A Textbook of Chemical Engineering Thermodynamics, 2e
A Textbook of Chemical Engineering Thermodynamics, 2e
A Textbook of Chemical Engineering Thermodynamics, 2e
Properties of Solutions
We have seen in Chapter 6 that the thermodynamic properties of homogeneous pure substances
depend only on the state of the system. The relationships developed for pure fluids are not applicable
to solutions and need modification. The thermodynamic properties of solutions and heterogeneous
systems consisting of more than one phase are influenced by the addition and removal of matter. The
term solution includes homogeneous mixtures of two or more components in the gas, liquid or solid
phase. The pressure, temperature and the amount of various constituents present determine the
extensive state of a solution; and pressure, temperature and composition determine the intensive state.
In this chapter, we discuss how the thermodynamic properties of a solution are determined and
introduce certain concepts that are essential to the study of phase equilibria and chemical reaction
equilibria.
In Eq. (7.1), n is the total number of moles and M is the molar property of the solution. ni denotes the
number of moles of component i in solution, so that n = Sni.
In general, any partial molar property is the increase, in the property Mt of the solution resulting
from the addition at constant temperature and pressure, of one mole of that substance to such a large
quantity of the system that its composition remains virtually unchanged. It is an intensive property and
its value depends only on the composition at the given temperature and pressure. The subscript njπi
indicates that the number of moles of all components in the solution other than the number of moles of
i are kept constant.
We have seen that the effective molar volume of water added to the ethanol–water solution, i.e. the
partial molar volume in the solution is less than the molar volume VW of pure water at the same
temperature and pressure. To be specific, when pure water is added to an ethanol water solution of
volume Vt and allowed sufficient time for heat exchange so that temperature remains the same as that
before addition, the increase in volume of the solution DVt π DnWVW, where DnW is the moles of
water added. The increase in volume is given by
In this process, a finite drop of water was added which may cause a finite change in composition. If
were to represent a property of the solution, it must be based on data for the solution at this
composition. For an infinitesimal amount of water added, Eq. (7.3) becomes
and it denotes the incremental change in mixture volume which occurs when a small quantity of
component i is added at constant pressure and temperature. The amount of i added is so small that no
detectable change in composition occurs. While the molar volume is always positive, the partial
molar volume may even be negative. The partial molar volume of MgSO4 in water at infinite dilution
(i.e. in the limit of zero concentration) is –1.4 10–6 m3/mol which means that the addition of one
mole of MgSO4 to a large volume of water results in a decrease in volume of
1.4 10–6 m3. The contraction may be due to the breaking up and subsequent collapse of the open
structure of water as the ions become hydrated.
Though different from molar properties of the pure components, to get a physical picture of the
concept of partial molar properties, we can treat them as the molar properties of the components in
solution. However, it is to be borne in mind that the components of a solution are intimately
intermixed and cannot have individual properties of their own. The partial molar properties in fact,
represent the contribution of individual components constituting the solution to the total solution
property as described in the following section.
7.1.2 Partial Molar Properties and Properties of Solution
Consider any thermodynamic extensive property (such as volume, free energy, heat capacity, etc.), its
value for a homogeneous system being completely determined by the temperature, pressure and the
amounts of various constituents present. Let M be the molar property of a solution and Mt be the total
property. Then, Mt = nM, where n is n1 + n2 + n3 + . . . . Here, n1, n2, n3, . . . are the number of
moles of the respective components 1, 2, 3, . . . of the system. The solution property is a function
represented by
Mt = f(P, T, n1, n2, . . ., ni, . . .)………(7.6)
If there is a small change in the pressure, temperature and the amounts of various constituents, then
It is evident that the partial molar properties are not extensive properties, but are intensive
properties of the solution. They depend, therefore, not upon the total amount of each constituent, but
only upon the composition, or the relative amounts of the constituents. If we add several constituents
simultaneously to a given solution at constant temperature and pressure, keeping the ratio of the
various constituents constant, the partial molar properties are not changed. Then, the change in
property
This equation along with Eq. (7.9), which can be written in the following form serves as the
relationship between partial molar properties and total solution property.
We see that the partial molar property of any constituent may be regarded as the contribution of
one mole of that constituent to the total value of the property under the specified conditions. In other
words, the partial molar properties may be treated exactly as though they represent the molar
properties of the components in the solution.
EXAMPLE 7.1 Give an alternative derivation for Eqs. (7.12) and (7.14) starting from Eq. (7.9)
Solution Equation (7.9) gives
n represents the total amount of various constituents and dn the changes in the total number of moles.
One is free to choose any value for n as well as dn. In short, n and dn can be independently changed.
For all possible values of n and dn, the above equation is to be satisfied. This is possible only if the
terms in brackets reduce to zero.
Method 3 (The tangent-intercept method). This is also a graphical method widely used for
the determination of partial molar properties of both components in a binary solution. The molar
volume V is plotted against mole fraction of one of the components (say, x2, the mole fraction of
component 2). To determine the partial molar volumes, draw the tangent to the curve at the desired
mole fraction. The intercept that this tangent makes with the vertical axis at x2 = 1 gives and the
intercept on the vertical axis at x2 = 0 (or x1 = 1) gives . In Fig. 7.3, BD = and AC = .
To prove this result, consider a binary solution containing n1 moles of component 1 and n2 moles of
component 2. Let the total volume be Vt and let V be the molar volume. Then
In Fig. 7.3, the length BD = BE + ED, where BE is the slope of the tangent at P times the length PE.
That is,
BE = (1 – x2) (∂V/∂x2)
and ED = V, the molar volume at the mole fraction x2. Thus
BD = V + (1 – x2) (∂V/∂x2)
which, by Eq. (7.22) is . Similarly, the length
AC = FC – FA = V – x2 (∂V/∂x2) =
The above methods are applicable for the determination of various other partial molar properties
also. Of the various mixture properties, only the volume can be determined absolutely. For the
determination of other properties like , etc., it becomes necessary to work with property changes
on mixing (discussed later in this chapter) like DG, DH, etc. The method of tangent intercept for the
determination of, say , requires the plot of DG per mole versus x2.
EXAMPLE 7.5 At 300 K and 1 bar, the volumetric data for a liquid mixture of benzene and
cyclohexane are represented by V = 109.4 10–6 – 16.8 10–6x – 2.64 10–6x2, where x is the
mole fraction of benzene and V has the units of m3/mol. Find expressions for the partial molar
volumes of benzene and cyclohexane.
Solution The molar volume of the solution as a function of composition is given:
V = 109.4 10–6 – 16.8 10–6x1 – 2.64 10–6x12………(7.23)
where x1 = mole fraction of component 1 (in this case, benzene). By Eq. (7.20),
EXAMPLE 7.6 The enthalpy at 300 K and 1 bar of a binary liquid mixture is
H = 400x1 + 600x2 + x1x2(40x1 + 20x2)
where H is in J/mol. For the stated temperature and pressure, determine:
(a) Expressions for and in terms of x1
(b) Numerical values for the pure component enthalpies H1 and H2
(c) Numerical values for the partial molar enthalpies at infinite dilution .
Solution The molar enthalpy of the solution as function of concentrations of the constituents is given:
H = 400x1 + 600x2 + x1x2(40x1 + 20x2)………(7.26)
(a) Differentiating Eq. (7.26), we get
Since x1 = 1 – x2, dx1 = – dx2 and ∂x2/∂x1 = –1, the above equation simplifies to
EXAMPLE 7.7 The volume of an aqueous solution of NaCl at 298 K was measured for a series of
molalities (moles of solute per kg of solvent) and it was found that the volume varies with molality
according to the following expression.
V = 1.003 10–3 + 0.1662 10–4m + 0.177 10–5m1.5 + 0.12 10–6m2
where m is the molality and V is in m3. Calculate the partial molar volumes of the components at m =
0.1 mol/kg.
Solution The partial molar volume of NaCl:
Differentiating Eq. (7.29) with reference to m,
………(7.30)
The total free energy Gt of a solution is a function of pressure, temperature and number of moles of
various components.
Gt = f(P, T, n1, n2, . . ., ni, . . .)………(7.31)
By reasoning analogous to that used in the derivation of Eq. (7.12), we have, at constant temperature
and pressure,
Gt = S mini
For a binary solution, the molar free energy of the solution is
G = x 1m 1 + x 2m 2
The chemical potential of a component is thus seen to be the contribution of that component to the
Gibbs free energy of the solution. The chemical potential is an important property of solution
extensively used in the study of phase and chemical equilibria.
7.2.1 Effect of Temperature and Pressure on Chemical Potential
Effect of temperature. Consider Eqs. (7.30) and (7.34). Differentiate Eq. (7.30) with respect to
temperature. Then
where is the partial molar entropy of the component i in the solution. This result, though gives the
variation of chemical potential with temperature, can be put in a more useful form [compare with Eq.
(6.73)] as follows: Since
Equation (7.50) is an alternative definition of chemical potential. But it should be understood that
(∂Ut/∂ni)S,V,njπi is not partial molar internal energy, for it refers to constant entropy and volume and
not to constant temperature and pressure. Partial molar internal energy is not equal to chemical
potential.
EXAMPLE 7.9 Show that for an ideal gas,
which states that the fugacity of a component in a mixture of ideal gases is equal to the partial
pressure of that component in the mixture. However, this is not true for real gases. Equation (7.57)
provides the means for computing fugacities in the real gaseous solution. But this requires the
evaluation of as a function of pressure, which in turn requires the knowledge of how the solution
volume varies with composition at each pressure. These types of data are rarely available, and hence
rigorous calculation of fugacities in gaseous mixtures using Eq. (7.57) is rarely done.
Note that the right-hand side of Eq. (7.57) reduces to the same result as that given by Eq. (6.128)
where the residual volume for the pure component is given by a = Vi – RT/P. That is, for pure
components at a temperature T and pressure P,
Subtracting Eq. (7.62) from Eq. (7.63),
which is commonly known as Lewis–Randall rule or Lewis fugacity rule. It states that fugacity of a
component in an ideal solution is directly proportional to the mole fraction of the component in the
solution. In Eq. (7.65), is the fugacity of the species i in an ideal gaseous solution, and fi is the
fugacity of pure i evaluated at the temperature and pressure of the mixture. Thus, we have now,
for ideal gaseous solution and, for ideal (perfect) gases.
For a gas mixture to behave as an ideal solution, it requires only that the molar volume in the pure
state and the partial molar volume in the solution be the same, or . For the mixture to be an ideal
gas it requires that , which means that the molar volumes of all the components are the
same whether in the mixture or in the pure state. For ideal solutions, the volumes of components may
differ from one another. In short, the concept of an ideal gaseous solution is less restrictive than that
of a mixture of ideal gases.
The Lewis–Randall rule is a simple equation and is therefore widely used for evaluating fugacities of
components in gas mixtures. It allows the fugacity of a component in the mixture to be calculated
without any information about the solution except its composition. However, it is not reliable because
of the severe simplification inherent in Amagat’s law of additive volumes. But at high pressures it is
often a very good assumption, because, at liquid like densities, fluids tend to mix with little or no
change in volume (J.M. Prausnitz et al., 1986). Lewis fugacity rule is valid for systems where the
intermolecular forces in the mixture are similar to those in the pure state. Thus, it can be said that this
rule is valid
1. At low pressures when the gas phase behaves ideally
2. At any pressure if the component is present in excess
3. If the physical properties of the components are nearly the same
4. At moderate and high pressures, the Lewis–Randall rule will give incorrect results if the
molecular properties of the components are widely different and the component under
consideration is not present in excess.
7.3.3 Fugacities in Liquid Solutions
Calculation of fugacity of a component in a liquid solution using Eq. (7.57) is not practical because
the volumetric data at constant temperature and composition are rarely available. These data are
required for the integration over the entire range of pressures from the ideal gas state to the pressure
of the solution including the two-phase region. For calculation of fugacities in liquid solutions,
another approach is used. We define an ideal solution whose fugacity can be easily calculated
knowing its composition and measure the departure from ideal behaviour for the real solution. A
quantitative measure of the deviation from ideality is provided by the function known as the activity
coefficient which will be discussed in Section 7.6.
7.3.4 Ideal Solutions and Raoult’s Law
A solution in which the partial molar volumes of the components are the same as their molar volumes
in the pure state is called an ideal solution. There is no volume change when the components are
mixed together to form an ideal solution. That is, for an ideal solution V = S xi = S xiVi, where V
is the molar volume of the solution, Vi and are the molar volume and partial molar volume
respectively of the component i, and xi is the mole fraction of component i in the solution. If a mixture
of two liquids is to behave ideally, theoretical considerations reveal that the two types of molecules
must be similar. The environment of any molecule and hence the force acting on it is then not
appreciably different from that existing in the pure state. We have shown that for ideal gaseous
solutions, the Lewis–Randall rule is applicable which states that fugacity of each constituent is
directly proportional to the number of moles of the constituent in the solution. The Lewis–Randall
rule is applicable to ideal liquid solutions also. It can be written as
………(7.66)
where is the fugacity of component i in the solution, fi is the fugacity of i in the pure state, and xi is
the mole fraction of component i in the solution.
While the ideal solution model is adequate for many gas mixtures for reasonable temperature and
pressure, the same is not true for the case of liquid solutions. Very few solutions follow
Eq. (7.66) over the entire composition range. Ideal liquid solution behaviour is often approximated
by solutions comprised of molecules not too different in size and chemical nature. Thus a mixture of
isomers (e.g. ortho- , meta- and para-xylene), adjacent members of homologous series of organic
compounds (e.g. n-hexane and n-heptane, ethanol and propanol, benzene and toluene, ethyl bromide
and ethyl iodide) etc., are expected to form ideal solutions.
Raoult’s Law. The criterion of phase equilibria permit us to replace the liquid phase fugacities
and fi with fugacities in the gas phase with which the liquid is in equilibrium. Thus, under
equilibrium. Here superscripts V and L refer to the vapour phase and the liquid phase respectively.
Thus, fugacity in Eq. (7.66) is equal to the fugacity of constituent i in the vapour phase. If the vapour
phase is assumed to be ideal gas, which is true if the pressure is not too high, the vapour phase
fugacity is the same as partial pressure of component i in the vapour. If the liquid phase is pure
i, the fugacity of pure i in the vapour phase can be replaced with the vapour pressure . Under these
conditions the Lewis–Randall rule, Eq. (7.66), becomes
………(7.67)
This expression is known as Raoult’s Law. This is a simplified form of the Lewis–Randall rule.
Whereas the Lewis–Randall rule is obeyed by all ideal solutions, the Raoult’s law is applicable to
ideal solutions if the vapour phase with which it is in equilibrium is an ideal gas.
Raoult’s law provides a very simple expression for calculating the fugacity of a component in the
liquid mixture which is the same as the partial pressure of the component in the vapour. It says that the
partial pressure is directly proportional to the mole fraction in the liquid solution. Ideal solutions
which conform to Raoult’s law over the entire range of concentrations are rare. A frequently cited
example for ideal solutions is mixtures of optical isomers of organic compounds. Raoult’s law
applies as fair approximation to mixtures of hydrocarbons showing a reasonable similarity in
molecular structure such as are encountered in petroleum industry. In most other cases Raoult’s law
applies only over a limited concentration range.
Often, the solute portion of the non-ideal liquid solution can be assumed to follow Henry’s law. is
the partial pressure of the solute over the solution, xi is its mole fraction in the solution and Ki is a
proportionality constant known as Henry’s law constant. Ki may be greater or less than , the
vapour pressure of the solute at the temperature and total pressure in question. When Ki and are
equal, Henry’s law and Raoult’s law are identical. Henry’s law may be thought of as a general rule of
which Raoult’s law is a special case. Henry’s law is obeyed in all solutions by the solute at
extremely low concentrations. Essentially all liquids will obey Henry’s law close to mole fraction
zero, but many will deviate from the law above 0.01–0.02 mole fraction. And almost all liquids
deviate above 0.1 mole fraction. But in some exceptional cases, Henry’s law is found to be obeyed
quite well up to xi = 0.5.
For ideal solutions, the partial fugacity (or partial pressure) of a component is proportional to its
mole fraction. For a real solution it has been found experimentally that as the mole fraction of the
component approaches unity, its fugacity approximates to the value for an ideal solution, though at
lower mole fractions, the behaviour departs markedly from ideal behaviour.
In Fig. 7.4, the fugacity curve becomes asymptotic to the straight line showing ideal behaviour as
mole fraction approaches unity. In a dilute solution, the component present in larger proportions
designated as solvent, obeys Raoult’s law even though it may depart from ideal solution behaviour in
a more concentrated solution. As the mole fraction of the solute—the component present in smaller
proportions—approaches zero, it will conform to the ideal behaviour predicted by Henry’s law.
Thus, we can generalise by saying that the solute in a dilute solution obeys Henry’s law and the
solvent obeys Raoult’s law. It can be shown that over the range of compositions where the solvent
obeys Raoult’s law, the solute obeys Henry’s law (see Example 7.15).
Therefore, the solubility of oxygen is 0.0568 10–4 moles per mole of water. In mass units, it can
be written as
0.0568 10–4 32 1/18 = 0.10110–4 kg oxygen per kg water
EXAMPLE 7.11 The partial pressure of acetone (A) and chloroform (B) were measured at 298 K and
are reported below:
(a) Confirm that the mixture conforms to Raoult’s law for the component present in excess and
Henry’s law for the minor component.
(b) Determine the Henry’s law constants.
Solution The partial pressures are plotted against mole fraction xA as shown in Fig. 7.5.
From the data given, it can be seen that = 0.457 and = 0.386 bar. The dotted line representing
the ideal behaviour (Raoult’s law) of component A is drawn by joining the origin and (x = 1, p =
0.457) by a straight line. Raoult’s law for component B is also drawn. The dotted lines PA and QB
represent the ideal behaviour. The Henry’s law line PR is drawn tangential to the curve versus xA
as xA tends to 0 and the line QS is drawn tangential to the versus xA curve as xA tends to 1.
(a) We see that the partial pressure curve for component A coincides with the Raoult’s law line in
the region where mole fraction of component A approaches unity and in this region, the partial
pressure of component B coincides with the Henry’s law line. Thus, in the region where
Raoult’s law is obeyed by A, Henry’s law is obeyed by B, and vice versa.
(b) The slopes of the Henry’s law line PR gives KA, the Henry’s law constant for A. KA = 0.23 bar.
Similarly slope of QS is KB. KB = 0.217 bar.
For ideal solutions as , the activity ai = xi. For real solutions, the activity can be shown to be
equal to the product of activity coefficient and mole fraction. The activity coefficient is discussed
later in this chapter.
The term activity is a ratio without dimensions. It is a widely used function in solution
thermodynamics, particularly in dealing with property changes of mixing. The relationship between
property change of mixing and activity is discussed later in this chapter.
That is, the activity of a component in a mixture of gases is equal to its fugacity, numerically. If
the mixture behaves as an ideal gas at the given conditions the activity and partial pressure are
the same. This standard state is used in the study of chemical reaction equilibrium.
2. The pure component gas at the pressure of the system. With this choice the activity of each
component in ideal gas solution becomes equal to its mole fraction.
This standard state becomes hypothetical at temperatures where the total pressure exceeds the
saturation pressure of the component gas in the pure state. Vapour–liquid equilibrium studies
conventionally use this standard state.
Liquids. Two standard states are common for liquids also.
1. The pure component liquid at a pressure of 1 bar. This state is hypothetical if the vapour
pressure of the pure liquid exceeds 1 bar.
2. The pure liquid at the pressure of the system. This state becomes hypothetical at temperatures
above the critical or saturation temperature of the pure liquid. This standard state is used in
vapour–liquid equilibrium studies.
Solids. The standard state chosen for solid is usually the pure component in the solid state at a
pressure of 1 bar.
where is the fugacity in the standard state. For ideal solutions gi = 1, and we have
which is same as the Lewis–Randall rule [Eq. (7.66)] with the pure liquid at the system pressure as
the standard state.
Two types of ideal behaviour are observed; the first conforms to Lewis–Randall rule (or Raoult’s
law) in which case , the fugacity of the pure species at the system pressure and the second type
conforms to an ideal dilute solution behaviour (the Henry’s law), in which case , the Henry’s
law constant. Depending upon the standard states on which they are based, the activity coefficients
can take different numerical values. For the standard state in the sense of Lewis–Randall rule or
Raoult’s law,
where ai is the activity of i in the solution. Equation (7.77) is, in fact, Lewis fugacity rule modified by
the factor gi to correct for deviation from ideality. This equation should reduce to Raoult’s law as x
approaches unity and to Henry’s law as x approaches zero. For this to be possible, g must equal unity
as mole fraction approaches unity (Raoult’s law region) and Ki/fi, as mole fraction, approaches zero
(Henry’s law region). In terms of partial pressures, Eq. (7.77) may well be written as
Then the activity coefficient approaches unity as x approaches zero. In Eqs. (7.81) and (7.82), is
the activity coefficient referred to infinite dilution.
When activity coefficients are defined with reference to an ideal solution in the sense of Raoult’s
law, then for each component i,
gi 1…as…xi 1
On the other hand, if activity coefficients are defined with reference to an ideal dilute solution, then
g1 1…as…x1 1 (solvent)
1…as…x2 0 (solute)
Activity coefficients with reference to ideal dilute solution would be useful when dealing with liquid
mixtures that cannot exist over the entire composition range as happens, for example, in a liquid
mixture containing gaseous solute. If the critical temperature of the solute is lower than the
temperature of the mixture, then a liquid phase cannot exist as x2 1, and the relations based on an
ideal mixture in the sense of Raoult’s law can be used only by introducing a hypothetical standard
state for component 2. However, relations based on an ideal dilute solution eliminate this difficulty.
Activity coefficients are very strong functions of concentration of solution. The variation of g with x
over the entire range of composition is usually complex, but can often be roughly approximated in
binary solutions by the empirical equations such as the one proposed by Porter:
where b is an empirical constant. These relationships apply best when the components are not too
dissimilar in structure and polarity.
7.6.1 Effect of Pressure on Activity Coefficients
The effect of pressure on fugacity was derived in Chapter 6 [Eq. (6.126)].
The molar volumes and Vi correspond to the particular phase under consideration. For liquid
solutions, the effect of pressure on activity coefficients is negligible at pressures below atmospheric.
For gaseous mixtures, activity coefficients are nearly unity at reduced pressures below 0.8.
7.6.2 Effect of Temperature on Activity Coefficients
The effect of temperature on fugacity of a pure substance was given by Eq. (6.125) as
Equation (7.88) gives the effect of temperature on activity coefficients. The term ( – Hi) is the
partial heat of mixing of component i from its pure state to the solution of given composition both in
the same state of aggregation and pressure. For gaseous mixtures, this term is negligible at low
pressures.
EXAMPLE 7.12 The partial pressures of acetone (A) and chloroform (B) were measured at 298 K
and are reported below:
xA 0 0.2 0.4 0.6 0.8 1.0
, bar 0 0.049 0.134 0.243 0.355 0.457
, bar 0.386 0.288 0.187 0.108 0.046 0
The above equations are used to calculate the activity and activity coefficients for different
concentrations. A sample calculation is provided below for the second set where
xA = 0.2, xB = 0.8, = 0.049 bar, = 0.288 bar, KB = 0.217 bar, = 0.386 bar
The above calculations are repeated for other concentrations and the results are given below:
xB 0 0.2 0.4 0.6 0.8 1.0
a 0 0.12 0.28 0.48 0.75 1.0
a 0 0.21 0.50 0.86 1.33 1.78
g 0.60 0.70 0.80 0.94 1.0
g 1.0 1.05 1.25 1.43 1.66 1.78
EXAMPLE 7.13 The fugacity of component 1 in binary liquid mixture of components 1 and 2 at 298
K and 20 bar is given by
The second terms on both sides of the above equation vanish, as they are equal to unity. Therefore,
As the activity coefficients directly measure the departure from the ideal solution behaviour,
Eq. (7.101) is the most useful form of the Gibbs–Duhem equation.
The various forms of Gibbs–Duhem equations are rigorous thermodynamic relations that are valid for
conditions of constant temperature and pressure. They tell us that the partial molar proper-ties of a
mixture cannot change independently; in a binary mixture, if the partial molar property of one of the
component increases, the partial molar properties of the other should decrease.
Gibbs–Duhem equations find wide applications in solution thermodynamics. These include:
(a) In the absence of complete experimental data on the properties of the solution, Gibbs–Duhem
equations may be used to calculate additional properties. For example, if experimental data are
available for the activity coefficient of one of the components in a binary solution over certain
concentration range, the activity coefficient of the other component over the same composition
range can be estimated using Gibbs–Duhem equations. This is particularly useful wherever the
volatilities of the two components differ markedly. The measurements usually give the activity
coefficient of the more volatile component whereas that of the less volatile component is
calculated using Eq. (7.101). Thermodynamic properties of some high-boiling liquids (e.g.
polymers) dissolved in a volatile liquid (say, benzene) can be computed by measuring the
partial pressure of the latter in the solution.
(b) Thermodynamic consistency of experimental data can be tested using Gibbs–Duhem equations.
If the data on the partial molar property of each component measured directly in experiments
satisfy the Gibbs–Duhem equation, it is likely that they are reliable, but if they do not satisfy the
Gibbs–Duhem equation, it is certain that they are incorrect.
(c) Gibbs–Duhem equations can be used for the calculation of partial pressure from isothermal
total pressure data. Suppose that in an experimental investigation of vapour–liquid equilibrium,
the total pressures are measured as a function of composition of one of the phases (usually the
liquid phase) and the composition of the other phase is not measured. The Gibbs–Duhem
equation facilitates the calculation of the composition of other phase thereby reducing the
experimental work considerably.
(d) Partial pressure data can be obtained from isobaric boiling point data using Gibbs–Duhem
equations. The isobaric T-x data can be easily converted to x-y data.
In the sections that follow the application of Gibbs-Duhem equation is illustrated in the
derivation of the relationship between Henry’s law and Raoult’s law for a real solution (see Example
7.15), in proving the essential criterion that the vapour and liquid compositions are the same for an
azeotropic mixture (see Example 8.11) etc.
EXAMPLE 7.14 Show that in a binary solution, if the molar volume of one of the components
increases with concentration, the molar volume of the other must decrease.
Solution When Eq. (7.94) is written for one mole of the solution with M replaced by V, we get
It means that if is positive, must be negative. That is, if partial molar volume of component 1
increases the partial molar volume of component 2 must decrease.
EXAMPLE 7.15 Prove that if Henry’s law is obeyed by component 1 in a binary solution over
certain concentration range, Lewis–Randall rule (Raoult’s law) will be obeyed by component 2 over
the same concentration range.
Solution Equation (7.99) gives Gibbs–Duhem equations in terms of fugacities
where a, b, c are constants independent of concentrations. Obtain an expression for g2 in terms of x1.
Solution Using the Gibbs–Duhem equation [Eq. (7.101)], we get
where C is a constant of integration. Integrating and using the boundary condition that when
x2 = 1 (or x1 = 0 ), g2 = 1 we get C = 0. Therefore, we get the required expression:
The above example illustrates how the activity coefficient of one of the species in a binary mixture
can be evaluated if the activity coefficient of the other is known as an analytical equation in x. Now,
suppose that g1 is determined experimentally and is reported as a function of x in a tabular form.
How is g2 evaluated? Rearrange Gibbs–Duhem equation, Eq. (7.101) in the form
The integral in Eq. (7.106) is to be evaluated graphically. For this, plot a graph taking x1/x2 along the
y-axis and ln g1 on the x-axis. The area under the curve from ln g1 at x1 = 0 to the
ln g1 value at the desired concentration x1 will give the integral in Eq. (7.106). The negative of this
is the value of ln g2 at x1.
The quantity can be treated as the change in the property of component i when one mole of
pure i in its standard state is brought to the solution of given composition at the same temperature and
pressure. Using Eq. (7.111), the volume change of mixing and free energy change of mixing can be
written as
DV = S xi ………(7.112)
DG = S xi ………(7.113)
are the partial molar volume and molar volume of component i in the standard state
respectively. Replacing in Eq. (7.112) in terms of using the preceding two
relations, we get
Enthalpy change of mixing, DH. The Gibbs–Helmholtz equation [see Eqs. (6.73) and (7.41)]
relates the free energy of the substance in the pure state or in the solution to the corresponding
enthalpies of the substance as
Substituting Eq. (7.114) into this, we get
Entropy change of mixing, DS. The partial molar entropy and molar entropy of component i
in the standard state are related to the free energy of i in the solution and free energy of pure i as
given below.
Equations (7.115), (7.117), (7.119) and (7.120) reveal that all property changes of mixing can be
written in terms of activity of the components in solution.
7.8.2 Property Changes of Mixing for Ideal Solutions
For ideal solutions fugacity is given by Lewis–Randall rule, . With reference to the pure
component standard state, the activity is given as ai = /fi = xi. Replacing ai with xi in the results in
the previous section and noting that the concentration is independent of pressure and temperature, we
have the following results for the property changes of mixing of ideal solutions.
DG = RT S xi ln xi………(7.121)
DV = 0
DH = 0
DS = – R S xi ln xi………(7.122)
Thus, we see that the volume change of mixing, and enthalpy change of mixing of ideal solution are
zero. This is true for internal energy change and heat capacity change of mixing as well. But, the free
energy change of mixing and entropy change of mixing are not zero. As an ideal gas is a special case
of an ideal solution, the above equations are applicable for ideal gas mixtures also.
7.9 HEAT EFFECTS OF MIXING PROCESSES
Since the energy of interaction between like molecules is different from that between unlike
molecules, the energy of a solution is different from the sum of the energies of its constituents. This
difference between the energy of the solution and the energy of the constituents leads to the absorption
and evolution of heat during the mixing process. The heat of mixing, DH (or the enthalpy change of
mixing), is the enthalpy change when pure species are mixed at constant pressure and temperature to
form one mole (or unit mass) of solution. For binary mixtures,
DH = H – (x1H1 + x2H2)………(7.123)
Knowing the enthalpies of the pure constituents H1 and H2 and the heat of mixing at the given
concentration, the enthalpy of the solution can be computed as
H = (x1H1 + x2H2) + DH………(7.124)
When solids or gases are dissolved in liquids, the accompanying enthalpy change is usually measured
as heats of solution, which is defined as the enthalpy change when one mole of the solute dissolves in
the liquid. Thus
………(7.125)
where DHS is the heat of solution per mole of solute (component 1). When the constituents are all
liquids and solutions of all proportions are possible, the heat effect is usually termed as heat of
mixing. Figures 7.7 and 7.8 illustrate the two types of presentation of heat of mixing.
