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This Physical Chemistry lecture uses graphs from the
following textbooks:
P.W. Atkins, Physical Chemistry, 6. ed., Oxford University
Press, Oxford 1998
G. Wedler, Lehrbuch der Physikalischen Chemie, 4. ed.,
Wiley-VCH, Weinheim 1997
PHYSICAL CHEMISTRY:
An Introduction
static phenomena
macroscopic
phenomena
equilibrium in macroscopic
systems
THERMODYNAMICS
ELECTROCHEMISTRY
dynamic phenomena
change of concentration as a
function of time
(macroscopic) KINETICS
(ELECTROCHEMISTRY)
STATISTICAL THEORY
OF MATTER
microscopic
phenomena
stationary states of particles
(atoms, molecules, electrons,
nuclei) e.g. during translation,
rotation, vibration
• bond breakage and
formation
• transitions between quantum
states
STRUCTURE OF MATTER
CHEMICAL BOND
STRUCTURE OF MATTER
(microscopic) KINETICS
CHEMICAL BOND
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=6.0221023 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 and molality
 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).
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.
Homogeneous system:
The macroscopic properties are identical in
all parts of the system.
Heterogeneous system:
Phase:
The macroscopic properties jump at the
phase boundaries.
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:
stable
unstable
Equilibria in Thermodynamics:
metastable
H2O (water)
25°C
1 bar
H2 + ½ O2
25°C
1 bar
stable
metastable
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):
A
+
B
A
B
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.
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 higher
temperature to one at a lower temperature if the
two are in contact through a diathermic wall, as
in (a) and (c). However, if the two regions have
identical temperatures, there is no net transfer of
energy as heat even though the two regions are
separated by a diathermic wall (b). The latter
condition corresponds to the two regions being
at thermal equilibrium.
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 pressure
p of a gas at constant volume V, or the volume V of the gas for
constant pressure p:
(x)  a  x  b
x  x 0C
 (x)  100 C 
x100C  x 0C
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 (-273.15 C).
For the pressure p, this transforms to:


p0C

p
 100  


C
p

p
p

p
100C
0C 
 100C 0C
Observation:
For all (ideal) gases one finds
100 
p0C
 273.15  0.01
p100C  p0C
 Introduction of the thermodynamic temperature scale (in ‘Kelvin’):
T
p
 100 
K
p100C  p0C
and
T


 273.15
K C
Work, heat, and energy:
 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)
 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.
 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.
 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.
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 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.
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.
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.
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.:
p = f(T) for V=const.:
p = f(V) for T=const.:
“isobars”
“isochors”
“isotherms”
1
2
V = const.  ( + 273.15°C)
= const.’  T
(Charles, 1798; Gay-Lussac, 1802)
p = const.  ( + 273.15°C)
= const.’  T
3
p  V = const.
(Boyle-Mariotte; 1664/1672)
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
T0 ,p0 , v 0  v 
v0
T
T0
Step 2: Isothermal change
T,p0, v  p  v  p0  v
 pv
v0
 p0  T
T0
}
= const. !
‘perfect gas equation’
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.
pv=RT
pV=nRT
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.
R : ‘gas constant’
(= 8.31434 J K-1 mol-1)
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
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.
Step 1: Isothermal change
T0,p0,v0  p  v ''  p0  v 0
Step 2: Isobaric change
T0 ,p,v ''  v 
 pv
v ''
T
T0
v0
 p0  T
T0
}
= const. !

same result !!!
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):
 z 
 z 
 z   z 
dz    dx    dy ;   ,   : partial differentials
 x  y
 x  y  y  x
 y  x
(ii) The mixed derivatives are identical (Schwarz’s
theorem):
  2z    2z 

 

 xy   yx 
(iii) Upon variation of x, y for z=const (Euler’s theorem):
z
 z 
x y
 z 
 y 
dz  0    dx    dy     
z
 x  y
 x  z
 y  x
y x


 
A more general approach to thermal expansion and compression:
 V = f(T) for p=const.:
(Gay-Lussac)
V = V0  (1 + )

 p = f(T) for V=const.:
1  p 
1  p 
    
p0    V p0  T  V
pV = const.

