram of silver – copper. The components are completely miscible in the liquid state, and partially miscible in the solid state. Eutectic behaviour is observed.

The temperature-composition diagram of potassium chloride –lithium chloride. The components are completely miscible in the liquid state, but not miscible in the solid state. Eutectic behaviour is observed.

Additional scenarios, with e.g. reacting systems, or incongruent melting where a compound is not stable as a liquid, are possible.

Ge

Si Ag

Cu KCl

LiCl

80

Chemical equilibrium: The Gibbs energy minimum

As the reaction advances (represented by motion from left to right along the horizontal axis) the slope of the Gibbs energy changes. Equilibrium corresponds to zero slope, at the foot of the valley.

The reaction Gibbs energy, rG, is defined as the slope of the graph of the Gibbs energy plotted against the extent of the reaction :

Although normally signifies a difference in values, here r

signifies a derivative, the slope of G with respect to . However, there is a close relationship with the normal usage. Assume e.g. the equilibrium reaction

A + B C + D

The corresponding change in Gibbs energy is

dG = µAdnA + µBdnB + µCdnC + µDdnD

= (AµA + BµB + CµC + DµD)d

That is, rG can be interpreted as the difference between the chemical potentials of the reactants and products at the composition of the reaction mixture.

Spontaneity of a reaction at constant temperature and pressure:

rG<0: the forward reaction is spontaneous (exergonic reaction)

rG>0: the reverse reaction is spontaneous (endergonic reaction)

rG=0: the reaction is at equilibrium

r

p,T

GG

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81

Chemical equilibrium: The description of equilibrium

Chemical reactions take place in real mixtures, so

and

The reaction quotient, Q, has the form with each species raised to the power given by its stoichiometric coefficient.

0i i i(p,T) (p,T) RTlna

r i iip,T

GG

i

i

0 0r i i i i r i

i i i

0 0r i r

i

G RT lna G RT lna

G RT ln a G RT lnQ

activities of productsQ

activities of reac tants

At equilibrium, the slope of G is zero: rG = 0. The activities then have their equilibrium values and one can write

An equilibrium constant expressed in terms of activities (or fugacities) is called a thermodynamic equilibrium constant. Note that, because activities are dimensionless numbers, the thermodynamic equilibrium constant is also dimensionless.It shows the position of equilibrium.

In elementary applications, the activities in this so-called law of chemical equilibrium (or law of mass action) are often replaced by the numerical values of molalities or molar concentrations, and fugacities are replaced by pressures. In either case, the resulting expressions are only approximations.

0r

i

G

RTi

i equlibrium

K a e ( const. for given p,T!)

82

Chemical equilibrium: The relation between equilibrium constants

Expression of the equilibrium constant in terms of partial pressure pi, mole fraction xi, concentration ci

or molality mi, and their correlation to the thermodynamic equilibrium constant:

All ci0 are 1 mol dm-3, and all mi

0 1 mol solute (kg solvent)-1. The activity coefficients I, yi and yi” approach unity in dilute solutions.

If pure solid or liquid substances take part in the equilibrium reaction, their activity is 1, and they do not appear in the law of chemical equilibrium:

i

i i

ii i

i i

ii equilibrium

x i x ii iequilibrium equilibrium

0p i p i

i iequilibrium equilibrium

c i c i ii iequilibrium equilibrium

K a

K x K K

K p K K (p )

K c K K y (c

i

ii i

i

0

" 0m i m i

i iequilibrium equilibrium

)

K m K K y (m )

e.g. Boudouard reaction C graphite + CO2 ⇌ 2 CO : K =a(CO)2

a(C)∙a(CO2)equilibrium

= a(CO)2

a(CO2)equilibrium

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83

The response of equilibria to temperature and pressure

Le Chatelier’s principle:

A system at equilibrium, when subjected to a disturbance, responds in a way that tends to minimize the effect of the disturbance (‘escape from constraint’).

Effect of temperature

Exothermic reactions: increased temperature favours the reactants

Endothermic reactions: increased temperature favours the products

Effect of pressure

Reactions with rV>0: increased pressure favours the reactants

Reactions with rV<0: increased pressure favours the products

When a reaction at equilibrium is compressed (from a to b), the reaction responds by reducing the number of molecules in the gas phase (in this case by producing the dimers represented by the ellipses).

e.g. ammonia synthesis

rH298 = -45.94 kJmol-1, rV < 0

for thermodynamic reasons the reaction should be conducted at low temperatures and high pressures.