In Fig. 7.7, the heat of mixing of ethanol–water is shown from which it is clear that mixing process at
lower temperatures and concentration of ethanol is exothermic and at higher temperatures and high
concentrations it is endothermic. In Fig. 7.8, heats of solution of various substances in water are
plotted with moles of water per mole of solute as abscissa.
Using the heat of mixing at one temperature and heat capacity data of pure species and the solution,
the heat of mixing at any temperature can be calculated. The method of calculation is similar to the
one employed for the calculation of standard heat of reaction at any temperature from the values at
298 K.
EXAMPLE 7.17 The enthalpy change of mixing for a binary liquid solution at 298 K and 1 bar is
given by the equation DH = x1x2(40x1 + 20x2), where DH is in J/mol and x1 and x2 are the mole
fractions of components 1 and 2 respectively. The enthalpies of the pure liquids at the same
temperature and pressure are 400 and 600 J/mol respectively. Determine numerical values of the
partial molar enthalpies at infinite dilution at 298 K and 1 bar.
Solution Refer Eqs. (7.20) and (7.21) and replace V by H.
EXAMPLE 7.18 At 300 K and 1 bar, the volumetric data for a liquid mixture of benzene and
cyclohexane are represented by V = 109.4 10–6 – 16.8 10–6x – 2.64 10–6x2, where x is the
mole fraction of benzene and V has the units of m3/mol. Determine the expression for volume change
of mixing for the standard state based on Lewis–Randall rule.
Solution The expressions for the partial molar volumes of benzene (1) and cyclohexane (2) were
derived in Example 7.5. They are
EXAMPLE 7.19 A vessel is divided into two compartments. One contains 100 moles nitrogen at 298
K and 1 bar and the other contains 100 moles of oxygen at the same conditions. The barrier separating
them is removed and the gases are allowed to reach equilibrium under adiabatic conditions. What is
the change in entropy of the contents of the vessel?
Solution As the gases are ideal, the temperature and pressure before and after mixing will be the
same. For one mole of the mixture the change in entropy on its formation from pure components is
given by Eq. (7.122).
DS = – R S xi ln xi
Since xi = 0.5,
DS = – R ln 0.5 = 8.314 ln 2 = 5.763 J/mol K
For 200 moles of the mixture, DS = 1152.57 J/K
EXAMPLE 7.20 The heat of formation of LiCl is – 408.610 kJ/mol at 298 K. The heat of solution for
1 mol LiCl in 12 moles water is –33.614 kJ at 298 K. Calculate the heat of formation of LiCl in 12
moles water at 298 K.
Solution The chemical reaction of formation of LiCl and the physical change of dissolution of LiCl in
water can be represented by the following:
EXAMPLE 7.21 A container is divided into two compartments. One contains 3.0 moles hydrogen at
298 K and 1.0 bar and the other contains 1.0 mol nitrogen at 298 K and 3.0 bar. Calculate the free
energy of mixing when the partition is removed.
Solution Assume that the gases behave ideally. The volume occupied by hydrogen is
(nRT)/P = 3RT
Volume occupied by nitrogen is
(nRT)/P = RT/3
Therefore, the total volume occupied after the partition is removed is
3RT + RT/3 = (10/3)RT
The final pressure attained by the mixture is
It is assumed that the process is taking place in two steps. In the first, the individual gases are
separately brought to the final pressure at constant temperature and in the second, the gases are mixed
at constant pressure and temperature. For the first step, the change in free energy is due to change in
the pressure and is equal to
EXAMPLE 7.22 Calculate the mean heat capacity of a 20 mole per cent solution of alcohol in water
at 298 K, given the following:
Heat capacity of water: 4.18 103 J/kg K; Heat capacity of ethanol: 2.18 103 J/kg K; Heat of
mixing for 20 mole per cent ethanol–water at 298 K: – 758 J/mol; Heat of mixing for 20% (mole)
ethanol–water at 323 K: – 415 J/mol. Assume that the heat capacities of pure liquids are constant
between 298 and 523 K.
Solution The enthalpy change when 0.8 moles of water and 0.2 moles of ethanol both at
323 K are mixed together is given by the heat of mixing at 323 K which is equal to – 415 J/mol of
solution. 0.8 mol water at 323 K + 0.2 mol ethanol at 323 K 1.0 mole 20 per cent ethanol–water;
DH = – 415 J/mol. This change can be assumed to be taking place in four steps as detailed below:
Step 1: 0.8 mol water is cooled from 323 K to 298 K. Let DH (1) be the enthalpy of cooling. Then
DH(1) = 0.8 18 4.18 (298 – 323) = – 1504.8 J
Step 2: 0.2 mol ethanol is cooled from 323 K to 298 K. Let DH(2) be the enthalpy of cooling.
DH(2) = 0.2 46 2.58 (298 – 323) = – 593.4 J
Step 3: 0.8 mol water and 0.2 mol ethanol at 298 K are mixed together. Heat of mixing is
DH(3) = –758 J/mol
Step 4: 20 per cent ethanol–water solution is heated to 323 K. The enthalpy of heating is
DH(4) = CPm(323 – 298)
where CPm is the mean specific heat of solution.
DH = DH(1) + DH(2) + DH(3) + DH(4)
– 415 = – 1504.8 – 593.4 – 758 + CPm (323 – 298)
Thus the mean heat capacity of a 20 per cent solution is CPm = 97.65 J/mol K
EXAMPLE 7.23 What temperature will be attained when a 20% mole ethanol–water mixture is
adiabatically formed from the pure liquids at 298 K? Heat of mixing for 20% mole ethanol–water at
298 K: –758 J/mol. The mean heat capacity of a 20% mole solution of alcohol in water at
298 K: 97.65 J/mol K.
Solution The adiabatic mixing process involves no change in enthalpy or DH = 0. It means the
temperature of the solution will increase or decrease on mixing depending upon whether heat is
absorbed or evolved during the process. Let T be the temperature attained by the solution on mixing.
Then its enthalpy above 298 K is CP m (T – 298), where CPm is its mean heat capacity. The
temperature T can be evaluated by assuming that the process is occuring in steps as shown in
Fig. 7.10.
DH = DHS + CPm(T – 298)
0 = – 758 + 97.65(T – 298)
Therefore, T = 305.8 K.
Equation (7.136) says that the molar excess property ME of a solution is the average of the partial
molar excess property of each component weighted according to its mole fractions.
7.10.1 Excess Gibbs Free Energy
For phase equilibrium studies the most useful excess property is the partial molar excess Gibbs free
energy which can be directly related to the activity coefficient. Excess Gibbs free energy is defined
as
GE = G – Gid………(7.137)
Using Eq. (7.136), we can write the excess Gibbs free energy as
In Eq. (7.140) the fugacity is related to xi,, i and as , so that Eq. (7.140) becomes
Dmi = RT ln xigi………(7.143)
Substituting Eq. (7.142) and Eq. (7.143) into Eq. (7.139), the result is
EXAMPLE 7.24 The two-suffix-Margules equation is the simplest expression for excess Gibbs free
energy that is obeyed by chemically similar materials.
GE = Ax1x2………(7.147)
where A is an empirical constant independent of composition. Derive the expressions for the activity
coefficients that result from this expression.
Solution Write Eq. (7.146) for components 1 and 2. Then
SUMMARY
A proper understanding of the thermodynamic properties of solutions is essential for the analysis of
many chemical engineering problems such as the phase equilibria and chemical reaction equilibria.
New concepts were found necessary to deal with the solutions, the concept of partial molar
properties being the most important among them. The general definition of partial molar property was
given by Eq. (7.1). The partial molar property of a substance in a solution is an intensive property
strongly dependent on the concentration. It gives the increase, in the property of the solution resulting
from the addition at constant temperature and pressure, of one mole of the substance, to such a large
quantity of the system that its composition remains virtually unchanged. The method of tangent
intercepts was found to be most suitable for the determination of partial molar properties. The partial
molar free energy of a substance was designated as its chemical potential. It is the contribution that
the component makes towards the total free energy of the solution. It is a widely used thermodynamic
property and serves as an index of chemical equilibrium in the same manner as temperature and
pressure are used as indices of thermal and mechanical equilibrium.
The concept of fugacity introduced in Chapter 6 was extended to take care of mixtures through Eqs.
(7.51) and (7.52). For the evaluation of fugacity of mixtures, data on the variation of the solution
volume with composition at different pressures is necessary [Eq. (6.57)]. Since such data are scarce,
the fugacity in the solution is to be evaluated by devising an ideal solution model and by measuring
the extent to which the real solutions deviate from it. The Lewis–Randall rule allows the fugacity of a
component in the mixture to be calculated without any information about the solution except its
composition. It states that the fugacity of a component in an ideal solution is directly proportional to
the mole fraction of the component in the solution [Eq. (7.65)]. For ideal liquid solutions, if the
vapour phase with which it is in equilibrium is assumed to behave as an ideal gas, the Lewis–Randall
rule may be simplified to the Raoult’s law [Eq. (7.67)]. The Raoult’s law states that the partial
pressure of a component in the vapour phase is directly proportional to the mole fraction of that
component in the liquid, which is in equilibrium with the vapour. Even for non-ideal solutions, the
fugacity (or the partial pressure) was found to be directly proportional to the mole fraction in the
liquid as the mole fraction approaches zero (Henry’s law). Often, the solute portion of a dilute non-
ideal liquid solution can be assumed to follow the Henry’s law and the solvent portion, the Raoult’s
law. The activity of a component in a solution (Section 7.6) can be related to compositions directly
and hence plays an important role in solution thermodynamics. Activity coefficients (Section 7.6)
measure the extent to which real solutions deviate from ideal behaviour. Two types of ideal
behaviour are observed; one conforming to the Lewis–Randall rule and the other conforming to the
Henry’s law, thus giving rise to two types of activity coefficients.
In Section 7.7, the Gibbs–Duhem equations were developed relating the partial molar properties of
the components to one another. They tell us that the partial molar properties of a mixture cannot
change independently; in a binary mixture, if the partial molar property of one of the component
increases, that of the other should decrease. Various forms of the Gibbs–Duhem equations applicable
for binary solutions were also developed.
The property changes of mixing were defined (Section 7.8) as the difference in the property of the
solution and the sum of the properties of the pure components constituting the solution, all at the same
temperature and pressure as the solution. The property changes were presented in terms of the activity
of the constituents as shown by Eq. (7.115) for the free energy, Eq. (7.117) for the volume, Eq.
(7.119) for the enthalpy and Eq. (7.120) for the entropy change on mixing. The difference between the
energy of the solution and the energy of the constituents leads to absorption and evolution of heat
during the mixing process. The heat effects of mixing process were dealt with in detail in Section 7.9.
The excess property (Section 7.10) was defined as the difference between an actual property and the
property that could be calculated for the same temperature, pressure and composition by the equations
for an ideal solution. For phase equilibrium studies the most useful excess property is the partial
molar excess Gibbs free energy which can be directly related to the activity coefficient. A simple
relationship was shown to exist between activity coefficient and excess chemical potential which
makes it possible to express the activity coefficient as a function of the composition.
REVIEW QUESTIONS
1. Distinguish between molar volume and partial molar volume. Does the partial molar volume of a
substance vary with the concentration of the substance in the solution?
2. Express the partial molar property as the partial derivative of the total property of the solution.
Is it an intensive property or an extensive property?
3. How are the partial molar volumes of the constituents of a binary mixture related to their mole
fractions of the constituents and the molar volume of the solution? Explain how these equations
are useful for the determination of partial molar volumes by the tangent-intercept method.
4. Define chemical potential. What is its physical significance?
5. Chemical potential can be equated to the partial derivatives of U, A, H or S under certain
constraints. However, it cannot be treated as the partial molar internal energy, partial molar
enthalpy, etc. Explain.
6. Show that the rate of change of chemical potential of a substance with pressure is equal to its
partial molar volume in the solution.
7. What are the characteristics of an ideal solution? What is Lewis–Randall rule?
8. “The concept of an ideal gaseous solution is less restrictive than the concept of an ideal gas
mixture.” Explain.
9. State Raoult’s law. Show that it is a simplified form of the Lewis–Randall rule.
10. State Henry’s law and show that the Raoult’s law is a special case of the Henry’s law.
11. Given the Henry’s law constant, how would you determine the solubility of a gas in a liquid?
12. Define activity and show that the activity and mole fraction in an ideal solution are identical.
13. The activities in a gas mixture may be numerically equal to the fugacities or the mole fractions
in the mixture. Explain.
14. Define activity coefficient. How do you distinguish between the activity coefficient based on
the Lewis–Randall rule and that based on the ideal dilute solution?
15. Do the activity coefficients vary with composition or not? What is the effect of temperature and
pressure on the activity coefficient?
16. Discuss the Gibbs–Duhem equation and its various forms. What are the major fields of
application of the Gibb’s Duhem equations?
17. What do you mean by property changes of mixing? How are these related to the activities of the
components in the mixture?
18. “All property changes of mixing are zero for ideal solutions”. Do you agree? Explain.
19. Define excess property. Under what circumstance the property change of mixing and the excess
properties are identical?
20. How is the activity coefficient related to the excess free energy?
EXERCISES
7.1 Prove the following:
(a)
(b)
(c)
7.2 Discuss the method for the calculation of entropy of solutions.
7.3 Discuss the variable pressure and variable temperature modifications of Gibbs–Duhem
equations.
7.4 Derive an expression for partial molar volumes using the following relation for the
molar volume of the binary liquid mixture of components 1 and 2.
V = x1V1 + x2V2 + x1x2[B + C(x2 – x1)]
where x1 and x2 are the mole fractions and V1 and V2 are the molar volumes in the pure state.
7.5 Describe schematically an experimental technique for the determination of volume change and
enthalpy change on mixing.
7.6 The activity coefficients in a binary mixture based on the Lewis–Randall rule standard state are
given by
Derive expressions for activity coefficients based on Henry’s law in terms of composition.
7.7 Show that the Henry’s law constant varies with pressure as
where V is the volume in m3/mol at 1.0 bar and 300 K. Find the expressions for .
7.10 If the partial molar volumes of species 1 in a binary liquid solution at constant temperature
and pressure is given by
derive the equat3ion for . What equation for V is consistent with this?
7.11 The molar enthalpy of a binary mixture is given by
H = x1(a1 + b1x1) + x2(a2 + b2x2)
Derive an expression for .
7.12 Using the method of tangent intercepts plot the partial molar volume of HNO3 in aqueous
solution at 293 K using the following data where w is the mass percentage of HNO3.
w 2.162 10.98 20.80 30.00 39.20 51.68 62.64 71.57 82.33 93.4 99.60
r 10–3, kg/m3 1.01 1.06 1.12 1.18 1.24 1.32 1.38 1.42 1.46 1.49 1.51
7.13 On addition of chloroform to acetone at 298 K, the volume of the mixture varies with
composition as follows:
x 0 0.194 0.385 0.559 0.788 0.889 1.000
where x is the mole fraction of chloroform. Determine the partial molar volumes of the
components and plot against x.
7.14 The partial molar volumes of acetone and chloroform in a mixture in which mole fraction of
acetone is 0.5307 are 74.166 10–6 m3/mol and 80.235 10–6 m3/mol respectively. What
is the volume of 1 kg of the solution?
7.15 The volume of a solution formed from MgSO4 and 1.0 kg of water fits the expression
7.17 The standard enthalpy of formation of HCl (in kJ/mol) from the elements at 298 K are given
below:
nw 1.0 2.0 3.0 4.0 5.0 6.0 8.0 10.0 50.0 100.0
92.66 119.0 141.67 149.73 156.96 158.81 161.16 162.42 166.22 166.79 205.9
Calculate the partial molar enthalpies of HCl and water in a solution containing 10 kmol
HCl/m3 of solution.
7.18 The following table gives the molality and density of aqueous solutions of KCl at
298 K. Determine the partial molar volume of KCl at m = 0.3.
m, mol/kg 0.0 0.1668 0.2740 0.3885 0.6840 0.9472
7.19 Calculate the concentration of nitrogen in water exposed to air at 298 K and 1 bar if Henry’s
law constant for nitrogen in water is 8.68 104 bar at this temperature.
Express the result in moles of nitrogen per kg water (Hint: Air is 79 per cent nitrogen by
volume).
7.20 The partial pressure of methyl chloride in a mixture varies with its mole fraction at
298 K as detailed below:
x 0.0005 0.0009 0.0019 0.0024
, bar 0.27 0.48 0.99 1.24
where R, A and B are constants. Derive expression for the activity coefficient of
component 2.
7.25 For a mixture of acetic acid and toluene containing 0.486 mole fraction toluene, the partial
pressures of acetic acid and toluene are found to be 0.118 bar and 0.174 bar respectively at 343
K. The vapour pressures of pure components at this temperature are 0.269 bar and 0.181 bar for
toluene and acetic acid respectively. The Henry’s law constant for acetic acid is 0.55 bar.
Calculate the activity and activity coefficient for acetic acid in the mixture
(a) Based on Lewis–Randall rule
(b) Based on Henry’s law.
7.26 Calculate the activity and activity coefficients for toluene for the conditions in Exercise 7.24
assuming pure liquid standard state.
7.27 Partial pressure of ammonia in aqueous solutions at 273 K varies with concentration as:
x 0.05 0.10 0.15 0.50 1.00
, bar 0.0179 0.0358 0.062 1.334 4.293
Calculate
(a) The activity coefficient of ammonia in 10 mole per cent solution using pure liquid standard
state
(b) The Henry’s law constant if the system obeys Henry’s law.
7.28 The activity coefficient of n-propyl alcohol in a mixture of water (A) and alcohol (B) at 298 K
referred to the pure liquid standard state is given below:
xB 0 0.01 0.02 0.05 0.10 0.20
gB 12.5 12.3 11.6 9.92 6.05 3.12
Obtain expressions for and show that they satisfy Gibbs–Duhem equations.
7.33 Water at a rate of 54 103 kg/h and Cu(NO3)2 ◊ 6H2O at a rate of 64.8 103 kg/h are
mixed together in a tank. The solution is then passed through a heat exchanger to bring the
temperature to 298 K, same as the temperature of the components before mixing. Determine the
rate of heat transfer in the exchanger. The following data are available. Heat of formation at 298
K of Cu(NO3)2 is –302.9 kJ and that of Cu(NO3)2 ◊ 6H2O is –2110.8 kJ. The heat of solution
of Cu(NO3)2 ◊ nH2O at 298 K is – 47.84 kJ per mol salt and is independent of n.
7.34 If pure liquid H2SO4 is added to pure water both at 300 K to form a 20 per cent (weight)
solution, what is the final temperature of the solution? The heat of solution of sulphuric acid in
water is H2SO4 (21.8 H2O) = –70 103 kJ/kmol of sulphuric acid. Standard heat of formation
of water = – 286 kJ/mol. Mean heat capacity of sulphuric acid may be taken from Chemical
Engineer’s Handbook.
7.35 LiCl H2O (c) is dissolved isothermally in enough water to form a solution containing
5 mol of water per mole of LiCl. What is the heat effect? The following enthalpies of formation
are given:
LiCl (c) = – 409.05 kJ, LiCl◊H2O (c) = – 713.054 kJ
LiCl (5H2O) = – 437.232 kJ, H2O (l) = – 286.03 kJ
7.36 Calculate the heat effects when 1.0 kmol of water is added to a solution containing
1.0 kmol sulphuric acid and 3.0 kmol of water. The process is isothermal and occurs at 298 K.
Data: Heat of mixing for H2SO4 (3H2O) = – 49,000 kJ per kmol H2SO4. Heat of mixing for
H2SO4 (4H2O) = – 54,100 kJ per kmol H2SO4.
7.37 A single effect evaporator is used to concentrate a 15% (weight) solution of LiCl in water to
40%. The feed enters the evaporator at 298 K at the rate of 2 kg/s. The normal boiling point of a
40% LiCl solution is 405 K and its specific heat is 2.72 kJ/kg K. For what heat transfer rate in
kJ/h, should the evaporator be designed? The heat of solution of LiCl in water per mole of LiCl
at 298 K are:
DHS for LiCl(13.35 H2O) = – 33.8 kJ, for LiCl (3.53 H2O) = – 23.26 kJ. Enthalpy of
superheated steam at 405 K, 1 bar = 2740.3 kJ/kg. Enthalpy of water at 298 K = 104.8 kJ/kg.
Molecular weight of LiCl = 42.39.
7.38 The excess Gibbs free energy of solutions of methyl cyclohexane (MCH) and tetrahydrofuran
(THF) at 303 K are correlated as
GE = RTx(1 – x)[0.4587 – 0.1077(2x – 1) + 0.0191(2x – 1)2]
where x is the mole fraction of methyl cyclohexane. Calculate the Gibbs free energy change on
mixing when 1 mol MCH and 3 mol THF are mixed.
7.39 Derive the relation between the excess Gibbs free energy of a solution based on the Lewis–
Randall rule and that based on the asymmetric treatment (Lewis–Randall rule for solvent and
Henry’s law for solute) of solution ideality.
7.40 The excess enthalpy of a solution is given by
HE = x1x2(40x1 + 20x2) J/mol
at 298 K and 1 bar. Determine for an equimolar mixture of components 1 and 2 given
that V1 = 0.12 m3/kmol and V2 = 0.15 m3/kmol.
8
Phase Equilibria
A system is said to be in a state of equilibrium if it shows no tendency to depart from that state either
by energy transfer through the mechanism of heat and work or by mass transfer across the phase
boundary. Since a change of state is caused by a driving force, we can describe a system at
equilibrium as one in which there are no driving forces for energy or mass transfer. That is, for a
system in a state of equilibrium, all forces are in exact balance. It may be noted here that the state of
equilibrium is different from a steady state condition. Under steady state there exist net fluxes for
material or energy transfer across a plane surface placed anywhere in the system. Under equilibrium
the net flux is zero.
Transfer of material or energy across phase boundaries occurs till equilibrium is established between
the phases. In our daily experience, we come across a number of processes in which materials are
transferred from one phase to another. During breathing we take oxygen from the air through the lungs
and dissolve it in the blood. During the preparation of tea or coffee we extract the soluble
components in the powder into boiling water. Dilute aqueous solution of alcohol is concentrated by
distillation in which a vapour rich in alcohol is produced from the boiling solution. The phase
equilibrium thermodynamics is of fundamental importance in many branches of science, whether
physical or biological. It is particularly important in chemical engineering, because majority of
manufacturing processes involve transfer of mass between phases either during the preparation of the
raw materials or during the purification of the finished products. Gas–liquid absorption, distillation,
liquid–liquid extraction, leaching, adsorption, etc., are some of the important separation techniques
employing mass transfer between phases. In addition to these, many industrial chemical reactions are
carried out under conditions where more than one phase exist. A good foundation in phase
equilibrium thermodynamics is essential for the analysis and design of these processes.
In this chapter due emphasis is given to the development of the relationship between the various
properties of the system such as pressure, temperature and composition when a state of equilibrium is
attained between the various phases constituting the system. The temperature–pressure-composition
relationships in multiphase system at equilibrium form the basis for the quantitative treatment of all
separation processes. The two types of phase equilibrium problems that are frequently encountered
are:
1. The determination of composition of phases which exist in equilibrium at a known temperature
and pressure
2. The determination of conditions of temperature and pressure required to obtain equilibrium
between phases of specified compositions.
The present chapter tries to provide solutions to these problems.
In this equation, the equality sign refers to a reversible process which can be treated as a succession
of equilibrium states and the inequality refers to the entropy change for a spontaneous process whose
ultimate result would be an equilibrium state. The first law of thermodynamics expressed
mathematically by Eq. (2.5) can be rewritten as
dQ = dU + dW………(8.1)
Substituting Eq. (8.1) into Eq. (4.44), we get
T dS ≥ dU + dW
dU T dS – dW………(8.2)
dW in Eq. (8.2) may be replaced by P dV so that
dU T dS – P dV………(8.3)
Equation (8.3) is valid for cases where external pressure is the only force and the work is, therefore,
the work of expansion only. By this, we exclude other effects like those due to gravitational and
electromagnetic fields and surface and tensile forces. Equation (8.3) can be treated as the combined
statement of the first and second law of thermodynamics applied to a closed system which interact
with its surroundings through heat transfer and work of volume displacement. This equation is utilised
for deriving the criteria of equilibrium under various sets of constraints, each set corresponding to a
physically realistic or commonly encountered situation. These different criteria are discussed now.
Constant U and V. An isolated system does not exchange mass, heat or work with the
surroundings. In Eq. (8.1), d Q = 0, d W = 0 and hence d U = 0. A well-insulated vessel of constant
volume would closely approximate this behaviour. Thus in Eq. (8.3) dU = 0 and dV = 0 so that
The entropy is constant in a reversible process and increases in a spontaneous process occurring in a
system of constant U and V. Since an irreversible process leads the system to an equilibrium state, the
entropy is maximum at equilibrium when no further spontaneous processes are possible.
Constant T and V. Helmholtz free energy is defined by Eq. (6.1).
A = U – TS
Rearranging Eq. (6.1), we get
U = A + TS
dU = dA + T dS + S dT
Substitute this result in Eq. (8.3) and rearrange the resulting expression to the following form
dA – P dV – S dT………(8.5)
Under the restriction of constant temperature and volume, the latter implying no work, the equation
simplifies to
Equation (8.6) means that the spontaneous process occurring at constant temperature and volume is
accompanied by a decrease in the work function and consequently, in a state of thermodynamic
equilibrium under these conditions the Helmholtz free energy or the work function is a minimum.
Constant P and T. Equation (6.6) defines Gibbs free energy as
G = H – TS
Since H = U + PV we can write Eq. (6.6) as
G = U + PV – TS
Taking the differentials
dG = dU + P dV + V dP – T dS – S dT
rearranging these as
dU = dG – P dV – V dP + T dS + S dT
and combining this result with Eq. (8.3), we obtain
dG V dP – S dT………(8.7)
At constant temperature and pressure, Eq. (8.7) reduces to
Equation (8.8) means that the free energy either decreases or remains unaltered depending upon
whether the process is spontaneous or reversible. It implies that for a system in equilibrium at a given
temperature and pressure the free energy must be minimum.
Since most chemical reactions and many physical changes are carried out under conditions of constant
temperature and pressure, Eq. (8.8) is the commonly used criterion of thermodynamic equilibrium. It
also provides a very convenient and simple test for the feasibility of a proposed process. No process
is possible which results in an increase in the Gibbs free energy of the system, because according to
Eq. (8.8) the Gibbs free energy always decreases in a spontaneous process and in the limit of the
reversible process, the free energy doesn’t change at all.
In the equilibrium state, differential variations can occur in the system at constant temperature and
pressure without producing any change in the Gibbs function. Thus, the equality in Eq. (8.8) can be
used as the general criterion of equilibrium or as a thermodynamic statement that characterises the
equilibrium state.
dG = 0 (at constant T and P)………(8.9)
To apply this criterion for phase equilibrium problems we need formulate an expression for dG as
function of the number of moles of the components in various phases and set it equal to zero. This
equation along with the mass conservation equations provides the solutions to phase equilibrium
problems.
DGid = RT S xi ln xi
Combining these three equations we find that
DG = RT S xi ln (gixi)
The last two equations imply that fugacity of components in a stable solution always increase with
increase in concentration.
The left-hand side of the above equation is the entropy change accompanying the phase change of one
mole of the substance (DS), and the numerator on the right-hand side represents the enthalpy change
for the phase change of one mole of the substance or the latent heat of phase change (DH). That is,
EXAMPLE 8.3 Deduce the Clapeyron equation using the criterion of equilibrium, Eq. (8.9).
Solution In Chapter 6 we have derived the Clapeyron equation, Eq. (6.25), using Maxwell’s
relations.
The criterion of equilibrium provides an alternate route for its derivation. Consider any two phases a
and b of the same substance under equilibrium. Since Ga and Gb are both functions of temperature
and pressure, and these functional relationships are different for different phases, the two phases can
coexist only at such values of the temperature and pressure that Ga = Gb. If the temperature and
pressure are altered infinitesimally without disturbing the equilibrium, the change in the free energy
must be the same in each phase.
dGa = dGb………(8.25)
In a phase change there is no work other than the work of expansion, so that
dG = V dP – S dT
Using this in Eq. (8.25),
Va dP – Sa dT = Vb dP – Sb dT………(8.26)
V and S are the molar volume and molar entropy of the fluid with the superscript representing the
phase for which the properties correspond to. Equation (8.26) can be rearranged to the following
form.
This relation gives the increase in pressure that is necessary to maintain the equilibrium between
phases for a pure substance when the temperature is increased. By using the following simplifications
Eq. (8.29) can be modified to yield the Clausius–Clapeyron equation applicable for vapour–liquid
equilibria.
1. The latent heat of vaporisation is constant and independent of temperature.
2. The molar volume of liquid is negligible compared to that of vapour.
3. The vapour behaves as ideal gas.
The Clausius–Clapeyron equation was derived [Eq. (6.28)] in Chapter 6 and is reproduced below.
The variations in the number of moles dni are independent of each other. However, they are subject to
the constraints imposed by Eq. (8.33). For all possible variations , Eq. (8.32) is to be satisfied.
This is possible only if
Equation (8.34) means that when a system consisting of several components distributed between
various phases is in thermodynamic equilibrium at a definite temperature and pressure, the
chemical potential of each component is the same in all the phases. If they are different, the
component for which such a difference exists will show a tendency to pass from the region of higher
to the region of lower chemical potential. Thus the equality of chemical potential along with the
requirement of uniformity of temperature and pressure serves as the general criterion of
thermodynamic equilibrium in a closed heterogeneous multicomponent system. In short, we can write
where C is a constant, an alternative and equally general criterion of equilibrium can be written in
terms of fugacities as
Fugacity is a more useful property than chemical potential for defining equilibrium since it can be
expressed in absolute values, whereas chemical potential can be expressed only relative to some
arbitrary reference state. Equation (8.36) is therefore widely used for the solution of phase
equilibrium problems.
