CHEMICAL THERMODYNAMICS
Introduction
Thermodynamics is concerned primarily with the study of heat and its effects. In more
general sense thermodynamics is the study of the transformation (inter-conversion) of energy.
The fundamental principles of thermodynamics are the basis of applied subjects such as
chemical & process engineering and process metallurgy. Chemical thermodynamics is
thermodynamics with emphasis on problems that chemists and chemical engineers are likely
to encounter. Thermodynamic reasoning can be used to find out under what conditions a
process can be possible. Thus it can save us the vain work of trying to carry out a process that
is impossible in principle.
Thermodynamics relates the properties of bulk matter to the behaviour in physical/chemical
processes without making any assumptions about the atomic/molecular nature of the matter.
Basic terms used in thermodynamics
A system: Is part of the universe or part of the world that is selected for consideration (or
what we focus our attention on). It may be a reaction vessel, an engine, an electrochemical
cell, etc.
The region around the system or whole of the universe outside the system is called the
surroundings (or the rest of the universe beyond the system is described as the
surroundings). The surface separating the system from the surroundings is called the
boundary.
Types of systems:
Based on the exchange of heat and matter between the system and the surroundings, the
following types of thermodynamic systems (shown in fig. 1) can be identified:
1.
An isolated system: This is a system where neither heat nor matter can be exchanged
with the surroundings. For example a sealed, well insulated reaction vessel.
2.
A closed system: This is a system which can only exchange heat with the surroundings
but no matter can be transferred across the boundary to the surroundings or vice versa.
For example liquid in stopped flask or a car tyre filled with air and no air may leave or
enter the tyre if the valve is closed. However energy may leave or enter the system as
heat when it is warmed in a strong sunlight.
3.
An open system: This system can exchange both heat and matter with the
surroundings. A good example is a beaker of water open to the atmosphere where
matter can leave the system by evaporation and energy can enter the system as heat by
placing a lighted Bunsen burner underneath the beaker.
Figure 1: Types of thermodynamic systems
Another distinction of systems is based on uniformity of material making up the system.
Homogeneous system is one where the chemical composition and physical properties are
identical throughout the system. Such a system may consist of only one component or a
mixture of components in one phase.
Heterogeneous system consists of two or more distinct phases.
State of a System
By the state of a system we refer to the equilibrium state of the system. The state of a system
is described by its properties, including composition, pressure and temperature. A simple
example of a state description is the equation of state for an ideal gas, PV = nRT (the ideal gas
equation).
State Function
A state function is a property of a system that depends only on the current state of the system,
and not on the way in which the system acquired that state (i.e. independency of path).
A state function is employed in describing the state of the system. Thus for an ideal gas,
pressure, volume and temperature are state functions. Because of relations between state
2
functions, it is sufficient to specify only a certain number of state functions then all other
state functions are prescribed.
For a system that is changed from an initial state to a final state, a state function is one whose
change on going from the initial to the final state is independent of the route/path taken.
So, the chief criterion of a SF is that in changing from a given initial state to a given final
state, the change in SF is the same regardless of the method or pathway by which the change
is made. Familiar SFs are V, P, and T. For example if we change the volume of a gas from V 1
to V2, the change in volume, V = V2 – V1, is independent of any intermediate change in
volume.
Other important characteristic of state functions is that once some state functions have been
specified, the values of all other state functions are fixed.
Path Function
A path function is one which depends upon how change in the system takes place. Two
examples of path functions are heat, q, and work-done, w. The amount of heat or work
transferred during the expansion of a gas depends on whether the expansion is reversible or
irreversible and on how (whether isothermal, adiabatic etc) the expansion occurred. Changes
in path functions are not represented by the symbol ‘’.
Types of State Functions
State functions are divided into two classes: extensive and intensive state functions.
1.
An extensive state function (or state property) is directly proportional to the mass of the
system or amount of the substance(s) present in the system. Thus, if the mass is doubled
then the magnitude of the SF also doubles. Examples are volume, internal energy,
Gibbs energy, entropy etc.
2.
