3 thermodynamics of pharmaceutical systems

Outline of the Learning Topics:
• Introduction to practical thermodynamics Laws
• Zero & First Law
• Second Law
• The concept of Gibbs Free Energy
• Free Energy and Equilibrium Constant
• Kinetics
• Heat Energy
• Thermodynamics & Material Properties Relationships
• Thermodynamics and Spontaneity
• Thermodynamic Principles in Pharmacy Education &
Practice

Thermodynamics
(Introduction)
• Thermodynamics describes the overall properties, behavior, and
equilibrium composition of a system;
• Thermodynamics: all about “if”
 tells whether or not a process or a reaction can occur (is there a decrease in
free energy?)
 applicable to systems in stable or metastable equilibrium
 sufficient driving force is needed to enforce a favorable transformation,.

Thermodynamics (continued)
(Introduction)
• Kinetics describes the rate at which a particular process will
occur and the pathway by which it will occur
• Kinetics: all about “how”
 how fast or slow a process can occur, i.e., determining the rate
 applicable to systems in transition from non equilibrium to equilibrium, or
between two equilibrium states
 kinetics of a process is generally about how to overcome the energy barrier
to finish the transformation from the starting (reactant) state to the final
(product) state.

• A description of any thermodynamic system employs the four laws of
thermodynamics that form an axiomatic basis:
• The zeroth law of thermodynamics states that if two systems are each
in thermal equilibrium with a third, they are also in thermal
equilibrium with each other.
• The first law specifies that energy can be exchanged between physical
systems as heat and work. The first law of thermodynamics states that
in a process without transfer of matter, the change in internal energy,
ΔU, of a thermodynamic system is equal to the energy gained as
heat, Q, less the thermodynamic work, W, done by the system on its
surroundings.
Thermodynamics
(Zero & First Law)

• The second law defines the existence of a quantity called entropy, that
describes the direction, thermodynamically, that a system can evolve and
quantifies the state of order of a system and that can be used to quantify the
useful work that can be extracted from the system,
• It states that Heat cannot spontaneously flow from a colder location to
a hotter location - In thermodynamics, interactions between large
ensembles of objects are studied and categorized.
• Central to this are the concepts of the thermodynamic system and
its surroundings - A system is composed of particles, whose average motions
define its properties, and those properties are in turn related to one another
through equations of state.
Thermodynamics
(Second Law)

• The third law of thermodynamics states that as the temperature of a system
approaches absolute zero, all processes cease and the entropy of the system
approaches a minimum value. This law of thermodynamics is a statistical law
of nature regarding entropy and the impossibility of reaching absolute zero of
temperature. This law provides an absolute reference point for the
determination of entropy. The entropy determined relative to this point is the
absolute entropy. Alternate definitions include "the entropy of all systems and
of all states of a system is smallest at absolute zero," or equivalently "it is
impossible to reach the absolute zero of temperature by any finite number of
processes".
• Absolute zero, at which all activity would stop if it were possible to achieve, is
−273.15 °C (degrees Celsius), or −459.67 °F (degrees Fahrenheit), or 0 K
(kelvin), or 0° R (degrees Rankine).
(Second Law)

The Gibbs Free Energy
J. Willard Gibbs used the ideas of enthalpy, entropy, and spontaneity in a concept called free
energy (G). Free energy refers to the maximum amount of energy free to do useful work.
It is related to enthalpy (H), temperature (T), and entropy (S) by Equation
G = H – T S
Free energy is also a measure of spontaneity. Negative values of G indicate a spontaneous
or forward (reactants make products) reaction. Positive values of G indicate a
nonspontaneous or reverse (products make reactants) system. If G = 0, the system is in
equilibrium. At equilibrium, the composition of the system (amount of products and
reactants) is constant.
The free energy of a sum of a series of equations is the sum of the free energies of those
equations. One form of this is Equation
G°rxn =  n G°f,product –  m G°f,reactant
where G°f refers to the free energy of the formation reaction.
(Second Law)

Exercise 1
Calculate G° for the following reaction:
1/2 O2 (g) + Mn2+ + H2O (l) = MnO2 (s, Pyrolusit) + 2 H+
G°f (H+) = 0 kJ mol-1
G°f (O2) = 0 kJ mol-1
G°f (MnO2, s) = -465.1 kJ mol-1
G°f (H2O, l) = -237.18 kJ mol-1
G°f (Mn2+) = -228.0 kJ mol-1
G° = -465.1 - (-237.18 + (-228.0)) = +0.08 kJ mol -1
Exercise 2
MnCO3 (s) = Mn2+ + CO3
2-

Homework: Exercise 3
SO4
2- + 9 H+ + 8 e- = HS- + 4 H2O (l)
G° = -194.2 kJ mol-1

The Law of Mass Action
• In a chemical reaction not all the reactants become products due to some reversible
reactions.
• At the point where the rate of the forward reaction is the same as the reverse reaction,
the concentrations of products and reactants are constant - This point is chemical
equilibrium. At equilibrium, the concentrations of reactant and product are constant,
but not equal. Individual molecules of reactants and products still react, but the overall
amount does not change.
• The law of mass action states that any reaction mixture eventually reaches a state
(equilibrium) in which the ratio of the concentration terms of the products to the
reactants, each raised to a power corresponding to the stoichiometric coefficient for
that substance in the balanced chemical equation, is a characteristic value for a given
temperature. For the reaction
aA + bB <--> cC + dD
the lowercase letters represent stoichiometric coefficients, A and B represent reactants,
and C and D represent products. The ratio described by law of mass action is a
constant, called the equilibrium constant (K):
KC = [C]c[D]d
[A]a[B]b

For gaseous reactions, partial pressures can be used instead of concentration values:
KP = PC
c PD
d
Pa
a PB
b
The relationship between KC and KP can be derived from the ideal gas law.
KP = KC(RT) n
where n is the difference in the number of moles of products (sum of their stoichiometric
coefficients) and moles of reactants, T is the temperature, and R is the universal gas
constant = 8.31 J mol-1 K-1.
The concentration of a solid or pure liquid is regarded as a constant. This is normally
combined with the equilibrium constant rather than being included as part of the
equilibrium constant expression.
The way the reaction is written affects the value of the equilibrium constant. For example,
the equilibrium constant of the reverse reaction is the reciprocal of the equilibrium
constant of the forward reaction.