EXAMPLE 8.4 Using the criterion of phase equilibrium show that the osmotic pressure over an ideal
solution can be evaluated as
where xA is the mole fraction of solute and VB is the molar volume of the solvent.
Solution Consider a vessel which is divided into two compartments by a semi-permeable membrane.
Pure solvent (say, water) is taken in one of the compartments and a solution (say, sucrose in water) is
taken in the other. Let T be the temperatures on both sides of the membrane and P be the pressure.
While the membrane is impermeable to the flow of the solute, it permits the flow of solvent into the
solution. This phenomenon of a solvent diffusing through a membrane which is permeable to it, but is
impermeable to the solute, is known as osmosis.
Osmosis is caused by the difference in the chemical potentials of the solvent on the two sides of the
membrane. At a given pressure, the chemical potential of a pure solvent is greater than that of the
solvent in the solution. By increasing the pressure at the solution side of the membrane, the chemical
potential of the solvent in the solution can be increased. When the pressure is increased to P
keeping the temperature constant, the chemical potential of the solvent in the solution would become
equal to that of the pure solvent at pressure P, and the diffusion would stop. If the pressure is
increased above P , the direction of diffusion would be reversed. In that event, the solvent would
diffuse from the solution to the pure solvent. This process is known as reverse osmosis. The excess
pressure P – P to be applied over the solution at constant temperature to arrest the process of
osmosis is known as the osmotic pressure. Thus, osmotic pressure is
Posmotic = P – P
Let the mole fraction of the solutes constituting the solution be represented by xA and the mole
fraction of the solvent be represented by xB. Let denotes the chemical potential of the solvent at
pressure P . Equation (7.51) relates the chemical potential of a component in a solution to its
fugacity. Thus
In this equation, is the chemical potential of pure solvent at pressure P , fB is its fugacity and
is its fugacity in the solution. Since the solution is ideal, the above equation may be simplified
utilising the Lewis–Randall rule which relates fB and that . Now we get the following
result for the chemical potential of the solvent in the solution at pressure P .
Since volume of a liquid is not affected by change in pressure, the integral in this equation can be
easily determined in terms of the molar volume VB. Thus
VB (P – P) = – RT ln xB
Noting the definition of the osmotic pressure, the preceding equation may be written as
where, represents the fugacity of component i in the solution and the superscripts V and L represent
the vapour and liquid phases, respectively. Using this equation, the problem of determining the
composition of the liquid and vapour phases in equilibrium is quite simple: it is necessary only to
evaluate the compositions so that the fugacity of each component be the same in both phases. For
example, for a binary mixture of ethanol and water in vapour-liquid equilibrium, at a definite
temperature and pressure, the mole fractions in the liquid and vapour must be such that the fugacity of
ethanol is the same in both phases. That is, . Here, is the fugacity of ethanol in the mixture.
To evaluate quantitatively the equilibrium compositions, the fugacity of a component should be
expressed in terms of its mole fraction in the mixture. Using the definition of activity coefficient, the
fugacity of a substance in the vapour phase can be written in terms of its mole fraction yi in the
mixture, the fugacity of pure i as a vapour at the system temperature and pressure.
If the stable state for i at T and P is not a vapour, evaluating requires the introduction of a
hypothetical state. The use of the concept of fugacity coefficient helps to overcome this difficulty. The
fugacity of a component in a gas mixture can be written as
where is the fugacity coefficient of i in the mixture. The fugacity coefficient may be evaluated
from an equation of state for the mixture.
For the liquid phase, the fugacity of a component can be expressed as the product of its mole fraction
xi in the solution, the activity coefficient gLi and the fugacity of the component in the standard state.
Equations (8.45) and (8.46) are the fundamental relationships for estimating the vapour–liquid
equilibrium by two different approaches—the equations of state approach and the activity coefficient
model approach. Equation (8.45) forms the basis of estimating VLE by the equation of state
approach, which will be discussed under Section 8.12.1.
Equation (8.46) is the fundamental relationship in the study of vapour–liquid equilibrium based on
activity coefficient model for the liquid phase fugacity. The liquid-phase activity coefficients in Eq.
(8.46) can be estimated by any of the models described in the following sections. Estimation of the
vapour–liquid equilibria using activity coefficients is useful when polar components are present in
the system. For most substances at low pressures, can be assumed to be unity. If the pressure is
very high, must be evaluated using an equation of state. Using the activity coefficient approach, the
vapour–liquid equilibrium problems may be attacked, by dividing them into the following grouping
for convenience.
Case 1: Ideal gas-phase, ideal liquid solution. For mixtures of ideal gases, = 1. For ideal
liquid solutions, at low pressures, gLi = 1, and the fugacity is equal to the saturation pressure of
pure liquid at the temperature of interest. Equation (8.46) becomes
Case 2: Low-pressure VLE problems. If the liquid phase is not an ideal solution so that g π 1,
but the pressure is low enough that the assumption of ideal gas behaviour for the gas phase would not
introduce any significant errors in practical calculations, Eq. (8.46) can be simplified as
Case 3: High-pressure VLE problems. In the general case where ideal behaviour cannot be
assumed for the gas and liquid phases, the fugacity coefficient and the activity coefficient gLi
should first be determined for solving vapour–liquid equilibrium problems using Eq. (8.46). These
are normally complex functions of temperature, pressure and compositions and can be written as
The fugacity in the reference state is the fugacity of pure i at the same T, P and state of aggre-
gation as the mixture. To calculate this, it is convenient to determine first the fugacity of pure i in the
liquid state at T under its equilibrium vapour pressure and then apply a correction term for the fact
that . The fugacity of the liquid under its equilibrium vapour pressure is equal to the vapour
pressure times the fugacity coefficient . The fugacity coefficient is the ratio of the fugacity of
the component i under its saturation conditions to the saturation pressure of the substance and tends to
unity if the vapour behaves as an ideal gas. Using Eq. (6.31), we can write
The exponential in the above equation is known as the Poynting correction and it is approximately
unity when pressure is low. Also at low pressures when the gas behaves ideally, and
the above equation reduces to Eq. (8.48).
To make these points clearer, consider a mixture whose temperature and composition ( x1) are such
that it is represented by point A in Fig. 8.2. Since the point A lies below the bubble-point curve, the
solution is entirely liquid. The mixture is taken in a closed container and the pressure over the system
is maintained at a constant value by a piston. The mixture is heated slowly so that its temperature
increases along the vertical line passing through point A till point B on the bubble-point curve is
reached. The temperature T1 corresponding to point B, is the bubble point of the original mixture. The
first bubble of the vapour is produced at this temperature and it will have the composition (y1)
represented by point C on the upper curve. The vapour is richer in the more volatile component.
Therefore y1 > x1, and the dew-point curve lies above the bubble-point curve. The mixtures at points
B and C are the liquid and vapour at equilibrium at the system pressure and temperature T1. Since
both are at the same temperature, they can be joined by a horizontal line BC, known as a ‘tie line’.
Further heating will result in the vaporisation of more liquid, and at temperature T2 the system will
consist of saturated liquid represented by point D and saturated vapour represented by point E, which
are in equilibrium. Since the vapour formed is not removed from the system, the overall composition
of the combined mixture of liquid and vapour will be same as x1, the composition of the original
mixture. However, the relative amounts of the liquid and the vapour change as the temperature is
changed.
These relative amounts are given by the ratio in which the point representing the combined mixture (in
this case, point F) divides the tie line DE. By material balance consideration, it can be easily verified
that
If heating is continued, eventually a temperature T3 is reached when almost all liquid is vaporised.
The last drop of liquid getting vaporised at this temperature has a composition denoted by point G
and the equilibrium vapour has the composition at H same as the original mixture. Temperature T3 is
the dew point of the original mixture. The mixture temperature increases along the vertical line HJ on
further heating. On cooling the superheated mixture at point J, the first drop of condensate appears
when the temperature drops to T3, the dew point of the mixture and the composition of the liquid is
given by point G.
We have seen that the mixture at point A has vaporised over a temperature range from T1 (the bubble
point) to T3 (the dew point), unlike a pure substance, which vaporises at a single temperature known
as the boiling point of the substance. For a solution, the term ‘boiling point’ has no meaning, because,
at a given pressure the temperature during vaporisation of a solution varies from the bubble point to
the dew point.
Equilibrium diagram. The vapour–liquid equilibrium data at constant pressure can be represented
on a x versus y plot or an equilibrium distribution diagram. If the vapour composition is taken as the
ordinate and the liquid composition is taken as the abscissa, a tie line such as line BC on the boiling
point diagram gives rise to a point such as point P on the distribution diagram
(Fig. 8.3). Since the vapour is richer in the more volatile component, the curve lies above the
diagonal on which x = y.
A liquid–vapour equilibrium curve very close to the diagonal means that the composition of the
vapour is not much different from the composition of the liquid with which it is in equilibrium; when
the curve coincides with the diagonal, x and y are equal.
Effect of pressure on VLE. On the boiling point diagram the temperatures corresponding to
x = 0 and x = 1 are the boiling points of pure substances B and A respectively. The boiling points of
pure substances increase with pressure. This is true for the bubble and dew points of a mixture of
given composition. Consequently the boiling point diagrams at higher pressures will be above the
boiling point diagrams at lower pressures as shown in Fig. 8.4. Since the relative volatility decreases
as pressure is increased, the closed loop formed by the dew-point and bubble-point curves become
narrow at high pressures. Figure 8.5 indicates the effect of pressure on the distribution diagram. In
Fig. 8.4, P3 is the critical pressure for component A and above this pressure, the looped curves are
shorter.
8.8.2 Constant-temperature Equilibria
Vapour–liquid equilibrium data at constant temperature are represented by means of P-x-y diagrams;
Fig. 8.6 shows a typical P-x-y diagram.
The pressure at x = 0 is the vapour pressure of pure B( ) and the pressure at x = 1 is the vapour
pressure of pure A( ). Since component A is assumed to be more volatile, and therefore,
the P-x-y diagram slopes upwards as shown in the figure. The P-y curve lies below the P-x curve so
that for any given pressure, y > x. A solution lying above the P-x curve is in the liquid region and that
lying below the P-y curve is in the vapour region. In between P-x and P-y curves the solution is a
mixture of saturated liquid and vapour. A horizontal line such as AB connects the liquid and vapour
phases in equilibrium and is therefore, a tie line. Assume that a liquid mixture whose conditions may
be represented by the point C in Fig. 8.6, is taken in a closed container. When the pressure over this
system is reduced at constant temperature, the first bubble of vapour forms at point D, and
vaporisation goes to completion at point E. Further reduction in pressure leads to the production of
superheated vapour represented by point F. The effect of temperature on P-x-y diagram is shown in
Fig. 8.6(b). When the temperature is less than the critical temperature of both components, the looped
curve such as the one shown at the bottom of Fig. 8.6(b) results. The other two curves refer to
temperatures greater than the critical temperature of A.
8.9 VAPOUR–LIQUID EQUILIBRIA IN IDEAL SOLUTIONS
It is possible to determine the vapour–liquid equilibrium (VLE) data of certain systems from the
vapour pressures of pure components constituting the system. If the liquid phase is an ideal solution
and the vapour behaves as an ideal gas, the VLE data can be estimated easily without resorting to
direct experimentation. A solution conforming to the ideal behaviour has the following
characteristics, all interrelated.
1. The components are chemically similar. The average intermolecular forces of attraction and
repulsion in the pure state and in the solution are of approximately the same order of magnitude.
2. There is no volume change on mixing (DV = 0) or the volume of the solution varies linearly with
composition.
3. There is neither absorption nor evolution of heat on mixing the constituents that form an ideal
solution (DH = 0); that is, there is no temperature change on mixing.
4. The components in an ideal solution obey Raoult’s law, which states that the partial pressure in
the vapour in equilibrium with a liquid is directly proportional to the concentration in the liquid.
That is, , where is the partial pressure of component i and xi is its mole fraction in
the liquid. is the vapour pressure of pure i. This criterion also
implies that the total vapour pressure over an ideal solution is a linear function of its
composition.
For an experimental test of an ideal solution, the last criterion is the safest one to use. For example,
the solution formed by two chemically dissimilar materials like benzene and ethyl alcohol should
definitely be non-ideal. It is found that for an equimolar mixture of benzene and ethyl alcohol, there is
no change in volume during mixing at room temperature. This peculiar behaviour is because of the
fact that when this solution is formed from its constituents, there is increase in volume up to certain
concentration and thereafter the volume decreases as shown in Fig. 8.7. When the solution volume is
plotted against the composition, the curve will intersect the broken line representing the volume of an
ideal solution at a particular concentration represented in the figure by point P.
If the volume were measured for the concentration of the solution corresponding to point P, no change
in volume would be observed. This may be the case for enthalpy change of mixing also at some
particular composition. The conclusion to be drawn is that negligible volume change or temperature
change for one particular composition of the mixture is not a safe criterion of an ideal solution. If
these are to be used as the tests for ideal behaviour, then these tests should be done for more than one
concentration of the solution. In contrast, the criterion that the total vapour pressure over an ideal
solution varies linearly with composition is safe and reliable.
It should be understood that there exists no ideal solution in the strict sense of the word; but actual
mixtures approach ideality as a limit. Ideality requires that the molecules of the constituents are
similar in size, structure and chemical nature; only optical isomers of organic compounds meet these
requirements. Thus a mixture of ortho- , meta- and para-xylene conforms very closely to the ideal
solution behaviour. Practically, adjacent or nearly adjacent members of the homologous series of
organic compounds can be expected to form ideal solutions. Thus mixtures of benzene and toluene,
n-octane and n-hexane, ethyl alcohol and propyl alcohol, acetone and acetonitrile, paraffin
hydrocarbons in paraffin oils, etc., can be treated as ideal solutions in engineering calculations.
Consider an ideal binary solution made up of component 1 and component 2. We have shown in
Chapter 7 that all ideal solutions obey Lewis–Randall rule.
Here is the fugacity of the component i in the liquid and fi is the fugacity of pure i. Using the
criterion of equilibrium and noting that if pressure is not too high, the vapour would not
depart too greatly from ideal gas behaviour, it is possible to write
Equation (8.51) shows that at a given temperature, the total pressure over an ideal solution is a linear
function of composition thus establishing the fourth criteria given above. When the partial pressures
and total pressure are plotted against mole fraction x1, we get according to Eq. (8.50) and Eq. (8.51)
the straight lines shown in Fig. 8.8. The broken lines give the partial pressures and the continuous line
gives the total pressure.
The P-x-y diagram can be easily constructed. At any fixed temperature, the total pressure can be
calculated using Eq. (8.51) for various x values ranging from 0 to 1. The corresponding equilibrium
vapour phase compositions are obtained by applying Dalton’s law according to which the partial
pressure in the vapour is equal to the mole fraction in the vapour (y) times the total pressure (P). That
is
Thus Eq. (8.51) is used to calculate the total pressure at given x and Eq. (8.54) is used to calculate
the corresponding equilibrium vapour phase composition y. The P-x-y diagram can now be plotted as
shown in the Fig. 8.8.
To prepare the T-x-y diagrams at a given total pressure P we can again use Eqs. (8.51) and
(8.54). Assume temperatures lying between the boiling points of pure liquids at the given pressure.
For the temperature assumed, find the vapour pressures of the pure liquids and calculate x from
Eq. (8.51). Use these in Eq. (8.54) and calculate the vapour composition y. Instead, if we attempt to
find the equilibrium temperature for the solution of known concentration x, the temperature may be
estimated by trial, such that the sum of the partial pressures is equal to the given total pressure. Once
the temperature is thus known, the vapour phase composition is determined as before. The T-x curve
is the lower curve in the figure and is called the bubble-point curve. The T-y curve is the upper curve
and is called the dew-point curve.
The y-x diagram is also prepared from the constant total pressure data. It can be constructed from the
boiling point diagram by drawing horizontal tie lines. The intersections of these lines with the
bubble-point curve give x and the intersections with the dew-point curve give y. Figure 8.10 shows a
typical equilibrium diagram.
There is an approximate method for the construction of the equilibrium diagram, and it is based on the
assumption that the ratio of vapour pressures of the components is independent of temperature. This
assumption may not introduce much error, as it is possible that the vapour pressures of both
components vary with temperature and these variations are to the same extent that their ratio remains
unaltered. Thus
which can be written in the following form.
Although Eq. (8.55) is not exact over a wide range of temperatures, the effect of variation in a is so
small that an average a value can be used in Eq. (8.55) and the whole y-x data required for the
preparation of the equilibrium curve can be evaluated.
EXAMPLE 8.5 Prove that if Raoult’s law is valid for one constituent of a binary solution over the
whole concentration range, it must also apply to the other constituent.
Solution Assume that Raoult’s law is obeyed by component 1 in a binary mixture. Then
As pointed out earlier, Raoult’s law is obeyed by ideal solutions when the vapour phase behaves as
an ideal gas whereas Lewis–Randall rule is obeyed by ideal solutions irrespective of whether the gas
phase is ideal or not. So for component 1, we can write
This equation is sometimes referred to as Duhem–Margules equation. Comparing Eq. (8.56) with
Eq. (8.57) we see that
This is Raoult’s law for component 2. The conclusion to be drawn from the above derivation is that if
Raoult’s law is applicable to one of the constituents of a liquid mixture at all concentrations, it must
be applicable to the other constituent as well.
EXAMPLE 8.6 n-Heptane and toluene form ideal solution. At 373 K, their vapour pressures are 106
and 74 kPa respectively. Determine the composition of the liquid and vapour in equilibrium at 373 K
and 101.3 kPa.
Solution Refer Eq. (8.51). Then,
where x is the mole fraction of heptane in the liquid. On solving this, we get x = 0.853
From Eq. (8.54),
y = 0.853 106/101.3 = 0.893
The liquid and the vapour at the given conditions contain respectively 85.3% (mol) and 89.3% (mol)
heptane.
EXAMPLE 8.7 An equimolar solution of benzene and toluene is totally evaporated at a constant
temperature of 363 K. At this temperature, the vapour pressures of benzene and toluene are 135.4 and
54 kPa respectively. What are the pressures at the beginning and at the end of the vaporisation
process?
Solution Put x = 0.5 in Eq. (8.51). Then P = 94.7 kPa. This is the pressure at the beginning of
vaporisation. Equation (8.51) can be written as
Put y = 0.5 in this. Thus, we get P = 77.2 kPa. (This is the pressure at the end of vaporisation).
EXAMPLE 8.8 A mixture of A and B conforms closely to Raoult’s law. The pure component vapour
pressures in kPa at T K are given by
If the bubble point of a certain mixture of A and B is 349 K at a total pressure of 80 kPa, find the
composition of the first vapour that forms.
Solution At 349 K, the vapour pressures of the pure components are:
At a given temperature and pressure, the composition of the liquid and vapour phases in equilibrium
is calculated using Eqs. (8.51) and (8.54), respectively. The composition of the vapour so calculated
at the bubble-point temperature is the composition of the first vapour produced from a liquid on
boiling.
Using Eq. (8.51),
The vapour formed contains 92.5% A.
EXAMPLE 8.9 The vapour pressures of acetone (1) and acetonitrile (2) can be evaluated by the
Antoine equations
where T is in K and P is in kPa. Assuming that the solutions formed by these are ideal, calculate
(a) x1 and y1 at 327 K and 65 kPa
(b) T and y1 at 65 kPa and x1 = 0.4
(c) P and y1 at 327 K and x1 = 0.4
(d) T and x1 at 65 kPa and y1 = 0.4
(e) P and x1 at 327 K and y1 = 0.4
(f) The fraction of the system that is liquid and the composition of the liquid and vapour in
equilibrium at 327 K and 65 kPa when the overall composition of the system is 70 mole per cent
acetone.
Solution (a) From the Antoine equations, at 327 K,
Assume a temperature and calculate the vapour pressures using Antoine equations. Substitute the
vapour pressure values in the above equation. See whether the Left-hand side = 0.4. This is repeated
till the left-hand side of the above equation becomes equal to 0.4.
At T = 334 K, = 107.91 kPa and = 51.01 kPa.
(e) At 327 K, we have = 85.12 kPa and = 39.31 kPa. Here y1 = 0.4
Equation (8.54) relates y to x. When P in Eq. (8.54) is eliminated using Eq. (8.51) we get
(f) The composition of the vapour and liquid in equilibrium at P = 65 kPa and T = 327 K were
determined in part (a). They are x1 = 0.5608 and y1 = 0.7344. Let f be the fraction of the mixture that
is liquid. Then an acetone balance gives
1 0.7 = f 0.5608 + (1 – f) 0.7344
Solving this, we get f = 0.1982. That is, 19.82% (mol) of the given mixture is liquid.
EXAMPLE 8.10 Mixtures of n-Heptane (A) and n-Octane (B) are expected to behave ideally. The
total pressure over the system is 101.3 kPa. Using the vapour pressure data given below,
(a) Construct the boiling point diagram and
(b) The equilibrium diagram and
(c) Deduce an equation for the equilibrium diagram using an arithmetic average a value.
T, K 371.4 378 383 388 393 398.6
Solution Sample calculation: Consider the second set of data. T = 378 K; PSA = 125.3 kPa; PSB =
Solution Sample calculation: Consider the second set of data. T = 378 K; P A = 125.3 kPa; P B =
55.6 kPa.
Using Eq. (8.51),
101.3 = 55.6 + xA(125.3 – 55.6)
Therefore, xA = 0.656.
Using Eq. (8.54), we see
yA = 0.656 125.3/101.3 = 0.811
Relative volatility is
a = PSA/PSB = 125.3/55.6 = 2.25
These calculations are repeated for other temperatures. The results are tabulated below:
T, K 371.4 378 383 388 393 398.6
xA 1.000 0.656 0.487 0.312 0.157 0
yA 1.000 0.811 0.674 0.492 0.279 0
a 2.28 2.25 2.17 2.14 2.08 2.02
Comparing Eq. (8.58) with Eq. (8.59) we see that /RT2 < 0, which means . The total
enthalpy of the solution is , whereas the enthalpy of the system before mixing is
S niHi. Since the former is greater than the latter, there is absorption of heat during mixing. Examples
of solutions showing positive deviation from ideality are oxygen–nitrogen, ethanol–ethyl ether,
water–ethanol, carbon disulphide–acetone, benzene–cyclohexane, acetonitrile–benzene,
n-hexane–nitroethane, etc.
For solutions exhibiting negative deviation from ideal behaviour, the partial pressures are less than
those given by Raoult’s law. By a derivation similar to the one presented in the preceding paragraph,
it can be shown that when solutions showing negative deviation are formed from pure constituents
there is evolution of heat. At the molecular level, appreciable negative deviation reflects stronger
intermolecular forces between unlike than between like pairs of molecules. Examples are
chloroform–ethyl ether, chloroform–benzene, hydrochloric acid–water, phenol–cyclohexanol,
chloroform–acetone, etc.
The general nature of the vapour pressure curves showing positive and negative deviation are shown
in Fig. 8.12. Figures 8.12(a) and (b) refer to constant temperature conditions. The uppermost curves
give the total vapour pressure as function of liquid composition. The corresponding curves, as a
function of the vapour composition lie below it, so that the vapour is rich in the more volatile
component.
8.10.1 Azeotropes
Azeotropes are constant boiling mixtures. The word ‘azeotrope’ is derived from Greek word meaning
‘boiling without changing’. When an azeotrope is boiled, the resulting vapour will have the same
composition as the liquid from which it is produced. Whereas, the equilibrium temperature of an
ordinary solution varies from the bubble point to the dew point, the boiling point of an azeotrope
remains constant till the entire liquid is vaporised. The azeotropes are formed by solution showing
large positive or negative deviation from ideality. If the vapour pressures of the constituents of a
solution are very close, then any appreciable positive deviation from ideality will lead to a maximum
in the vapour pressure curve and negative deviations from ideality under the same conditions leads to
a minimum in the vapour pressure curve. Even if an appreciable difference exists in the vapour
pressures of the pure components, the chances for the occurrence of maximum or minimum in the
vapour pressures should not be overruled if the deviation from ideal behaviour is quite high. At the
composition at which there exists a maximum or minimum in the vapour pressure curve, a minimum or
maximum, as the case may be, exists in the boiling point diagrams. The mixture is said to form an
azeotrope at this composition under the given temperature and pressure and it will distil without
change in composition, because the vapour produced has the same composition as the liquid.
Minimum-boiling azeotropes. Solutions showing positive deviation from ideality in certain
cases may lead to the formation of azeotropes of the minimum-boiling type. The P-x-y, T-x-y and x-y
curves for the minimum-boiling azeotropes are shown in Fig. 8.13.
In the boiling point diagram, the liquid and vapour curves are tangent at point M, the point of
azeotropism at this pressure. The temperature at M is the minimum temperature of boiling for the
system. For all mixtures of composition less than M, the equilibrium vapour is richer than the
liquid in the more volatile component. For all mixtures richer than M, the vapour is less rich than the
liquid in the more volatile component. A mixture of composition M boils producing a vapour of
identical compositions and consequently at a constant temperature and without change in composition.
If solutions either at P or Q are boiled in an open vessel with continuous escape of vapours, the
temperature and composition move along the lower curve away from M and towards the pure
substances. Solutions like these cannot be distilled by usual distillation methods. One
of the most important azeotropes in this category is ethanol–water which forms azeotrope at
89.4% (mol) ethanol at 351.4 K and 101.3 kPa. Other examples are benzene–ethanol (341.2 K, 55%
benzene), carbon disulphide–acetone (312.5 K, 61% carbon disulphide), isopropyl ether–isopropyl
alcohol (345.1 K, 39.3% alcohol), all at 101.3 kPa.
Maximum-boiling azeotropes. When the total pressure of the system at equilibrium is less than
the ideal value, the system is said to exhibit negative deviation from ideality. When the difference in
vapour pressures of the components is not too great, and in addition, the negative deviations are large,
the curve for total pressure against composition passes through a minimum. This condition results in a
maximum in the boiling temperature and a condition of azeotropism as at point M in Fig. 8.14.
The vapour is leaner in the more volatile component for liquids whose concentration is less than the
azeotropic concentration. Solution on either side of the azeotrope, if boiled in an open vessel with
escape of vapours will ultimately leave a residual liquid of the azeotropic composition in the vessel.
Maximum-boiling azeotropes are less common than the minimum-boiling type. Hydrochloric acid–
water system forms an azeotrope at 11.1% (mol) HCl at 383 K and 101.3 kPa. Other examples are
chloroform–acetone (337.7 K and 65.5% acetone), phenol–cyclohexanol
(455.65 K, 90% phenol), all at 101.3 kPa.
The plot of activity coefficients versus mole fraction in the liquid phase for ideal solutions, minimum-
boiling azeotrope and maximum-boiling azeotrope are given in Fig. 8.15.
Effect of pressure on azeotropes. The azeotropic composition shifts continuously with change
in pressure or temperature. In some cases, changing the pressure may eliminate azeotropism from the
system. Azeotropism disappears in ethanol–water system at pressures below 9.33 kPa. The table
below illustrates the effect of pressure on this system. The last row gives the mole per cent of alcohol
in the azeotrope.
P, kPa 13.3 20.0 26.6 53.2 101.3 146.6 193.3
T, K 307.4 315.2 321 336 351.3 361 368.5
%(mol) 99.6 96.2 93.8 91.4 89.43 89.3 89.0
EXAMPLE 8.11 Prove that at the azeotropic composition, the vapour and liquid have the same
composition.
Solution Refer the Duhem–Margules equation, Eq. (8.57).
For a maximum or minimum on the total pressure curve, (dP/dx1) is zero. Then Eq. (8.64)
necessitates that either the terms in the parenthesis should be zero or is zero. If the latter
term is zero the partial pressure would be unaffected by concentration changes, which is not true.
Hence, at the point of azeotropism, we have, the following relation:
Since x2 = 1 – x1 and y2 = 1 – y1, the above result means that at azeotropic condition, x1 = y1 or the
vapour composition and liquid composition are the same.
The minimum requirement to be met by an equation for activity coefficient is that it should conform to
the restriction imposed by the above relationship. Among a number of equations between g and x that
are available, as far as phase equilibrium problems are concerned, some equations have got wide
acceptance. They are discussed in the following sections.
Equation (8.65) is known as Wohl’s three-suffix equation . It involves three parameters, A, B and
(q1/q2) which are characteristics of the binary system.
Margules equation. When the term (q1/q2) is unity in Eq. (8.65), we get the following expression,
which is known as the Margules three-suffix equation.
The constant A in the above equation is the terminal value of ln g1 at x1 = 0 and the constant B is the
terminal value of ln g2 at x2 = 0. The three-suffix Margules equation adequately represents the VLE
data of systems like acetone–methanol, acetone–chloroform, chloroform–methanol, etc.
When A = B in Eq. (8.66), the Margules equation takes the following simple form:
………(8.67)
Equation (8.67) is called the Margules two-suffix equation. It represents sufficiently and accurately
the activity coefficients of simple liquid mixtures, i.e. mixtures of molecules, which are similar in
size, shape and chemical nature. The constant A may be positive or negative. While in general, the
constant depends on temperature, for many systems it is a weak function of temperature. Vapour–
liquid equilibrium data of argon–oxygen, benzene–cyclohexane, etc., are well represented by the
Margules equation [Eq. (8.67)].
van Laar equation. Let (q1/q2) = (A/B) in Eq. (8.65). The resulting two-parameter equation is
known as the van Laar equation. The van Laar equations can be written as
Strictly speaking, van Laar equations are applicable only for solutions of relatively simple,
preferably non-polar liquids. But empirically, it has been found that these are applicable for more
complex mixtures. The van Laar equations are widely used for vapour–liquid equilibrium
calculations because of their flexibility and mathematical simplicity. Activity coefficients in benzene–
isooctane system, n-propanol–water system, etc., are accurately represented by the van Laar
equations.