Modern ammonia plants: conversion of synthesis gas (N2, H2) at ~250 atm and 450°C with catalyst of iron containing potassium and aluminum oxide promoters (Haber-Bosch process).

1

2N2+

3

2H2 ⇌ NH3

84

The response of equilibria to temperature and pressure: Exact treatment

(i) Temperature dependence

From

one obtains

(van’t Hoff’s isobar)

Integration over small temperature ranges, i.e. with rH0 almost independent of temperature:

0r

0r

p

GG lnK 1 Tln K and

RT T R T

0 0r r

2p p

lnK H lnK H(a) and (b)

T RT 1/ T R

When ln K is plotted against l/T, a straight line is expected with slope equal to -rH0/R. This is a non-calorimetric method for the measurement of reaction enthalpies.

2

1

1/ T0

2 1 r

1/ T

1ln K(T ) ln K(T ) H d(1/ T)

R

0r

2 12 1

H 1 1ln K(T ) ln K(T )

R T T

(ii) Pressure dependence

0r

T

lnK V

p RT

• “escape from constraint”

• K is almost independentof pressure in condensed phases

0r

T T

lnK 1 G

p RT p

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85

The physical liquid interface: Surface tension

Molecules at a liquid-vapour interface have less nearest neighbours than molecules in the bulk liquid, and will thus have different thermodynamic properties. These surface effects may be expressed in the language of Helmholtz and Gibbs energies:

dA is the work needed to change the surface area, A, by an infinitesimal amount. The constant of proportionality, , is called the surface tension. Its dimensions are energy/area and its units are typically Jm-2 or Nm-1. The surface tension is responsible for the spherical shape of droplets, which have the smallest surface/bulk ratio. (Note: Sometimes the symbol is still in use for the surface tension!)

Comparison of different liquids is eased by normalization with respect to the particle density in the interface:

mol = v2/3 ,

where mol is the molar surface tension and v the molar volume of the liquid. mol as well as depend on temperature, and must become zero at the critical temperature Tc:

mol = a((Tc – 6 K) – T)

This empirical rule was found by the Hungarian physicist Eötvös. The constant a is about 2.110-4 mJK-1mol-2/3

for non-associated liquids, like C5H10, C6H12, C6H6, or O2 ( principle of corresponding states).

mo

l

dG= ∂G

∂Tp,A

dT + ∂G

∂pT,A

dp + ∂G

∂Ap,T

dA = -SdT + Vdp + γdA

86

The physical liquid interface: Vapour pressure above curved surfaces

The vapour pressure of a liquid depends on the curvature of its surface. The pressure on the concave side of an interface, pin, is always greater than the pressure on the convex side, pout. This relation is expressed by the Laplace equation:

Thus the vapour pressure of a liquid which is dispersed as droplets (a small volume of liquid at equilibrium surrounded by its vapour) of radius r is

where p* is the vapour pressure above a flat surface. One implication of this Kelvin equation is the growth of larger droplets at the expense of smaller ones. A similar phenomenon is found for crystal growth, and known as ‘Ostwald ripening’. These effects are important for e.g. meteorology (cloud formation) and technical applications (supersaturated vapour phases, superheated and supercooled liquids)

in out

2p p

r

The dependence of the pressure inside a curved surface on the radius of the surface, for two different values of the surface tension.

2 v

rRTp p * e

44

87

equilibrium in macroscopic systems

THERMODYNAMICSELECTROCHEMISTRY

change of concentration as a function of time

(macroscopic) KINETICS(ELECTROCHEMISTRY)

stationary states of particles (atoms, molecules, electrons, nuclei) e.g. during translation, rotation, vibration

STRUCTURE OF MATTERCHEMICAL BOND

•bond breakage and formation

•transitions between quantum states

STRUCTURE OF MATTER(microscopic) KINETICS

CHEMICAL BOND

STATISTICAL THEORY OF MATTER

PHYSICAL CHEMISTRY: An Introduction

macroscopicphenomena

microscopicphenomena

static phenomena dynamic phenomena

88

The rates of chemical reactions

The definition of rate

Consider a reaction of the form AA + BB CC + DD.

A unique rate of reaction, v, can be defined as the rate of change of the extent of reaction, :

(Remember that i is negative for reactants and positive for products).