Exact differential of V=f(p,T):
 Due to
 : (thermal) expansion coefficient
p = p0  (1 + )

 p = f(V) for T=const.:
(Boyle-Mariotte)
1  V 
1  V 



V0   p V0  T p
and
d(pV) = pdV + Vdp = 0
1  V 

V  p  T
 : (isothermal) compressibility
 V 
 V 
dV  
dT


 p  dp

T

p

T
generally valid!
   V  dT   V  dp
 V 
 V   T 
 p     T    p  (Euler)

p 

T
V


1 
generally valid!

p 
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
p  pA  pB  ...
where, for each substance J,
pJ 
nJ  R  T
V
The mole fraction, xJ, is the amount of J
expressed as a fraction of the total amount of
molecules, n, in the sample:
xJ 
nJ
n
n  nA  nB  ...
When no J molecules are present, xJ=0;
when only J molecules are present, xJ=1.
Thus the partial pressure can be defined as:
pJ  x J  p
and
p A  pB  ... 
 (x A  xB  ...)  p  p
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.
Real gases: An introduction
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.
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.
Compression factor
Z
pv
RT
For a perfect gas, Z=1 under all conditions.
Deviation of Z from 1 is a measure of departure
from perfect behaviour.
At very low pressures, all the gases have Z1 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.
The variation of the
compression factor Z = pv/RT
with pressure for several gases
at 0C. 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.
Real gases: The virial equation of state
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:
p  v  R  T  (1 Bp  Cp2  ...)
 B C

p  v  R  T   1  2  ... 
 v v

The third virial coefficient, C, is usually less
important than the second one, B, 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.
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.
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
dZ
 B  2pC  ...  B
dp
as p  0, and
dZ
B
d(1/ v)
as v  , corresponding to p  0.
Real gases: The Boyle temperature
Because the virial coefficients depend on the
temperature (see table above), there may be a
temperature at which Z1 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 coinicide with those of a perfect
gas as p 0, and B=0. It then follows that pvRTB 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.
Real gases: Condensation and critical point
Reconsider 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.
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.
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
a
(p  2 )  (v  b)  R  T
v
or
a  n2
(p  2 )  (V  n  b)  n  R  T
V
van der Waals equation
 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
Comparison to the virial equation of state:
a
Bb 
RT
The surface of possible
states allowed by the van
der Waals equation.
Analysis of the van der Waals equation of gases
(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:
dp
RT
2a


0
2
3
dv
v
 v  b
d2p
2RT
6a


0
dv 2  v  b 3
v4
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.
(1) Perfect gas isotherms are obtained at high
enough temperatures and large molar volumes.
(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 presure
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’)
at the critical point. The solution is
v c  3b
pc 
a
27b2
Tc 
8a
27Rb
and the critical compression factor, Zc, is
predicted to be equal to
Zc 
for all gases.
pc v c 3

RTc
8
Van der Waals constants of selected gases
The principle of corresponding states
Idea
b) Reduced melting temperature
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
pr 
p
pc
vr 
v
vc
Tr 
0.44 at 1.013 bar
c) Reduced boiling temperature
T
Tc
Tb
Tc
and obtain the reduced van der Waals equation:
pr 
Tm
Tc
0.64 at 1.013 bar
d) Trouton’s rule (or: Pictet-Troutons’s rule)
8Tr
3
 2
3v r  1 v r
A wide range of liquids gives approximately the
same standard entropy of vaporization of
S  85 J K-1 mol-1
Examples
a) Compression factors
But:

approximation!

works best for gases composed of spherical
particles

fails, sometimes badly, when the particles are
non-spherical or polar
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.
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 First Law of Thermodynamics:
The internal energy of an isolated system is constant.
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.
or:
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
Rationale for the path independency of U, and the path dependencies of Q and W
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 dU
characterizes an exact, i.e. path-independent differential, while the changes Q and W are inexact, or
path-dependent, differentials.
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 nonadiabatic 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.
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’:
 U 
 U 
dU  
dT  

 dV 

T

V

 V,n

T,n
 U 
  n 

i
i
T,V,nji
dni
Closed system (dni = 0):
a)
isochoric changes of state (dV = 0):
dU = Q
(UII – UI) = Q
 U: exchanged heat
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:
 H 
 H 
dH  
dT  

 dp 
 T p,n
 p T,n
 H 
i  n  dni
 i T,p,nji
The temperature dependence of internal energy and enthalpy: Heat capacities
assume: 1 phase, 1 mol, pure substance
 u 
 V  const.: du  Q    dT  c V dT
 T  V
 h 
 p  const.: dh  Q    dT  cpdT
 T p
cv: molar heat capacity at constant volume
cp: molar heat capacity at constant pressure
uT2  uT1 
T2
c
dT
p
dT
T1
hT2  hT1 
T2
c
T1
experimentally, cp is often easier to determine than cv  calculate cv from cp
 U 
  v 
cp  c v  

p
  T 


v


p
T


v 2
 p   v 
 T
 T




 T  v  dT p
Perfect gas: the internal energy is independent of the volume v
 U 
 p 

T
 v 
 T   p  0

T
 v
v
 v 
cp  c v  p    R
 T p
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
dnA  A

dnB
B
dnC C

dnD D

dnA dnB dnC dnD ! dni




 d
A
B
C
D
i
 : ‘extent of reaction’
note: stoichiometric coefficients of reactants have negative, of products positive sign
it is:
dni = ni – nistart = id and I = const.
 U=f(V,T,n1start,n2start, … nkstart,) and H=f(p,T,n1start,n2start, … nkstart,)
 U 
 U 
 U 
dU  
dT  
dV  


 d

T

V


 V,

 T,

T,V
since all nistart = const.:
 H 
 H 
 H 
dH  
dT  
dp  

 d


p

 T p,

T,

T,p
Reaction Enthalpy and Internal Reaction Energy (cont’d)
 U 
 H 
What is the meaning of   and   ?
  T,V
  T,p
a) isothermal and isochoric processes
 U 
dU  QT,V  
d