The selection of a proper equation for VLE data correlation depends on the molecular complexity of
the system and the precision of the experimental data. When an equation is selected that fits the
experimental data well, the constants for the constant pressure conditions will be different from those
applicable for constant temperature conditions. The effect of pressure on the constants is usually
negligibly small, whereas the effect of temperature is appreciable and cannot be neglected. The van
Laar constants vary with temperatures unless the temperature range involved is small. However, in
vapour–liquid equilibrium calculations, the effect of temperature on the activity coefficient is usually
ignored (Prausnitz, 1985).
The Margules three-suffix equation is suited for symmetrical systems, i.e. where the constants A and B
are nearly the same. The van Laar equations can be used for unsymmetrical solutions, where the ratio
A/B does not exceed 2. Though many systems follow van Laar equations, they cannot represent
maxima or minima in the ln g curve. Margules three-suffix equation should be used in such cases. For
choice of an appropriate equation, a rule of thumb usually employed is this: When the ratio of molar
volumes is close to unity, the Margules equation is preferred. When the ratio is quite different from
unity, as is the case when water is one of the constituents, the van Laar equations are found to be
satisfactory. For example, the chloroform–ethyl alcohol system, which shows a maximum and a
minimum on the ln g curves and whose ratio between pure component molar volumes is 1.38, is
accurately represented by the Margules equation. For n-propanol–water system this ratio is 4.16 and
the van Laar equations are found to represent the behaviour accurately.
It is to be remembered that in equations having only two constants, determination of g1 and g2 at a
single known composition permits the evaluation of the constants and the complete g curve. Equation
(8.47) permits the evaluation of g1 and g2 when it is rearranged to the following form.
The data required are a single set of equilibrium vapour–liquid composition values and the vapour
pressures of the pure components. When an azeotrope is formed, only the azeotropic composition
need be known, because it represents the composition of both the liquid and the vapour phases. The
activity coefficients can be evaluated by putting x = y in Eq. (8.70).
Wilson equation. All the activity coefficient equations discussed so far can be deduced from the
original Wohl’s equation under proper simplifying assumptions. However, there are many equations
that cannot be derived from the Wohl’s general equation. Among such equations, the Wilson equation,
the NRTL equation and the UNIQUAC equation are important from practical point of view. All these
are based on the concept of local compositions, which are different from the overall mixture
compositions. Based on molecular considerations, Wilson (1964) proposed the following equations
for activity coefficients in a binary mixture.
Wilson equations have two adjustable positive parameters L12 and L21. These are related to the pure
component molar volumes and to the characteristic energy differences by
where V1 and V2 are the molar volumes of pure liquids and l’s are the energies of interaction
between the molecules designated in the subscripts. The differences in the characteristic energies
(aij) are assumed to be temperature independent and this introduces no serious error in practical
calculations. Wilson equation provides a good representation of VLE of a variety of miscible
mixtures. It is particularly suitable for solutions of polar or associating components like alcohols in
non-polar solvents for which the Margules and van Laar equations are generally inadequate.
Wilson equation suffers from two disadvantages, though not serious for many applications. Firstly, it
is not suitable for systems showing maxima or minima on the ln g versus x curves. Secondly, it is not
useful for systems exhibiting limited miscibility. The use of Wilson equation is therefore
recommended only for liquid systems that are completely miscible, or for partially miscible systems
in the region where only one liquid phase exists.
Non-random two-liquid (NRTL) equation. The NRTL model, proposed by Renon and Prausnitz
(1968), also is based on the local composition concept. The activity coefficients are
The constants b12 and b21 are similar to the constants representing characteristic energy
differences appearing in the Wilson equation. These, as well as the constant a12 are independent of
composition and temperature. The parameter a12 is related to the non-randomness in the
mixture. If a1 2 is zero, the mixture is completely random and the NRTL equation reduces to the
Margules equation. It is found from fitting of experimental data that a12 varies from about 0.20
to 0.47. In the absence of the experimental data, the value of a12 is arbitrarily set, a typical
choice being a12 = 0.3. When a12 is arbitrarily fixed, NRTL equation becomes a two-parameter
model.
NRTL equation is applicable to partially miscible as well as totally miscible systems. For
moderately non-ideal systems, it offers no advantage over the van Laar and Margules equations. But,
for strongly non-ideal solutions and especially partially miscible systems, the NRTL equations
provide a good representation.
Universal quasi-chemical (UNIQUAC) equation. Abrams and Prausnitz (1975) extended the
quasi-chemical theory of liquid mixtures to solutions containing molecules of different sizes. This
extension is called the UNIQUAC theory. The UNIQUAC model consists of two parts—the
combinatorial part, which describes the prominent entropic contribution and a residual part, which is
due primarily to the intermolecular forces that are responsible for the enthalpy of mixing. The
combinatorial part is determined by the sizes and shape of the molecules and requires only pure-
component data. The residual part depends on the intermolecular forces and involves two adjustable
binary parameters.
The UNIQUAC equations for activity coefficients are
z is the coordination number, r, q and q are pure-component molecular structure constants. The
molecular size and surface area are given by r and q respectively. For fluids other than water or
lower alcohols, q = q . For alcohols, the surface of interaction q is smaller than the geometric
surface q. The adjustable binary parameters t12 and t21 are related to the characteristic energies Du
as follows.
The UNIQUAC equation satisfies a large number of non-electrolyte mixtures containing non-polar
fluids such as hydrocarbons, alcohols, nitriles, ketones, aldehydes, organic acids, etc., and water,
including partially miscible mixtures. The main advantages of this equation are its wide applicability
and simplicity arising primarily from the fact that there are only two adjustable parameters.
Universal functional activity coefficient (UNIFAC) method. In the UNIFAC method, the
activity coefficients are estimated through group contributions. The liquid is treated as a solution of
different structural groups from which the molecules are formed, rather than a solution of molecules
themselves. This method is based on the UNIQUAC model where the activity coefficient is divided
into two parts—the molecular size contribution (the combinatorial part) and the interaction
contributions (the residual part).
The combinatorial contribution can be estimated from pure-component properties and the size and
shape of the molecules, whereas for the estimation of the second part, group areas and group
contributions are needed. A large number of group interaction parameters are already reported.
UNIFAC has been successfully used for the design of distillation columns, involving even azeotropic
and extractive distillation (Prausnitz et al., 1986).
EXAMPLE 8.12 Liquids A and B form an azeotrope containing 46.1 mole per cent A at 101.3 kPa and
345 K. At 345 K, the vapour pressure of A is 84.8 kPa and that of B is 78.2 kPa. Calculate the van
Laar constants.
Solution Let the material A be component 1 and B be component 2. The activity coefficients at the
azeotropic concentration can be evaluated by Eq. (8.71)
EXAMPLE 8.13 The azeotrope of the ethanol–benzene system has a composition of 44.8% (mol)
ethanol with a boiling point of 341.4 K at 101.3 kPa. At this temperature the vapour pressure of
benzene is 68.9 kPa and the vapour pressure of ethanol is 67.4 kPa. What are the activity coefficients
in a solution containing 10% alcohol?
Solution Let benzene be component 1 and alcohol component 2. For the azeotrope
EXAMPLE 8.14 Water (1)–hydrazine (2) system forms an azeotrope containing 58.5% (mol)
hydrazine at 393 K and 101.3 kPa. Calculate the equilibrium vapour composition for a solution
containing 20% (mol) hydrazine. The relative volatility of water with reference to hydrazine is 1.6
and may be assumed to remain constant in the temperature range involved. The vapour pressure of
hydrazine at 393 K is 124.76 kPa.
Solution The vapour pressure of water at 393 K = 1.6 vapour pressure of hydrazine at
393 K = 1.6 124.76 = 199.62 kPa.
To evaluate the vapour compositions using these equations, we should know the vapour pressure
values at the new equilibrium temperature. Taking the ratio of the last two equations, we get
The composition of the vapour in equilibrium with the liquid containing 20% hydrazine is 5.28%
hydrazine and 94.72% water.
EXAMPLE 8.15 At 318 K and 24.4 kPa, the composition of the system ethanol (1) and toluene (2) at
equilibrium is x1 = 0.3 and y1 = 0.634. The saturation pressure at the given temperature for the pure
EXAMPLE 8.16 The activity coefficients in a mixture of components A and B at 313 K are given by
At 313 K, A and B form an azeotrope containing 49.4 mol percent A at a total pressure of 27 kPa. If
the vapour pressures of pure A and pure B are 25.0 and 24.3 kPa, respectively, calculate the total
pressure of the vapour at temperature 313 K in equilibrium with a liquid mixture containing 12.5 mol
percent A.
Solution At the azeotropic composition, Eq. (8.71) is applicable, so that the activity coefficients are:
EXAMPLE 8.17 Using van Laar constants and the vapour pressures of the pure substances how
would you prove whether a given binary system forms an azeotrope or not?
Solution If the mixture does not exhibit azeotropic behaviour, the ratio (y1/x1) will be greater than
(y2/x2) for the entire concentration range 0 to 1. Denoting the ratio of (y1/x1) to (y2/x2) by a, then a
> 1 for 0 < x < 1. However, if the mixture forms an azeotrope, then the value of a will be greater than
1 over some concentration range and will be less than 1 over the remaining portion. Since a varies
continuously with x, a should have a value equal to 1 at some x which is the azeotropic composition.
Writing Eq. (8.47) for both components and rearranging the result, we get
A and B are the van Laar constants. If the mixture forms an azeotrope, one of the above values will be
greater than 1 and the other less than 1.
EXAMPLE 8.18 For the binary system methanol (1) and benzene (2), the recommended values of the
Wilson parameters at 341 K are L12 = 0.1751 and L21 = 0.3456. The vapour pressures of pure
species are = 68.75 kPa and = 115.89 kPa. Show that the given system can form an azeotrope
at 341 K. Assume that the vapour behaves like an ideal gas.
Solution Wilson equations [Eq. (8.72)] provide the activity coefficients in a binary mixture as:
Equation (8.48) gives the relationship between activity coefficient and equilibrium phase
compositions as
If the mixture forms an azeotrope, then the value of a will be greater than 1 over some concentration
range and will be less than 1 over the remaining portion. Relative volatilities are calculated at x1 = 0
and x1 = 1. If the system forms an azeotrope, one of these values will be greater than unity and the
other less than unity.
At x1 = 0,
At x2 = 0,
Since the relative volatility at x1 = 0 is greater than unity, and that at x2 = 0 is less than unity, it is
clear that the system forms an azeotrope.
EXAMPLE 8.19 A stream of isopropanol–water mixture is flashed into a separation chamber at 353
K and 91.2 kPa. A particular analysis of the liquid product showed an isopropanol content of 4.7%
(mol), a value that deviated from the norm. It is suspected that an air leak into the separator might
have caused this. Do you agree? The vapour pressures of the pure propanol and water are 91.11 kPa
and 47.36 kPa respectively, and the van Laar constants are A = 2.470 and B = 1.094.
Solution For x1 = 0.047, x2 = 0.953, A = 2.47 and B = 1.094, Eq. (8.68) gives
g1 = 7.388 and g2 = 1.011
The total pressure corresponding to this equilibrium composition is
This is less than the total pressure. This error must have been caused by an air-leak.
EXAMPLE 8.20 Construct the P-x-y diagram for the cyclohexane (1)–benzene (2) system at 313 K
given that at 313 K the vapour pressures are = 24.62 kPa and = 24.41 kPa. The liquid-phase
activity coefficients are given by
The results are plotted taking P on the y-axis and x1 and y1 on the x-axis.
EXAMPLE 8.21 From vapour–liquid equilibrium measurements for ethanol–benzene system at 318
K and 40.25 kPa it is found that the vapour in equilibrium with a liquid containing 38.4% (mol)
benzene contained 56.6% (mol) benzene. The system forms an azeotrope at 318 K. At this
temperature, the vapour pressures of ethanol and benzene are 22.9 and 29.6 kPa respectively.
Determine the composition and total pressure of the azeotrope. Assume that van Laar equation is
applicable for the system.
Solution Let benzene be component 1 and ethanol component 2. Using Eq. (8.47) the activity
coefficients are determined.
For any liquid composition, the activity coefficients are calculated using these equations. If the
mixture forms an azeotrope at any composition, then as per Eq. (8.71), the following relations also
give the activity coefficients.
Thus, activity coefficients calculated using the van Laar equations should also satisfy the relation
at the azeotropic composition. The azeotropic composition is obtained by trial
assuming values of x. For x1 = 0.6, g1 = 1.3830 and g2 = 1.7806.
These values are so close that it can be assumed that the composition corresponds to an azeotrope.
Thus, the liquid mixture forms an azeotrope containing 60% benzene which boils at 318 K and
approximately 40.86 kPa, (mean of 40.937 and 40.775 kPa).
EXAMPLE 8.22 The activity coefficients in a binary system are given by ln .
Show that if the system forms an azeotrope, then and the azeotropic composition
is given by
At the azeotropic composition, the P-x curve has a maximum or minimum. Therefore, the above result
is equated to zero and after rearrangement, we get
Since the value of x1 lies between 0 and 1 it is essential that . Noting that x1x2 for
a binary mixture is less than or equal to 0.25, the second root leads to A ≥ 2. This condition usually
results in partial miscibility. Thus, condition for homogeneous azeotropy is and A
< 2.
EXAMPLE 8.23 The following values refer to the Wilson parameters for the system acetone(1)–
water(2):
a1 2 = 1225.31 J/mol, a21 = 6051.01 J/mol, V1 = 74.05 10–6 m3/mol, V2 = 18.07 10–6
m3/mol.
The vapour pressures are given by
where P is in kPa and T is in K. Calculate the equilibrium pressure and composition of
(a) Vapour in equilibrium with a liquid of composition x1 = 0.43 at 349 K.
(b) The liquid in equilibrium with a vapour of concentration y1 = 0.8 at 349 K.
Solution Using the Antoine equations, at 349 K, the vapour pressures are calculated as
The vapour composition is found out from the relation y1P = g1x1
(b) Since the liquid composition is not known, the activity coefficients cannot be calculated. Assume
g1 = g2 = 1. The relation Pyi = xi can be written for component 1 and component 2, which on
rearrangement gives
Activity coefficients can now be determined for this composition using Wilson equations.
Now the pressure is recalculated incorporating the activity coefficient values. The equation
This is same as the previous value calculated for P. Therefore, P = 164.48 kPa and x1 = 0.4568.
EXAMPLE 8.24 The system methanol–methyl ethyl ketone forms an azeotrope containing 84.2%
(mol) methanol at 337.5 K and 101.3 kPa. The vapour pressures of the pure species are given by the
Antoine equation
The equations of state provide thermodynamic models for evaluating in Eq. (8.45) from
volumetric properties. The fugacity coefficients for the vapour and liquid phases are to be obtained
from appropriate equations of state. Equations of state widely used in engineering calculations were
discussed in Chapter 3. The most generally used equations of state for VLE calculations are given in
Table 8.1.
The fugacity coefficients in Eq. (8.45) can be calculated using a suitable equation of state. The
relationship between fugacity coefficient and the volumetric properties can be written as
In Eq. (8.83a), is the fugacity coefficient in the saturation state, Vi is the molar volume of pure i as
saturated liquid. are evaluated using an equation of state. Assuming the Poynting factor to
be unity, Eq. (8.83a) can be rearranged as
which is nothing but Raoult’s law which is applicable for ideal solution where the vapour behaves as
ideal gas. We see that for ideal solution, the vaporisation equilibrium constant is the ratio of vapour
pressure to total pressure. The K-factors for ideal solutions depend only on the temperature and
pressure and are readily correlated as a function of these two variables.
For mixtures of light hydrocarbons, Eq. (8.83a) can be simplified using two assumptions:
1. Intermolecular forces are weak, so that vapour phase behaves as an ideal solution, so that ,
where fi is the fugacity coefficient of the pure components.
2. Liquid phase behaves as an ideal solution so that =1.0.
The resulting equation is very convenient as it involves properties of pure component only.
Equation (8.83e) reveals that for mixtures of light hydrocarbons, the vaporisation equilibrium
constants are independent of the composition of the liquid and vapour phases in equilibrium
and can be evaluated as pure component properties. Since can be determined from
equations of state or generalised correlations, it is possible to provide correlations for K-values of
substances as functions of temperature and pressure. DePriester nomographs [C.L. DePriester, Chem.
Eng. Progr., Symposium Ser. 7, 49 (1953)] provide such correlations for many
hydrocarbons. These nomographs are available in standard references such as Chemical Engineers
Handbook. Figure 8.16 gives the K-factor for light hydrocarbons in the high temperature
range.
8.12.3 Bubble-point Equilibria
The bubble-point temperature is the one at which the first bubble of vapour is produced from the
liquid on heating at constant pressure. At the bubble point the liquid has the same composition as the
original mixture. Therefore, in problems where bubble-point temperature is to be determined, the xi
are known. Assume a temperature and get the Ki values at this temperature. Calculate yi using yi =
Kixi. If the assumed temperature is correct then
S yi = S Kixi = 1………(8.84)
Otherwise, repeat the calculations with another temperature. To find the bubble-point pressure , a
similar procedure as above is adopted by assuming various values of pressure until S Kixi = 1.
Otherwise, repeat the calculation by assuming another temperature till this equation is satisfied.
Determination of the dew-point pressure involves a similar procedure assuming pressure instead of
temperature.
8.12.5 Flash Vaporisation
The general flash vaporisation problem can be stated as: Given a mixture of known overall
composition zi at temperature T and pressure P, what is the fraction that is vapour (V) and what are
the composition of the liquid and vapour phases in equilibrium? The overall material balance for the
system is
F = V + L………(8.86)
where F is the total number of moles of the initial mixture. The component-i balance for the
system is
Fzi = Vyi + Lxi………(8.87)
Since yi = Kixi, it can be eliminated from Eq. (8.87) to get the following:
Equation (8.91) can also be utilised in an iterative procedure to estimate T, P or the fraction of the
initial mixture that is vaporised.
EXAMPLE 8.25 A mixture contains 45% (mol) methanol (A), 30% (mol) ethanol (B) and the rest
n-propanol (C). Liquid solution may be assumed to be ideal and perfect gas law is valid for the
vapour phase. Calculate at a total pressure of 101.3 kPa.
(a) The bubble point and the vapour composition
(b) The dew point and the liquid composition.
The vapour pressures of the pure liquids are given below:
Solution The vapour pressures of the components are plotted against temperature so that interpolation
of vapour pressure can be done easily.
(a) If the vapour phase can be treated as an ideal gas and liquid phase, an ideal solution, the K-values
can be written as Ki = yi/xi = . Equation (8.84) can be written as
Now temperatures are assumed till the above equality is satisfied. It is seen that at 344 K,
The bubble-point lies between 344 and 345 K. By interpolation, the bubble-point is obtained
as 344.6 K. At this temperature the vapour pressures are obtained from the P vs T plots. =
137.3 kPa, = 76.20 kPa and = 65.40 kPa.
Component xi Ki = /P yi = Ki xi
Methanol 0.45 137.30 1.355 0.610
Ethanol 0.30 76.20 0.752 0.226
Propanol 0.25 65.40 0.646 0.162
S Ki xi 0.998
The equilibrium vapour contains 61% methanol, 22.6% ethanol and 16.2% propanol.
(b) Equation (8.85) for the present case becomes
The dew-point temperature is to be determined by trial such that the above relation is satisfied. By
trial, it can be seen that at 347.5 K, = 153.28 kPa, = 85.25 kPa and = 73.31 kPa.
Component yi Ki = /P xi = yi /Ki
The values in the last column are the liquid composition at the dew point. Thus, liquid contains
29.7% methanol, 35.7% ethanol, and 34.5% propanol.
EXAMPLE 8.26 A hydrocarbon mixture contains 25% (mol) propane, 40% (mol) n-butane and 35%
(mol) n-pentane at 1447.14 kPa. Assume ideal solution behaviour and calculate
(a) The bubble-point temperature and composition of the vapour
(b) The dew-point temperature and the composition of the liquid
(c) The temperature and the composition of the liquid and vapour in equilibrium when 45% (mol)
of the initial mixture is vaporised. (The values of Ki can be obtained from Fig. 13.6 of Chemical
Engineer’s Handbook, 5th ed.)
Solution (a) Assume temperature, say 355.4 K, and the Ki values are found out from the nomograph
[Fig. 13.6(b) in Chemical Engineer’s Handbook]. The products of Ki and xi are calculated and their
sum S xiKi is found out. The results for two temperatures 355.4 K and
366.5 K are shown below.
T = 355.4 K T = 366.5 K
Component xi Ki Ki xi Ki Ki xi
Propane 0.25 2.000 0.500 2.30 0.575
n-Butane 0.40 0.780 0.312 0.90 0.360
n-Pentane 0.35 0.330 0.116 0.40 0.140
S Ki xi 0.928 1.075
The bubble-point temperature lies between 355.4 K and 366.5 K. By interpolation, the temperature is
found out to be 361 K. The calculations are carried out at this temperature and the results are as
follows:
Component xi Ki Ki xi
Propane 0.25 2.12 0.530
n-Butane 0.40 0.85 0.340
n-Pentane 0.35 0.37 0.130
S Ki xi 1.000
Since S xiKi is approximately 1.00, the bubble-point temperature is 361 K. The values in the last
column are the mole fraction of various components in the vapour. At the bubble-point, the vapour
contains 53% propane, 34% butane and 13% pentane.
(b) At the dew-point temperature, S yi/Ki = 1. At 377.6 K, this value is 1.1598 and at 388.8 K it is
0.9677.
T = 377.6 K T = 388.8 K
Component yi Ki yi /Ki Ki yi /Ki
Propane 0.25 2.6 0.0962 2.9 0.0862
n-Butane 0.40 1.1 0.3636 1.3 0.3077
n-Pentane 0.35 0.5 0.7000 0.61 0.5738
Syi /Ki 1.1598 0.9677
By interpolation, the dew-point temperature is found to be 387 K. The calculations for this
temperature is given below.
Component yi Ki yi /Ki
Propane 0.25 2.85 0.0877
n-Butane 0.40 1.25 0.3200
n-Pentane 0.35 0.59 0.5932
S yi /Ki 1.0009
The last column in the above table is the liquid compositions. The equilibrium liquid at the dew point
contains 8.77% propane, 32.0% butane and 59.32% pentane.
(c) In the following calculations, temperature is assumed so as to satisfy Eq. (8.91). For a basis of
100 mol of the initial mixture, F = 100 mol, V = 45 mol and L = 55 mol. Equation (8.91) becomes
T = 366.5 K T = 377.6 K
Component zi Ki zi /[1 + L/(VKi )] Ki zi /[1 + L/(VKi )]
Propane 0.25 2.30 0.1632 2.6 0.1701
n-Butane 0.40 0.90 0.1696 1.1 0.1895
n-Pentane 0.35 0.40 0.0863 0.5 0.1016
S zi /[1 + L/(VKi )] 0.4191 0.4612
From the calculations given above, we see that the equilibrium temperature is between 366.5 K and
377.6 K. By interpolation, T = 374.6 K.
T = 374.6 K
Component zi Ki zi /[1 + L/(VKi )]
These are calculated using the values in the last column. Corresponding xi values are found out using
the material balance [Eq. (8.87)].
Fzi = Vyi + Lxi
The results of the calculation are given below:
Component yi xi
Plot the logarithm of the activity coefficients against mole fraction x1 of component 1 in a binary
solution as shown in Fig 8.17(a) and measure the slopes of the tangents drawn to the resulting curves
at any selected composition x1.
Equation (8.92) tells us that if the Gibbs–Duhem equation is to be satisfied, both slopes must have
opposite sign. Otherwise, the data are inconsistent. If the slopes are of opposite sign, substitute the
values in Eq. (8.92) and if it is satisfied reasonably well, then the data is consistent at the selected
composition. For a complete test, the slopes determined at other compositions are substituted into Eq.
(8.92) to see whether the equality is satisfied or not.
In addition to the above observations, we can make the following generalisations with the help of Eq.
(7.101).
1. If one of the ln g curves has a maximum (or minimum) at certain concentration, the other curve
should have a minimum (or maximum) at the same composition.
2. If there is no maximum or minimum point, then both curves must be positive or both must be
negative over the entire range. Or in other words, if one component has g values always greater
than unity and has no maximum, the g values of the other component must likewise be greater
than unity. This is a consequence of the fact that Raoult’s law is to be obeyed by the component
as its mole fraction tends to unity.
From the above discussion, it is clear that Fig. 8.17(b) represents plots of consistent data whereas
Figs. 8.17(c) and (d) are plots of thermodynamically inconsistent data. In Fig. 8.17(c), though there is
maximum on one curve and minimum on the other, these are shown at different compositions. In Fig.
8.17(d), the slopes have the same sign and the data are thermodynamically inaccurate.
From the experimental values of activity coefficients, ln (g1/g2) values are calculated and plotted
against x1 taken on the x-axis. The net area of the diagram should equal zero if the data are
thermodynamically consistent. That is, the area above the x-axis will be equal to the area below it as
shown in Fig. 8.19.
8.13.4 Using the Coexistence Equation
The coexistence equation can be used for testing the consistency of vapour–liquid equilibrium data. If
the vapour in equilibrium with a binary liquid mixture behaves as an ideal gas, Eq. (8.47) can be used
to describe the equilibrium. Rearranging Eq. (8.47), the activity coefficients can be written as
Substitute this in the Gibbs–Duhem Equations [Eq. (8.92)] written in the following form
Equation (8.99) is known as the coexistence equation. It can be used to calculate any one of the three
variables P, x or y if experimentally measured values of the other two variables are available. If all
the three variables are experimentally determined, then Eq. (8.99) can be used to test the consistency
of the measured data.
or
The partial pressures of both components are plotted against mole fraction x1 as in Fig. 8.20. The
slopes, are determined at any selected composition. and
are calculated. Then, according to Eq. (8.100), the absolute values of these
quantities should be the same if the data are thermodynamically consistent.
EXAMPLE 8.27 The following results were obtained by experimental VLE measurements on
the system, ethanol (1)–benzene (2) at 101.3 kPa. Test whether the data are thermodynamically
consistent or not.
Solution Assuming that the gas phase behaves ideally, the activity coefficients are calculated as
ln (g1/g2) values are plotted against x1. The net area is found out. Since this is not equal to zero, the
given experimental measurements do not satisfy the Redlich–Kister criterion [Eq. (8.97)] for
consistency.
EXAMPLE 8.29 The following data gives the composition versus total pressure for the system
chloroform (1)–ethyl alcohol (2) at 328 K are:
Vapour pressures of chloroform and acetone at 328 K are 82.35 and 37.30 kPa respectively. Estimate
the constants in the Margules equation [Eq. (8.66)].
Solution The Margules equations are:
(Note: Equations (8.103) and (8.104) are known as Carlson and Colburn relations for activity
coefficient.)
EXAMPLE 8.30 The following table gives the partial pressure of acetone versus liquid composition
for acetone (1)–water (2) system at 333 K.
x1 0 0.033 0.117 0.318 0.554 0.736 1.000
The vapour pressure of water at 333 K is 19.91 kPa. Calculate the partial pressure of water in the
vapour phase.
Solution Equation (8.100) is the Gibbs–Duhem equation in terms of partial pressures.
Using the given data calculate x1/[(1 – x1) ] and this is plotted against . The area under the curve
between the limits 0 and gives the integral in Eq. (8.106) from which the partial pressure can be
calculated. The results are given below:
x1 0 0.033 0.117 0.318 0.554 0.736 1.000
, kPa 19.91 19.31 18.27 16.99 15.42 13.90 0
On further addition of heat, the temperature remains constant at T* and more vapour of the same
composition as given by point E is formed. This continues till one of the components disappears from
the liquid and the system becomes a two-phase mixture either LA – V or LB – V depending upon the
initial composition.
Now let us consider the cooling of a vapour of initial composition and temperature indicated by point
J. When the temperature is lowered to that corresponding to point K the vapour pressure of pure
liquid B will be equal to the partial pressure of B in the vapour. Pure liquid B gets condensed and the
vapour composition changes along the line KE. When the temperature T* is reached, the partial
pressure of A in the vapour will be equal to the vapour pressure of A and at this condition, pure
liquids A and B and the vapour are present in equilibrium. Further cooling results in the elimination of
the vapour phase and the system now consists of two immiscible liquids. If the initial vapour were at
point P, pure liquid A would have condensed out first, instead of B.
EXAMPLE 8.31 A high boiling organic liquid is purified from non-volatile impurities by allowing it
to mix with steam directly at a total pressure of 93.30 kPa. The vapour pressure data are given as
follows:
Temperature, K 353 373
Vapour pressure of water, kPa 47.98 101.3
Vapour pressure of liquid, kPa 2.67 5.33
Assume that water and the organic liquid are immiscible and the impurities do not affect the
vaporisation characteristics. The vapour pressures vary linearly with temperature. Calculate under
three-phase equilibrium
(a) The equilibrium temperature and
(b) The composition of the resulting vapour.
Solution (a) At 353 K, sum of the vapour pressures is 50.65 kPa and at 373 K it is 106.63 kPa. Since
the vapour pressures vary linearly, the temperature at which the sum of vapour pressures is 93.3 kPa
is obtained by interpolation.
(b) At 368.2 K, the vapour pressure of water is 88.50 kPa and that of the liquid is 4.80 kPa. Since at
three-phase equilibrium, the partial pressure is equal to the vapour pressure, the ratio of mole
fractions of the components will be same as the ratio of vapour pressures. Let y be the mole fraction
of water in the vapour. Then
The point P gives the critical solution temperature. Outside the dome the mixture is homogeneous.
8.16.2 Ternary Equilibrium Diagrams
Liquid–liquid equilibria involving three components are important in the analysis of extraction
operations. The extraction process involves bringing a binary mixture of components A and C into
intimate contact with a solvent B. The solvent B is either partially soluble in liquid A or is immiscible
with it. The component C gets distributed in different proportions between the two insoluble phases
known as the ‘raffinate’ and the ‘extract’. The A-rich phase is known as the raffinate and the B-rich
phase is known as the extract. When the solvent added is only partially miscible with A, the extract
and raffinate phases contain three components. The ternary liquid–liquid equilibrium diagrams are
usually represented on equilateral triangular coordinates. On the equilateral triangle the length of the
altitude is allowed to represent 100% composition and the length of the perpendiculars from any
point to the bases represent the percentages of the three components. The apexes of the triangle
represent the pure componen ts A, B, and C and points on the sides represent binary mixtures.