For a homogeneous system these expression are often divided by the (constant) volume of the system, and the reaction rate expressed in terms of concentrations:

For heterogeneous reactions division by the surface area of the species i leads to

where i is the surface density of i.

The definition of (instantaneous) rate as the slope of the tangent drawn to the curve showing the variation of concentration with time. For negative slopes, the sign is changed when reporting the rate, so all reaction rates are positive.

CA B D

A B C D

dn1 dn 1 dn 1 1 dn d molv

dt dt dt dt dt s

3i

1 d[i] 1 d molv

dt V dt dm s

i2

i

1 d molv

dt m s

45

89

The rates of chemical reactions

Rate laws, rate constants, and reaction order

• In virtually all chemical reactions that have been studied experimentally, the reaction rate depends on the concentration of one or more of the reactants. In general, the rate may be expressed as a function of these concentrations: v = f ([A], [B], … ).

• The most frequently encountered functional dependence is the rate's being proportional to a product of algebraic powers of the individual concentrations, i.e.,

v [A]a[B]b

The exponents a and b may be integer, fractional, or negative. This proportionality can be converted to an equation by inserting a proportionality constant k, thus:

v = k [A]a[B]b

This equation is called a rate law, rate equation or rate expression. The exponent a is the order of the reaction with respect to reactant A, and b is the order with respect to reactant B. The proportionality constant k is called the rate constant. The overall order of the reaction is simply p = a + b.

• The rate law of a reaction is determined experimentally, and cannot in general be inferred from the chemical equation of the reaction. The reaction of hydrogen and bromine, for example, has a very simple stoichiometry, H2(g) + Br2(g) 2 HBr(g), but its law is complicated:

• In certain cases the rate law does reflect the stoichiometry of the reaction; but that is either a coincidence or reflects a feature of the underlying reaction mechanism.

1/ 22 2

2

k [H ] [Br ]v

[HBr]1 k '

[Br ]

90

The rates of chemical reactions (cont’d)

Elementary reactions and molecularity

• Most reactions occur in a sequence of steps called elementary reactions, each of which involves only a small number of atoms, molecules or ions. An elementary reaction itself proceeds in a single step.

• The molecularity of an elementary reaction is the number of molecules coming together to react in it. Common are unimolecular reactions, in which a molecule shakes itself apart or its atoms in a new arrangement, as in the isomerization of cyclopropane to propene. In a bimolecular reaction, a pair of molecules collide and exchange energy, atoms, or groups of atoms, or undergo some other kind of change (e.g. F + H2 HF + H). Three reactants that come together to form products constitute a termolecular reaction. Reactions with four, five, etc. reactants involved in an elementary reaction have not been encountered in nature.

• It is important to distinguish molecularity from order: reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism.

• Molecularity and overall order of an elementary reaction are the same!

• If the elementary steps of a complex mechanism are known, the overall rate law can be deduced!

Isomerization as a typical unimolecular elementary reaction.

46

91

Simple integrated rate laws

First-order reactions, half-lives and time constants

• e.g. decomposition reactions, isomerizations, radioactive decay

exponential decay

• The half-life t1/2 of a substance is the time taken for the concentration of a reactant to fall to half its initial value:

(Only) for a first-order reaction, the half-life of a reactant is independent of its initial concentration.

• The time-constant of a first-order reaction is thetime required for the concentration of a reactant to fall to 1/e of its initial value:

The exponential decay of the reactant in a first-order reaction. The larger the rate constant, the more rapid the decay: here klarge

= 3ksmall.

0

[A] t

[A] 0

d[A] d[A]v k [A] k dt

dt [A]

kt0

0

[A]ln kt [A] [A] e

[A]

The determination of the rate constant of a first-order reaction: a straight line is obtained when ln [A] (or, as here, ln p) is plotted against t; the slope gives k.

1/ 2

ln2t

k

1

k

92

Simple integrated rate laws (cont’d)

Second-order reactions

(i) A + A products

0

[A] t2

2[A] 0

d[A] d[A]v k [A] k dt

dt [A]

0

0 0

[A]1 1kt [A]

[A] [A] 1 kt[A]

The variation with time of the concentration of a reactant in a second-order reaction. The grey line is the corresponding decay in a first-order reaction with the same initial rate. For this illustration, klarge

= 3ksmall.