• Reactions move toward equilibrium from either the products or the reactants. If no
equilibrium concentrations (or pressures) are used in the mass action expression, the
value is called the reaction quotient (Q). If the value of Q is smaller than K, the
reaction must go in a forward or spontaneous (–G) direction to reach the final value
(K). If the value of Q is larger than K, products must react to reach the final value (K).
The reaction goes in a reverse or no spontaneous (+G) direction. The relationship
between free energy and the equilibrium constant is
G = G° + RT ln(Q) R = gas constant = 8.314 J/mol•K
T = temperature (standard conditions 298 K)
• At equilibrium, the rate is neither forward nor reverse, so G is zero. However, the
equilibrium constant K can be determined from the free energy at standard state.
G° = –RT ln(K)
• A system at equilibrium can be perturbed by changing conditions. Le Chatelier's
principle states that if a stress (perturbation) is applied to a system at equilibrium, the
system will adjust to minimize that stress. Consequently, if reactant is added, the
reaction must go in a forward reaction to use up that reactant and minimize the
stress.

Exercise 4
Calculate the equilibrium constant K for the following reaction:
1/2 O2 (g) + Mn2+ + H2O (l) = MnO2 (s, Pyrolusit) + 2 H+
Exercise 5
Calculate K (equilibrium constant = solubility product) for the following reaction;
what is K for the respective precipitation reaction?
MnCO3 (s) = Mn2+ + CO3
2-
ln K = -G° / RT = -0.08 x 1000 / (8.314 x 298) = -0.0323
K = 0.97
K = 3 x 10-11 or ln K = -24.2
K (precipitation) = 1/K (dissolution) = 3.5 x 1010

Calculate K for the following reaction:
SO4
2- + 9 H+ + 8 e- = HS- + 4 H2O (l)
An aqueous solution contains 10-4 M CO3
2- and 10-3 M Ca2+. The concentration or activity,
respectively, of a solid is defined as 1. Will the reaction
Ca2+ + CO3
2- = CaCO3 (s) with K = 108.1 take place?
K = 1034 or ln K = 78
Q = 1/(10-3 x 10-4) = 107
Q < K --> CaCO3 will precipitate.

Review:
Free Energy and Equilibrium Constant

Kinetics
• Kinetics is a term that relates to how fast a reaction occurs. Whereas thermodynamics
is concerned with the ultimate equilibrium state and not concerned with the pathway to
equilibrium, kinetics concerns itself with the reaction pathway. Very often, equilibrium in
the Earth is not achieved, or achieved only very slowly, which naturally limits the
usefulness of thermodynamics. Kinetics helps to understand why equilibrium is
occasionally not achieved.
• While the rate of the forward reaction is equal to the rate of the reverse reaction at
equilibrium state, equilibrium constant expressions are not a measurement of rate. The
expression is determined from the overall reaction rather than from the rate-
determining step. The concentrations at equilibrium give no information on how long it
takes to reach that equilibrium. Catalysts will help the reaction reach equilibrium faster
but will not affect the equilibrium concentration. Instead, equilibrium concentrations
(and equilibrium constants) are related to thermodynamic parameters like G and H.
• The rate of reaction is measured as the change in concentration of a product or
reactant ([X]) over a given time (t). The rate of reaction for reactants is negative, since
reactants are disappearing, and positive for products, which are appearing. Rate can
be measured as average rate using the equation

Kinetics
• Rate decreases over time. Therefore instantaneous rate, the rate at any given time, is
sometimes used. The instantaneous rate can be determined from a tangent line at the
relevant instant of time on a graph of concentration versus time. The instantaneous
rate at the start of the reaction (t = 0) is called the initial rate.
• The relationship between concentration and rate is called the rate law. The rate is
proportional to the product of the concentration of reactants raised to some exponent:
rate = k[A]m[B]n
• The proportionality constant (k) of this equation is called the rate constant. The
exponents on the reactant concentration are called the order. With the form given, m
is the order in A and n is the order in B. The sum of the exponents (m + n) is called the
overall order. The order of the reaction is normally an integer or simple fraction.

Kinetics

Kinetics
Reaction order
•To characterize the affect that changing a particular reactant has on the rate of a
reaction, kineticists use the term "reaction order." When the rate of a reaction is directly
related to the concentration of a substance, it is said to be "first order" in that substance.
This is the case for radioactive decomposition.
•The reaction of chlorine atoms and ozone, which has the rate law rate = k[Cl][O3] is first
order in chlorine atoms and first order in ozone. The order of the entire rate law, called
the reaction order, is the sum of all the exponents of the concentrations in the rate law.
For the above reaction, the overall order is 2.
•A reaction is of zero order when the rate of reaction is independent of the concentration
of materials. The rate of reaction is a constant. When the limiting reactant is completely
consumed, the reaction stops abruptly.
A zero order reaction obeys the rate law:
-d[A]/dt = k
This type of reaction is important in enzyme catalyzed reactions.
Zero order reactions are also typically found when a material required for the reaction to
proceed, such as a surface or a catalyst, is saturated by the reactants.