Figure 8.26(a) shows the equilibrium diagram of type-I systems in which one pair is partially soluble.
The pairs A-C and B-C are miscible in all proportions and the pair A-B is miscible only partially.
Examples are water (A)–chloroform (B)–acetone (C), water (A)–benzene (B)–acetic acid (C), water
(A)–methyl isobutyl ketone (B)–acetone (C), etc. Liquid C dissolves completely in A and B whereas A
and B dissolve only to a limited extent in each other. In the region below the mutual solubility curves
the two liquid phases exist under equilibrium. The compositions of the equilibrium phases are
obtained at the ends of the tie line passing through the point representing the overall composition of
the mixture. For example, the mixture, whose combined composition is represented by point M
separates into a raffinate R and an extract E at equilibrium. Thus, RE is a tie line for the system. The
weight fraction of C in the raffinate is denoted by xR, and that in the extract is denoted by yE. Several
tie lines can be drawn and each gives rise to a set of equilibrium xR and yE values, which can be
used to plot the equilibrium diagram shown in Fig. 8.26(b).
The curve DRPF is the binodal solubility curve, which shows the change in the solubility of A-rich
and B-rich phases upon addition of C at a fixed temperature. Any mixture outside this curve will be a
homogeneous solution of a single liquid phase. There is one point on the binodal curve P, which will
represent the last of the tie lines where the A-rich and B-rich phases become identical. It is known as
the plait point. With increase in temperature the mutual solubilities of A and B increase and as a
result the heterogeneous area shrinks. Above the critical solution temperature of the binary A-B they
dissolve completely and the heterogeneous area vanishes completely. Extraction is not possible under
this condition.
The ternary equilibrium diagram for type II systems is shown in Fig. 8.27. In this type of systems, two
pairs are partially soluble. Examples of type II systems are chlorobenzene (A)–water (B)–Methylethyl
ketone (C) , n-heptane (A)–aniline (B)–methylcyclohexane(C), etc. Here A and C are completely
miscible while A-B and B-C pairs show only limited solubility. Points D and F represent the mutual
solubility of A and B and points H and G those of B and C at the prevailing temperature. Curves DRH
and FEG are the ternary solubility curves. Mixtures such as at M inside the heterogeneous area form
two liquid phases in equilibrium at E and R. As temperature is increased the mutual solubilities
increase and above the critical solution temperature of the binary pair B-C, the system becomes
identical to type I system.
SUMMARY
The phase equilibrium thermodynamics is of fundamental importance in chemical engineering,
because, majority of chemical process industries employ transfer of mass between phases either
during the preparation of the raw materials or during the purification of the finished products. The
major thrust of the present chapter was the development of the relationship between the various
properties of the system such as pressure, temperature and composition when a state of equilibrium
was attained between the different phases constituting the system. For a system to be in mechanical
equilibrium, the pressure and temperature should be uniform throughout the system. Since, the
uniformity of temperature and pressure do not eliminate the possibility of transfer of mass between
the phases, to describe the state of thermodynamic equilibrium, additional criteria are developed
(Section 8.1). They are:
dSU, V ≥ 0, dAT, V 0, dGT, P 0
Since, most chemical reactions and physical changes are carried out at constant T and P, the last
criterion formed the basis for phase equilibrium calculations. This criterion of equilibrium also led to
the criterion of stability as given by Eq. (8.12). The criterion of stability requires that at constant
temperature and pressure the free energy change on mixing DG, its first and second derivatives are all
continuous functions of the concentration x, and the second derivative should be positive.
For single-component systems in thermodynamic equilibrium under a given temperature and pressure,
the molar free energy should be the same in each phase (Section 8.3). Its logical extension to
multicomponent multi-phase systems reveals that if a system consisting of several components
distributed between various phases is in thermodynamic equilibrium at a definite temperature and
pressure, the chemical potential of each component will be the same in all the phases. Since absolute
values of fugacities are known, it was found convenient to use fugacities in phase equilibrium
calculations, rather than the chemical potentials. Accordingly, the general criterion of phase
equilibrium was expressed as the equality of fugacities [Eq. 8.36]. The Gibbs Phase rule follows
from the criterion of equilibrium (Section 8.5). The phase rule allows us to determine the number of
independent variables that must be arbitrarily fixed so as to establish uniquely the intensive state of
the system. The Duhem’s theorem helps in establishing the extensive state of the system (Section 8.6).
Vapour–liquid equilibrium problems essentially involve the calculation of the composition of the
liquid and vapour phases such that the fugacities of the components are the same in both phases. To
evaluate quantitatively the equilibrium compositions, the fugacity need be expressed in terms of the
mole fractions in the mixture. The fundamental relationship for a general VLE problem was derived
[Eq. (8.45)] and the various possible simplifications were described (Sections 8.7–8.12). For
evaluating the liquid phase fugacity, the activity coefficients should be known as a function of the
composition. Several equations were used for estimating the activity coefficients as function of
composition of the liquid. The Wohl’s equations, the Margules equations and the van Laar equations,
the local composition models for activity coefficients such as the Wilson equations, the NRTL
equations and the UNIQUAC equations, and the UNIFAC group contribution model are some of the
widely used activity coefficient equations.
Thermodynamics provides tests for consistency of experimental VLE data (Section 8.13). Almost all
these tests are based on the Gibbs–Duhem equations, the Redlich–Kister method
[Eq. (8.97)] being the most reliable among them. The discussion on the vapour–liquid equilibrium for
systems of limited miscibility (Section 8.15) and the liquid–liquid equilibrium (Section 8.16) would
be helpful for the analysis of many important separation processes in chemical engineering.
REVIEW QUESTIONS
1. How would you state the criterion of equilibrium in terms of the entropy, the work function and
the Gibbs free energy?
2. What do you know about the free energy change of mixing and its partial derivatives, for stable
liquid phases?
3. Show that for equilibrium between phases of a pure substance, the fugacities in both phases
should be equal.
4. How do you obtain the Clapeyron equation from the criterion of phase equilibrium? What
simplifications are used in the derivation of the Clausius–Clapeyron equation?
5. For a heterogeneous multicomponent system, what is the general criterion of phase equilibrium?
6. What do you understand by the number of degrees of freedom? How is it determined using the
phase rule for a non-reacting system?
7. State the Duhem’s theorem. What is its significance in establishing the state of the system?
8. What are the available degrees of freedom in the following non-reactive equilibrium systems?
(a) Two partially miscible liquid phases, each containing the same three liquid phases.
(b) A vapour phase containing ammonia in air and a liquid phase containing ammonia in water
at a specified temperature.
(c) A mixture of benzene and toluene undergoing a simple distillation operation.
9. Write down the equation for solving a general VLE problem. How does this equation get
simplified for (a) ideal gas phase, ideal liquid phase and (b) low-pressure equilibrium?
10. What is Poynting correction?
11. Distinguish between the bubble-point and dew-point temperatures.
12. What is meant by a ‘tie line’? How does the tie line help in determining the amount of liquid
and vapour in equilibrium?
13. Why does the boiling point diagram at a higher pressure lie above that at a lower pressure?
14. What are the salient features of an ideal liquid solution? How does the total pressure over an
ideal solution vary with composition?
15. How would you calculate the constant pressure y-x data of a binary mixture using an average
value of the relative volatility?
16. Component 1 in a binary non-ideal solution is found to obey the Raoult’s law over a certain
concentration range. What do you know about the behaviour of component 2 over the same
range?
17. What do you mean by positive and negative deviation from ideality? “A solution exhibiting
positive deviation from ideality is formed accompanied by an absorption of heat and a solution
exhibiting negative deviation from ideal behaviour is formed accompanied by an evolution of
heat”. Explain.
18. What are azeotropes? With proper phase diagrams, distinguish between minimum and maximum
boiling azeotropes. What is the effect of pressure on the azeotropic composition?
19. Discuss the suitability of different activity coefficient equations for VLE data correlation.
20. What is vaporisation equilibrium constant? How do you estimate the bubble-point temperature
and the bubble-point pressure of a multicomponent system?
21. A multicomponent liquid mixture of known composition is flash vaporised at a given pressure
and temperature. How would you estimate the fraction of the liquid vaporised?
22. How are the Gibbs–Duhem equations helpful in testing the consistency of the VLE data?
23. What is the zero area method for testing the consistency of VLE data?
24. What is coexistence equation? What are its major applications?
25. The activity coefficients of one of the components in a binary solution are known as function of
concentration. How would you evaluate the activity coefficients of the other component as a
function of composition?
26. What are the critical solution temperature and the three-phase temperature with reference to
partially miscible liquid systems?
27. Why does immiscibility occur in liquid solutions?
28. How would you estimate the composition of the vapour phase in equilibrium with two
immiscible liquid phases?
EXERCISES
8.1 Show that the following equations provide the criteria of equilibrium under certain constraints.
(a) dUS,V = 0 (b) dSH,P = 0 (c) dHS,P = 0
8.2 For each of the following non-reactive equilibrium systems, determine the number of available
degrees of freedom.
(a) Two miscible materials in vapour–liquid equilibrium with vapour composition specified at
a given temperature and pressure.
(b) A mixture of methane and air in contact with a solid adsorbent at atmospheric pressure and
a specified temperature.
(c) Liquid water in equilibrium with a mixture of water vapour and nitrogen.
(d) Two partially miscible liquid phases and a vapour phase in equilibrium with them at a
constant pressure.
(e) A liquid mixture of benzene and toluene in equilibrium with its vapour at 1 bar.
(f) A vapour phase consisting of ammonia and air and a liquid phase consisting of ammonia and
water at a given temperature.
(g) A liquid mixture of components A and C in equilibrium with a liquid solvent B in which
only C is soluble at a given temperature and pressure.
8.3 Determine the mole fraction of methane, xi, dissolved in a light oil at 200 K and 20 bar.
Henry’s law is valid for the liquid phase, and the gas phase may be assumed to be an ideal
solution. At these conditions, Henry’s law constant for methane in oil = 200 bar, fugacity
coefficient of pure methane gas = 0.90 and mole fraction of methane in the gas phase, y1 = 0.95.
8.4 The vapour pressures of benzene and toluene are given below.
Calculate the equilibrium data for the system at 101.3 kPa and formulate an equation for the
equilibrium diagram in terms of average relative volatility.
8.5 At 303 K, the vapour pressures of benzene (A) and toluene (B) are 15.75 kPa and 4.89 kPa
respectively. Determine the partial pressures and weight composition of the vapour in
equilibrium with a liquid mixture consisting of equal weights of the two components.
8.6 An equimolar mixture of benzene and toluene is contained in a piston/cylinder arrangement at a
temperature T. What is the maximum pressure below which the mixture exists as a vapour phase
alone? At the given T, the vapour pressures of benzene and toluene are 203.9 kPa and 85.3 kPa,
respectively. Assume that Raoult’s law is valid.
8.7 Two substances A and B are known to form ideal liquid solutions. A vapour mixture containing
50% (mol) A and 50% (mol) B is at 311 K and 101.3 kPa. This mixture is compressed
isothermally until condensation occurs. At what pressure does condensation occur and what is
the composition of the liquid that forms? The vapour pressures of A and B are 142 kPa and 122
kPa respectively.
8.8 Air is cooled to 80 K at 101.3 kPa. Calculate the composition of the liquid and vapour phases
at this condition assuming that the mixture behaves ideally. The vapour pressure of nitrogen and
oxygen at 80 K are 135.74 kPa and 30.04 kPa respectively.
8.9 The binary system, acetone (1)–acetonitrile (2) conforms closely to Raoult’s law. Using the
vapour pressure data given below plot the following
(a) P-x1 and P-y1 curves at 323 K
(b) T-x1 and T-y1 curves at 53.32 kPa
8.10 Assuming Raoult’s law to be valid for the system benzene (1)–ethyl benzene (2) and the
vapour pressures are given by the Antoine equations
The normal boiling points of acetone and chloroform are respectively 329.5 K and 334.1 K.
(Hint: The ratio of vapour pressures remains almost constant. Use the method employed in
Example 8.13 for calculating y for arbitrarily chosen x values.)
8.16 Show that the van Laar equation and Margules equation are consistent with the Gibbs–Duhem
equations.
8.17 The toluene–acetic acid mixture forms an azeotrope containing 62.7% (mol) toluene and
having a minimum boiling point of 378.6 K at 101.3 kPa. The vapour pressure data are:
The normal boiling point of toluene and acetic acid are respectively 383.9 K and 391.7 K.
(a) Calculate the van Laar constants A and B
(b) Plot ln g1 and ln g2 as ordinate against mole fraction of toluene.
8.18 Under atmospheric pressure, the acetone–chloroform azeotrope boils at 337.8 K and contains
33.5% (mol) acetone. The vapour pressures of acetone and chloroform at 337.8 K are
respectively 132.62 kPa and 113.96 kPa.
(a) Calculate the composition of the vapour in equilibrium with a liquid analysing 11.1%
(mol) acetone. How does it compare with the experimental value of 6.5% acetone in the
vapour?
(b) What is the total pressure at this condition?
8.19 Ethyl alcohol and hexane form an azeotrope at 33.2% (mol) ethanol. It boils at 331.9 K at
101.3 kPa. At 331.9 K, the vapour pressures are 44.25 kPa for ethanol and 72.24 kPa for
hexane. Determine:
(a) The van Laar constants
(b) The vapour composition for a solution containing 50% (mol) hexane boiling at 331.9 K
(c) The total pressure for the conditions in part (b).
8.20 At atmospheric pressure, ethyl acetate and ethyl alcohol form an azeotrope containing 53.9%
(mol) acetate boiling at 345 K. Determine:
(a) The van Laar constants
(b) The azeotropic composition and the total pressure if the mixture forms an azeotrope boiling
at 329.5 K
(c) The composition of the vapour in equilibrium with a liquid of composition 60% (mol)
alcohol and boiling at 329.5 K
8.21 An organic liquid solution containing A (molecular weight 46) and B (molecular weight 78)
form an azeotrope containing 52% by weight A at 333 K and 101.3 kPa. Vapour pressures of A
and B are 69.31 kPa and 68 kPa respectively. Determine the van Laar constants.
8.22 For the acetone (1)–diethylamine (2) system the activity coefficients values as function of
concentration are given below:
Using the above data estimate the van Laar constants for the system.
(Hint: A = ln g1 as x1 0 and B = ln g2 as x2 0)
8.23 Find the van Laar constants for the binary system benzene (1)–ethanol (2) using the following
data
8.24 The T-x-y data for the system metaxylene (1)–propionic acid (2) at 101.3 kPa is given below:
Calculate the total pressure and vapour composition in equilibrium with a liquid containing 31%
(mol) acetone at 333 K.
8.31 For the 2-propanol (1)–water (2) system, the following Wilson parameters are reported.
a12 = 1833.74 J/mol,……a21 = 5183.26 J/mol
V1 = 76.92 10–6 m3/mol,……V2 = 18.07 10–6m3/mol
The vapour pressures can be calculated by the Antoine equations, which are given below:
8.40 Calculate the constants A and B in the van Laar equation from the following data. Check
whether the data are consistent.
8.41 The following data were reported for vapour–liquid equilibrium for ethanol–water system at
298 K. Test whether the data are thermodynamically consistent.
8.42 The following vapour-liquid equilibrium data were obtained for water (1)–nitric acid (2)
system at 293 K.
8.44 The partial pressure of ether at 303 K for the ether (1)–acetone (2) system is given as follows:
Use the area test to determine the thermodynamic consistency of the data.
8.46 The activity coefficient of thallium in amalgams at 293 K are given below:
What is the activity coefficient of water in a 10% (mol) n-propyl alcohol solution?
8.48 At 323 K, the vapour pressures of pure ether and pure ethyl alcohol are 170.13 and 29.47 kPa
respectively. The total pressures versus liquid composition data are given below:
Using Gibbs–Duhem equation compute from these data the partial pressures of ether and alcohol
over liquid solutions of various compositions at 323 K.
8.49 The data given below refer to the boiling points of ethanol (1)–benzene (2) system at
100 kPa and the vapour pressures of pure ethanol and benzene at these temperatures.
Calculate the van Laar constants from these data assuming g to be independent of temperature.
Also, find g1 and g2 from the van Laar equations.
8.50 The total pressure versus solution concentration data for the system dioxane (1)–water (2) at
353 K is given below:
The vapour pressures of pure water and dioxane at this temperature are 47.33 and 51.05 kPa.
Calculate:
(a) The van Laar constants
(b) The constants in the Margules equation
(c) The vapour composition in equilibrium with a liquid containing 60% water by weight and
the total pressure over this solution using van Laar method.
8.51 Benzene (1)–cyclohexane (2) form an azeotrope at 0.525 mole fraction benzene at a
temperature of 350.8 K and 101.3 kPa. At this temperature, the vapour pressure of benzene is
99.3 kPa and that of cyclohexane is 98 kPa. Using the van Laar model estimate the activity
coefficients at x1 = 0.2 and 0.9. Using this activity coefficient information calculate the
equilibrium pressure and the vapour compositions at 350.8 K for the two liquid compositions.
8.52 The azeotrope of the n-propanol–water system has a composition 56.83% (mol) water with a
boiling point of 360.9 K at a pressure of 101.3 kPa. At this temperature, the vapour pressures of
water and propanol are respectively 64.25 kPa and 69.71 kPa. Evaluate the activity coefficients
for a solution containing 20% water through the van Laar equations.
8.53 The pressure exerted over the binary system ethanol–methylcyclohexane containing 40.5%
(mol) ethanol at 308 K is 20.31 kPa. The vapour phase contained 54.7% (mol) ethanol. The
vapour pressures at 308 K are 13.74 kPa for ethanol and 9.81 kPa for methylcyclohexane. What
are the total pressure and composition of the vapour in equilibrium with a liquid containing 60%
(mol) ethanol at 308 K?
8.54 A binary liquid mixture of components A and B containing 80% (mol) A is in equilibrium with
a vapour containing 84.3% (mol) A at 101.3 kPa and 339 K. Estimate the pressure and
composition of the vapour in equilibrium with a liquid containing 50% A at 339 K. The vapour
pressures of A and B at this temperature are 106.6 kPa and 79.97 kPa respectively.
8.55 At 333 K, compounds A and B each has vapour pressures of 106.63 kPa. The mixture of A and
B forms an azeotrope at 333 K and 133.29 kPa and has a composition of 50% A.
(a) Calculate the equilibrium pressure and vapour composition over a liquid solution
containing 25% A.
(b) If A and B have equal latent heats of vaporisation, how do you expect the azeotropic
composition to respond to an increase in temperature?
8.56 At 353 K, compounds A and B each has vapour pressures of 93.30 kPa. At this temperature
mixtures of A and B form azeotrope containing 50% (mol) A and exerts a pressure of 127.96 kPa.
Calculate the equilibrium pressure and vapour composition at 353 K over a liquid solution
containing 25% (mol) A.
8.57 For the binary mixture of A and B the activity coefficients are given by
The vapour pressures of A and B at 353 K are 119.96 kPa and 79.97 kPa respectively. Does an
azeotrope exist at 353 K? If so, what is the azeotropic pressure and composition for A = 0.6?
8.58 It is proposed to purify benzene from small amounts of non-volatile impurities by subjecting it
to distillation with saturated steam at 99.3 kPa. Calculate the temperature at which distillation
will proceed and the weight of steam accompanying 1 kg benzene. The vapour pressure data is
given in Example 8.28.
8.59 At 383 K, saturated solution of aniline in water contains 7.95% aniline by weight and a
saturated solution of water in aniline contains 88.05% aniline by weight. The vapour pressures
of pure aniline and of water at 383 K are 9.22 kPa and 143.10 kPa respectively. Construct the P-
x-y diagram for the mixture at 383 K.
8.60 Construct T-x-y diagram for the ether (1)–water (2) system at 101.3 kPa from the following
data.
Assume that Raoult’s law is valid for ether in ether phase and for water in the water phase.
8.61 Dimethylaniline is distilled with steam at 90 kPa to free it from non-volatile impurities.
Assuming it to be completely immiscible with water determine
(a) The distillation temperature
(b) The composition of the vapour produced.
The vapour pressure data are following:
Plot of ln PS versus 1/T may be assumed linear.
8.62 A stream contains 30% (mol) toluene, 40% (mol) ethyl benzene and 30% (mol) water.
Assuming that mixtures of ethylbenzene and toluene obey Raoult’s law and they are completely
immiscible in water, calculate the following for a total pressure of 101.3 kPa:
(a) The bubble-point temperature and the composition of the vapour
(b) The dew-point temperature and the composition of the liquid.
The vapour pressure data are given below:
8.63 n-Heptane (1) and water (2) are essentially immiscible as liquids. A vapour mixture
containing 65% (mol) water at 373 K and 101.3 kPa is cooled slowly at constant pressure until
condensation is complete. Construct a plot for the process showing temperature versus
equilibrium mole fraction of heptane in the residual vapour. For n-heptane,
8.66 An experimental determination of vapour–liquid equilibrium state of ether (1) and acetone (2)
gave the following results:
x1 = 0.3, y1 = 0.42, T = 313 K and P = 105 Pa
The saturation vapour pressures of the pure components at 313 K are: ether = 1.21 105 Pa
and acetone = 0.56 105 Pa. The vapour phase can be assumed ideal.
(a) Calculate the liquid-phase activity coefficients.
(b) What is the value of excess Gibbs free energy GE/RT for the liquid phase?
9
Chemical Reaction Equilibria
The chemical process industries are concerned with the transformation of raw materials into useful
products. Such transformation in most cases is achieved by means of chemical reactions. The design
and operation of reaction equipment are therefore quite an important field in the chemical engineering
profession. To be successful in this profession, the chemical engineer should be versatile with the
thermodynamics and kinetics of chemical reactions. Thermodynamics predicts the equilibrium
conversion that would be achieved in a chemical reaction and also the effect of operating conditions
on it, whereas the kinetics deals with the rate or speed with which the desired conversion is attained
in practice. Thermodynamic analysis can also give information about the feasibility of chemical
reactions.
The progress and extent of a chemical reaction are affected by changes in the reaction conditions like
temperature, pressure, composition of the reactants, etc. For example, in the synthesis of methanol
from carbon monoxide and hydrogen, the equilibrium conversion as well as the rate of reaction are
affected by changing the pressure, temperature or the relative amounts of carbon monoxide and
hydrogen in the reactant stream. The influence of these controllable variables on the thermodynamics
of reaction, or to be specific, on the equilibrium conversion, in some situation may be in conflict with
the influence of these variables on the kinetics of the reaction. This can be illustrated by considering
the effect of temperature on the oxidation of sulphur dioxide to sulphur trioxide. The rate of this
reaction increases with temperature and from the point of view of rate alone it is better to operate the
reactor at as high a temperature as permissible. However, the equilibrium conversion to sulphur
trioxide falls off sharply with increase in temperature. The conversion is above 90% at temperatures
near 800 K, but it is only 50% at 950 K. It is clear that both the kinetics (the rate) and
thermodynamics (the equilibrium) of the reaction must be considered in the choice of reaction
conditions in the commercial process for any chemical reaction. The purpose of the present chapter is
to identify the role of thermodynamics in the design and operation of chemical reaction systems.
Equilibrium conversion of a reaction sets a limit and provides a goal by which we measure
improvement in the process. It is impossible at a given set of conditions to attain a conversion that is
better than the equilibrium value calculated from thermodynamic principles. Even if this conversion
is not attainable in practice within a reasonable time, its knowledge is valuable because it represents
the best that can be expected from the reaction. It tells us whether or not an experimental investigation
of a proposed new process is worthwhile. There is no point in trying improvement in the process by
improving the rate by introducing suitable catalysts, if thermodynamics predicts an equilibrium yield,
of say, only 20% whereas a 50% yield is necessary for the process to be economically viable. The
choice of an appropriate catalyst may give a better reaction rate, but it will not alter the equilibrium
yield of the product. The emphasis in this chapter is on determining the conversion at equilibrium and
on predicting the effect of controllable variables like temperature and pressure on the conversion.
where A is the chemical symbol for the various species taking part in the reaction and n is the
stoichiometric number. Consider the reaction
2A + 3B L + 2M
This is a special case of the general form of Eq. (9.1), with nL = 1, nM = 2, nA = – 2, and nB = – 3.
In the general form, this reaction may be represented as
0 = L + 2M – 2A – 3B
The stoichiometric numbers are positive for products, negative for reactants and zero for inert
species. The changes in the number of moles of various species taking part in the reaction are in
direct proportion to their stoichiometric numbers. Let Dni denote the change in the number
of moles of component i due to the reaction. For one mole of A disappearing in the reaction
DnA = –1, DnB = –1.5, DnL = 0.5 and DnM = 1. We see that
For the thermodynamic analysis of chemical reactions the concept of ‘extent of reaction,’ also called
‘reaction coordinate’ is useful. It is denoted by e. The reaction coordinate measures the progress of a
reaction and is defined as the degree to which a reaction has advanced. It has the advantage that the
change in the extent of reaction de is the same for each component, whereas the changes in the number
of moles are different for different species taking part in the reaction. The extent of reaction and the
number of moles taking part in the reaction are related as
For the initial state of the system, that is, before the reaction, the value of e is zero.
EXAMPLE 9.1 Derive the relationship between the mole fraction of the components taking part in the
reaction and the extent of the reaction.
Solution Let ni0 be the number of moles of the species initially present in the system and
ni the number of moles present after the reaction. Then ni = ni0 + Dni where Dni is the change in the
number of moles of i due to the reaction. Integration of Eq. (9.2) yields
EXAMPLE 9.2 A gas mixture containing 2 moles nitrogen, 7 moles hydrogen and 1 mole ammonia
initially, is undergoing the following reaction:
N2 + 3H2 2NH3
(a) Derive expressions for the mole fractions of various components in the reaction mixture in
terms of the extent of reaction.
(b) Explain how the conversion of limiting reactant is related to the extent of reaction.
Solution (a) Equations (9.3) and (9.4) relate the mole fraction of various constituents in the system to
the extent of reaction.
(b) The limiting reactant here is nitrogen. Let the fractional conversion of nitrogen be z. Then
Moles of nitrogen in the reaction mixture is =
Moles of nitrogen in the mixture in terms of the extent of reaction is =
Comparing the two results, we see that
EXAMPLE 9.3 Derive the relationship between mole fraction of species in multiple reactions and
the extent of reactions.
Solution When two or more reactions occur simultaneously, the number of moles of each component
changes because of several reactions. Equation (9.2) can be modified as
Let , the sum of the stoichiometric numbers in the jth reaction. Then the above equation can
be written as
EXAMPLE 9.4 A gas mixture containing 3 mol CO2, 5 mol H2 and 1 mol water is undergoing the
following reactions:
CO2 + 3H2 CH3OH + H2O
CO2 + H2 CO + H2O
Develop expressions for the mole fraction of the species in terms of the extent of reaction.
Solution The total moles initially present,
n0 = 3 + 5 + 1 = 9
For the first reaction,
n1 = – 1 – 3 + 1 + 1 = – 2
For the second reaction,
n2 = – 1 – 1 + 1 + 1 = 0
The mole fractions are calculated using Eq. (9.9)
where mi is the chemical potential of component i. For the reaction under consideration, Eq. (9.12)
takes the form
where – a, – b, l and m are the stoichiometric numbers which are positive for products and negative
for the reactants and e is the extent of reaction. In general, for an infinitesimal change in a reacting
system, we can write Eq. (9.13) as
The left-hand side of Eq. (9.16) is the free energy change DG accompanying the complete reaction
under equilibrium conditions. Hence, DG = 0 under equilibrium.
The physical significance of the criterion of chemical equilibrium can now be examined. Consider a
simple chemical reaction equilibrium: A D B. Let the extent of the reaction be e. The change in the
number of moles of A = – de and the change in the number of moles of B = de. The change in free
energy at constant temperature and pressure is found out by Eq. (9.14)
dGt = (mB – mA) de………(9.17)
This equation can be written in the following form.
Equation (9.18) gives the slope of the curve obtained when the Gibbs free energy is plotted against
extent of reaction as in Fig. 9.1.
The slopes given by Eq. (9.18) are not constant because the chemical potentials are functions of
composition, which varies as the extent of reaction changes. Since the reaction proceeds in the
direction of decreasing Gibbs free energy G, the forward reaction (A B) takes place if mA > mB
and the backward reaction (A B) proceeds if mA < mB. When mA = mB, the slope of the curve
is zero. This occurs at the minimum of the curve and corresponds to the position of chemical
equilibrium. The composition of the reaction mixture at the point where the Gibbs free energy is the
minimum is the equilibrium composition at the specified temperature and pressure. Thus the criterion
of equilibrium, Eq. (9.10), means that differential displacement of chemical reaction can occur at the
equilibrium state, but without changing the total Gibbs free energy. If the system is not in chemical
equilibrium, the reaction occurring must be irreversible and the total Gibbs free energy must decrease
at constant temperature and pressure.
where ai is the activity of component i in the reaction mixture and ni is the stoichiometric number of
i. Activities of the species appearing in Eq. (9.19) are raised to the respective stoichiometric
numbers. Since the activity is defined as the ratio of the fugacity of the component in the solution to
the fugacity in the standard state,
Thus the equilibrium constant is determined by the standard free energy change and the temperature.
The standard free energy change depends on the temperature, the specification of standard state for
each component and the number of moles involved in the stoichiometric equation under consideration.
The numerical values of the equilibrium constant will be of no significance unless accompanied by
the specifications for these three factors. However, it is independent of pressure at equilibrium. The
effect of the reaction stoichiometry on the equilibrium constant has already been discussed. The
choice of standard state is being dealt with in the following section.
9.4.1 Choice of Standard State
Though the choice of standard state in Eq. (9.31) is arbitrary and is left to our convenience, certain
conventions are followed in this choice. The choice of pure component standard state will be
convenient in many situations, as this requires only the specification of temperature and pressure for
defining the state completely. The temperature in the standard state is the same as that of the reaction.