(ii) A + B products

it follows from the reaction stoichiometry that, when the concentration of A has fallen to [A]0-x, the concentration of B will have fallen to [B]0-x:

The integral on the left is evaluated by using the method of partial fractions and by using [A]=[A]0 and [B]=[B]0 at t=0 to give:

Therefore, a plot of the expression to the left against t should be a straight line from which k can be obtained.

For [A]0=[B]0 the solutions are identical to those given to the left.

d[A]v k [A] [B]

dt

x t

0 00 0

dxk dt

[A] x [B] x

0

0 0 0

[B] /[B]1ln kt

[B] [A] [A] /[A]

47

93

Simple integrated rate laws (cont’d)

Reversible first-order reactions

Net change of concentration:

If the initial concentration of A is [A]0 and no B is presentinitially, than [A] + [B] = [A]0 at all times

Solution:

Equilibrium concentrations:The approach of concentrations to their equilibrium values as predicted for a reaction A B that is first-order in each direction, and for which k = 2k’.

d[A]k [A] k ' [B]

dt

0 0

d[A]k [A] k ' [A] [A] (k k ')[A] k '[A]

dt

(k k ')t0

(k k ')t0

[A][A] k ' k e

k k 'k [A]

[B] 1 ek k '

0 0eq eq 0

k ' [A] k [A][A] [B] [A] [A]

k k ' k k '

Ak⇌k'

B

94

Acceleration of reaction rates: Temperature

Rule of thumb

The rate constants of most reactions increase as the temperature is raised. For many reactions in solution a temperature increase by 10°C causes an increase in reaction rate by a factor of 2-4.

The Arrhenius equation

In 1889 Svante Arrhenius found on the basis of numerous experimental rate measurements that rate constants varied as the negative exponential of the absolute temperature:

The parameter A, which corresponds to the intercept of the line in the plot ln(k)=f(1/T) at 1/T=0 (at infinite temperature), is called the pre-exponential factor or the frequency factor. The parameter Ea, which is obtained from the slope of the line (-Ea/R), is called the activation energy. Collectively, the two quantities are called the Arrhenius parameters.

aE /RT

a

k A e

Elnk lnA

RT

The Arrhenius plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the Arrhenius equation. The slope gives Ea/R and the intercept at 1/T = 0 gives ln A.

k

Temperature T

A

Plot of the rate constant k against temperature. Starting from k=0 at T=0, it approaches A for T .

48

95

Acceleration of reaction rates: Interpretation of the Arrhenius parameters

A potential energy profile for an exothermic reaction. The horizontal axis is the reaction coordinate, and the vertical axis is potential energy. The activated complex is the region near the potential maximum, and the transition state corresponds to the maximum itself. The height of the barrier between the reactants and the products is the activation barrier of the reaction.

• Consider how the potential energy changes in the course of a chemical reaction that begins with a collision between molecules of A and molecules of B. As the reaction proceeds, A and B come into contact, distort, and begin to exchange or discard atoms. The reaction coordinate is the collection of motions, such as changes in interatomic distances and bond angles, that are directly involved in the formation of products from reactants.

• The potential energy rises to a maximum and the cluster of atoms that corresponds to the region close to the maximum is called the activated complex. After the maximum, the potential energy falls as the atoms rearrange in the cluster, and it reaches a value characteristic of the products. The climax of the reaction is at the peak of the potential energy, which corresponds to the activation energy Ea. Here two reactant molecules have come to such a degree of closeness and distortion that a small further distortion will send them in the direction of products. This crucial configuration is called the transition state of the reaction.

• The activation energy is the minimum kinetic energy that reactants must have in order to form products. For example, in a gas-phase reaction there are numerous collisions each second, but only a tiny proportion are sufficiently energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess of an energy Ea is given by the Boltzmann distribution as exp(-Ea/RT). Hence, the exponential factor can be interpreted as the fraction of collisions that have enough kinetic energy to lead to reaction.

• The pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Hence, the product of A and the exponential factor, exp(-Ea/RT), gives the rate of successful collisions.

Ea

96

Acceleration of reaction rates: Catalysis

Definition

A catalyst is a substance that accelerates a reaction, but undergoes no net chemical change (Ostwald, 1907). Usually only small quantities of the catalyst are required for significant effects.

• A catalyst lowers the activation energy of the reaction by providing an alternative path that avoids the slow, rate-determining step of the uncatalysed reaction.

• The acceleration occurs without alteration of the general energy relations. A reaction, which is impossible without a potential catalyst for thermodynamic reasons, is still impossible in its presence.