Kinetics

Kinetics
Homework:
Exercise 12

Kinetics
The rate law can be integrated to get a relationship between time (t) and concentration. For
a first–order reaction with a single reactant (rate = k[X]), the integrated rate law is
ln[X] = –kt + ln[X]0
where [X]0 is the initial concentration of X. The integrated rate law for a second–order
reaction with a single reactant (rate = k[X]2) is
A reaction that is first order in two reactants (rate = k[X][Y]) can be expressed as a
pseudo–first–order reaction if the concentration of one reactant is significantly
greater than that of the other. For example, if [Y] is much greater than [X], the rate
law can be expressed as
rate = k'[X], where k' = k[Y]
It is also possible to have a zero–order reaction (rate = k). For zero–order reactions, the
integrated rate law is
[X] = –kt + [X]0

Kinetics
Another way to express the rate of reaction is with the half–life. Half–life is the time
required for the reactant concentration to decrease to half its initial value ([X] =
1/2[X]0. The integrated rate laws can be used to relate the half–life (t1/2) to rate
constant (k) and initial concentration ([X]0). For a first–order reaction,
A first–order reaction is not dependent on concentration of reactant. All nuclear reactions
are first order reactions and the rates of nuclear reactions are commonly designated
by the half–life.
The half–life of a second–order reaction is
and that for a zero–order reaction is

Kinetics
• The rate law is determined experimentally, rather than from the chemical reaction.
This is because the overall chemical reaction does not necessarily reflect the way in
which the reaction occurs. A mechanism is the step–by–step sequence by which a
chemical reaction occurs. Each of these elementary steps goes at a specific rate.
The rate law is determined by the slowest, rate–determining, elementary step rather
than by the overall reaction.
• Reactions occur when bonds are broken and formed. The substance formed during
this process, as bonds are breaking and forming, is called an activated complex. In
some steps, an unstable substance (intermediate product) that later undergoes
further reaction is formed.

Kinetics
•Bonds breaking and forming usually occur
as a result of a collision. For bond
breakage to occur in the collision, the
molecules must have sufficient kinetic
energy.
•The energy required to get a reaction
going is called the activation energy (Ea).
The energy difference between products
and reactants is the H (or G) for the
reaction.
• The relationship between temperature (T ) and rate constant (k) is described by the
Arrhenius equation
• where Ea is the activation energy, R is the gas constant (8.314 J/mol•K) and A is the
frequency factor. The frequency factor is related to how successful the collisions
between molecules are.
• One way to increase the rate of a reaction is to add a catalyst. A catalyst increases
the rate of reaction without itself being consumed. It does this by lowering the
activation energy, often by directing the orientation of the colliding molecules.

Kinetics
For a repetition and/or more information, look at:
http://www.louisville.edu/a-s/chemistry/Chapter7.ppt

•Heat energy is present in every object above absolute zero.
•How do we know that heat is present in every object?
Temperature of the body is an indication. (Absolute zero is the
lowest temperature possible and is equal to -273oC).
•Are heat and temperature one and the same? No.
•Heat is a form of energy and temperature represents the degree
of hotness of a body.
•We can say, 'Heat is the cause and temperature is the effect'.

State function
A state function or state property refers to a property of the system that
depends only on its present state
It does not depend on the systems past (or future)
In other words, the value of a state function does not depend on how the
system arrived at the present state: it depends only on the characteristics
of the present state.
An important characteristic of a state function is that a change in this
function in going from one state to another state is independent of the
particular pathway.
Energy is a state function, but work and heat are not state functions.

Chemical energy
•The mechanical examples discussed earlier are applicable to chemical
systems too.
CH4(g) + 2O2(g) ------- CO2 + 2H2O (g) + energy (heat)
•This reaction is used to heat homes in winter
•The SYSTEM is the part of the universe on which we wish to focus our
attention
•The SURROUNDINGS include everything else in the universe
•Heat flows out of the system-EXOTHERMIC
•Reactions that absorb energy from the surroundings are ENDOTHERMIC

Some Terminology
• System – part of the universe under
investigation
• Surroundings – everything outside the
system
• Three types of systems: open, closed,
isolated
– Open systems – can exchange energy and mass
with their surroundings

• State of a system – set of macroscopic
properties which completely define a
system:
100 g of H2O(l) at 25 °C, 1 atm
• Thermodynamic process – change in state
of a system
100 g H2O(l) at 25 °C, 1 atm  100 g H2O(s) at -10 °C, 1 atm
100 g H2O(l) at 25 °C, 1 atm
 11.1 g H2(g) + 88.9 g O2(g) both at 25°C, 1 atm

CH4(g) + 2O2(g) ------- CO2 + 2H2O (g) + energy (heat)
Where does the energy, released as heat, coming from?
Here the PE of the products is less than the PE of the reactants
The potential energy stored in the chemical bonds is being converted to
thermal energy.
Delta PE represents the change in PE stored in the bonds of the products
as compared with the bonds in the reactants
This quantity represents the difference between the energy required to
break the bonds in the reactants and the energy released when the
bonds in the product are formed. Bonds weaker
In exothermic reactions the bonds in the
products are stronger than those of the
reactants. More energy is released in
forming the new bonds in the products
than is consumed in breaking
the bonds in the reactants
Bonds stronger

1/2N2 + 3/2H2
N + 3H
472.7 + 3 (216) =
1120.7 kJ/mol 1166. 8 kJ/mol
-46.1
kJ/mol NH3

First Law of Thermodynamics
For any change in state
ΔE = q + w
the change in
internal energy
of the system
the heat flowing
between the
surroundings
and the system
the work done
on the system by
the surroundings
or vice versa

Types of Energy
• Internal energy – total energy of a particles
in system
– Sum of their kinetic and potential energy
• Kinetic energy – energy due to motion
• Potential energy – energy due to position,
or composition

State Functions
• Energy is a state function
• State function – define the
state of a system
– Independent of the history of
a system
• Other important state
functions
pressure volume
temperature
Potential energies of
hiker 1 and hiker 2
are equal regardless
of path taken

Heat: A Chemist’s View
• Heat (q) – transfer of thermal energy
between a system and its surrounding as a
result of temperature difference
– Flows from hot to cold
– Heat “lost” equals heat “gained”
• Misconception:
– Heat is not a property of a system
– Heat appears at the boundary of a system only
when a system is undergoing a change of state
qsystem = - qsurroundings