If the standard state chosen for a substance is a solution, the composition must also be specified.
For gases, as has been pointed out earlier, the standard state chosen is the pure component at the
temperature of the reaction and at unit fugacity. Fugacity will be unity at 1 bar (or 1 atm) if the gas
behaves as an ideal gas at this condition. For ideal gases, therefore, the standard state pressure
approaches 1 bar and DG0 can be easily evaluated at this pressure. By this choice, K = Kf and Eq.
(9.31) becomes
DG0 = – RT ln Kf………(9.32)
The standard state of unit fugacity may not be convenient for reactions involving solids, liquids or
solutions. By convention, the standard state chosen for solids and liquids is the pure solid or liquid as
the case may be, at a pressure of 1 bar (or 1 atm), the temperature being the same as the temperature
of the reaction.
9.4.2 Feasibility of a Reaction
From the values of standard free energy change, we can formulate an approximate criterion for the
feasibility of a chemical reaction, which will be useful in preliminary exploratory work. It would be
worthwhile to have some idea about whether or not the equilibrium is favourable, before we search
for catalysts and other conditions necessary to cause the reaction. If the reaction is not
thermodynamically feasible, there is no point in pursuing a long and expensive experimental
investigation on improving the rate of reaction.
Any reaction starting with pure reactants uncontaminated with any of the products will have a
tendency to proceed to some extent, though this may be infinitesimally small. It is the value of the
equilibrium constant, which, in turn, is related to the standard free energy of the reaction that gives the
necessary information on the thermodynamic possibility of the reaction. Even the decomposition of
water vapour to hydrogen and oxygen will proceed to some extent under atmospheric temperature and
pressure. From the value of the standard free energy change, DG0, the equilibrium constant for the
reaction
at 298 K is found to be about 1 10–40. This means that the extent of decomposition of water
vapour is infinitesimally small at equilibrium and the reaction is not thermodynamically feasible.
I f DG0 for a reaction is zero, then K = 1, the reaction proceeds to a considerable extent before
equilibrium is reached. If DG0 is negative, then K > 1, the reaction is quite favourable. But the
situation becomes less favourable as DG0 increases in the positive direction. It should be borne in
mind that many reactions with positive values of DG0 are certainly feasible from the standpoint of
industrial operation. For example, the methanol synthesis reaction with DG0 = 46,200 kJ/kmol at 600
K is found to be feasible. This reaction is carried out at high pressure to overcome the unfavourable
free energy change. In short, there is no well-defined demarcation to separate favourable and
unfavourable reactions. The following guide may be useful as an approximate criterion for
ascertaining the feasibility of chemical reactions:
DG0 < 0, the reaction is promising.
0 < DG0 < 40,000 kJ/kmol, the reaction may or may not be possible and needs further study.
DG0 > 40,000 kJ/kmol, the reaction is very unfavourable.
EXAMPLE 9.5 Device a series of hypothetical steps for carrying out the gas-phase reaction
aA + bB lL + mM
when the reactants and the products are at their standard state. Show that the free energy changes
calculated for these series of steps add up to give the same result as the one provided by Eq. (9.31).
Solution The free energy change accompanying the process in which the reactants at their standard
state are converted to products also at their standard state may be calculated via any convenient path.
Let us assume the following computational path for carrying out the reaction, which is represented in
Fig. 9.2.
Step 1: The reactants are initially in their pure form and are at their standard state of unit fugacity and
at the temperature of the reaction. Then they are compressed to a fugacity of the reaction mixture at
equilibrium. The free energy change for this process is DG1.
Step 2: The pure reactants are introduced to the reaction system through membranes permeable only
to single species. Since the fugacities of the components before and after this step are the same, the
free energy change DG2 for this process is zero. DG2 = 0.
Step 3: The introduction of the reactants disturbs the state of equilibrium prevailing in the reaction
system. To bring the system back to the equilibrium condition the forward reaction occurs at the given
temperature and pressure. According to the criterion of equilibrium, this reaction proceeds without
any change in the free energy of the system. Therefore, DG3 = 0.
Step 4: The product gases are separated by means of membranes into pure components at the reaction
temperature and pressure. As in step 2, the free energy change in this process is zero. That is, DG4 =
0.
Step 5: The pure components with fugacities equal to are expanded to standard state fugacities
. The free energy change for this step,
which is same as Eq. (9.31).
Equation (9.36), known as van’t Hoff equation, predicts the effect of temperature on the equilibrium
constant and hence on the equilibrium yield. DH0 in Eq. (9.36) is the standard heat of reaction. It is
apparent that if DH0 is negative, i.e. if the reaction is exothermic, the equilibrium constant decreases
as the reaction temperature increases. Alternatively, for an endothermic reaction, the equilibrium
constant will increase with increase in temperature.
If DH0, the standard heat of reaction, is constant, Eq. (9.36) on integration yields
K and K1 are the equilibrium constant values at temperatures T and T1 respectively. Equation (9.37)
may be used to evaluate the equilibrium constant with good results over small temperature ranges.
The equation is exact if DH0 is independent of temperature. A reasonably accurate method of
interpolation or extrapolation of equilibrium constant is provided by plotting ln K versus reciprocal
of temperature, which leads to a straight line according to Eq. (9.37).
The variation of the standard heat of reaction with temperature may be taken into account if the molal
heat capacities of the various species taking part in the reaction are known as functions of
temperature. Suppose that the specific heats at constant pressure are expressed as a power function in
T.
CP = a + bT + gT2………(9.38)
Then the effect of temperature on the standard heat of reaction may be developed as follows: Since
heat of reaction is the enthalpy change between the given initial and final states, it may be evaluated
by devising any convenient path between these terminal states for which the enthalpy changes are
readily available. Assume that the standard heat at temperature T1, , is known and it is desired to
calculate the standard heat at temperature T.
The actual reaction occurring at temperature T for which the heat of reaction is may be treated
as occurring along the three paths as depicted in Fig. 9.3.
1. The reactants are cooled from temperature T to T1. The enthalpy change for this step is
3. The temperature of the products is raised from T1 to T in this step. The enthalpy change is
The standard heat of reaction at temperature T, is obtained by adding the preceding three equations.
The constant DH in the above equation can be evaluated if the heat of reaction at a single
temperature is known. Equation (9.46) can then be used for the evaluation of the heat of reaction at
any temperature T.
Substitute Eq. (9.46) into Eq. (9.36) and integrate the resulting expression. The result is
A in Eq. (9.47) is a constant of integration, which may be evaluated from the knowledge of the
equilibrium constant at one temperature. Equation (9.31) relates the equilibrium constant to the
standard free energy change. Using this relationship, we get
9.5.1 Evaluation of Equilibrium Constants
Equation (9.47) can be used for the evaluation of the equilibrium constant, provided, we know the
dependence of heat capacities on temperature and we also have enough information for the evaluation
of the constants DH and A. Assuming that the heat capacity data are available, the general methods
used for the evaluation of the constants DH and A are listed below.
M ethod 1. K may be calculated from the experimentally measured composition of the equilibrium
mixture using Eq. (9.19). If K values are thus known at two different temperatures, they may be
substituted into Eq. (9.47). The resulting two equations are solved for the constants DH and A.
M ethod 2. Standard heat of reaction at one temperature and one value for the equilibrium constant
that is determined by direct experimental measurements are available. The former is used in Eq.
(9.46) for the evaluation of the constant DH and the latter in Eq. (9.47) for evaluating the constant
A.
M ethod 3. This method involves no direct experimental measurements for the equilibrium constant
and therefore this is the most convenient and most widely used method. The method makes use of
thermal data only, usually in the form of standard heat of reaction DH0, and a standard free energy
change of reaction DG0. Then the constants DH and A are evaluated using Eq. (9.46) and
Eq. (9.48) respectively.
DH0 for a reaction may be evaluated from the standard heat of formation, , that are tabulated for
most of the compounds. The standard free energy of a reaction can be estimated from the values of
standard free energy of formation, of the various species participating in the reaction and their
respective stoichiometric numbers as
That is, the standard free energy of a reaction is the algebraic sum of the free energies of formation of
the products minus the algebraic sum of the free energies of formation of the reactants. When an
element enters into a reaction, its standard free energy of formation may be taken to be zero.
9.5.2 Giauque Functions
Data for calculation of standard free energy of reactions are sometimes tabulated as Giauque
functions. These are Gibbs free energy functions that vary very slowly with temperature. Two such
functions are in general use—the first is referred to 0 K and the second referred to 298 K. These are
written as
where are respectively the free energy in the standard state at temperature T, the enthalpy
in the standard state at T and the enthalpy in the standard state at 298 K. Because only standard state
properties are involved, these functions depend only on temperature. This temperature dependence is
found to be very weak which makes these functions suitable for tabular interpolation. Using the
definition of free energy, we can show that
The difference in enthalpy values, the terms in brackets in Eq. (9.53), needed for applying
Eq. (9.53) also are listed in tables along with 0. The standard free energy change of a reaction may
be calculated from the Gibbs free energy functions. Equation (9.51) can be rearranged as
The Gibbs free energy at the standard state for each of the species taking part in the reaction as given
by Eq. (9.54) or Eq. (9.55) multiplied by the respective stoichiometric numbers add together to give
the standard free energy of the reaction.
Note that the enthalpy of a compound in the standard state, , is the same as its standard enthalpy of
formation, . The standard free energy of a reaction determined using Eqs. (9.56) or (9.57) may
be used in Eq. (9.31) to calculate the equilibrium constant.
EXAMPLE 9.6 Calculate the equilibrium constant at 298 K of the reaction
N2O4 (g) 2NO2 (g)
given that the standard free energies of formation at 298 K are 97,540 J/mol for N2O4 and
51,310 J/mol for NO2.
Solution Using Eq. (9.50) for the dissociation of N2O4,
Therefore, K = 0.1287.
EXAMPLE 9.7 The standard heat of formation and standard free energy of formation of ammonia at
298 K are –46,100 J/mol and –16,500 J/mol respectively. Calculate the equilibrium constant
for the reaction
N2 (g) + 3H2 (g) 2NH3 (g)
at 500 K assuming that the standard heat of reaction is constant in the temperature range 298 to 500 K.
Solution The standard free energy of reaction is estimated from Eq. (9.50).
The standard heat of reaction at 298 K = 2 – 46,100 = – 92,200 J/mol. This is assumed constant
within the temperature range involved. Now use Eq. (9.37) to evaluate the equilibrium constant.
Assuming that activities are equal to the mole fractions, calculate the standard free energy of the
reaction at 317 K and 391 K and average value of heat of reaction over this temperature range.
Solution Since activities are equal to mole fractions, K = yib/ynb, where yib is the mole fraction of i-
butane and ynb the mole fraction of n-butane in the equilibrium mixture. Therefore,
Assuming that the heat of reaction is independent of temperature we can use Eq. (9.37) for calculating
it.
EXAMPLE 9.9 Estimate the standard free energy change and equilibrium constant at 700 K for the
reaction
N2 (g) + 3H2 (g) 2NH3 (g)
given that the standard heat of formation and standard free energy of formation of ammonia at
298 K to be – 46,100 J/mol and –16,500 J/mol respectively. The specific heat (J/mol K) data are
given below as function of temperature (K):
CP = 27.27 + 4.93 10–3T for N2
EXAMPLE 9.10 Evaluate the equilibrium constant at 600 K for the reaction
CO (g) + 2H2 (g) CH3OH (g)
given that the Gibbs free energy function
for CO, H2 and methanol at 600 K are respectively –203.81, –136.39 and –249.83 J/mol K. The
heats of formation at 298 K of CO (g) and CH3OH (g) at 298 K are –110,500 J/mol and –200,700
J/mol.
Solution The standard free energy of formation at 600 K is evaluated by means of Eq. (9.57).
EXAMPLE 9.11 Calculate the equilibrium constant for the reaction
N2 (g) + 3H2 (g) 2NH3 (g)
at 500 K, given that the free energy function
at 500 K for nitrogen, hydrogen and ammonia are respectively –177.5, –116.9 and –176.9 J/mol K.
The function for nitrogen, hydrogen and ammonia are respectively 8669, 8468 and
9920 J/mol. The free energy of formation of ammonia at 298 K is –46,100 J/mol.
Solution Equation (9.53) gives
The equilibrium constant defined above is independent of the pressure. By Eq. (9.31), the equilibrium
constant is known if the standard free energy of the reaction and the reaction temperature are known.
The standard free energy of a reaction is determined by the free energies of the substances in their
standard states. The standard states are defined by specifying a pressure and are in no way affected
by the reaction pressure. That is, the standard free energy of a reaction, and hence, the equilibrium
constant are not affected by changes in the equilibrium pressure.
9.6.2 Effect of Pressure on Equilibrium Composition
Though the equilibrium constant is unaffected by pressure, it does affect the equilibrium composition
in gas-phase reactions. This effect is explained qualitatively by Le Chatelier’s principle. Consider for
example, the equilibrium in the gas-phase reaction A 2B. When pressure is applied to this system,
it responds in such a way as to minimise the effect of the increase in pressure. This is achieved by
decreasing the number of moles in the system, which in turn is achieved by the reaction A 2B.
Thus, increase in pressure decreases the number of B molecules and increases the number of A
molecules. By the same reasoning we can deduce that in the case of the reaction equilibrium for N2 +
3H2 2NH3 formation of ammonia will be favoured by an increase in pressure as there is a
reduction in the number of moles due to this reaction. It should be remembered that when the
composition of the system changes in this manner in response to increase or decrease in pressure, it
does so without changing the equilibrium constant.
Except at very high pressures, properties of solids, liquids or solutions are not affected appreciably
by pressure. Therefore, the equilibrium concentrations in reactions involving solids, liquids or
solutions are not affected significantly by changes in pressure.
To predict the effect of pressure quantitatively, the relationship between equilibrium constant and
equilibrium composition must be established. Equation (9.19) defines the equilibrium constant as a
function of activities of the species in the reacting system. The activities of the components are
affected by changes in pressure, temperature and composition. As K is independent of pressure, and
activities are not, it requires that the activities of the components change with pressure in such a way
that the complex function of activities, which we have defined as equilibrium constant, remains
unaltered. The equilibrium constant written in terms of activities, K, and the equilibrium constant Kf,
which is written in terms of the fugacities of the components were shown to be equal for gaseous
systems employing ideal-gas standard state through Eq. (9.21).
Since, K is independent of pressure, the variation in the Pn term in the above equations must be
balanced by a corresponding change in the value for Ky. The change in Ky means the change in the
equilibrium compositions. If there is a decrease in the number of moles during the reaction as in the
case of ammonia synthesis reaction, n will be negative. An increase in pressure in this case will
decrease Pn and as a result, Ky and the equilibrium yield would increase. On the other hand, if the
reaction results in an increase in the number of moles, n will be positive and the equilibrium yield
would decrease with increase in pressure.
The above observations are in agreement with the Le Chatelier’s principle. In addition,
Eq. (9.62) can be used to explain the effect of pressure on reactions where n is zero, which cannot be
explained by Le Chatelier’s principle. One would expect pressure to have no effect on reaction such
as the water-gas shift reaction
CO (g) + H2O (g) CO2 (g) + H2 (g)
because there is no change in the number of moles during the reaction. The effect of pressure on the
equilibrium composition in this case can be explained by the effect of pressure on Kf. Kf measures the
deviation from ideal-gas behaviour, and its value may change with change in pressure. If Kf
decreases in any reaction, then Ky and the equilibrium yield would increase even when n is zero. The
effect of pressure on Kf can be calculated from fugacity coefficients. It is seen that when the
compressibility of the products is greater than the compressibility of the reactants, Kf decreases with
pressure, thereby increasing the conversion.
EXAMPLE 9.12 Industrial grade methanol can be produced according to the reaction
EXAMPLE 9.13 A compound M polymerises in the gas phase at low pressure to Mn, where n > 1.
(a) Show that the mole fraction of the polymer at equilibrium increases with increase in pressure at
constant temperature
(b) The mole fraction of the polymer in the equilibrium mixture at 300 K is 0.15 at 1 bar and 0.367
at 2 bar. Find the value of n.
Solution (a) The reaction is nM Mn. There is a decrease in the number of moles during the
forward reaction. The increase in pressure therefore favours the polymerisation reaction and as a
result, the mole fraction of the polymer at equilibrium increases with pressure.
(b) From Eq. (9.62), Ky = (K/Kf)P–n. Assuming ideal gas behaviour, Ky = KP–n. Here, n = 1 – n
and at 1 bar,
Ky = mole fraction of Mn/(mole fraction of M)n = 0.15/0.85n
Ky at 2 bar = 0.367/0.633n
Therefore,
0.15/0.85n = KPn–1 = K
0.367/0.633n = KPn–1 = K 2n–1
Dividing the second equation by the first,
On solving, we get n = 4.
EXAMPLE 9.14 In the synthesis of ammonia, stoichiometric amounts of nitrogen and hydrogen are
sent to a reactor where the following reaction occurs
N2 + 3H2 2NH3
The equilibrium constant for the reaction at 675 K may be taken equal to 2 10–4.
(a) Determine the per cent conversion of nitrogen to ammonia at 675 K and 20 bar.
(b) What would be the conversion at 675 K and 200 bar?
Solution Basis: 1 mol nitrogen and 3 mol hydrogen are in the reactant mixture. Let e be the extent of
reaction. Then the number of moles of various species at equilibrium are calculated using Eq. (9.3) as
ni = ni0 + nie. Thus the moles of nitrogen, hydrogen and ammonia at equilibrium are, respectively, 1
– e, 3 – 3e and 2e. Total moles at equilibrium is = 4 – 2e. The mole fractions of nitrogen, hydrogen
and ammonia are, respectively,
Therefore, e = 0.5375. So, conversion of nitrogen = 53.75%.
We see that the increase in pressure favours the formation of ammonia as this reaction is
accompanied by a decrease in the number of moles.
where i is any species taking part in the reaction and ni is the number of moles of i. N represents the
total number of moles in the reaction mixture, and if any inert material is present in the system, N
includes nI moles of inert material also.
N = S ni + nI
Combining Eqs. (9.65) and (9.66) we obtain
Any changes in the reaction conditions that results in an increase in the right-hand side of
Eq. (9.67) leads to an improved conversion.
of various components as . Equation (9.4) gives the relationship between mole fractions
and extent of reaction at equilibrium as
(a) The mole fractions of the constituents in the equilibrium mixture are expressed in terms of the
extent of reaction as given in the table below:
EXAMPLE9.17 The reaction takes place in the gas phase at 2975 K and 2025
kPa. The reaction mixture initially comprises 15 mol percent oxygen, 77 mol percent nitrogen and the
rest inerts. The standard Gibbs free energy change for the reaction is 113.83 kJ/mol at this
temperature. Assuming ideal gas behaviour, calculate the partial pressures of all species at
equilibrium. How is the conversion of oxygen affected when the initial mixture were free of inerts?
Solution Basis: 15 mol oxygen, 77 mol nitrogen and 8 mol inert in the reaction mixture. Let e be the
extent of reaction. The mole fractions of the components under equilibrium are:
Oxygen: (15 – e)/100, Nitrogen: (77 – e)/100, NO: 2e/100
Partial pressures are obtained by multiplying the mole fractions by the total pressure. The values are
given below:
O2: 271.6 kPa, N2: 1527.1 kPa, NO: 64.4 kPa, Inerts: 162 kPa
If the initial mixture were free of inerts, the mole fractions of the components under equilibrium will
be Oxygen: (15 – e)/92, Nitrogen: (77 – e)/92, NO: 2e/92 and the equilibrium constant will be given
by the same expression as before:
(b)
We see that the equilibrium conversion of water vapour was 50% when the reactants were in
stoichiometric proportions, it increases to 66.7% when CO was present 100% in excess and falls to
33.3% when CO becomes the limiting reactant.
EXAMPLE9.19 Ethanol is produced by the vapour phase hydration of ethylene according to the
reaction:
The reactor operates at 400 K and 2 bar and the feed is a gas mixture of ethylene and steam in the
ratio 1:3. The equilibrium constant is 0.25. Estimate the composition (mol %) of the equilibrium
mixture. Assume ideal gas behaviour. How is the conversion of ethylene affected when the initial
reactant stream contains stoichiometric quantities of the reactants?
Solution Basis: 1 mole of ethylene and 3 moles of N2 in the reactant stream.
Solving this, we get e = 0.183. That is, conversion of ethylene is 18.3%. The conversion was 26.8%
in the first case when excess of water vapour was present in the reactant stream.
9.7.3 Presence of Products
If the initial reaction mixture contained any of the products of the reaction, then the number of moles
of that product formed by the reaction so as to establish equilibrium will decrease as indicated by Eq.
(9.67). Therefore, the addition of the products to the original reactant stream decreases the
equilibrium conversion.
EXAMPLE 9.20 A gas mixture which contained 1 mol CO, 1 mol water vapour and 1 mol CO2 is
undergoing the following reaction at a temperature of 1100 K and a pressure of 1 bar.
CO (g) + H2O (g) CO2 (g) + H2 (g)
The equilibrium constant for the reaction is K = 1. Assume that the gas mixture behaves as ideal gas.
Calculate the fractional dissociation of steam and discuss the effect of the presence of the products on
the equilibrium conversion.
Solution The mole fractions of the different species in the equilibrium mixture is expressed in terms
of the extent of reaction as below:
Solving the above equation, we get e = 0.333, which means that the conversion of water vapour gets
reduced to 33.3% due to the presence of CO2, the product of the reaction in the reactant stream.
EXAMPLE 9.21 A gas mixture containing 25% CO, 55% H2 and 20% inert gas is to be used for
methanol synthesis. The gases issue from the catalyst chamber in chemical equilibrium with respect to
the reaction
CO (g) + 2H2 (g) CH3OH (g)
at a pressure of 300 bar and temperature of 625 K. Assume that the equilibrium mixture forms an
ideal solution and Kf and Kf are 4.9 10–5 and 0.35 respectively. What is the per cent conversion
of CO?
Solution Basis: 100 moles of initial gas mixture.
Let e be the extent of reaction at equilibrium. n0 = 100.
n=–1–2+1=–2
The mole fractions in the equilibrium mixture are calculated using Eq. (9.4)
The equilibrium constant Kp = 1.25 10–2. The pressure is maintained at 50 bar. Assume ideal
gas behaviour for the gas mixture. Determine the composition of the gases leaving the reactor.
Solution Basis: 100 moles of the reactant gases.
Solving this, we get e = 8.71. Mole fraction of nitrogen is obtained as
[20 – (e/2)]/(100 – e) = 0.1714
and mole fraction of hydrogen is
[60 – (3e/2)]/(100 – e) = 0.5141
Mole fraction of ammonia:
e/(100 – e) = 0.0954
Mole fraction of inert gas:
1 – 0.1714 – 0.5141 – 0.0954 = 0.2191
Analysis of exit gases from the reactor:
N2 = 17.14%, H2 = 51.41%, NH3 = 9.54% and inert gas = 21.91%
For liquid-phase reactions, the evaluation of equilibrium constant using this equation requires a
relationship between activity and composition. Since, activity is the ratio of the fugacity to the
fugacity in the standard state, such a relationship can be established once the standard state is
specified. The standard state for liquid-phase reactions may be the pure liquid at 1 bar and the
reaction temperature. The fugacity in this state is not much different from the fugacity of pure liquid at
the pressure and temperature of the reaction fi. This is because, pressure has very negligible effect on
the properties of liquids. With this choice, the equilibrium constant becomes
The fugacity of a component in the solution is related to the fugacity in the pure state by = gixifi,
where i is the activity coefficient in the solution. Using this in Eq. (9.68), we get
For components present in low concentration, the standard state of the solute is usually the fictitious
or hypothetical state which would exist if the solute obeyed Henry’s law over a concentration range
extending up to a molality of unity. This hypothetical state is illustrated in Fig. 9.4.
where Ki is the Henry’s law constant and mi is the molality. Using the hypothetical standard state, it
can be shown that the standard state fugacity is equal to the Henry’s law constant and the activity and
the molality are equal. That is
ai = mi………(9.72)
With this choice for the standard state, a very simple relationship exists between the activity and the
concentration for cases where Henry’s law is applicable.
All the above methods give the same results for equilibrium compositions, but the values for
equilibrium constant depend on the choice of the standard state.
9.9.2 Equilibria involving Pure Solids and Liquids
When a pure liquid or a pure solid is involved in a heterogeneous reaction with gases, its activity
may be taken as unity provided the pressure of the system is not much different from the standard
state. Activity as we know, is defined as the ratio of the fugacity to the fugacity in the standard state.
The fugacity in the standard state is almost equal to that in the equilibrium state, as these two states
differ only in their pressures and not in their temperatures. Pressure, unless extremely high, has only a
negligible effect on the properties of liquids and solids. Where the standard state for solids and
liquids is taken at 1 bar or at low equilibrium vapour pressures, the activities of pure solids and pure
liquids may be taken as unity at moderate pressures. Therefore, the composition of the gaseous phase
at equilibrium is not affected by the presence of the solid or liquid.
9.9.3 Pressures of Decomposition
Many solid compounds decompose to give another solid and a gas, as in the calcination of calcium
carbonate to CO2 and lime.
CaCO3 (s) CaO (s) + CO2 (g)
The equilibrium constant for this reaction is
The activities of the solid components present at equilibrium are close to unity provided the pressure
is moderate and both solids are present at equilibrium. Since the standard state for gases is the ideal-
gas state at 1 bar, the standard state fugacity is equal to unity and therefore, the activity of CO2 in the
equilibrium mixture is equal to its fugacity, . But fugacity of a component is equal to its partial
pressure at low pressures and, therefore, Eq. (9.74) reduces to
This is the equilibrium partial pressure exerted by CO2 and its value depends only on temperature. If
the partial pressure is lowered below this equilibrium value, CaCO3 will decompose and will
eventually disappear. On the other hand, if the pressure on the system is kept above the equilibrium
partial pressure, CaO will combine with CO2 resulting in the formation of CaCO3.
For a general solid decomposition reaction represented by
aA (s) lL (s) + mM (g)
the above treatment can be generalised as
In the above equation, m is the stoichiometric coefficient; DH0 and DS0 are the standard heat of
reaction and standard entropy of reaction respectively.
EXAMPLE 9.23 Ethylene gas reacts with water forming aqueous solution of ethanol.
C2H4 (g) + H2O (l) C2H5OH (aq)
Equilibrium measurements at 530 K and 85 bar showed that the aqueous phase contained
1.5% (mole) ethanol and 95.0% (mole) water. The vapour phase analysed 48% ethylene. The fugacity
coefficient for ethylene is estimated to be 0.9. Evaluate the equilibrium constant.
Solution Equation (9.73) may be used for evaluating K.
The standard state for aqueous solution is 1 molal solution; for water, it is pure liquid water at
1 bar; and for gaseous ethylene, it is the pure ethylene at 1 bar.
The molality of aqueous solution = moles ethanol/kg water
= 1.5/(95.0 18 10–3) = 0.8772 mol/kg water
EXAMPLE 9.24 Calculate the decomposition pressure of limestone at 1000 K.
CaCO3 (s) CaO (s) + CO2 (g)
The standard free energy of this reaction as function of temperature is
DG0 = 1.8856 105 – 243.42T + 11.8478T ln T – 3.1045 10–3T2 + 1.7271 10–6 T3
– 4.1784 105/T
Also calculate the decomposition temperature at 1 bar.
Solution From Eq. (9.75), the decomposition pressure is = K, where K can be calculated by
EXAMPLE 9.25 Solid calcium oxalate dissociates at high temperatures into solid calcium carbonate
and carbon monoxide:
Solution By Eq. (9.75),
Therefore,
EXAMPLE 9.26 Iron oxide is reduced to iron by passing over it a mixture of 20% CO and 80% N2
at 1200 K and 1 bar.
FeO (s) + CO (g) Fe (s) + CO2 (g)
The equilibrium constant for this reaction is 0.403. Assuming that equilibrium is attained, calculate
the weight of metallic iron produced per 100 m3 of gas admitted at 1200 K and 1 atm. Gas mixture
may be assumed to behave as ideal gas.
Solution Basis: 100 mol of gas entering.
The activities of solid components can be taken to be unity.
9.10 SIMULTANEOUS REACTIONS
With a given set of reactants many reactions may be possible. When we consider the equilibrium
yield of methanol in the reaction
CO + 2H2 CH3OH………(9.78)
by the methods already discussed, we are in fact ignoring the presence of intermediate product,
formaldehyde in the reaction mixture. The above reaction proceeds in two steps in series as:
CO + H2 HCHO………(9.79)
HCHO + H2 CH3OH………(9.80)
For the thermodynamic analysis of a reaction that proceeds in two or more steps, the presence of
intermediate products can sometimes be ignored on the assumption that they are very unstable and
their concentrations at equilibrium are negligible in comparison with that of the main product. The
above assumption is implicit in treating the equilibrium mixture in the methanol synthesis as
consisting of only CO, H2 and CH3OH. In this case, this assumption is a valid one as formaldehyde is
very unstable, but in many other situations, the presence of intermediate products in the reaction
mixture at equilibrium also should be taken into account as explained below:
The free energy change for a reaction is equal to the sum of the free energy changes in the individual
step reactions. Thus,
where are the free energy changes in the two step reactions that occur and DG0 is the
standard free energy change in the overall reaction. Since DG0 = – RT ln K, the above equation gives
K = K1K2
K1 and K2 are the equilibrium constants for the individual steps and K is the equilibrium constant for
the combined reaction. For a given value of K, an infinite number of combinations of K1 and K2 are
possible such that K = K1K2. For example, let us take K = 10–4 and consider the cases where (a) 10–
4 = 10–10 106, (b) 10– 4 = 10–2 10–2, and (c) 10–4 = 106 10–10. For case (a), the
concentration of intermediate products at equilibrium would be negligible and correct result would
be obtained by considering only the overall reaction. For case (b), there would be considerable
amounts of intermediates at equilibrium and their presence cannot be ignored. For case (c), the
equilibrium mixture would be mostly intermediates. The use of an overall equilibrium constant for the
calculation of equilibrium compositions is limited to cases where the intermediate products are not
present in significant quantities.