An intensive state function is one which is independent of the amount of the material
present in a system. Examples are temperature, density, viscosity and pressure.
Types of processes
Depending on which particular variable is maintained constant when the process is carried
out, the following are typical thermodynamic processes:
3
An isothermal process: this is a process in which the temperature of the system remains
constant. Isothermal processes can be made to occur in systems that are exposed to and
surrounded by large constant temperature baths known as thermostats.
An adiabatic process: is the process in which no heat is allowed to enter or leave the system
(i.e. q = 0). Systems with such processes are thermally insulated from the surroundings.
An isobaric process: is a process in which the pressure of the system remains unchanged
(constant). A good example is a reaction taking place in an open beaker always at
atmospheric pressure.
An isometric (isochoric) process: is a thermodynamic process in which the volume of the
system remains constant. ∆V = 0. This implies that under isometric process, the system does
no p-v work.
ENERGY
Energy is the most fundamental concept in thermodynamics. To begin with we shall deal
with three forms of energy: Mechanical work, Heat and Internal energy.
1.
Mechanical work is defined as the product of force and displacement. If we raise a
mass m to a height h against the gravitational field, g, we apply a force mg over a
distance h and perform the work w = mgh on the mass. Units of work, w: Nm or J or
kg m2/s2).
Pressure-Volume work (PV-work): This is an important mechanical work in
chemistry. This work results from compression or expansion of gases under the
influence of external pressure.
Work is a measurable quantity.
2.
Heat, q: This is the form of energy that is transferred as a result of temperature change.
It is a measurable quantity. q  mCs T  nCm T , whereby Cs and Cm are specific heat
capacity and molar heat capacity, respectively.
4
3.
Internal Energy, U: This is the energy content of the system. In molecular dimentions,
the internal energy of the system can be attributed to:
i. molecular motion - translational, rotational and vibrational (i.e. the kinetic energy of
the individual molecules (or atoms)) and,
ii. The chemical bonds and other interaction between the molecules or atoms (the
potential energy of the molecules or atoms).
The sum of these contributions is equal to the internal energy, U.
N.B. The internal energy (absolute values of the internal energy) can not be measured.
We can only deduce changes in the internal energy of a system from the heat and work
received from the surroundings.
THE FIRST LAW OF THERMODYNAMICS
1st law of thermodynamics is the statement of the conservation of energy: “energy of the
universe is constant”. It deals with the conversion of energy as well as the exchange of
energy between a system and the surroundings. It requires that the total energy of the
universe be the same before and after exchange or conversion.
For a closed system, in which energy can be exchanged with the surroundings but not matter,
the first law of thermodynamics postulates that changes in internal energy, is due to energy
received as heat and work from the surroundings.
Thus, changes in the internal energy of the system, dU, result from the sum of energy added
in form of heat, q, from the surroundings and work performed on the system, w, by the
surroundings.
dU  dq  dw .
The above equation applies for infinitesimal changes in the system.
For finite changes in the system then the equation becomes: U  w  q .
This equation is a postulate, it can not be derived.
5
dw
dU
dq
Figure 2:
The internal energy is changed by dU due to the supply of heat, dq , and work, dw, from the
surroundings.
Note: Our reference is always the system. Therefore:
the work, w, done on the system by the surroundings is positive, while
the work, w, done by the system on the surroundings is negative
the heat, q, added to the system from the surroundings is positive, while
the heat, q, evolved/released from the system to the surroundings is negative
Internal energy, U, is a state function, hence change in internal energy, U, is independent of
path.
U2
 dU  U
2
 U 1  U
U1
A direct consequence of U being a state function is that for a cyclic process, U =  dU  0 .
Note that q and w are not state functions and can be substantiated as follows:
Consider isothermal (constant temperature) expansion of an ideal gas to double its volume.
Assume in the initial state we have two equal volumes, one empty and another one is filled up
with gas and in the final state we have the gas filled completely on both volumes. This can be
achieved in two ways:
(1) The first way is by making a hole in the separation wall and let the gas expand freely
and fill up the empty part. In this type of expansion, the external pressure is zero. That
is, the system expands against a vacuum. The work done then equals to zero.