• Heat Capacity (C) – quantity of heat
required to raise the temperature of a
substance at 25 °C by 1 °C
q = mc∆T = C∆T
• Specific heat capacity (c) – heat capacity
of 1 g of substance
c = q/m∆T
cwater = 4.184 J•g-1•°C-1
• Molar heat capacity – heat capacity of 1
mol of substance
VERY IMPORTANT

Measuring Specific Heat
Heated to
100 °C

ΔH = Δ E when Δn = 0 PV = nRT
ΔH is enthalpy change for a process-
heat absorbed by the system when
the process is carried out at constant
pressure
ΔH =
P ΔV = 0 when V does not change
ΔE = E2 – E1 and PΔV = P( V2 – V1)

Important relation PV = RT, R = PV/T

Note this definition
ΔH = Δ E + P ΔV
This term is
important for
gases but not
significant for
liquids

Practice – Calculate the amount of heat released when 7.40
g of water cools from 49° to 29 °C
q = m ∙ Cs ∙ T
Cs = 4.18 J/gC (Table 6.4)
T1= 49 °C, T2 = 29 °C, m = 7.40 g
q,J
Check:• Check
Solution:• Follow the
concept plan to
solve the
problem
Conceptual
Plan:
Relationship
s:
• Strategize
Given:
Find:
• Sort
Information
Cs m, T q
65
the unit is correct, the sign is reasonable as the
water must release heat to make its temperature
fall
Tro: Chemistry: A Molecular Approach, 2/e

Hess’s Law
• Hess’s Law - If a process occurs in steps,
the enthalpy change for the overall process
is the sum of the enthalpy changes for the
individual steps
• Steps can be carried out hypothetically
• Only possible because enthalpy (H) is a
state function

Standard Enthalpies of Formation
• Standard enthalpies of formation may be
used in conjunction with Hess’ Law to
predict ΔH° for virtually any reaction
CaO(s) + H2O(l)  Ca(OH)2(s) ΔH° = ?
Ca(s) + H2(g) + O2(g)  Ca(OH)2(s) ΔH°f = -986.09 kJ/mol
CaO(s)  Ca(s) + ½O2(g) -ΔH°f = +635.09 kJ/mol
H2O(l)  H2(g) + ½O2(g) -ΔH°f = +285.83 kJ/mol
CaO(s) + H2O(l)  Ca(OH)2(s) ΔH° = -65.17 kJ/mol

Standard Enthalpies of Formation
In general, for a reaction
a A + b B  c C + d D
ΔH°rxn = cΔH°f,C + dΔH°f,D - aΔH°f,A - bΔH°f,B
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
ΔH°rxn= 6ΔH°f,CO2
+ 6ΔH°f,H2O - ΔH°f,C6H12O6
- 6ΔH°f,O2
= 6(-394 kJ) + 6(-286 kJ) – (-1272 kJ) – 6(0 kJ)
= -2808 kJ
= -3.726 kcal/g sugar = -3.726 Cal/g sugar

21 February 2020 70
Thermodynamics
&
Material Properties Relationships

1-Introduction
• One of the significant achievements of classical
thermodynamics is its ability to provide connections
between various properties, so that only a few
measurements are needed for a complete description of a
substance.
• For a more accurate description, particularly for nonideal
gases and condensed phases under extreme conditions,
the relationships between thermodynamic properties need
to be developed.
71 21 February 2020

72 21 February 2020
g = h - Ts
a = u – Ts
• Few of thermodynamic properties can be directly measured
by laboratory experiments. Pressure, temperature and
volume are obviously among the measurable properties.
• There is no instrument to measure entropy or any of the
properties related to energy (u, h, a, and g).
• These quantities cannot be assigned absolute values; only
changes in them as a result of a process have quantitative
meaning.

73 21 February 2020
2-Mathematical Relationships - Partial
Derivatives and Associated Relations
• Derivative of a function f(x)
• Total differential of a function z = f (x, y)

74 21 February 2020
2-i Mathematical Relationships in Thermodynamics
• z = f(x,y)
– z = dependent
– x = independent
– y = independent
• P= P(v,T)

Example:
• If a fluid is heated in a constant-volume container from T to T +  T, what is the
pressure rise  P?
For this problem, the starting function is the equation of state v(T,P), for which the
total differential is:
Dividing by dT and holding v constant:
where  and  are the coefficients of thermal expansion and isothermal
compressibility, respectively. If these coefficients are substantially constant over the
range of T and P involved, the above equation can be directly integrated to yield:
75 21 February 2020

76 21 February 2020
2-ii Partial Differential Relations

77 21 February 2020
2-iii Reciprocity and Cyclic Relations
“divide-and-hold-constant” method

78 21 February 2020
• Reciprocity relation of partial derivatives
• Cyclic relation of partial derivatives

79 21 February 2020
2-iv Mathematical Relationships
M
y




x

N
x




y

2
z
xy

2
z
yx
x
y




z
y
z




x
z
x




y
 1

80 21 February 2020
3-The Maxwell Relations and other Useful
Formulas
• 3-i Maxwell Relations
• The four most common Maxwell relations
• The four most common Maxwell relations are the equalities of the
second derivatives of each of the four thermodynamic potentials, with
respect to their thermal natural variable (temperature T or
entropy S ) and their mechanical natural variable (pressure P or
volume V ).