  T,V
b) isothermal and isobaric processes
 H 
dH  QT,p  
d

  T,p
 U 
 H 
d

and
  
   d correspond to the heat q produced or absorbed by a

T,V

T,p
chemical reaction or a phase transition for one formula conversion.
The expression / is usually abbreviated by the Greek   U, H
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)
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
The correlation between H and U
for dT = dp = 0 holds:
 U 

H  U  
 p  V

 V T,


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 + pV
= U + pivi
= 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.
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.
The temperature dependence of reaction enthalpies H:
Kirchhoff’s law
Standard reaction enthalpies at different
temperatures may be estimated from heat capacities
and the reaction enthalpy at some other
temperature.
It is:
 H 
 H 
dH  
dT  

 dp

p
 T p

T
and:
  2H 
  2H 
 Cp 
 H 








  Cp 
 T 

p  T p  T p   p
 c
i p,i
Integration yields Kirchhoff’s law for the standard
reaction enthalpy change from rH°(T1) to
rH(T2 )  rH(T1) 
T2
  C dT
r
o
p
T1
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.
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.
A similar expression is found for rU°:
e.g. ammonia synthesis (1/2 N2 + 3/2 H2  NH3):
rCp° = cp(NH3) – 1/2 cp(N2) – 3/2 cp(H2)
rU(T2 )  rU(T1) 
T2
  C dT
r
T1
o
v
The pressure dependence of reaction enthalpies H
It is: rH(p2 )  rH(p1) 
 for perfect gases H is independent of
pressure, since T(V/T)p = V

 V  

V

T
 T   dp
p 

p 
1
p2
 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 pathindependence 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
2. CO + ½ O2  CO2
3. C + O2  CO2
3
I
C+O2
1?

H1, immeasurable
H2 = -283.1 kJ mol-1
H3 = -393.7 kJ mol-1
II
CO+ ½O2
III
CO2
2
H1 = H3 - H2 = -110.6 kJ mol-1
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) + 4fH°(N2,g)}
- {2fH°(HN3,l) + 2fH°(NO,g)}
= {-187.78 + 4(0)} – {2(264.0) + 2(90.25)}
= -892.3 kJ mol-1
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.
Expansion work:
Assume a system with massless, frictionless, rigid, perfectly fitting
piston of area A und an external pressure pex.
dw   pex dV
Vf
w    pex dV
Vi
i: initial state
f: 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:
 U 
 U 
dU  Q  pdV  
dT  

 dV
 T  V
 V T
=0 for a
perfect gas
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.
- Free expansion
i.e. expansion against zero opposing force, or pex = 0
 w=0
- 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:
V
f
w   pex  dV   pex (Vf  Vi )
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.
Vi
w   pex V
This type of expansion is irreversible.
- 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:
Vf
w   nRT 
Vi
V 
dV
  nRT ln  f 
V
 Vi 
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.
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.
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.
cp
Starting point: no exchange with surroundings, i.e.
q = 0
 U 
 dU  w   pdV  
 dT
 T  V
i.e.:  expansion work at the expense of U
 temperature change of the perfect gas
compression  heating
expansion
 cooling
it is:
1
 Vi  c v
 Tf 
Since R = cp – cv:


 
 
 Vf 
 Ti 
With T = pV/nR and the heat capacity ratio
 = cp/cv we find:
Tf  Vf 1  Ti  Vi  1  const.
pf  Vf
 pi  Vi 
 const.
‘Poisson’s law’
p1f    Tf  p1i    Ti   const
pdV  nc V dT
RT
dV  c V dT
V
R dln V  c V dlnT

for a change of state from Vi, pi, Ti  Vf, pf, Tf:
V
T
R ln f  c V ln f
Vi
Ti

 Vi 
 
 Vf 
R
cv
T 
  f
 Ti 
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.
The work of (reversible) adiabatic changes
Starting point:
W   pdV  dU  nc VdT
Provided the heat capacity is independent of temperature,
the adiabatic expansion work is
W
Vf
 pdV  U  nc
V
(Tf  Ti )
Vi
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.
The Carnot Cycle
Starting point: - ideal gas
- alternating isothermal and adiabatical 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
Determination of T2 (Tc):
 V 
T
T2
 c  1
T1 Th  V 4 
1
V 
 2
 V3 
1
The Carnot Cycle (cont’d)
Step
Gas
1
Isothermal, reversible
expansion
2
Adiabatic, reversible
expansion
3
Isothermal, reversible
compression
4
Adiabatic, reversible
compression
Hot
Source
Cold sink
Work storage
+ Q(T1)
-nRT1ln(V2/V1)
- Q(T1)
-
+nRT1ln(V2/V1)
-ncv(T1 – T2)
-
-
+ncv(T1 – T2)
- Q(T2)
+nRT2ln(V3/V4)
-
+ Q(T2)
-nRT2ln(V3/V4)
+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
W  nR(T1  T2 )ln
V2
V1
+
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
The Carnot Cycle (cont’d)
work output
supply of heat from the hot reservoir
Efficiency of the Carnot-Cycle:
W