• For the same reason, a catalyst does not disturb the final equilibrium composition of the system, only the rate at which that equilibrium is approached.

A catalyst provides a different path with a lower activation energy. The result is an increase in the rate of formation of products.

• Homogeneous catalysis: Catalyst and reactants are in the same phase (e.g. decomposition of H2O2 in aqueous solution, catalysed by bromide ions or catalase, which is an enzyme, i.e. a biological catalyst)

• Heterogeneous catalysis: Catalyst and reactants are in different phases (e.g. hydrogenation of ethene to ethane in the presence of a solid catalyst such as palladium, platinum or nickel)

• Autocatalysis: Catalysis of a reaction by the products.

49

97

Heterogeneous catalysis

• About 20 % of the value of all commercial products manufactured in the USA are derived from processes involving catalysis, the vast majority involving heterogeneous catalysis.

• Estimated heterogeneous catalyst market: $6.5 billion in 2000.

• Estimated costs of catalysts: about 0.1% of the value of fuels produced, about 0.22% of chemicals.

An example: The three-way catalyst

Conversion of the main pollutants CO, NOx, and hydrocarbons in a relatively narrow window of the air-to-fuel ratio:

CO + ½O2 CO2

NO + CO ½ N2 + CO2

CmHn + (m+n/4)O2

m CO2 + n/2 H2O

Elementary steps of heterogeneous catalysis:

• (dissociative) adsorption• reaction• desorption

CO O2

desorption

reaction

diffusion

adsorptionprecursor

dissoziation

CO2

Pt

98

Adsorption isotherms

The Langmuir isotherm

Simplest physically plausible isotherm, based on three assumptions:

• Adsorption cannot proceed beyond monolayer coverage.

• All sites are equivalent and the surface is uniform.• The ability of a molecule to adsorb at a given site

is independent of the occupation of neighbouring sites (i.e. no interactions between adsorbed molecules).

Dynamic equilibrium:

Rate of change of surface coverage due to adsorption (p: partial pressure of A, N: total number of surface sites, N(1-): number of vacant sites):

Rate of change due to desorption:

a

d

k

kA(g) M(surface) AM(surface)

a

dk pN(1 )

dt

d

dk N

dt

The Langmuir isotherm for non-dissociative adsorption for different values of K.

The Langmuir isotherm for dissociative adsorption (X22X) for different values of K.

a

d

kKpK

1 Kp k

Langmuir isothermfor non-dissociative adsorption

for dissociative adsorption:

Kp

1 Kp

50

99

Adsorption and catalysis

The Langmuir-Hinshelwood mechanism

In the Langmuir-Hinshelwood (LH) mechanism of surface-catalysed reactions, the reaction takes place by encounters between molecular fragments and atoms adsorbed on the surface. A second-order rate law is expected:

If A and B follow Langmuir isotherms:

and

LHkLH A BA B Pr v k

A A B BA B

A A B B A A B B

K p K p

1 K p K p 1 K p K p

LH A A B B

2

A A B B

k K p K pv

1 K p K p

Variation of the product formation rate d[Pr]/dt in a LH mechanism for fixed partial pressure of B.

The Eley-Rideal mechanism

In the Eley-Rideal (ER) mechanism of surface-catalysed reactions, a gas-phase molecule collides with another molecule already adsorbed on the surface. It follows that the rate law should be

If A follows a Langmuir isotherms in the pressure range of interest:

ERkER B AA B Pr v k p

ER A A B

A A

k K p pv

1 K p

Variation of the product formation rate d[Pr]/dt in a ER mechanism for fixed partial pressure of B.

• Almost all thermal surface-catalysed reactions are thought to take place by the LH mechanism.

• Distinction between LH and ER mechanisms e.g. via molecular beam studies, or via dependence of the rate on the partial pressures (see figures).

100

More complicated reaction mechanisms: Consecutive first-order reactions

Reaction from the reactant through an intermediate to the product:(e.g. decay of a radioactive family, such as

where the times are half-lives)

Approach: Solution: Suppose that initially only A is present, and thatits concentration is [A]0.