Similarly no heat is gained by the system to effect expansion; hence it is also equal to
zero. w  0 and q  0 . This is what we call free expansion (Pext = 0).
6
(2) The same final state can be obtained by an arrangement where the gas performs
mechanical work on the surroundings (w < 0). Heat must be supplied to the gas to
keep the temperature constant (q > 0).
So although the final state is the same in both cases, heat, q, and work, w, supplied to the
system are different. Hence, q and w are not SF but path functions since they depend on the
path used.
Pressure – Volume Work (or PV work)
Many chemical reactions result in the generation or consumption of gases. This can lead to
changes in volume of the reaction system and performance of expansion (pv) work.
PV work: Isothermal expansion
We will derive expressions for work done under isothermal expansion. Consider a gas
enclosed in a cylinder fitted with a frictionless piston of area A as shown in the Fig. 3 below.
Vacuum
q
A
Pin
l
Figure 3: Reversible isothermal expansion of an ideal gas.
Suppose the pressure inside the system, Pin, is very close to being balanced by the weights or
force on the piston area of A, corresponding to an external pressure Pex = Force/A. If Pex <
Pin by an infinitesimal value, the piston will move upwards an infinitesimal distance, dl,
hence, performing an infinitesimal amount of work, dw, to the surroundings. Note that in the
case of expansion the system does work so dw is negative.
 Force 
 dw   Force  dl  
  A  dl   Pex Adl   Pex dV
 A 
where dV is the change in volume. Since Pex  Pin then we can write
dw = -PindV or dw =-PdV.
7
PV work: Reversible Isothermal Expansion
During isothermal expansion, the pressure, Pin will decrease. If Pex is adjusted to be close to
Pin at all times (by removing small weights above the system one by one) then a reversible
process is attained.
The total work done, w, as the gas expands isothermally and reversibly from the initial
volume, V1, to the final volume, V2 (this is the volume after expansion) is given by:
w

V2
 dw    PdV
V1
Note: when dV is positive (expansion), w is negative and when dV is negative (compression),
w is positive.
The equation above can be used when the relation between P and V is specified.
For the ideal gas system, the equation of state: PV = nRT is valid. Then, one can evaluate the
integral as shown below.
P
nRT
V
V2
V
2
V
nRT
dV
dV  nRT 
 nRT ln 2 where, V1 and V2 are the initial volume and the
V
V
V1
V1
V1
w  
final volume, respectively.
The above expression can also be given in terms of pressures by use of Boyle’s law (PV =
constant).
w  nRT ln
P1
P2
 nRT ln
P2
P1
Example: Calculate the work done when one mole of an ideal gas reversibly expands from 5
dm3 to 10 dm3 at 25 oC.
Solution:
w = - (1 mol) (8.314 J/mol-K)(298 K) ln (10/5) = -1718 J
8
PV work: Irreversible Isothermal Expansion
For irreversible isothermal expansion, the gas expands against a constant external pressure,
Pex. Since there is a considerable difference between the gas pressure inside the cylinder and
the external pressure, there is an instant drop in the internal pressure from P1 to P2 = Pex
Therefore the work done depends on the final pressure, P2 = Pex.
V2
w    Pex dV   Pex V2  V1    Pex V
V1
Example: Calculate the work done when one mole of a gas expands from 5 dm3 to 10 dm3
against a constant pressure of 1 atmosphere.
Solution:
w = -PexV = -(101325 N/m2) [(10-5) dm3 x 10-3 m3/dm3] = -506.6 J
Figure 4: The area under the P-V curve during a reversible isothermal expansion. The area
of P-V curve during an irreversible isothermal expansion (shaded yellow) is also shown.
Comparison of work done in reversible and irreversible isothermal expansion
Figure 4 above indicates clearly that more work is done when the expansion occurs reversibly
than when it occurs against a constant pressure (irreversibly). i.e., wirr < wrev.