• Maxwell relations are equations that relate the
partial derivatives of properties P, v, T, and s of a
simple compressible system to each other.
81 21 February 2020
du  Tds  Pdv dh  Tds  vdP
u  u(s,v) h  h(s,P)
Helmholtz Function Gibbs Function
A  U  TS a  u  Ts G  H  TS g  h  Ts
da  du  Tds  sdT dg  dh  Tds  sdT
du  Tds  Pdv dh  Tds  vdP
da  sdT  Pdv dg  sdT  vdP
a  a(T,v) g  g(T,P)

82 21 February 2020
Maxwell Relations are:
du  Tds  Pdv 
T
v




s
 
P
s




v
dh  Tds  vdP 
T
P




s

v
s




P
da  sdT  Pdv 
P
T




v

s
v




T
dg  sdT  vdP 
v
T




P
 
s
P




T

83 21 February 2020
3-ii Other Useful Formulas
du  Tds  Pdv 
u
s




v
 T &
u
v




s
 P
h
s




P
 T &
h
P




s
 v
da  sdT  Pdv 
a
T




v
 s &
a
v




T
 P
dg  sdT  vdP 
g
T




P
 s &
g
P




T
 v

85 21 February 2020
4-The Clapeyron Equation
• Clapeyron equation after the French engineer and physicist E. Clapeyron (1799-1864),
relates the enthalpy change associated with a phase change (such as the enthalpy of
vaporization hfg) from knowledge of P, v, and T data alone.
• If a phase change is occurring:
• The 3rd Maxwell relationship:
• Therefore, the combination of the two equation:

• The Clapeyron equation (also called the Clausius-Clapeyron equation)
relates the slope of a reaction line on a phase diagram to fundamental
thermodynamic properties.
• The form of the Clapeyron equation most often used is:
dP/dT = ΔS/ΔV
• This equations states that the slope (rise/run) of an univariant
equilibrium plotted on a P-T diagram is equal to the entropy change
(ΔS) of the reaction divided by the volume change (ΔV) of the
reaction.
86 21 February 2020

87 21 February 2020
For isothermal liquid–vapor phase–change process:
ds
f
g
 
P
T




v
dv
f
g

sg  sf 
P
T




sat
vg  vf 
sfg
vfg

P
T




sat
dPsat
dT

sfg
vfg

hfg
Tvfg

88 21 February 2020
Clapeyron-Clausius Equation
Ideal gas Pv=RT
Low pressure vg-vf ~ vg vg=RT/P
Small temperature differences hfg linear with temperature

90 21 February 2020
5-Thermodynamic Relations for
Nonideal Behavior
• The objective of this section is to express the deviations from
ideality in terms of the equation of state of the substance and its
heat capacity. Basically, we seek to write total differentials of du,
dh and ds in terms of p, v, and T, and of CP or CV.
• It is required to write total differentials of:
– du, dh, ds
And the values of:
– CP , CV
in terms of p, v, and T.

9121 February 2020
5-i General Relation for internal energy
• Internal energy is a property, it may be a function of temperature and
volume.
• Maxwell relation
• General equation
P
T




v

s
v




T
P
v
s
T
v
u
PdvTdsdu
dv
v
u
dT
T
u
duvTuu
TT
Tv

































 ),(
du  CvdT  T
P
T




v
 P





dv
duv  CvdTv
duT  T
P
T




v
 P





dvT

92 21 February 2020
5-ii General Relation for enthalpy
• Enthalpy is a property, it may be a function of temperature
and pressure.
h  h(T,P)  dh 
h
T




P
dT 
h
P




T
dP
h
P




T
 T
s
P




T
 v
v
T




P
 
s
P




T
dh  CPdT  v  T
v
T




P





dP
dhP  CPdTP
dhT  v  T
v
T




P





dPT

93 21 February 2020
h2  h1  CPdT 
1
2
 v  T
v
T




P





dP
1
2

h2  h1  CPdT 
T1
T2
 v  T
v
T




P





dP
P1
P2

Path
1– x – 2
1– y – 2
1– m – n – 2

94 21 February 2020
• Two relationships of enthalpy and internal energy are:
• Relation of internal energy and enthalpy:
dh  CPdT  v  T
v
T




P





dP
du  CvdT  T
P
T




v
 P





dv
u2  u1  h2  h1  P2v2  P1v1 

9521 February 2020
5-iii General Relation for entropy(1)
• Entropy is a property, it may be a function of temperature and
pressure.
• Recall
s  s(T,P)  ds 
s
T




P
dT 
s
P




T
dP
CP 
h
T




P
 T
s
T




P
v
T




P
 
s
P




T
ds  CP
dT
T

v
T




P
dP

9621 February 2020
5-iii General Relation for entropy(2)
• Entropy is a property, it may be a function of temperature and
volume.
• Recall
s  s(T,v)  ds 
s
T




v
dT 
s
v




T
dv
Cv 
u
T




v
 T
s
T




v
P
T




v

s
v




T
dv
T
P
T
dT
Cds
v
v 









97 21 February 2020
Departure
• Enthalpy departure is the difference between the
enthalpy of a real gas and the enthalpy of the gas at an ideal
gas state and it represents the variation of the enthalpy of a
gas with pressure at a fixed temperature.
• Entropy departure is the difference between the entropy
of a real gas at a given P and T and the entropy of the gas at
an ideal gas state at the same P and T.

98 21 February 2020
ds  CP
dT
T

v
T




P
dP
dv
T
P
T
dT
Cds
v
v 








5-iv General Relations for Specific Heats Cp and Cv

9921 February 2020
• Mayer relation
• Volume expansivity
• Isothermal Compressibility
 




vT
v
vTCC VP
2
2 1








PT
v
v









1

TP
v
v









1


Example:
• liquid water at 20oC and 10 MPa
• The pertinent properties are:
 = 2.0x10-4 K-1;
= 4.4x10-4 MPa-1; and
v = 1.04x10-3 m3/kg
• With these values,
• CP - CV = 0.5 J/mole-K
• By way of comparison, CP of water is 75 J/mole-K. The difference in
the heat capacities of this substance is clearly negligible.
100 21 February 2020
 




vT
v
vTCC VP
2
2 1









101
21 February 2020
dv
T
P
T
dT
Cds
v
v 








ds  CP
dT
T

v
T




P
dP

6- Relations for nonideal gases with
special process restraints
• Isentropic process
dv
T
P
T
dT
Cds
v
v 









• Joule Expansion (constant internal energy process)
Joule’s experiment can be modified to more clearly reveal the
thermal effects of expansion of a nonideal gas when its internal energy
is held constant.
du  CvdT  T
P
T




v
 P





dv

Example:
What is the change in temperature of nitrogen when the specific
volume initially at 20oC and 10 MPa is doubled at constant internal
energy?