Substitute this equation into the approach, andsolve the differential equation:

a bk kA I P 2.35 min 2.35 d239 239 239U Np Pu

a

a b

b

d[A]k [A]

dtd[I]

k [A] k [I]dtd[P]

k [I]dt

The concentrations of A, I, and P in the consecutive reaction scheme A I P, for ka = 10kb. If the intermediate I is in fact the desired product, it is important to be able to predict when its concentration is greatest.

ak t0[A] [A] e

ak tb a 0

d[I]k [I] k [A] e

dt

a b

b a

k t k ta0

b a

k t k ta b

0b a

k[I] e e [A]

k k

k e k e[P] 1 [A]

k k

51

101

The role of the rate-determining step

One feature of the above calculation is evident: There is a considerable increase in mathematical complexity as soon as the reaction mechanism has more than only one step. A reaction scheme involving many steps is nearly always insolvable analytically. Two approaches are possible:

- integrate the rate laws numerically, or- make approximations.

The latter is still widely used because it often leads to convenient expressions and more readily digestible results.

Suppose now that kb>>ka, so whenever an I molecule is formed it decays rapidly into P. Because

the above equation reduces to

Here, the rate of formation of the final product depends on only the smaller of the two rate constants. That is, the rate of formation of Pdepends on the rate at which I is formed, not on the rate at which I is changed into P. For this reason, the step A I Is called the rate-determining step of the reaction.

Comparison of the concentration profiles for the first (left) and the second (right) step being the rate-determining one. Compare also the steady-state approximation on the next page.

ak t0[P] 1 e [A]

In these diagrams of reaction schemes, heavy arrows represent fast steps, and light arrows represent slow steps. (a) The first step is rate-determining; (b) the second step is rate-determining; (c) although one step is slow, it is not rate-determining because there is a fast route that circumvents it.

e-kbt ≪ e-kat kb- ka≈ kb

102

The steady-state approximation

Similar remarks apply to more complicated reaction mechanisms, and in general the rate-determining step is the slowest step in a reaction and controls its overall rate. However, the rate-determining step is not just the slowest step: it must be slow and must be a crucial gateway for the formation of the products. If a faster reaction can also lead to the products, than the slowest step is irrelevant because the reaction can then be side-stepped.

The steady-state approximation assumes that, after an initial inductionperiod, an interval during which the concentration of intermediates, I, rises from zero, and during the major part of the reaction the rates of change of concentration of all reaction intermediates are negligible small:

When this approximation is e.g. applied to the consecutive first-ordermechanism:

and finally:

This is the same (approximate) result as before, but much quicker obtained.

d[I]0

dt

a b

a b

b a

0 k [A] k [I]

[I] k / k [A]

d[P]k [I] k [A]

dt

The basis of the steady-state approximation. It is supposed that the concentrations of intermediates remain small and hardly change during most of the course of the reaction.

P ≈ka[A]0 e-katdt = 1 - e-kat [A]0

t

0

52

103

Pre-equilibria

Assume a slightly more complicated reaction mechanism in which an intermediate I reaches a so-called pre-equilibrium with the reactants A and B:

A pre-equilibrium arises when the rates of formation of the intermediate and its decay back into reactants are much faster than its rate of formation of products; thus, the condition is possible when ka’ >> kb but not when kb >> ka’. Because we assume A, B, and I to be in equilibrium, we can thus write

Under the assumption that the rate of reaction of I to form P is too slow to affect the maintenance of the pre-equilibrium, the rate of formation of P may now be written as:

This rate law has the form of a second-order law.

It is easy to show that a pre-equilibrium mechanism of the type

results in an overall third-order reaction:

This type of mechanism is e.g. given in the reaction

2 NO + O2 2 NO2

a

a

k[I]K K

[A][B] k '

b b

a bb

a

d[P]k [I] k K[A][B]

dtk kd[P]

k[A][B] k k Kdt k '

2b

d[P]k K[A] [B]

dt

A + B ka⇌ka

' I

kb⟶ P 

A + B ka⇌ka

' I

I + B kb⟶ P

104

Chain reactions: The hydrogen-bromine reactionIn a chain reaction, a reaction intermediate produced in one step generates an intermediate in a subsequent step, then that intermediate generates another intermediate, and so on. The intermediates in a chain reaction are called chain carriers. In many cases, a chain reaction leads to a complicated rate law. An example is the hydrogen-bromine reaction, which was first experimentally investigated by Bodenstein and Lind in 1906:

The mechanism which can account for this reaction rate isgiven by the following steps (Christiansen, Herzberg andM. Polanyi, 1919/1920):

Initiation:

Propagation:

Retardation:

Termination:

The initiation is an example of a thermolysis, a reaction initiatedby heat or vigorous intermolecular collisions. In the termination step, the third body removes the energy of recombination. It turns out that only Br atom recombination is important.