However for isothermal compression, the reversible process requires less amount of work
compared to the irreversible process. i.e. wirr > wrev
9
ENTHALPY
The enthalpy (denoted by H) of a system whose internal energy, pressure and volume are U,
P and V, respectively, is defined as:
H = U + PV; since U, P and V are State Functions, H is a State Function. Note from dw = PdV that the product of P and V has the dimension of work, energy. Naturally H has units of
energy.
For any change of state, the enthalpy change is
H  H 2  H1  U 2  P2V2  (U1  P1V1 )  U  ( PV )
For a constant pressure process
P2  P1 and ( PV )  PV2  PV1  PV
 H  U  PV
constant P
We have shown that q is not a SF. But it can be replaced by a SF under specified
conditions.
Most chemical reactions are carried out not at constant volume but constant pressure.
Chemical reactions are normally conducted in flasks/test tubes or beakers under constant
pressure of the atmosphere.
In this case, w = -PV
Using 1st law of thermodynamics
U = q + w = q + (-PV)  q = U + PV
(i)
But from above, the enthalpy change is given by the following relation:
H = U + (PV)
At constant pressure
H = U + P V
(ii)
When (i) and (ii) are compared
qp = H
The enthalpy change of a reaction is the value of the total amount of heat a chemical reaction
can provide to the surroundings or absorb from the surroundings at constant pressure.
Also, consider a process taking place at constant volume (calorimetric studies). Volume
changes (dV) equals to zero. Thus, -PdV = 0, i.e. no work performed (w = 0).
U = q + w (1st law of therm.); U = q (w = 0).
Therefore, U = qv.
10
Internal energy and enthalpy for an ideal gas: Joule’s experiment
Joule performed an experiment to study the internal energy of a gas. The set of his
experiment included a closed container consisting of two bulbs connected together by a valve
and immersed in a calorimeter. One bulb is filled with gas while the other bulb is evacuated.
The set-up ensures that no work can be performed or supplied when the valve is opened,
(expansion into vacuum), i.e. w = 0. The temperature of the calorimeter is measured by
means of a thermometer (cf. figure 5 below). After reading the temperature of the
calorimeter, the valve is opened and allowed to fill up the other empty bulb. When the
equilibrium has been established in these two bulbs the temperature of the calorimeter is
recorded again.
Thermometer
Gas
Vacuum
Calorimeter
Figure 5: Apparatus for the Joule experiment: The gas is allowed to expand irreversibly into
the evacuated side of the apparatus
Results of Joule experiment: Joule found no temperature change for an ideal gas, i.e., T =
0, which means q = 0 and since no work was performed (free expansion),
U  q  w  0
 U 
This shows that for an ideal gas, 
  0.
 V T
Thus the internal energy change is a function of temperature only. ( U = 0 at constant
temperature).
That the internal energy change for an ideal gas is a function of temperature only can also be
inferred from the following line of argument:
For an ideal gas of fixed mass, the internal energy U may be taken to be a function of any two
variables of T, V, P because the three variables are interconnected. It is however most
11
convenient to take U as a function of T and V i.e., U = f(T,V). Since U is a state function, the
total differential, dU, is given by:
U 
U 
dU    dT    dV
T V
V T
 U 
But Joule experiment for an ideal gas showed that 
 0
 V T
dU  U 
 U 
Hence, dU  

 dT 
 . This means that internal energy of an ideal gas
dT  T V
 T V
depends only on one function and that function is the temperature.
By definition; H = U + (PV). But from ideal gas equation, PV = nRT the enthalpy change
equals to
H = U + (nRT) = U + nR(T)
Since Joules experiment showed that for an ideal gas if T = 0 then U = 0, therefore also
H = 0 for a change of pressure at constant temperature.
Alternatively: The enthalpy of a system is taken to be a function of pressure and temperature.
i.e., H = f(P,T)
The total differential of H is given by:
 H 
 H 
dH  
 dP  
 dT
 P T
 T  P
 H 
But Joule experiment for an ideal gas showed that 
 0
 P T
dH  H 
 H 
Hence, dH  

 dT 
 Thus, H depends on T only.
dT  T  P
 T  P
12