• Throttling process
• In fact, the main practical purpose of these devices is to produce an
abrupt reduction in pressure.
• If the fluid is a gas, the change in temperature across the device may
be positive, negative, or zero, depending on the equation of state and
the upstream temperature.
Y
X
h = c

 Joule-Thomson coefficient (JT) is a measure of the
change in temperature with pressure during a constant-
enthalpy process.
 < 0, temperature increases under throttling
 = 0, temperature constant
 > 0, temperature decreases under throttling

Example:
If N2 at 20oC is reduced in pressure in a throttling device
from 10 MPa to 0.1 MPa, what is the temperature change?

Thermodynamics and Spontaneity
– processes that will occur are called spontaneous
• nonspontaneous processes require energy input to go
– if the system after reaction has less potential energy
than before the reaction, the reaction is
thermodynamically favorable.
• Spontaneity ≠ fast or slow
109

Reversibility of Process
• Any spontaneous process is irreversible
– it will proceed in only one direction
– reversible process is at equilibrium
– no change in free energy
• If a process is spontaneous in one direction, it must be
nons-pontaneous in the opposite direction
110

Thermodynamics vs. Kinetics
111

Diamond → Graphite
• Graphite is more stable than diamond
• Graphite into graphite is spontaneous – but don’t worry,
it’s so slow that your ring won’t turn into pencil lead in
your lifetime (or through many of your generations)
112

Spontaneous Processes
• Spontaneous processes occur -release energy from the
system
• higher potential energy to lower potential energy
– exothermic
• some spontaneous processes -proceed from lower
potential energy to higher potential energy
– endothermic
• How can something absorb potential energy, yet have a
net release of energy?
113

Melting Ice
114
Melting is an
Endothermic
process, yet ice will
spontaneously melt
above 0 °C.
When a solid melts, the
particles have more
freedom of movement.
More freedom of motion
increases the
randomness of the
system. When systems
become more random,
energy is released. We
call this energy, entropy

Factors Affecting Whether a
Reaction Is Spontaneous
• There are two factors that determine whether a
reaction is spontaneous. They are the enthalpy
change and the entropy change of the system
• The enthalpy change, H, is the difference in
the sum of the internal energy and PV work
energy of the reactants to the products
• The entropy change, S, is the difference in
randomness of the reactants compared to the
products
115

Enthalpy Change
 H generally measured in kJ/mol
• Stronger bonds = more stable molecules
• A reaction is generally exothermic if the bonds in the
products are stronger than the bonds in the reactants
– exothermic = energy released, H is negative
• A reaction is generally endothermic if the bonds in the
products are weaker than the bonds in the reactants
– endothermic = energy absorbed, H is positive
• The enthalpy change is favorable for exothermic reactions
and unfavorable for endothermic reactions
116

Tro: Chemistry: A
Molecular Approach
118
ENTHALPY OF REACTION
[2ΔHf(NH3 (g))] - [1ΔHf(N2 (g)) + 3ΔHf(H2 (g))]
[2(-46.11)] - [1(0) + 3(0)] = -92.22 kJ
-92.22 kJ (exothermic)
ENTROPY CHANGE
[2ΔSf(NH3 (g))] - [1ΔSf(N2 (g)) + 3ΔSf(H2 (g))]
[2(192.34)] - [1(191.5) + 3(130.59)] = -198.59 J/K
-198.59 J/K (decrease in entropy)
FREE ENERGY OF REACTION (AT 298.15 K)
From ΔGf° values:
[2ΔGf(NH3 (g))] - [1ΔGf(N2 (g)) + 3ΔGf(H2 (g))]
[2(-16.48)] - [1(0) + 3(0)] = -32.96 kJ
-32.96 kJ (spontaneous)
From ΔG = ΔH - TΔS:
-33.01 kJ (spontaneous)
EQUILIBRIUM CONSTANT, K (AT 298.15 K)
595200.760261229

Entropy
• Entropy is a thermodynamic function that
increases as the number of energetically
equivalent ways of arranging the components
increases,
• Random systems require less energy than
ordered systems
119

Changes in Entropy, S
 S = Sfinal − Sinitial
• Entropy change is favorable when the result is a more
random system
– S is positive
• Some changes that increase the entropy are
– reactions whose products are in a more random state
• solid more ordered than liquid - more ordered than gas
– reactions that have larger numbers of product
molecules than reactant molecules
– increase in temperature
– solids dissociating into ions upon dissolving
120

Increases in Entropy
121

S
• For a process where the final condition is more
random than the initial condition, Ssystem is
positive and the entropy change is favorable for
the process to be spontaneous
• For a process where the final condition is more
orderly than the initial condition, Ssystem is
negative and the entropy change is unfavorable
for the process to be spontaneous
 Ssystem  Sreaction = n(S°products) − n(S°reactants)
122

Entropy Change and State Change
123

Practice – Predict whether Ssystem is + or −
for each of the following
• A hot beaker burning your fingers
• Water vapor condensing
• Separation of oil and vinegar salad dressing
• Dissolving sugar in tea
• 2 PbO2(s)  2 PbO(s) + O2(g)
• 2 NH3(g)  N2(g) + 3 H2(g)
• Ag+(aq) + Cl−(aq)  AgCl(s)
S is +
S is −
S is −
S is +
S is +
S is +
S is −
124