Q(T1)
nR(T1  T2 )ln
nR T1 ln
V2
V1
V2
V1

T1  T2
T
 1 2
T1
T1
  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
Experience 2
All spontaneous processes in nature
proceed only in one direction.
All spontaneous processes in nature are
irreversible. They cause a loss of
useable work and lead to a gain in heat.
A
TA
+
B
TB
with TA < TB
A
B
TC
with TA < TC < TB
 not required by the First Law of Thermodynamics!
The Second Law of Thermodynamics
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.
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 (perpetuum
mobile) of the second kind is not feasible.
Question:
Which state function lets us
assess the direction of
spontaneous change?
Answer:
The Entropy S !
(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.
The Entropy S
V2
Q(T1)
V1
T


 1
Q(T2 ) nR T ln V2
T2
2
V1
nR T1 ln
Derivation via the Carnot process:
Q = 0 for adiabatic processes:
Extension to general cycles:
Each cyclic process can be
regarded as a sequence of two
lines of the Carnot cycle:



Qrev
 0
T

Qirrev
< 0
T


Q(T1) Q(T2 )

T1
T2
“reduced heat”;
state function!

Q(T1) Q(T2 )

 0 for reversible changes
T1
T2

Q(T1) Q(T2 )

< 0 incase of irreversible contributions
T1
T2
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.
The Entropy S (cont’d)
Introduction of the Entropy S as new state function, with dS 
dQrev
(Clausius)
T
Consider e.g. an isothermal cyclic process, carried out irreversible from 1 to 2, and reversible from 2 to 1:

dQ

T
2
dQirrev
1 T 
dQrev
2 T
2
1
dQirrev
 

T
1
Isolated system: dQ = 0
2
1
 dS

< 0
2
 S2 – S 1  0
S2 – S 1 = 0
dQirrev
1 T < S2  S1
or, more
general:
dQ
 dS
T
Clausius inequality
for irreversible processes
for reversible processes
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
Entropy changes accompanying specific processes
General expression:
dQrev
dU  pdV
 S 
 S 
dS  
dT

dV



 V 
T
T
 T  V

T
dQrev
dH  Vdp
 S 
 S 
dS  
dT

dp



 p 
T
T
 T p

T
Perfect gases:

dS 
dS 
cv

dT  dV
T

cp
T
dT  Vdp
dS  nc v dlnT  nR dln V  nc p dlnT  nR dlnp
S2  S1  nc v ln
Reactions:

T2
V
T
p
 nRln 2  nc p ln 2  nRln 2
T1
V1
T1
p1
 S 
 S 
 S 
dS    dT    dp    d
 T p,
 p T,
  T,p
Phase transitions:
 trsS 
 trsH
Ttrs
for cv, cp = const.
 S 
    S 
  T,p
 s
i
for p, T = const.
Caution: Chemical reactions are (in general) irreversible, and can not be described via dS 
dQrev
T
i
The variation of entropy with temperature: a more general approach
usually cp is not independent of temperature

Tf
dQrev
S(Tf )  S(Ti )  
 S(Ti ) 
T
Ti
Tf

Ti
Cp dT
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.
This leads to the
Third Law of Thermodynamics:
The entropy of all perfect crystalline substances is zero at T = 0.
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:
dQSurroundings   dQSystem
+
dS  dSSystem  dSSurroundings
 dSSystem 
1st law of thermodynamics
dQSurroundings
T
 0
2nd law of thermodynamics
dSSystem 
dQSystem
T
(i) Isothermal and isobaric heat transfer
Qsystem = dH  TdS – dH  0
from
G  H - T·S
Gibbs Energy
follows (dG)p,T  0
 0
Clausius inequality
(ii) Isothermal and isochoric heat transfer
Qsystem = dU  TdS – dU  0
from
A  U - T·S
Helmholtz Energy
follows (dA)V,T  0
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
The “characteristic functions” and the “Maxwell relations”
•
The ‘characteristic functions’ relate energies and enthalpies to other 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
•
The ‘Maxwell relations’ yield additional interrelations between state functions and state variables
 T 
 p 