In addition to the formation of intermediate products, which subsequently reacts to form the final
desired products, many side reactions may also occur within the system. For example, starting with
the pairs CO and H2 some of the possible reactions are:
CO + H2 HCHO
CO + 2H2 CH3OH
CO + 3H2 CH4 + H2O
2CO + 5H2 C2H6 + 2H2O
3CO + 6H2 C3H7OH + 2H2O
In dealing with methanol synthesis, it was assumed that the side reactions proceeded at a negligible
rate in comparison with the steps involved in the synthesis reaction. Theoretically, when the
equilibrium yield of a particular component is to be determined, we should consider simultaneous
equilibria in all possible reactions between the substances involved. However, for practical
calculations, it is possible to reduce the number of reactions that are to be considered.
In the general case when all intermediates and final products must be considered, it is necessary that
the equilibrium equations of all reactions must be satisfied by the compositions of the system at
equilibrium. Determination of the equilibrium compositions involves simultaneous solution of r
equilibrium equations where r is the number of independent reactions that can be written. After
determining the number of independent reactions as explained later, the equilibrium constant is
evaluated for each reaction by
Here the suffix j is used to represent the jth reaction under consideration. The above equation is
written for all r independent reactions. Assuming the equilibrium mixture to behave as ideal gases,
these lead to r equations relating the composition to the pressure and the equilibrium constant.
Let the equilibrium constants be K1 and K2 for the reactions indicated by Eqs. (9.84) and (9.85)
respectively and let the corresponding extent of reaction be e1 for reaction (9.84) and e2 for reaction
(9.85). The initial reactant mixture is assumed to consist of 1 mol A and x mol B. The mole fractions
in simultaneous reactions can be calculated using Eq. (9.9).
Note that K1 = 0.574, K2 = 2.21 and P = 1 bar. The resulting equations are solved for e1 and e2.
Assume a value for e1 and calculate e2 by each equation. These two e2 values are plotted against
e1. This is repeated for various assumed e1 values. The intersection of the two curves gives the
solution. e1 = 0.9124; and e2 = 0.623. The mole fractions are evaluated by supplying the values of
e1 and e2. The results are:
CH4: 0.0112,……H2O: 0.4415,……CO: 0.0357,……H2: 0.4307……and……CO2: 0.0804
This in fact is a relationship between T, yA and yB. Only two of these three variables are therefore
independent. As the degree of freedom is 4 it means that in addition to these three variables, two
more variables are to be specified to define the intensive state of the system uniquely.
O.A. Hougen, et al., define the number of independent reactions that must be considered as the least
number that includes every reactant and product present to an appreciable extent in all phases of
the equilibrium system, and accounts for the formation of each product from the original
reactants. It can be determined as follows:
1. For each chemical compound present in the system, equation for its formation reaction from its
elements is written.
2. The elements that are not present in the system are eliminated by properly combining the
equations written in step 1.
The number of equations, r, that results from the above procedure is equal to the number of
independent chemical reactions occurring.
EXAMPLE 9.28 Determine the number of degrees of freedom in a gaseous system consisting of CO,
CO2, H2, H2O and CH4 in chemical equilibrium.
Solution The number of independent chemical reactions occurring in the system is first determined.
The formation reactions for each of the compounds are written:
The elements C and O2 are not present in the system. C is eliminated first, from Eqs. (9.91), (9.92)
and (9.94). Combining Eq. (9.91) with Eq. (9.92) we get
The equations that remain after this elimination process are Eqs. (9.97) and (9.98) which represent
the independent chemical reactions occurring in the system. Therefore, r = 2. Equation (9.90) gives
the degrees of freedom as F = C – p – r + 2. Here C = 5; p = 1 and therefore, F = 4.
SUMMARY
Thermodynamics of chemical reactions is mainly concerned with the prediction of the equilibrium
conversion attainable in a chemical reaction and the effect of operating conditions on the degree of
completion of the reaction. The criterion of chemical equilibrium requires that for a chemical reaction
occurring at equilibrium, there should be no change in the Gibbs free energy of the system at constant
temperature and pressure. If the system is not in chemical equilibrium, the reaction occurring must be
irreversible and the total Gibbs free energy must decrease at constant temperature and pressure
(Section 9.2).
The equilibrium constant K for a reaction was defined in terms of the activities of the reactants and
the products as
where ai is the activity of component i in the reaction mixture and i is the stoichiometric number of i
(Section 9.3). The equilibrium constant was related to the standard free energy change by
Eq. (9.31). Thus the numerical value of the equilibrium constant depends upon the temperature,
the form of the stoichiometric equation and the definition of the standard state for each component.
However, it is independent of the pressure at equilibrium (Section 9.4). Equation (9.31) also
provided an approximate criterion for feasibility of reactions. If DG0 for a reaction is zero, then
K = 1, the reaction proceeds to a considerable extent before equilibrium is reached. If DG0 is
negative, then K > 1, the reaction is quite favourable.
The effect of temperature on the equilibrium constant was quantitatively expressed by
van’t Hoff equation [Eq. (9.36)]. For an exothermic reaction, the equilibrium constant decreases as
the reaction temperature increases and for an endothermic reaction, the equilibrium constant will
increase with increase in temperature (Section 9.5). Three methods for the evaluation of equilibrium
constant were discussed; the one which made use of thermal data in the form of standard heat of
reaction DH0, and a standard free energy change of reaction DG0 at a given temperature was found to
be the most convenient and widely used. The usefulness of the Giauque functions for tabulation of
standard free energy of reactions and calculation of the equilibrium constant was also established.
The equilibrium constant is independent of pressure whereas the composition at equilibrium varies
with pressure as evident from Eq. (9.62). If there is a decrease in the number of moles during the
reaction, the equilibrium yield would increase with increase in the pressure, whereas if the reaction
results in an increase in the number of moles, the equilibrium yield would decrease with increase in
pressure. It was also shown that the effect of the presence of inert gas in the reactant stream on the
equilibrium conversion was just the opposite of the effect of pressure (Section 9.7).
For liquid-phase reactions, the equilibrium constant may be written as K = KgKx. K is an
equilibrium constant in terms of activity coefficients, which may be assumed, equal to unity. For
components present in low concentration, the activity and the molality are equal (Section 9.8). Under
heterogeneous equilibrium (Section 9.9), a brief discussion on the reaction between a gas and liquid
resulting in the formation of a solution and reaction equilibria in which a solid or liquid reacted with
a gas, were provided. Also, it was seen that for reactions in which solid compounds decomposed to
give another solid and a gas, the equilibrium constant was equal to the partial pressure of the gas. If
the partial pressure was lowered below this equilibrium value the solid would decompose and if the
pressure on the system was maintained above this value, the formation of solid was favoured. For
simultaneous reactions in which all intermediate and final products in the equilibrium mixture were to
be considered for determining the composition, equilibrium equations were written for all the
independent reactions and these were solved simultaneously (Section 9.10).
REVIEW QUESTIONS
1. What do you mean by the ‘extent of reaction’? How is it related to the mole fraction of the
species in the reaction mixture?
2. What is the criterion of chemical reaction equilibria?
3. Define equilibrium constant K of a chemical reaction. How is it related to Kf and KP?
4. Does the numerical value of the equilibrium constant depend on the form of the stoichiometric
equation?
5. How is the equilibrium constant K related to the standard free energy change? Does K vary with
pressure?
6. What is the effect of temperature on the equilibrium constant? Using van’t Hoff equation predict
the effect of increasing the temperature on endothermic and exothermic reactions.
7. How would you predict the feasibility of a reaction from the value of the standard free energy
change?
8. How would the equilibrium yield in a gaseous chemical reaction be affected by increasing the
pressure, if there is a decrease in the number of moles during the reaction? How would you
explain the effect of pressure on reactions such as the water–gas shift reaction, where there is no
change in the number of moles?
9. How would the equilibrium yield of ammonia be affected if argon is present in the synthesis gas
fed to the ammonia converter?
10. Explain how the equilibrium constant for liquid-phase reactions is evaluated.
11. Show that the equilibrium constant in the decomposition of calcium carbonate into CO2 and
lime is equal to the partial pressure of carbon dioxide. Explain how would you estimate the
decomposition pressure? What would happen if the CO2 pressure is reduced below this value?
12. A reaction proceeds in two steps. The equilibrium constants for the individual steps are K1 and
K2. What would be the equilibrium constant for the overall reaction?
13. What do you mean by the number of independent reactions in a chemically reacting system?
How would you determine it?
14. What is phase rule as applicable to a reacting system?
EXERCISES
9.1 Water vapour decomposes according to the following reaction:
Derive expressions for the mole fraction of each species in terms of the extent of reaction
assuming that the system contained n0 moles of water vapour initially.
9.2 The following reaction occurs in a mixture consisting of 2 mol methane, 1 mol water,
1 mol carbon monoxide and 4 mol hydrogen initially.
CH4 + H2O CO + 3H2
Deduce expression relating the mole fractions of various species to the extent of reaction.
9.3 A system consisting of 2 mol methane and 3 mol water is undergoing the following reaction
CH4 + H2O CO + 3H2
CH4 + 2H2O CO2 + 4H2
Derive expressions for mole fractions in terms of the extent of reactions.
9.4 The following gas-phase reactions occur in a mixture initially containing 3 mol ethylene and 2
mol oxygen.
(a) Over what range of temperature is the reaction promising from a thermodynamic viewpoint?
(b) For reaction of pure butene at 800 K, calculate the equilibrium conversion for operation at
1 bar and 5 bar.
(c) Repeat part (b) if the feed consists of 50% (mol) butene and the rest inerts.
9.12 Calculate the equilibrium constant for the vapour-phase hydration of ethylene to ethanol at 600
K
C2H4 + H2O C2H5OH
The following data are available:
9.13 The equilibrium constant at 420 K for the vapour-phase hydration of ethylene to ethanol
according to the reaction
C2H4 + H2O C2H5OH
is 6.8 10–2 and standard heat of reaction at 298 K is –45.95 103 J. The specific heat data
are as follows:
Formulate general relationships for estimating the equilibrium constant and standard free energy
change as functions of temperature.
9.14 For the vapour-phase hydration of ethylene to ethanol according to
C2H4 + H2O C2H5OH
the equilibrium constants were measured at temperature 420 K and 600 K. They are
6.8 10–2 and 1.9 10–3 respectively. The specific heat (J/mol K) data are:
Develop general expressions for the equilibrium constant and standard free energy change as
functions of temperature.
9.15 The water–gas shift reaction
CO (g) + H2O (g) CO2 (g) + H2 (g)
takes place at 373 K. The equilibrium constant KP for this reaction at 537 K = 9.8 10–4. The
heats of formation at 298 K are: CO = –110,525 J/mol, CO2 = –393,509 J/mol,
H2O = –241,818 J/mol. Calculate the equilibrium constant at 1000 K.
9.16 Calculate the fraction of pure ethane that would dehydrogenate at 750 K and 5 atm, if the
following reaction goes to equilibrium.
DG0 for the reaction at 750 K is 42.576 kJ. Assume ideal behaviour.
9.17 Ethanol can be prepared by the following vapour-phase reaction from ethylene:
The value of DG0 for the above reaction at 1 bar and 398 K is 5040 J. Calculate the conversion
obtained if an isothermal reactor operating at 398 K and 2 bar is fed with a mixture containing
50 mol percent ethylene and 50 mol percent steam. Assume that equilibrium is reached at the
exit of the reactor and the gases behave ideally.
9.18 A gaseous mixture containing 30% CO, 50% H2 and the rest inert gas is sent to a reaction
chamber for methanol synthesis. The following reaction occurs at 635 K and 310 bar.
CO (g) + 2H2 (g) CH3OH (g)
Assuming that the gas mixture behaves as an ideal solution calculate the per cent conversion of
CO given that Kf = 5 10–5 and Kf = 0.35.
9.19 Estimate the maximum conversion of ethylene to alcohol by vapour phase hydration at
523 K and 34 bar.
C2H4 (g) + H2O (g) C2H5OH (g)
The equilibrium constant varies with temperature as
ln K = 4760/T – 1.558 ln T + 2.22 10–3T – 0.29 10–6T2 – 5.56
The steam–ethylene ratio in the initial mixture is 5.0. The fugacity coefficients for ethylene,
ethanol and water vapour are 0.98, 0.84 and 0.91.
9.20 Ethanol is manufactured by the vapour-phase hydration of ethylene to ethanol according to the
reaction,
C2H4 (g) + H2O (g) C2H5OH (g)
Starting with a gas mixture containing 25% ethylene and 75% steam, determine the composition
of the products if the reaction were carried out at 400 K and 1 bar. The standard free energy of
reaction at 400 K is 4548.3 J.
9.21 What would be the equilibrium yield of ethanol at 1 bar and 373 K in the following reaction?
C2H4 (g) + H2O (g) C2H5OH (g)
The reactant stream consists of an equimolar mixture of steam and ethylene. The standard free
energy change may be taken as = 1264 J/mol.
9.22 Calculate the equilibrium percentage conversion of nitrogen to ammonia at 700 K and
300 bar, if the gas enters the converter with a composition of 75% (mol) hydrogen and 25%
(mol) nitrogen. For the reaction
may be taken as K = 85. Calculate the composition of gases leaving the converter.
9.24 One mol carbon at 298 K reacts with 2 mol oxygen at 298 K to form an equilibrium mixture of
CO2, CO and O2 at 3000 K and 1 bar. If the equilibrium constant K = 0.328, determine the
equilibrium composition.
9.25 One mol carbon at 298 K and 1 bar reacts with 1 mol oxygen at 298 K and 1 bar to form an
equilibrium mixture of CO2, CO and O2 at 3000 K and 1 bar in a steady flow process.
Determine the equilibrium composition and heat transfer for this process if the equilibrium
constant K = 0.328. Standard heat of formation are 393.509 kJ/mol for CO2, 110.525 kJ/mol for
CO. The mean heat capacity of products = 45 J/mol K.
9.26 Pure N2O4 at a low temperature is diluted with air and heated to 298 K and 1 bar. The
following reaction occurs
N2O4 (g) 2NO2 (g)
If the mole fraction of N2O4 in the N2O4–air mixture before dissociation begins is 0.2,
calculate the extent of decomposition and mole fraction of NO2 and N2O4 present at
equilibrium. The standard free energy change for the reaction at 298 K = 4644.7 J/mol.
9.27 Methanol is manufactured according to the reaction
CO (g) + 2H2 (g) CH3OH (g)
The reaction is carried out at 400 K and 1 bar. The standard heat of reaction at this condition is
– 9.4538 104 J and the equilibrium constant is 1.52. Analysis of the equilibrium vapour
product from the reactor shows 40% hydrogen. Equilibrium gas mixture can be treated as an
ideal gas.
(a) Determine the concentrations of CO and CH3OH in the product.
(b) If the reaction occurred at 500 K and 1 bar starting with the same feed as in part (a) would
you expect the concentration of hydrogen in the equilibrium mixture to be greater or less than
40% mole? Why?
9.28 Determine the maximum percentage of ethane that may get dehydrogenated to ethylene at 750
K and 5 bar according to the reaction
C2H6 (g) C2H4 (g) + H2 (g)
in equilibrium at 775 K what pressure is required for a 90 per cent conversion of SO2 if the
initial mixture is equimolar in the reactants. Assume ideal gases. Take the free energy of the
reaction at 775 K to be –2.8626 104 J.
9.32 1-butene is dehydrogenated to 1,3-butadiene according to the reaction
C4H8 (g) C4H6 (g) + H2 (g)
Determine the extent of reaction at equilibrium at 900 K and 1 bar with
(a) 1 mol butene as the reactant
(b) a reactant mixture consisting of 1 mol butene and 10 mol steam.
The following free energy functions and heat of formation data are available:
where T is in K and K is in (bar)–1/2. A feed mixture containing 12% SO2, 9% O2 and 79% N2
is reacted at 749 K and 1 bar. Calculate the fractional conversion of SO2.
9.36 Ethanol is produced by vapour-phase hydration of ethylene:
C2H4 (g) + H2O (g) C2H5OH (g)
9.37 Acetic acid is esterified in the liquid phase with ethanol at 373 K and 1 bar to produce ethyl
acetate and water according to the reaction
CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l)
The feed consists of 1 mol each of acetic acid and ethanol, estimate the mole fraction of ethyl
acetate in the reacting mixture at equilibrium. The standard heat of formation and standard free
energy of formation at 298 K are given below:
Assume that the heat of reaction is independent of temperature and the liquid mixture behaves as
ideal solution.
9.38 The esterification of ethanol with acetic acid occurs in an aqueous solution as follows:
C2H5OH (aq) + CH3COOH (aq) CH3COOC2H5 (aq) + H2O (l)
The free energies of formation of acetic acid, ethanol and ethyl acetate in a hypothetical 1 molal
solution at 298 K are –3.9645 105 J, –1.8053 105 J and –3.3296 105 J respectively.
The free energy of formation of water at 298 K is –2.3735 105 J. What is the equilibrium
constant? Starting with a dilute equimolar mixture of ethanol and acetic acid, calculate the extent
of reaction and the molalities of ethyl acetate and acetic acid in the equilibrium solution.
Assume dilute solution behaviour.
9.39 Carbon dioxide is reduced by graphite according to the equation
C (s) + CO2 (g) 2CO (g)
Assuming that equilibrium is attained at 1000 K and 1 bar, calculate the degree of completion of
reduction of CO2. The following data are available:
9.42 Ammonium chloride decomposes upon heating to yield a gas mixture of ammonia and
hydrochloric acid. At what temperature does ammonium chloride exert a decomposition
pressure of 1 bar? The standard heat of formation and the standard free energy of formation are
as follows:
9.43 The following decomposition reaction occurs at 373 K in the liquid phase.
A B+C
The equilibrium constant based on pure liquid standard state is 2. The vapour pressures are PA
= 5 bar, PB = 20 bar and PC = 2 bar. Assume that all vapours are ideal, liquid B is immiscible
with A–C liquid mixture and the A–C mixture is ideal. Calculate the equilibrium pressure and
the composition of the liquid and vapour phases.
9.44 The equilibrium constant for the following reaction is found to be 2.
A (l) B (l) + C (l)
The vapour pressures are PA = 5 bar, PB = 20 bar and PC = 2 bar. A and C form ideal solution
and B is immiscible with either A and C or their mixtures. The system consisted of pure A
initially. Find the pressure below which only a gas phase exists.
9.45 Mixtures of CO and CO2 are to be processed at temperatures between 900 K and 1000 K.
Determine the conditions under which solid carbon might deposit according to the reaction
CO2 (g) + C (s) 2CO (g)
The equilibrium constants for this reaction are 0.178 at 900 K and 1.58 at 1000 K. (Hint: The
activity of solid carbon is less than unity if carbon is not present in the system.)
9.46 Acetylene is catalytically hydrogenated to ethylene at 1500 K and 1 bar. Starting with an
equimolar mixture of acetylene and hydrogen what will be the mole fractions at equilibrium?
Assume ideal gases.
C2H2 2C + H2; K = 5.2
2C + 2H2 C2H4; K = 0.1923
9.47 What would be the equilibrium conversion of ethyl alcohol to butadiene at 700 K and
1 bar given the following reactions?
C2H5OH C2H4 + H2O; DG0 = – 45,427 J/mol
A.
B.
C.
D.
16. The reversible work of expansion in a flow process under isothermal condition is equal to
A. –(DA)T
B. –(DG)T
C. –(DU)T
D. –(DH)T
17. The decrease in enthalpy accompanying a reversible expansion measures the shaft work in the
case of
A. A non-flow isothermal process
B. An isothermal flow process
C. An isentropic non-flow process
D. An isentropic flow process
18. For a reversible process occurring at constant temperature and pressure, the decrease in Gibbs
free energy measures
A. The maximum reversible work
B. The maximum reversible work, other than the electrical work
C. The maximum reversible work, other than the wok of expansion
D. The heat supplied
19. CP = CV when
A.
B.
C.
D.
20. At the triple point of water, the number of degrees of freedom is
A. zero
B. one
C. two
D. three
21. The canonical variables for H are:
A. P and T
B. V and T
C. P and S
D. V and S
22. Which one of the following is incorrect?
A. dU = T dS – P dV
B. dH = T dS – V dP
C. dA = – S dT – P dV
D. dG = – S dT + V dP
23. The volume coefficient of expansion b of an ideal gas equals
A. 1/T
B. 1/P
C. T
D. P
24. Fugacity has the same dimensions as that of
A. Gibbs free energy
B. Pressure
C. Temperature
D. Fugacity is dimensionless
25. The change in free energy when a real gas undergoes an isothermal change in state is
A. DG = RT ln (V2/V1)
B. DG = RT ln (P2/P1)
C. DG = RT ln (f2/f1)
D. DG = RT ln (g2/g1)
26. The difference between the heat supplied and the work extracted in a steady flow process in
which the kinetic and potential energy changes are negligible, is equal to
A. The change in internal energy
B. The change in enthalpy
C. The change in the work function
D. The change in the Gibbs free energy
27. Compressibility factor Z of a gas is
A. The ratio of fugacity in the given state to fugacity in the standard state
B. The ratio of actual volume to the volume of the gas if it were ideal
C. The change in volume with temperature at constant pressure
D. The difference between actual volume and ideal gas volume
28. The temperature at which a transition occurs from a compressibility factor less than 1.00 to that
greater than 1.00 is known as
A. The critical temperature
B. The critical solution temperature
C. The inversion point
D. The Boyle point
29. The coefficient of compressibility k is defined as
A.
B.
C.
D.
30. For any equation of state to be valid, at the critical point the critical isotherm should
have
A. A maximum
B. A minimum
C. A point of inflection
D. Negative slope
31. At constant temperature and pressure, the decrease in Gibbs free energy is a measure of
A. The maximum work
B. The maximum net work
C. The unavailable energy
D. The loss in capacity to do work
32. In thermodynamics, a phase means
A. A closed system
B. An open system
C. A homogeneous system
D. A heterogeneous system
33. The net change in a state function is zero for
A. A reversible process
B. An Irreversible process
C. A cyclic process
D. A non-cyclic process
34. The third law of thermodynamics deals with
A. Chemical reactions
B. Quantitative equivalence between heat and work
C. Rate of change of a process
D. Absolute entropy of perfect crystalline substances
35. DG = DA for a process occurring at
A. Constant pressure
B. Constant volume
C. Constant pressure and constant temperature
D. Constant pressure and constant volume
36. As pressure approaches zero, fugacity coefficient value tends to
A. Pressure
B. Zero
C. Unity
D. Infinity
37. For a gas obeying the van der Waals equation of state, at the critical temperature,
A. Both (∂P/∂V)T and (∂2P/∂V2)T are zero
B. The first derivative is zero, while the second derivative is non-zero
C. The second derivative is zero while the first derivative is non-zero
D. Both the derivatives are non-zero
(1991)
38. For an ideal gas, the slope of the pressure-volume curve at a given point will be
A. Steeper for an isothermal than for an adiabatic process
B. Steeper for an adiabatic than for an isothermal process
C. Identical for both the processes
D. Of opposite sign
(1991)
39. The shape of T-S diagram for Carnot cycle is
A. A rectangle
B. A rhombus
C. A trapezoid
D. A circle
(1991)
40. During Joule–Thomson expansion of gases,
A. Enthalpy remains constant
]B. Entropy remains constant
C. Temperature remains constant
D. None of the above
(1992)
41. For a single-component, two-phase mixture, the number of independent variable properties are
A. Two
B. One
C. Zero
D. Three
(1992)
42. Ideal gas law is applicable at
A. Low T, low P
B. High T, high P
C. Low T, high P
D. High T, low P
(1994)
43. The second law of thermodynamics states that
A. The energy change of a system undergoing any reversible process is zero
B. It is not possible to transfer heat from a lower temperature to a higher temperature
C. The total energy of the system and the surroundings remain constant
D. None of the above
(1994)
44. A solid is transformed into vapour without changing into the liquid phase
A. At the triple point
B. At the boiling point
C. Below the triple point
D. Always
(1995)
45. At the inversion point, the Joule–Thomson coefficient is
A. Positive
B. Negative
C. Zero
D. Cannot be generalised
(1995)
46. The kinetic energy of a gas molecule is zero at
A. 0∞C
B. 273∞C
C. 100∞C
D. –273∞C
(1995)
47. Assuming that CO2 obeys the perfect gas law, the density of CO2 in kg/m3 at 536 K and 202.6
kPa is
A. 1
B. 2
C. 3
D. 4
(1995)
48. A closed system is cooled reversibly from 373 K to 323 K. If no work is done on the system
A. Its internal energy (U) decreases and its entropy (S) increases
B. U and S both decrease
C. U decreases but S is constant
D. U is constant but S decreases
(1995)
49. The equation dU = T dS – P dV is applicable to infinitesimal changes occurring in
A. An open system of constant composition
B. A closed system of constant composition
C. An open system with changes in composition
D. A closed system with changes in composition
(1996)
50. A system undergoes a change from a given initial state to a given final state either by an
irreversible process or by a reversible process. Then
A. DSI is always > DSR
B. DSI is sometimes > DSR
C. DSI is always < DSR
D. DSI is always = DSR
where DSI and DSR are the entropy changes of the system for the irreversible and reversible
processes, respectively.
(1997)
51. The change in Gibbs free energy for vaporisation of a pure substance is
A. Positive
B. Negative
C. Zero
D. May be positive or negative
(1997)
52. A change in state involving a decrease in entropy can be spontaneous only if
A. It is exothermic
B. It is isenthalpic
C. It takes place isothermally
D. It takes place at constant volume
(1998)
53. A Carnot cycle consists of the following steps:
A. Two isothermals and two isentropics
B. Two isobarics and two isothermals
C. Two isochorics and two isobarics
D. Two isothermals and two isochorics
(1998)
54. It is desired to bring about certain change in the state of a system by performing work on the
system under adiabatic conditions
A. The amount of work needed is path-dependent
B. Work alone cannot bring about such a change of state
C. The amount of work needed is independent of path
D. More information is needed to conclude anything about the path-dependence or otherwise of
the work needed.
(1998)
55. Chemical potential is
A. An extensive property
B. An intensive property
C. A path property
D. A reference property
56. According to the phase rule, the triple point of a pure substance is
A. Invariant
B. Univariant
C. Bivariant
D. None of the above
57. Which one of the following is incorrect with reference to partial molar properties?
A. They are intensive properties
B. They are always positive
C. They represent the contribution of individual components to the total solution property
D. They vary with composition of the solution
58. All but one of the following represent the chemical potential of component i in solution. Find
the odd man out.
A.
B.
C.
D.
59. Which one of the following statements is not valid for an ideal solution?
A. There is no volume change on mixing
B. There is no enthalpy change on mixing
C. There is no entropy change on mixing
D. Fugacity is directly proportional to concentration.
60. For the standard state of pure component at the solution pressure, the activity of a component in
an ideal solution is equal to
A. its fugacity in the solution
B. its mole fraction in the solution
C. its partial pressure
D. its chemical potential
61. Which one of the following is the correct form of Gibbs–Duhem equation for a binary solution?
A.
B.
C.
D.
62. Which one of the following is true for the excess property ME?
A. ME = M – S xiMi
B. ME = M – S xi
C. ME = M – Mid
D. ME = DM
63. One of the following statements is incorrect for a multicomponent system consisting of two
phases in thermodynamic equilibrium. Identify it.
A. The temperatures of both phases are the same
B. The pressure is uniform throughout
C. The concentrations of a component in both phases are equal
D. The chemical potentials of a component in both phases are equal
64. When an ideal binary solution is boiled at constant pressure in a closed container,
A. The boiling temperature remains constant at some value between the bubble point and the
dew point, till the entire liquid is vaporised
B. The boiling temperature varies between the bubble point and the dew point of the solution
C. The boiling occurs at a constant temperature known as the bubble point
D. The boiling occurs either at the bubble point or at the dew point
65. The value of activity coefficient for an ideal solution is
A. One
B. Zero
C. Equal to Henry’s law constant
D. Equal to the vapour pressure
66. A solution exhibiting positive deviation from ideality
A. Always forms a minimum boiling azeotrope
B. Always forms a maximum boiling azeotrope
C. Has a total vapour pressure that is less than that predicted by Raoult’s law
D. When formed from its constituents there is an absorption of heat
67. Which one of the following statements is true with reference to the minimum boiling
azeotropes?
A. There is a minimum on the vapour-pressure curve
B. The solution exhibits positive deviation from ideality
C. The dew point is greater than the bubble point
D. The activity coefficients are less than unity
68. The vaporisation equilibrium constant (K-factor) depends upon
A. Temperature only
B. Pressure only
C. Temperature and pressure only
D. Temperature, pressure and concentration
69. The vapour–liquid equilibrium data are thermodynamically inconsistent if
A. The slopes of the ln g1 vs x1 and ln g2 vs x1 curves have opposite signs
B. When plotted against x1, ln g2 and ln g1 curves pass through a maximum at the same
composition
C. Both g1 and g2 are greater than unity
D. ln g1 vs x1 curve has a maximum and ln g2 vs x1 curve has a minimum at a particular x1.
70. A mixture of two immiscible liquids A and B is in equilibrium with its vapour at temperature T
and pressure P. The vapour pressures of pure A and pure B are respectively. The
relation applicable to the system is
A.
B.
C.
D. T > TA + TB, where TA and TB are boiling points of pure A and pure B respectively
71. The mutual solubility of two partially miscible liquids increases with temperature. At what
temperature do the two liquid phases become identical?
A. At the critical point
B. At the three-phase temperature
C. At the upper critical solution temperature
D. At the dew point
72. Benzene and water may be considered immiscible. A mixture of benzene (20 g) and water (80
g) is taken in a vessel and boiled. It boils at 101.3 kPa and 342 K. At this temperature vapour
pressure of benzene is 71.18 kPa and that of water is 30.12 kPa. What is the concentration of
benzene in the vapour in mass per cent?