The 2nd Law of Thermodynamics
• The 2nd Law of Thermodynamics says that the
total entropy change of the universe must be
positive for a process to be spontaneous
– for reversible process Suniv = 0
– for irreversible (spontaneous) process Suniv > 0
• Suniverse = Ssystem + Ssurroundings
• If the entropy of the system decreases, then the
entropy of the surroundings must increase by a
larger amount
– when Ssystem is negative, Ssurroundings must be
positive and big for a spontaneous process
127

Heat Flow, Entropy, and the 2nd Law
According to the 2nd Law,
heat must flow from
water to ice because it
results in more
dispersal of heat. The
entropy of the universe
increases. 128
When ice is placed in
water, heat flows from
the water into the ice

Quantifying Entropy Changes in
Surroundings
• entropy change in the surroundings - amount of
heat gained or lost
– qsurroundings = −qsystem
• entropy change in the surroundings - inversely
proportional to its temperature
• At constant pressure and temperature, the overall
relationship is
129

Gibbs Free Energy and Spontaneity
very important……
• The Gibbs Free Energy, G, is the maximum amount of
work energy that can be released to the surroundings by a
system
– for a constant temperature and pressure system
– the Gibbs Free Energy is often called the Chemical Potential
because it is analogous to the storing of energy in a mechanical
system
 Gsys = Hsys−TSsys very important
• Because Suniv determines if a process is spontaneous, G
also determines spontaneity
 Suniv is + when spontaneous, so G is −
130

Gibbs Free Energy, G
• A process will be spontaneous when G is negative
 When G = 0 the reaction is at equilibrium
131

Δ H = 10, T = 5, ΔS = 10
T = 5 high
T = 0.5 low
1 Δ G = -10 – [5 x (+ 10)] = - 60
2. Δ G = -10 – [ 0.5 x (–10) ]= - 5 ( T is low)
2 Δ G = -10 – (5 x (–10) = + 40 (T high)
3. Δ G = +10 – (0.5 x (+10) = + 5 (T low
3. Δ G = +10 – (5 x (+10) = - 40 (T high)
4. Δ G = +10 – (5 x (–10) = + 60

Standard Conditions
• The standard state is the state of a material at a
defined set of conditions
• Gas = pure gas at exactly 1 atm pressure
• Solid or Liquid = pure solid or liquid in its most
stable form at exactly 1 atm pressure and
temperature of interest
– usually 25 °C
• Solution = substance in a solution with
concentration 1 M
133

The Standard Entropy Change,
S
Sºreaction = (∑npSºproducts) − (∑nrSºreactants)
– remember: though the standard enthalpy of
formation, Hf°, of an element is 0 kJ/mol, the
absolute entropy at 25 °C, S°, is always positive
134

Calculate S for the reaction
2 H2(g) + O2(g)  2 H2O(g)
135
S is −, as you would expect for a reaction with more
gas reactant molecules than product molecules
standard entropies from Appendix IIB
S, J/K
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
SSH2, SO2, SH2O
Substance S, J/molK
H2(g) 130.6
O2(g) 205.2
H2O(g) 188.8

Calculating G
• At 25 C
Go
reaction = nGo
f(products) - nGo
f(reactants)
• At temperatures other than 25 C
– assuming the change in Ho
reaction and
So
reaction is negligible
Greaction = Hreaction – TSreaction
• or
Gtotal = Greaction 1 + Greaction 2 + ...
136

Standard Free Energies of Formation
• The free energy of formation (Gf°) is the change
in free energy when 1 mol of a compound forms
from its constituent elements in their standard
states
• The free energy
of formation of
pure elements in
their standard
states is zero
137

G Relationships
• sum of the G values of the individual reaction is the
G of the total reaction
• reaction is reversed - sign of its G value reverses
• amount of materials is multiplied by a factor- value of
the G is multiplied by the same factor
138

The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K at 25 °C.
Calculate G and determine if it is spontaneous.
139
Because G is +, the reaction is not spontaneous
at this temperature. To make it spontaneous, we
need to increase the temperature.
H = +95.7 kJ, S = 142.2 J/K, T = 298 K
G, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
GT, H, S

The reaction SO2(g) + ½ O2(g)  SO3(g) has
H = −98.9 kJ and S = −94.0 J/K at 25 °C.
Calculate G at 125 C and determine if it is more or less
spontaneous than at 25 °C with G° = −70.9 kJ/mol SO3.
140
because G is −, the reaction is spontaneous at
this temperature, though less so than at 25 C
H = −98.9 kJ, S = −94.0 J/K, T = 398 K
G, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
GT, H, S

Gº and K
 Under nonstandard conditions, G = G + RTlnQ
Q is the reaction quotient
 At equilibrium G = 0
G = −RTlnK
• When K < 1, Gº is + - reaction spontaneous in the
reverse direction
• When K > 1, Gº is − - reaction is spontaneous in the
forward direction
• When K = 1, Gº is 0 - reaction is at equilibrium
141

Calculate K at 298 K for the reaction
N2O4(g)  2 NO2(g)
142
standard free energies of formation from Appendix IIB
K
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
Substance Gf°, kJ/mol
N2O4(g) +99.8
NO2(g) +51.3
GGf of prod & react K
Gº = −RTln(K)

Calculate G at 25 C for the reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
143
standard free energies of formation from Appendix IIB
G, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
GGf° of prod & react
Substance Gf°, kJ/mol
CH4(g) −50.5
O2(g) 0.0
CO2(g) −394.4
H2O(g) −228.6
O3(g) +163.2

van’t Hoff Equation – derivation from first principles
But from second law of
thermodynamics
bcomes zero
PV = RT, so V = RT/P

‘a’ moles of A
react with ‘b’
moles of B etc.
So we can write
cG(C) = cG0
(C) + cRT ln Pc

Very important
equation.
Measurement of
equilibrium
constant of a
reaction can lead to
the determination of
the thermodynamic
quantity
Reactants and products do the same amount of work in
different directions