 V 
 S 

S

V
 T 
 V 

 S 
 p 

p

S
 p 
 S 

 T 
 V 

V

T
 S 
 V 


 T 
 p 

p

T
The “Guggenheim scheme”
S U V
+ H
A
p G T
-
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 multiphase 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:
That is, the chemical potential is the slope of a plot of e.g. Gibbs
 G 
 A 
energy against the amount of substance of the component i, with
i  




the pressure and temperature (and the amounts of all other

n

n
 i p,T,nji
 i  V,T,nji
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 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 + …
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 ‘chemical potential’ (cont’d)
Change of G and A in chemical reactions:
 G 
G  


  p,T
It holds:
 i i
i
 A 
A  


   V,T
 
i
i
i
• In equilibrium (i.e. dG=0) the chemical potential i of substance i
must 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 = ii < 0.
The dependence of µ on temperature T, pressure p and composition x
(i) pure substances:
 G 
 n   µ

p,T

µg
and therefore
 µ 
 T    s

p
,
 µ 
 p   v
 T
The ‘chemical potential’ (cont’d)
thus for the “standard pressure” p° = 1 bar one finds for a pure and
µ(p)

a) perfect (ideal) gas:
dµ 
µ(p )
p
p
p
p
 vdp  RT  dln(p)

µideal  µ  RTln
p
p
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.

µreal  µ  RTln
f
p
(ii) simple mixtures
assume a perfect gas in an ideal gas mixture, and compare this to the pure gas:
A
pure gas i
p, T
µi* (p,T)
B
pure gas i
pi, T
µi* (pi,T)
C
gas mixture
pi, T, pj=p
µi (p,T)
*: pure compound
since
µi* (pi,T)  µi* (p,T)  RTln
and
µi (p,T)  µi* (pi,T)  0
pi
 RTlnx i
p
µi (p,T)  µ (p,T)  RTlnx i
*
i
also valid for condensed phases!
connection permeable only for gas i,
and equilibrium between B and C !
in case of real mixtures, xi has to be replaced by an
effective mole fraction, called the activity ai.
µi (p,T)  µi* (p,T)  RTlnai
xi and ai are related by the activity coefficient i: ai=ixi.
Simple mixtures: Partial molar quantities
example: the volume of a mixture
a) perfect (ideal) mixtures:
no interactions
V  V1*  V2*  ...  n1v1*  n2v *2  ... 
b) real mixtures:
nv
*
i i
(analogous: e.g. U, H, cp, …)
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).
Simple mixtures: Partial molar quantities (cont’d)
 V 
vi  

 ni p,T,nji
formal definition of the partial molar volume of a
substance i at some general composition:
e.g. binary mixture:
 V 
 V 
dV  
dn

dnB  v AdnA  vBdnB



A

n

n
 A p,T,nB
 B p,T,nA
and, after integration with fixed composition:
V  nAv A  nBvB 
nv
i i
  
a general approach:    nii   ni 

 ni nji
 : extensive property of the mixture
i : partial molar property, e.g. vi ,ui ,hi ,cp,i ...
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.
experimental determination: e.g. via mean molar quantities
 


n
i
x 
i
i
 x11  x 22
1    x 2
d
dx 2
for a binary mixture
2    (1  x 2 )
d
dx 2
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 and pressure. The Gibbs energy of
this initial system, Gi, is then
Gi  nAµA  nBµB
p
p


 nA  µA   RTln   nB  µB   RTln 
p 
p 


• after mixing, the partial pressures of the gases are
pA and pB, with p=pA+pB. The total Gibbs energy
changes to
p 
p 


Gf  nA  µA   RTln A   nB  µB   RTln B 
p 
p 


• the difference Gf – Gi, the Gibbs energy of mixing,
mixG, is therefore
mix G  nA RTln
pA
p
 nBRTln B
p
p
 nRT  x A lnx A  x B lnx B 
mix G  nRT  xi lnxi
i
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.
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
  G 
mix S    mix 
 T p,nA ,nB
  nR  x A lnx A  x B lnx B 
mix S   nR xi lnxi > 0
i
• in case of ideal mixtures, from G = H - TS
follows mixH=0. This is to be expected for a
system in which there are no interactions.
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.
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.
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):
- expansion of a gas through a barrier from one constant
pressure to another in an insulated apparatus, i.e. adiabatic
process
- observation of a 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.
The process:
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.
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.
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
with