A schematic representation of the mechanism of the reaction between hydrogen and bromine. Note how the reactants and products are shown as arms to the circle, but the intermediates (H and Br) occur only within the circle.

1/ 22 2

2 2

2

k [H ] [Br ]H (g) Br (g) 2HBr(g) v

[HBr]1 k '

[Br ]

1

2

3

4

5

k2

k2

k2

k2

k2

.

.

.

Br M 2Br M

Br H HBr H

H Br HBr Br

H HBr H Br

Br Br M

.

Br M *.

53

105

Chain reactions: the hydrogen-bromine reaction (cont’d)In the hydrogen-bromine reaction, the net rate of formation of the product HBr is:

This rate equation can be solved analytically when approximate solutions are used. Here, the steady-state approximation is applied to the intermediates H and Br, and the net rates of change of their concentrations are set equal to zero:

The steady-state concentrations of the intermediates are obtained by solving these two simultaneous equations and are

Note that [M] has cancelled. When these concentrations are substituted into the expression for d[HBr]/dt, we obtain

This equation has the same form as the empirical law, and the two empirical constants can be identified.

2 2 3 2 4

d[HBr]k [Br ][H ] k [H ][Br ] k [H ][HBr]

dt

2 2 3 2 4

21 2 2 2 3 2 4 5

d[H ]k [Br ][H ] k [H ][Br ] k [H ][HBr] 0

dt

d[Br ]2k [Br ][M] k [Br ][H ] k [H ][Br ] k [H ][HBr] 2k [Br ] [M] 0

dt

1/ 2 1/ 2 1/ 22 1 5 2 21

25 3 2 4

k (k / k ) [H ][Br ]k[Br ] [Br ] [H ]

k k [Br ] k [HBr]

1/ 2 1/ 21/ 22 1 5 2 2 1 4

24 5 3

3 2

2k (k /k ) [H ] [Br ] k kv k 2k ( ) , k '

k [HBr] k k1k [Br ]

106

Explosions

- Thermal explosions: A thermal explosion is a very rapid reaction arising from a rapid increase of reaction rate with increasing temperature. The temperature of the system rises if the energy releasedby an exothermic reaction cannot escape, and the reaction goes faster. The acceleration of the rateresults in an even faster rise of temperature, and the reaction becomes catastrophically fast.

- Chain-branching explosions: A chain-branching explosion occurs when the number of chain centres grows exponentially.

An example of both types of explosion is the reaction between hydrogen and oxygen:

2 H2(g) + O2(g) 2 H2O(g)

Although the net reaction is very simple, the mechanism is very complex and has not been fully elucidated (at least 8 main species, 16 reactions). Some of the major steps are:

Initiation:

Branching:

Propagation:

Termination:

A branching step is an elementary reaction that produces more than one chain carrier. Recall that an O atom has the ground-state configuration [He]2s22p4, with 2 unpaired electrons.

2

2

2

2 2

2

2 2

H 2H

H O O OH

O H OH H

OH H H H O

1H wall H M *

2

H O M

.

.

.

.

. M

.

HO *

54

107

Explosions (cont’d)

The occurrence of an explosion depends on the temperature and pressure of the system. The explosion regions for the hydrogen-oxygen reaction, the conditions under which explosion occurs, are shown in the graph.

The explosion limits for the H2 + O2

reaction. In the explosive regions the reaction proceeds explosively when heated homogeneously.

- At very low pressures the system is outside the explosive region and the mixture reacts smoothly. At these pressures the chain carriers produced in the branching steps can reach the walls of the container where they combine.

- Increasing the pressure along a vertical line in the illustration takes the system through the first explosion limit (provided the temperature is greater than about 730 K). The chain carriers react before reaching the walls and the branching reactions are explosively efficient.

- The reaction is smooth when the pressure is above the second explosion limit. The concentration of molecules in the gas is then so great that the radicals produced in the branching reactions recombine via termination steps. (In the last elementary step of the reaction mechanism, the third body (M) can remove excess energy. At low pressure three-particle collisions are unimportant and recombination is much slower. At higher pressures, when three-particle collisions are important, the explosive propagation of the chain slows down because the branching steps are diverted into simple propagation steps.)

- When the pressure is increased to above the third explosion limit, the reaction rate increases so much that a thermal explosion occurs.