By measuring the K at
two different temperatures
one can determine ΔH
using the equation
And Δ S from the
equation
The solubility product at 90 degrees
is only slightly higher than that at 25
degrees

• Combining these two equations
G° = H° − TS°
G° = −RTln(K)
• It can be shown that
• This equation is in the form y = mx + b
• The graph of ln(K) versus inverse T is a
straight line with slope
Why Is the Equilibrium
Constant Temperature
Dependent?
151

152
Very important
By measuring K one can find H and S for a reaction

Concentrati
on of H3O+
and A- are
same

Mg 2+ and O2- , Δ H is
greater
For NaCl
Δ G = Δ H - T ΔS
= 20 – 5 X 5
Δ G = Δ H - T ΔS
= -5
For MgO
= 50 – 5 X 8
= + 10
So for MgO Δ G is
+ ve while for NaCl
it is -ve
Why? Ionic interactions replaced by weaker interactions
So, the solubility of NaCl is due to
entropy effect

Reaction that has a higher K is a
better reaction for the synthesis

Thermodynamic Principles
in
Pharmacy Education & Practice

• Many physical and chemical concepts are highly applicable in both
pharmacy education and practice.
• Also, biological principles play a significant role in understanding the
fundamentals of drug action and its absorption, distribution, and elimination
from the body.
• Among the physical and chemical information that often taught in pharmacy
programs are those related to thermodynamics.
• In particular, courses in biochemistry, physiology, and pharmaceutics can
use thermodynamic concepts to explain the various phenomena in these
fields of study, for example, the diffusion of the drug through
a biological membrane is a spontaneous process (the word spontaneous
implies that the process is natural and irreversible).

• When drug molecules move from an area of high drug concentration (e.g., the intestine) to
an area with low drug concentration (the blood circulation), this pathway is spontaneous.
• It occurs because all natural systems change in such a way so that further changes are less
possible or not possible at all (i.e., they approach a state of equilibrium).
• These principles can explain the direction of the process, and often they are more applicable
nearby an equilibrium point instead of being far away from it.
• Also, they are more pertinent to isolated closed systems than those encountered in
physiological processes.
• So the dilemma that faces a pharmacy educator or practitioner is that how to bring these
scientific principles into actual practice.

• Therefore, it is perhaps best to explain at the students level about what the various
thermodynamic measures mean rather than how to calculate them from observed data, for
example, the enthalpy, as a thermodynamic measure, is defined by the amount of heat
absorbed or released by a system while the system undergoes a change.
• This must be useful in formation to a student-pharmacist who wants to understand how drug
molecules dissolve in a solvent system.
• During dissolution, the system initially absorbs heat from the environment to break down
molecular drug aggregates and in the final step the drug molecules thus released can be
deposited in the free Cavitation available in the solvent.
• During that deposition of drug molecules in the cavity of the solvent, the system gives up the
heat.
• Another important measure in thermodynamics is the entropy - It is the thermodynamic
measure for the amount of energy that is not useful to do work.

• The higher the entropy of a system, the closer that system to its point of equilibrium and the
lower its capacity to do work.
• The pharmacy educator may use for example the states of matter to illustrate the concept of
entropy to his students. In this case, a crystalline solid has the lowest entropy, followed by
the liquid solute, and then the gaseous/vapor phase of the solute where the entropy value is
found at its highest.
• Another illustration for a thermodynamic measure is the concept of free energy - Generally
speaking, the free energy of a system is the amount of energy available to a system to
perform work.
• This often is accompanied initially by the system absorbing heat from the environment.
Since the free energy is being utilized by the system to do work, it is expected therefore that
it decreases in value as the system is doing work.
• Thus, the difference in free energy must be negative if the system undergoes a spontaneous
reaction. If the change in free energy is positive, then no spontaneous response can
happen.

• Another pharmaceutical application of the free energy is in the interfacial phenomenon -
The ability of one liquid to spread on the surface of another liquid, when this occurs, an
amount of free energy is utilized by the system for the spreading to occur. This energy loss
by the system is quantified by the difference between the work of adhesion between the two
liquids, and the work of cohesion of the fluid which is spreading onto the surface.
• By convention, when the work of adhesion is greater in value than the work of cohesion,
spreading is possible.
• The principles of thermodynamics has been introduced in a pharmacy program at
the second year so as to make it continuous and relevant to scientific information as
gained from other course as related above such as biochemistry, pharmaceutics,
and physiology are all candidates where these concepts may be taught.
• Emphasis on understanding the measures and how they relate to each other is more
important than to show actual calculations for these measures.
• Perhaps the instructions may be supplemented with scientific articles where
students can read how these thermodynamic measures helped in delineating a
particular pharmaceutical process (e.g., distinguishing between the various
polymorphic forms of a medicinal substance and other scientific platforms)

Study Questions
• Define the following terms:
• [Thermodynamic, entropy, enthalpy, isentrope, coefficiency, spontaneous, kinetics, exponent, half-
life, catalyst, exotherm, endotherm, isotherm, microscope, macroscope, reciprocity, throttle,
spontaneity, reactant, etc]
• Respond to the following questions:
 State and explain the main three thermodynamics laws and how they are applied in
pharmaceutical processes
 Explain the Law of Mass in thermodynamically reactions
 State and explain some of the variables in a kinetic reaction
 State and explain some of the variable factors that have direct effects on spontaneity of
kinetic reactions
 Explain why equilibrium constant is temperature dependent during the kinetic reaction
process

• Group work discussional questions:
 Write on the laws of thermodynamics during the kinetic reaction process
 Write in details the varying measures and factors during the process of thermodynamically
influenced kinetic reaction
 Write on the thermodynamics and properties of pharmaceutical material substances as
examples exhibited during the kinetic reaction process
 Write on the core relationship of thermodynamics and spontaneity of chemical reaction
 What value do thermodynamic principles of chemical reactions in pharmaceutical procedures

3 thermodynamics of pharmaceutical systems

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