Hi = Hf
!
 H 
 H 
dH  
dT

dp

0

 p 
 T p

T
The isenthalpic Joule-Thomson coefficient µ is then defined
as:
 H 
 p 
 T 
T   
µ


cp
 H 
 p H
 T 

p
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
Simplest approach (Virial equation / van der Waals equation, via given pressure, the temperature must be below
a certain value if cooling is required but, if it
pv=RT+Bp and B=b-a/RT):
becomes too low, the boundary is crossed
again and heating occurs. Reduction of
2a
2a
b
pressure under adiabatic conditions moves the
Ti 
µ  RT
system along one of the isenthalps, or curves of
Rb
cp
constant enthalpy. The inversion temperature
curve runs through the points of the isenthalps
where their slopes change from negative to
For a perfect gas, µ = 0.
positive.
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).
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.
A gas typically has two inversion temperatures, one at high
temperature and the other at low.
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.
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 an T at which two phases coexist in equilibrium.
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 solid 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 374C and the vapour pressure is then 218 atm.
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.
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:
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).
Right: The experimentally
determined phase diagram for
water showing the different solid
phases. Note the change of
vertical scale at 2 atm.
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
Justification:
total number of variables (p, T, composition x1, x2, ... xC-1) in all P phases
minus (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
(phase rule)
 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.
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.
Phase stability and phase transitions: The location of phase boundaries
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:
dµ(p,T) = dµ(p,T)
With
dµ = -sdT + vdp
follows
-sdT + vdp = -sdT + vdp
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.
Negative slope of the water
solid-liquid boundary due to
trsv < 0
and
(v - v)dp = (s - s)dT
 trss
 trsh
 p 


 T 
 coex  trsv T trsv
Clapeyron equation
The Clapeyron equation is an exact expression and
applies to any phase equilibrium of any pure
substance!
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):
 trsh
 p 

 T 
 coex T(RT / p)
which arranges into the Clausius-Clapeyron equation for the variation of vapour pressure with
temperature:
 trsh
  lnp 

 T 
2

coex RT
  lnp 
 h
 
  trs

  1/ T  
R

coex
If temperature independence of the enthalpy of vaporization (sublimation) is assumed, this
integrates to
 h1
1 
linear relation ln(p)  1/T
p  p * e 
  trs  

R T
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.
Two-component systems: Ideal solutions and Raoult’s law
It was shown before, that for an ideal mixture the chemical potential is given by
µi (p,T)  µi* (p,T)  RTlnx i
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:
p A  x p , pB  x p
*
A A
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.
This relation is known as Raoult’s
*
B B
law.
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.
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)).
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:
p A  x AK A , pB  x BKB
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.
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).
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.
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.
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)
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:
µi (p,T)  µi* (p,T)  RTlnx i < µi* (p,T)
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 solidliquid 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!
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).
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
T 
R  Tb * 
2
 xB  Kb  b
 vapH
Kb is the ebullioscopic constant of the solvent, b its
molality (mol solute / kg solvent).
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
R  Tf * 
T ' 
 xB  K f  b
 fusH
2
Kc is the cryoscopic constant of the solvent, b its
molality (mol solute / kg solvent).
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.
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 but not to the
solute.
-
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+)
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 + .
-
Thermodynamic treatment:
V = nBRT
or
 = [B]RT
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!)
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.
Two-component systems: Vapour pressure diagrams
The composition of the vapour
-
Partial vapour pressures of the components of an ideal solution of
two volatile liquids

-
pA = xApA*
pB = xBpB*
Total vapour pressure of the mixture:

-
Raoult’s law
p = pA + pB = xApA* + xBpB* = pB* + (pA* - pB*)xA
Mole fractions y in the gas:


Dalton’s law
yA 
yA 
pA
p
x Ap A *
pB *  (p A *  pB *) x A
yB 
pB
p
yB  1  y A
 The gas is richer than the liquid in the more volatile component!
The general scheme of interpretation of a pressurecomposition diagram (a vapour pressure diagram).
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 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.
Two-component systems: Vapour pressure diagrams (cont’d)
Consider the effect of lowering of the pressure on a liquid mixture of
overall composition a:


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.
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 isopleth
to 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(ml = ml for balance).
Two-component systems: Temperature-composition diagrams and 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:
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.

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 nonvolatile 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.
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 to
form 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.
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:
The temperature-composition phase diagram
for two almost immiscible solids and their completely miscible liquids. The isopleth through e
corresponds to the eutectic composition, the
mixture with lowest melting point.

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.
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.
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)
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.
Chemical equilibrium: The Gibbs energy minimum
The reaction Gibbs energy, rG, is defined as the slope of the
graph of the Gibbs energy plotted against the extent of the
reaction :
 G 
r G  

  p,T
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
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.
Chemical equilibrium: The description of equilibrium
Chemical reactions take place in real mixtures, so
i (p,T)  i0 (p,T)  RTlnai
and
 G 
r G  
 


p,T
 
i i
i
i
0
0
r G   
i i  RT  i lnai  r G  RT  lnai
i
i
i
 r G0  RT ln  ai i  rG0  RT lnQ
i
activities of products
The reaction quotient, Q, has the form Q 
with each species raised to the power
activities of reac tants
given by its stoichiometric coefficent.
At equilibrium, the slope of G is zero: rG = 0. The activities then have their equilibrium values and one
can write
 G0
 r

i 
K    ai 
 e RT
 i
equlibrium
(  const. for given p,T!)
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 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.
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:


K    ai i 
equilibrium
 i


K x    x i i 
equilibrium
 i


K  K x     i i 
equilibrium
 i


K p    pi i 
equilibrium
 i


 
 (p0 )  i
K  K p    i i 
 equilibrium
 i


K c    ci i 
equilibrium
 i


 
 (c i0 )  i
K  K c    yi i 
 equilibrium
 i


K m    mi i 
equilibrium
 i


 
 (mi0 )  i
K  K m    y "i i 
 equilibrium
 i
All ci0 are 1 mol dm-3, and all mi0 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:
 a(CO) 2 
 a(CO) 2 
e.g. Boudouard reaction C(graphite)  CO2
2CO : K  
 


a(
C)

a(
CO
)
2  equilibrium

 a(CO2 ) equilibrium
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
Effect of pressure
increased temperature
favours the reactants
Reactions with rV>0:
increased pressure
favours the reactants
Endothermic reactions: increased temperature
favours the products
Reactions with rV<0:
increased pressure
favours the products
Exothermic reactions:
e.g. ammonia synthesis
1
3
N2  H2
2
2
NH3
rH298 = -45.94 kJmol-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).
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).
The response of equilibria to temperature and pressure: Exact treatment
(i) Temperature dependence
From
ln K  
r G
RT
0
and
one obtains
rH0
  lnK 
(a) 
 
2
 T p RT
 r G0
1 T
  lnK 
 T    R  T

p


and
(b)





rH0
  lnK 
 1/ T    R

p
(van’t Hoff’s isobar)
Integration over small temperature ranges, i.e. with rH0 almost
independent of temperature:
1
ln K(T2 )  ln K(T1)  
R
1/ T2

rH0 d(1/ T)
1/ T1
rH0  1
1
 ln K(T2 )  ln K(T1)  
 

R  T2 T1 
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.
(ii) Pressure dependence
1  rG0 
  lnK 
 p    RT  p 

T

T
• “escape from constraint”
r V
  lnK 
 p    RT

T
0
• K is almost independent
of pressure in condensed
phases
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:
 G 
 G 
 G 
dG  
dT  
dp  

 d   SdT  V dp   d


T

p


p,

p,T

T,
d is the work needed to change the surface area, , by an infinitesimal amount. The constant of
proportionality, , is called the surface tension. Its dimensions are energy/area and its units are typically
Jm-2 or Nm-1. The surface tension is responsible for the spherical shape of droplets, which have the
smallest surface/bulk ratio.
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 2.110-4 mJK-1mol-2/3
for non-associated liquids, like C5H10,
C6H12, C6H6, or O2 ( principle of
corresponding states).
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:
pin  pout 
2
r
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
p  p*  e
2 v
rRT
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)
The dependence of the pressure inside a
curved surface on the radius of the surface, for
two different values of the surface tension.
PHYSICAL CHEMISTRY:
An Introduction
static phenomena
macroscopic
phenomena
equilibrium in macroscopic
systems
THERMODYNAMICS
ELECTROCHEMISTRY
dynamic phenomena
change of concentration as a
function of time
(macroscopic) KINETICS
(ELECTROCHEMISTRY)
STATISTICAL THEORY
OF MATTER
microscopic
phenomena
stationary states of particles
(atoms, molecules, electrons,
nuclei) e.g. during translation,
rotation, vibration
• bond breakage and
formation
• transitions between quantum
states
STRUCTURE OF MATTER
CHEMICAL BOND
STRUCTURE OF MATTER
(microscopic) KINETICS
CHEMICAL BOND
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, :
v
1 dnA
1 dnB
1 dnC
1 dnD d




 A dt
B dt
C dt
D dt
dt
 mol 
 s 


(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:
v
1 d[i] 1 d

i dt
V dt
 mol 
 dm3  s 


For heterogeneous reactions division by the surface area of the
species i leads to
v
1 di
i dt
 mol 
 m2  s 


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.
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:
k  [H2 ]  [Br2 ]1/ 2
v
[HBr]
1 k '
[Br2 ]
• 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.
The rates of chemical reactions
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 cyclopropene 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.
Isomerization as a typical
unimolecular elementary
reaction.
• 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!
Simple integrated rate laws
First-order reactions, half-lives and time constants
• e.g. decomposition reactions, isomerizations,
radioactive decay
d[A]
v
 k  [A]
dt
[ A]
t
d[A]
  k   dt

[A]
[ A]0
0

 [A] 
ln 
   kt
[A]
 0
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.
[A]  [A]0 e kt
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:
ln2
t1/ 2 
k
(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 the
time required for the concentration of a reactant to
fall to 1/e of its initial value:
 
1
k
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.
Simple integrated rate laws (cont’d)
Second-order reactions
(i)
A + A  products
d[A]
v
 k  [A]2
dt
1
1

 kt
[A]
[A]0
(ii)
[ A]

A + B  products
t
d[A]
  k   dt
2

[A]
[ A]0
0
[A] 
[A]0
1 kt[A]0
v
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:
x

The variation with
time of the
concentration of a
reactant in a secondorder reaction. The
grey line is the
corresponding decay
in a first-order
reaction with the
same initial rate. For
this illustration, klarge
= 3ksmall.
d[A]
 k  [A]  [B]
dt
t
dx
0 [A]0  x [B]0  x   k  0 dt
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:
 [B] /[B]0 
1
 ln 
  kt
[B]0  [A]0
[A]
/[A]
0 

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.
Simple integrated rate laws (cont’d)
Reversible first-order reactions
A
k
B
k'
Net change of concentration:
d[A]
  k  [A]  k ' [B]
dt
If the initial concentration of A is [A]0 and no B is present
initially, than [A] + [B] = [A]0 at all times
d[A]
  k  [A]  k ' [A]0  [A]   (k  k ')[A]  k '[A]0
dt
Solution:

[A]0
 k '  k  e (k k ')t
k  k'
k  [A]0
[B] 
 1  e (k k ')t
k  k'
[A] 



Equilibrium concentrations:
[A]eq 
k ' [A]0
k  k'
[B]eq  [A]0  [A] 
k  [A]0
k  k'
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’.
Simple integrated rate laws (cont’d)
Consecutive first-order reactions
Reaction from the reactant through an intermediate to the product:
(e.g. decay of a radioactive family, such as
239
2.35 min
U 

239
2.35 d
Np 

239
ka
kb
A 
 I 
 P
Pu
where the times are half-lives)
Approach:
d[A]
  k a [A]
dt
d[I]
 k a [A]  kb [I]
dt
d[P]
 kb [I]
dt
The role of the rate-determining step:
a) ka kb: After an initial induction period, an
interval during which the concentrations of
intermediates rise from zero, the rates of
change of concentrations of all intermediates The basis of the steady-state
are negligible small (steady-state approxi- approximation. It is supposed
that the concentrations of
mation, d[I]/dt0).
intermediates remain small and
b) ka kb: Significant build-up of intermediate,hardly change during most of
the course of the reaction.
with noticeable product formation after an
initial induction period.
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.
The temperature dependence of reaction rates
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.
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:
k  A  e Ea / RT
E
lnk  ln A  a
RT
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.
A
k
The Arrhenius equation
Temperature T
Plot of the rate constant
k against temperature.
Starting from k=0 at
T=0, it approaches A for
T.
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.
The temperature dependence of reaction rates: Interpretation of the parameters
Ea
• 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.
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.
• 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.
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, ratedetermining 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.
• Homogeneous catalysis: Catalyst and reactants are in the
same phase (e.g. decomposition of H2O2 in aqueous solution,
catalysed by bromide ions or catalase (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.
A catalyst provides a different path with a
lower activation energy. The result is an
increase in the rate of formation of products.
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.
Elementary steps of heterogeneous catalysis:
•
•
•
(dissociative) adsorption
reaction
desorption
• 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.
desorption
CO O2
adsorption
CO2
precursor
reaction
The three-way catalyst
Conversion of the
main pollutants CO,
NOx, and
hydrocarbons in a
relatively narrow
window of the air-tofuel ration:
CO + ½O2  CO2
NO + CO
 ½ N2 + CO2
CmHn + (m+n/4)O2
 m CO2 + n/2 H2O
Pt
dissoziation
diffusion
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:
ka

 AM(surface)
A(g)  M(surface) 
kd
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):
d
 k a pN(1  )
dt
The Langmuir isotherm for
dissociative adsorption
(X22X) for different values
of K.
The Langmuir isotherm for
non-dissociative adsorption
for different values of K.
 
Kp
1  Kp
K
Langmuir isotherm
for non-dissociative adsorption
Rate of change due to desorption:
d
  kd N 
dt
ka
kd
for dissociative adsorption:
 
Kp
1  Kp
Adsorption and catalysis
The Langmuir-Hinshelwood mechanism
The Eley-Rideal 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 secondorder rate law is expected:
In the Eley-Rideal (ER) mechanism of surfacecatalysed reactions, a gas-phase molecule collides
with another molecule already adsorbed on the
surface. It follows that the rate law should be
A  B 
 Pr
kLH
v  kLH  A  B
If A and B follow Langmuir isotherms:
A 
K A pA
1  K A p A  KB pB
B 
KB pB
1  K A p A  K B pB
kER
A  B 
 Pr
v  kER  pB  A
If A follows a Langmuir isotherms in the pressure
range of interest:
v 
kER K A p ApB
1  K A pA
and
v 
kLH K A p AKB pB
1  K A pA  KB pB 
2
Variation of the
product formation
rate d[Pr]/dt in a
LH mechanism for
fixed partial
pressure of B.
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).
The End!