A. 70%
B. 91%
C. 20%
D. 80%
73. The necessary and sufficient condition for equilibrium between two phases is
A. Concentration of each component should be same in the two phases
B. The temperature of each phase should be the same
C. The pressure should be same in the two phases
D. The chemical potential of each component should be same in the two phases
(1992)
74. For a system in equilibrium, at a given temperature and pressure,
A. The entropy must be a minimum
B. The enthalpy must be a minimum
C. The internal energy must be a minimum
D. The Gibbs free energy must be a minimum
(1991)
75. To obtain the integrated form of Clausius–Clapeyron equation
A.
B.
C.
D.
(1999)
97. In a binary liquid solution of components A and B, if component A exhibits positive deviation
from Raoult’s law then component B
A. exhibits positive deviation from Raoult’s law
B. exhibits negative deviation from Raoult’s law
C. obeys Raoult’s law
D. may exhibit either positive or negative deviation from Raoult’s law
(2000)
98. Assume that benzene is insoluble in water. The normal boiling points of benzene and water are
353.3 K and 373.2 K, respectively. At a pressure of 1 atm, the boiling point of a mixture of
benzene and water is
A. 353.3 K
B. less than 353.3 K
C. 373.2 K
D. greater than 353.3 K but less than 373.2 K
(2000)
99. On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible
isothermal line at point A. Then at point A, the slope of the reversible adiabatic line
(∂P/∂V)S and the slope of the reversible isothermal line (∂P/∂V)T are related as
A.
B.
C.
D.
where g = CP/CV.
(2000)
100. The thermal efficiency of a reversible heat engine operating between two given thermal
reservoirs is 0.4. The device is used either as a refrigerator or as a heat pump between the same
reservoirs. The coefficient of performance as a refrigerator (COP)R and the coefficient of
performance as a heat pump (COP)HP are
A. (COP)R = (COP)HP = 0.6
B. (COP)R = 2.5; (COP)HP = 1.5
C. (COP)R = 1.5; (COP)HP = 2.5
D. (COP)R = (COP)HP = 2.5
(2000)
101. At a given temperature, K1, K2 and K3 are the equilibrium constants for the following
reactions 1, 2, 3 respectively:
where fi is a vapour fugacity coefficient, gi is the liquid activity coefficient and is the
fugacity of pure component i. The Ki value (yi = Kixi) is therefore, in general, a function of
A. temperature only
B. temperature and pressure only
C. temperature, pressure and liquid composition xi only
D. temperature, pressure, liquid composition xi, and vapour composition yi
(2001)
103. High pressure steam is expanded adiabatically and reversibly through a well insulated turbine
which produces some shaft work. If the enthalpy change and entropy change across the turbine
are represented by DH and DS respectively, for this process:
A. DH = 0 and DS = 0
B. DH 0 and DS = 0
C. DH 0 and DS 0
D. DH = 0 and DS 0
(2001)
104. For the case of a fuel gas undergoing combustion with air, if the air/fuel ratio is in-creased,
the adiabatic flame temperature will
A. increase
B. decrease
C. increase or decrease depending on the fuel type
D. not change (2001)
105. The Maxwell relation derived from the differential equation for the Helmholtz free energy
(DA) is
A.
B.
C.
D.
(2001)
106. At 373 K, water and methyl cyclohexane both have a vapour pressure of 1.0 atm. The latent
heats of vaporization are 40.63 kJ/kmol for water and 31.55 kJ/kmol for cyclohexane. The
vapour pressure of water at 423 K is 4.69 atm. The vapour pressure of methyl-cyclohexane at
423 K is expected to be
A. Significantly less than 4.69 atm
B. Nearly equal to 4.69 atm
C. Significantly more than 4.69 atm
D. Indeterminate due to lack of data
(2001)
107. Air enters an adiabatic compressor at 300 K. The exit temperature for a compression ratio of
3, assuming air to be an ideal gas (g = Cp/CV = 7/5) and the process to be reversible, is
A. 300(32/7)
B. 300(33/5)
C. 300(33/7)
D. 300(35/7)
(2001)
108. The extent of reaction is
A. different for reactants and products
B. dimensionless
C. dependent on the stoichiometric coefficients
D. all of the above
(2002)
109. An exothermic reaction takes place in an adiabatic reactor. The product temperature
the reactor feed temperature.
A. is always equal to
B. is always greater than
C. is always less than
D. may be greater or less than
(2002)
110. The number of degrees of freedom for an azeotropic mixture of ethanol and water in vapour-
liquid equilibrium is
A. 3
B. 1
C. 2
D. 0
(2002)
111. The partial molar enthalpy of a component in an ideal binary gas mixture of composition z, at
a temperature T and pressure P, is a function only of
A. T
B. T and P
C. T, P and z
D. T and z
(2002)
112. Which of the following identities can most easily be used to verify steam table data for
superheated steam?
A.
B.
C.
D.
(2002)
113. Steam undergoes isentropic expansion in a turbine from 5000 kPa and 673 K (entropy = 6.65
kJ/kg K) to 150 kPa (entropy of saturated liquid = 1.4336 kJ/kg K, entropy of saturated vapour =
7.2234 kJ/kg K). The exit condition of steam is
A. superheated vapour
B. partially condensed vapour with quality of 0.9
C. saturated vapour
D. partially condensed vapour with quality of 0.1
(2002)
114. A rigid vessel, containing three moles of nitrogen gas at 303 K is heated to 523 K. Assume the
average heat capacities of nitrogen to be CP = 29.1 J/mol K and CV =
20.8 J/mol K. The heat required, neglecting the heat capacity of the vessel, is
A. 13728 J
B. 19206 J
C. 4576 J
D. 12712 J
(2002)
115. One cubic metre of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its
volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is
21 J/mol K, the final temperature will be
A. 35 K
B. 174 K
C. 274 K
D. 154 K
(2002)
116. Ammonia is produced by the following reaction:
N2 + 3H2 2NH3
In a commercial process for ammonia production, the feed to an adiabatic reactor con-tains 1
kmol/s of nitrogen and stoichiometric amounts of hydrogen at 700 K. Assume the feed and
product streams to be ideal gas mixtures. The heat of reaction at 700 K for the above reaction is
calculated to be – 94.2 kJ/mol. The mean molar heat capacity in the range of 700–800 K are
0.03, 0.0289 and 0.0492 kJ/mol K for nitrogen, hydrogen and ammonia respectively. What is the
maximum allowable conversion in the reactor, if the adiabatic temperature rise across the
reactor should not exceed 100 K?
A. 87.9%
B. 12.1%
C. 25.8%
D. 74.2%
(2002)
117. In Joule’s experiments, an insulated container contains 20 kg of water initially at 25°C. It is
stirred by an agitator, which is made to turn by a slowly falling body weighing
40 kg through a height of 4 m. The process is repeated 500 times. The acceleration due to gravity
is 9.8 m/s2. Neglecting the heat capacity of the agitator, the temperature of water in (°C) is
A. 40.5
B. 34.4
C. 26.8
D. 25
(2003)
118. One mole of nitrogen at 8 bar and 600 K is contained in a piston-cylinder assembly. It is
brought to 1 bar isothermally against a resisting pressure of 1 bar. The work done (in joules) by
the gas is
A. 30554
B. 10373
C. 4988.4
D. 4364.9
(2003)
119. For water at 573 K, it has a vapour pressure of 8592.7 kPa and fugacity 6738.9 kPa. Under
these conditions, one mole of water in liquid phase has a volume of 25.28 cm3, and that in
vapour phase 391.1 cm3. The fugacity of water (in kPa) at 9000 kPa is
A. 6738.9
B. 6753.5
C. 7058.3
D. 9000
(2003)
120. The heat capacity of air can be approximately expressed as
CP = 26.693 + 7.365 10–3T
where CP is in J/mol K and T is in K. The heat given off by one mole of air when cooled at 1
atmospheric pressure from 500°C to –100°C is
A. 10.73 kJ
B. 16.15 kJ
C. 18.11 kJ
D. 18.33 kJ
(2003)
121. A solid metallic block weighing 5 kg has an initial temperature of 500°C; 40 kg of water
initially at 25°C is contained in a perfectly insulated tank. The metallic block is brought into
contact with water. Both of them come to equilibrium. The specific heat of the block material is
0.4 kJ/kg K. Ignoring the effect of expansion and contraction, and also the heat capacity of the
tank, the total entropy change in kJ/kg K is
A. –1.87
B. 0.0
C. 1.26
D. 3.91
(2003)
122. The following heat engine produces a power of 100,000 kW. The heat engine operates
between 800 K and 300 K. It has a thermal efficiency equal to 50% of that of the Carnot engine
for the same temperatures. The rate at which heat is absorbed from the hot reservoir is
A. 100,000 kW
B. 160,000 kW
C. 200,000 kW
D. 320,000 kW
(2003)
123. A steam turbine operates with a superheated steam flowing at 1 kg/s. The steam is supplied at
41 bar and 500°C, and discharges at 1.01325 bar and 100°C.
A.
B.
C.
D.
(2004)
134. A cyclic engine exchanges heat with two reservoirs maintained at 100°C and 300°C,
respectively. The maximum work (in J) that can be obtained from 1000 J of heat extracted from
the hot reservoir is
A. 349
B. 651
C. 667
D. 1000
(2004)
135. The vapour pressure of water is given by Psat = A – , where A is a constant, Psat is
vapour pressure in atm, and T is temperature in K. The vapour pressure of water
(in atm) at 50°C is approximately
A. 0.07
B. 0.09
C. 0.11
D. 0.13
(2004)
136. At standard conditions,
B.
C.
D.
(2005)
140. Which one of the following statements is true?
A. Heat can be fully converted into work.
B. Work cannot be fully converted to heat.
C. The efficiency of a heat engine increases as the temperature of the heat source is increased
while keeping the temperature of the heat sink fixed.
D. A cyclic process can be devised whose sole effect is to transfer heat from a lower
temperature to a higher temperature.
(2005)
141. A Carnot heat engine cycle is working with an ideal gas. The work performed by the gas
during the adiabatic expansion and compression steps, W1 and W2 respectively, are related as
A. |W1| > |W2|
B. |W1| < |W2|
C. W1 = W2
D. W1 = –W2
(2005)
142. The van Laar activity coefficient model for a binary mixture is given in the form
Given g1 = 1.40, g2 = 1.25, x1 = 0.25, x2 = 0.75, determine the constants A* and B*.
A. A* = 0.5, B* = 0.3
B. A* = 3, B* = 0.5
C. A* = 0.333, B* = 0.2
D. A* = 2, B* = 0.333
(2005)
143. A liquid mixture of benzene and toluene is in equilibrium with its vapour at 101.3 kPa and
373 K. The vapour pressures of benzene and toluene at 373 K are 156 and 63 kPa respectively.
Assuming that the system obeys Raoult’s law, the mole fraction of benzene in the liquid phase is
A. 0.65
B. 0.41
C. 0.065
D. 0.04
(2005)
144. A frictionless cylinder piston assembly contains an ideal gas. Initially at pressure (P1) = 100
kPa, temperature (T1) = 500 K and volume (V1) = 700 10–6 m3. This system is supplied
with 100 J of heat and pressure is maintained constant at 100 kPa. The enthalpy variation is
given by h (J/mol) = 30000 + 50T, where T is the temperature in K, and the universal gas
constant R = 8.314 J/mol K. The final volume of the gas (V2) in m3 is
A. 700 10–6
B. 866.32 10–6
C. 934.29 10–6
D. 1000.23 10–6
(2005)
145. A frictionless cylinder piston assembly contains an ideal gas. Initially pressure (P1) = 100
kPa, temperature (T1) = 500 K and volume (V1) = 700 10–6 m3. This system is supplied
with 100 J of heat and pressure is maintained constant at 100 kPa. The enthalpy variation is
given by h (J/mol) = 30000 + 50T, where T is the temperature in K, and the universal gas
constant R = 8.314 J/mol K. The change in internal energy of the gas is
A. 0
B. 100
C. 23.43
D. 83.37
(2005)
146. Heat and work are
A. intensive properties
B. extensive properties
C. point functions
D. path functions
147. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m3. It
expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output
(in kJ) during the process will be
A. 8.32
B. 12.0
C. 554.67
D. 8320.00
148. The contents of a well-insulated tank are heated by a resistor of 23 W in which 10 A current is
flowing. Consider the tank along with its contents as a thermodynamic system. The work done by
the system and the heat transfer to the system are positive. The rates of heat Q, work W and
change in internal energy DU during the process in kW are
A. Q = 0, W = –2.3, DU = +2.3
B. Q = +2.3, W = 0, DU = +2.3
C. Q = +2.3, W = 0, DU = –2.3
D. Q = 0, W = +2.3, DU = –2.3
149. A compressor undergoes a reversible steady-flow process. The gas at inlet and outlet of the
compressor is designated as state 1 and state 2, respectively. Potential and kinetic energy
changes are to be ignored. The following notations are used:
V specific volume and P pressure of the gas
The specific work required to be supplied to the compressor for this gas compression process is
A.
B.
C. V1(P2 – P1)
D. P2(V1 – V2)
150. A gas contained in a cylinder is compressed, the work required for compression being 5000
kJ. During the process heat interaction of 2000 kJ causes the surroundings to be heated. The
change in internal energy of gas during the process is
A. –7000 kJ
B. –3000 kJ
C. 3000 kJ
D. 7000 kJ
151. A mono-atomic ideal gas (g = 1.67, molecular weight = 40) is compressed adiabatically from
0.1 MPa, 300 K to 0.2 MPa. The universal gas constant is 8.314 kJ/kmol K. The work of
compression of the gas (in kJ/kg) is
A. 29.7
B. 19.9
C. 13.3
D. 0
152. A gas expands in a frictionless piston-cylinder arrangement. The expansion process is very
slow and is resisted by an ambient pressure of 100 kPa. During the expansion process, the
pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is
0.01 m3. The maximum amount of work that could be utilized from the above process is
A. zero
B. 1 kJ
C. 2 kJ
D. 3 kJ
153. One kilogram water at room temperature is brought into contact with a high temperature
thermal reservoir. The entropy change of the universe is
A. equal to entropy change of the reservoir
B. equal to entropy change of water
C. equal to zero
D. always positive
154. If a closed system is undergoing an irreversible process, the entropy of the system
A. must increase
B. always remains constant
C. must decrease
D. can increase, decrease or remain constant
155. Two moles of oxygen are mixed adiabatically with another 2 mol of oxygen in a mixing
chamber, so that the final total pressure and temperature of the mixture become equal to that of
the individual constituents at their initial states. The universal gas constant is given as R. The
change in entropy due to mixing per mole of oxygen is given by
A. –R ln 2
B. zero
C. R ln 2
D. R ln 4
156. Availability of a system at any given state is
A. a property of the system
B. the maximum work obtainable as the system goes to dead state
C. the total energy of the system
D. the maximum useful work obtainable as the system goes to dead state
157. Consider the following two processes:
I. A heat source at 1200 K loses 2500 kJ of heat to sink at 800 K
II. A heat source at 800 K loses 2000 kJ of heat to sink at 500 K
which of the following statements is true?
A. Process I is more irreversible than Process II
B. Process II is more irreversible than Process I
C. Irreversibility associated in both the processes is equal
D. Both the processes are reversible
158. An irreversible heat engine extracts heat from a high temperature source at a rate of
100 kW and rejects heat to a sink at a rate of 50 kW. The entire work output of the heat engine is
used to drive a reversible heat pump operating between a set of independent isothermal heat
reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high
temperature sink is
A. 50
B. 250
C. 300
D. 360
Common data for Questions 159 and 160.
In an experimental set-up, air flows between two stations P and Q adiabatically. The direction
of flow depends on the pressure and temperature conditions maintained at P and Q. The
conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The
following are the properties and relations pertaining to air:
Specific heat at constant pressure CP = 1.005 kJ/kg K, Specific heat at constant volume CV =
0.718 kJ/kg K, Universal gas constant R = 0.287 kJ/kg K, Enthalpy H = CpT, Internal energy U =
CVT.
159. If the air has to flow from station P to station Q, the maximum possible value of pressure in
kPa at station Q is close to
A. 50
B. 87
C. 128
D. 150
160. If the pressure at station Q is 50 kPa, the change in entropy (SQ – SP) in kJ/kg K is
A. – 0.155
B. 0 C. 0.160
D. 0.355
161. A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is
transferred to/from the cyclic device. It is assumed that heat transfer between each thermal
reservoir and cyclic device takes place across negligible temperature difference. Interactions
between the cyclic device and the respective thermal reservoirs that are shown in the figure are
all in the form of heat transfer.
A.
B.
C.
D. zero
Common data for Questions 164 and 165.
Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition
of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal
condition, until the volume becomes 2 m3. Heat exchange occurs with the atmosphere at 298 K
during this process.
164. The work interaction for the nitrogen gas is
A. 200 kJ
B. 138.6 kJ
C. 2 kJ
D. –200 kJ
165. The entropy change for the universe during the process in kJ/K is
A. 0.4652
B. 0.0067 C. 0
D. – 0.6711
166. A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature
reservoir is 727°C, what is the temperature of the low temperature reservoir?
A. 23°C
B. –23°C
C. 0°C
D. 250°C
167. A cyclic heat engine does 50 kJ of work per cycle. If the efficiency of the heat engine is 75%,
the heat rejected per cycle is
A. kJ
B. kJ
C. kJ
D. kJ
168. A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the
internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is used to
run a heat engine which rejects heat at 313 K. If the heat engine is to deliver 2.5 kW power, the
minimum area of the solar collector required would be
A. 8.33 m2
B. 16.66 m2
C. 39.68 m2
D. 79.36 m2
169. Considering the relationship TdS = d U + P d V between the entropy S, internal energy U,
pressure P, temperature T and volume V, which of the following statements is correct?
A. It is applicable only for a reversible process
B. For an irreversible process, TdS > dU + PdV
C. It is valid only for an ideal gas
D. It combines first and second laws for a reversible process
170. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The
balloon ruptures and the gas fills up the entire room. Which one of the following statements is
true at the end of the above process?
A. The internal energy of the gas decreases from its initial value but the enthalpy remains
constant.
B. The internal energy of the gas increases from its initial value but the enthalpy remains
constant.
C. Both internal energy and enthalpy of the gas remain constant.
D. The internal energy and enthalpy of the gas increase.
171. In a steady-state flow process taking place in a device with a single inlet and single outlet, the
work done per unit mass flow rate is given by , where V is the specific volume
and P is the pressure. The expression for W given above is
A. valid only if the process is both reversible and adiabatic
B. valid only if the process is both reversible and isothermal
C. valid for any reversible process
D. incorrect; it must be
172. The following four figures have been drawn to represent a fictitious thermodynamic cycle on
the P–V and T–S planes:
177. At the end of the process, which one of the following situations will be true?
A. Superheated vapour will be left in the system
B. No vapour will be left in the system
C. A liquid + vapour mixture will be left in the system
D. The mixture will exist at a dry saturated vapour state
178. The work done by the system during the process is
A. 0.1 kJ
B. 0.2 kJ
C. 0.3 kJ
D. 0.4 kJ
179. The net entropy generation (considering the system and the thermal reservoir together) during
the process is closest to
A. 7.5 J/K
B. 7.7 J/K
C. 8.5 J/K
D. 10 J/K
180. A gas having a negative Joule Thomson coefficient (m < 0), when throttled will
A. become cooler
B. become warmer
C. remain at the same temperature
D. either be cooler or warmer depending on the type of gas
181. An ideal Brayton cycle operating between the pressure limits of 1 bar and 6 bar has minimum
and maximum temperatures of 300 K and 1500 K. The ratio of the specific heats of the working
fluid is 1.4. The approximate final temperatures in kelvin at the end of the compression and
expansion processes are, respectively
A. 500 and 900
B. 900 and 500
C. 500 and 500
D. 900 and 900
182. The values of enthalpy of steam at the inlet and outlet of a steam turbine in a Rankine cycle are
2800 kJ/kg and 1800 kJ/kg respectively. Neglecting pump work, the specific steam consumption
in kg/kWh is
A. 3.6
B. 0.36
C. 0.06
D. 0.01
Statements for linked answer Questions 183 and 184.
The temperature and pressure of air in a large reservoir are 400 K and 3 bar, respectively. A
converging-diverging nozzle of exit area 0.005 m2 is fitted to the reservoir as shown in the
figure. The static pressure of air at the exit section, for isentropic flow through the nozzle, is 50
kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287 kJ/kg K and
1.4, respectively.
186. Nitrogen at an initial state of 10 bar, 1 m3 and 300 K is expanded isothermally to a final
volume of 2 m3. The P-V-T relation is , where a > 0. The final pressure.
A. will be slightly less than 5 bar
B. will be slightly more than 5 bar
C. will be exactly 5 bar
D. cannot be ascertained in the absence of the value of a
Common data for Questions 187 and 188
The following table of properties was printed out for saturated liquid and saturated vapour of
ammonia. The titles for only the first two columns are available. All that we know is that the
other columns (columns 3 to 8) contain data on specific properties, namely, internal energy
(kJ/kg), enthalpy (kJ/kg) and entropy (kJ/kg K).
attains equilibrium at 400 K and 3 bar. The standard Gibbs free energy change of reaction at
these condition is DG0 = 4000 J/mol. For two moles of an equimolar feed of ethylene and
steam, the equation in terms of the extent of reaction e (in moles) at equilibrium is
A.
B.
C.
D.
(2007)
210. A methanol-water vapour liquid system is at equilibrium at 333 K and 60 kPa. The mole
fraction of methanol in liquid is 0.5 and in vapour is 0.8. Vapour pressures of methanol and
water at 333 K are 85 kPa and 20 kPa, respectively. Assuming vapour phase to be an ideal gas
mixture, what is the activity coefficient of water in the liquid phase?
A. 0.3
B. 1.2
C. 1.6
D. 7.5
(2007)
211. For conditions in Question 210, what is the excess Gibbs free energy (GE, J/mol) of the liquid
mixture?
A. 9.7
B. 388
C. 422
D. 3227
(2007)
212. A perfectly insulated cylinder of volume 0.6 m3 is initially divided into two parts by a thin,
frictionless piston, as shown in the figure. The smaller part of volume 0.2 m3 has ideal gas at 6
bar pressure and 373 K. The other part is evacuated.
At certain instant of time t, the stopper is removed and the piston moves out freely to the other
end. The final temperature is
A. 124 K
B. 240 K
C. 306 K
D. 373 K
(2007)
213. The cylinder insulation is removed and the piston is pushed back to restore the system to the
initial state. If this is to be achieved only by doing work on the system (no heat addition, only
heat removal allowed), what is the minimum work required?
A. 3.4 kJ
B. 107 kJ
C. 132 kJ
D. 240 kJ
(2007)
214. For a Carnot refrigerator operating between 40°C and 25°C, the coefficient of performance is
A. 1
B. 1.67
C. 19.88
D. 39.74
(2008)
215. The work done by one mole of a van der Waals fluid undergoing reversible isothermal
expansion from initial volume Vi to final volume Vf is
A.
B.
C.
D.
(2008)
216. The standard Gibbs free energy change and enthalpy change at 298 K for the liquid phase
reaction CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) are given as DG0 = –
4650 J/mol and DH0 = – 3640 J/mol. If the solution is ideal and enthalpy change is assumed to
be constant, the equilibrium constant at 368 K is
A. 0.65
B. 4.94
C. 6.54
D. 8.65
(2008)
217. A binary mixture containing species 1 and 2 forms an azeotrope at 378.6 K and 1.013 bar. The
liquid phase mole fraction of component 1 (xl) of this azeotrope is 0.62. At 378.6 K, the pure
component vapour pressures for species 1 and 2 are 0.878 bar and 0.665 bar, respectively.
Assume that the vapour phase is an ideal gas mixture. The van Laar constants, A and B, are given
by the expressions
A.
B.
C.
D.
(2009)
221. An ideal gas with molar heat capacity CP = 5/2 R (where R = 8.314 J/mol K) is compressed
adiabatically from 1 bar and 300 K to pressure P2 in a closed system. The final temperature
after compression is 600 K and the mechanical efficiency of compression is 50%. The work
required for compression in (kJ/mol) is
A. 3.74
B. 6.24
C. 7.48
D. 12.48
(2009)
222. In the above problem, the pressure P2 (in bar) is
A. 23/4
B. 25/4
C. 23/2
D. 25/2
(2009)
223. A new linear temperature scale, denoted by °S, has been developed, where the freezing point
of water is 200°S and the boiling point is 400°S. On this scale, 500°S corresponds, in degree
Celsius, to
A. 100°C
B. 125°C
C. 150°C
D. 300°C
(2010)
224. An equimolar mixture of species 1 and 2 is in equilibrium with its vapour at 400 K. At this
temperature, the vapour pressures of the species are kPa and kPa. Assuming
Raoult’s law is valid, the value of y1 is
A. 0.30
B. 0.41
C. 0.50
D. 0.60
(2010)
225. A saturated liquid at 1500 kPa and 500 K, with an enthalpy of 750 kJ/kg, is throttled to a
liquid-vapour mixture at 150 kPa and 300 K. At the exit conditions, the enthalpy of the saturated
liquid is 500 kJ/kg and the enthalpy of the saturated vapour is 2500 kJ/kg. The percentage of the
original liquid, which vaporises, is
A. 87.5%
B. 67%
C. 12.5%
D. 10%
(2010)
226. At constant temperature and pressure, the molar density of a binary mixture is given by r = 1 +
x2, where x2 is the mole fraction of component 2. The partial molar volume at infinite dilution
of component 1, is
A. 0.75
B. 1.0
C. 2.0
D. 4.0
(2010)
227. Minimum work (W) required to separate a binary gas mixture at a temperature T0 and
pressure P0 is
where, y1 and y2 are mole fractions, fpure, 1, and fpure, 2 are fugacities of pure species at T0
and P0 and are fugacities of species in the mixture at To, Po and y1. If the mixture is
ideal, then W is
A. 0
B. W = – RT0[y1 ln y1 + y2 ln y2]
C. W = RT0[y1 ln y1 + y2 ln y2]
D. W = RT0
(2011)
228. The partial molar enthalpies of mixing (in J/mol) for benzene (component 1) and cyclohexane
(component 2) at 300 K and 1 bar are given by , where x1 and
x2 are the mole fractions. When 1 mol of benzene is added to 2 mol of cyclohexane, the enthalpy
change (in J) is
A. 3600
B. 2400
C. 2000
D. 800
(2011)
229. One mol of methane is contained in a leak-proof piston-cylinder assembly at 8 bar and 1000
K. The gas undergoes isothermal expansion to 4 bar under reversible conditions. Methane can
be considered as an ideal gas under these conditions. The value of universal gas constant is
8.314 J/mol K. The heat transferred (in kJ) during the process is
A. 11.52
B. 5.76
C. 4.15
D. 2.38
(2011)
230. Consider a binary mixture of methyl ethyl ketone (component 1) and toluene
(component 2). At 323 K, the activity coefficients g1 and g2 are given by
where x1 and x2 are the mole fractions in the liquid mixture, and Y1 and Y2 are parameters
independent of composition. At the same temperature, the infinite dilution activity coefficients,
are given by . The vapour pressures of methyl ethyl ketone
and toluene at 323 K are 36.9 and 12.3 kPa, respectively. Assuming that the vapour phase is
ideal, the equilibrium pressure (in kPa) of a liquid mixture containing 90 mol % toluene is
A. 19
B. 18
C. 16
D. 15
(2011)
231. In a throttling process, the pressure of an ideal gas reduces by 50%. If CP and CV are the heat
capacities at constant pressure and constant volume, respectively (g = CP/CV), the specific
volume will change by a factor of
A. 2
B. 21/g
C. 2g–1/g
D. 0.5
(2012)
232. If the temperature of saturated water is increased infinitesimally at constant entropy, the
resulting state of water will be
A. liquid
B. liquid-vapour coexistence
C. saturated vapour
D. solid
(2012)
233. In a parallel flow heat exchanger operating under steady state, hot liquid enters at a
temperature Th, in and leaves at a temperature Th, out. Cold liquid enters at a temperature Tc, in
and leaves at a temperature Tc, out. Neglect any heat loss from the heat exchanger to the
surrounding. If Th, in > > Tc, in, then for a given time interval, which one of the following
statements is true?
A. Entropy gained by the cold stream is greater than the entropy lost by the hot stream
B. Entropy gained by the cold stream is equal to the entropy lost by the hot stream
C. Entropy gained by the cold stream is less than the entropy lost by the hot stream
D. Entropy gained by the cold stream is zero.
(2012)
234. For an exothermic reversible reaction, which one of the following correctly describes the
dependence of the equilibrium constant (K) with temperature (T) and pressure (P)?
A. K is independent of T and P.
B. K increases with an increase in T and P.
C. K increases with T and decreases with P.
D. K decreases with an increase in T and is independent of P.
(2012)
235. An insulated, evacuated container is connected to a supply line of an ideal gas at pressure PS,
temperature TS and specific volume VS. The container is filled with the gas until the pressure in
the container reaches PS. There is no heat transfer between the supply line to the container, and
kinetic and potential energies are negligible. If CP
and CV are the heat capacities at constant pressure and constant volume, respectively
(g = CP/CV), then the final temperature of the gas in the container is
A. gTS
B. TS
C. (g – 1) TS
D. (g – 1)TS/g
(2012)
236. Consider a binary liquid mixture at constant temperature T and pressure P. If the enthalpy
change of mixing DH = 5x1x2, where xl and x2 are mole fraction of species 1 and 2,
respectively, and the entropy change of mixing DS = – R(x1 ln x1 + x2 ln x2) with
R = 8.314 J/mol K, then the minimum value of the Gibbs free energy change of mixing at 300 K
occurs when
A. x1 = 0
B. x1 = 0.2
C. x1 = 0.4
D. x1 = 0.5
(2012)
C.2 Fill in the blanks:
1. properties of a system do not depend on the quantity of matter contained in it.
2. An open system exchanges with the surroundings.
3. The maximum efficiency of a heat engine depends only on the between which it operates
and is independent of the nature of the cyclic process.
4. Gibbs free energy is defined as .
5. Mollier diagram is a plot of versus .
6. The efficiency of a Carnot engine working between 1000 K and 300 K is .
7. In the statement (DS)total ≥ 0 , the inequality refers to process.
8. P, V, T and S are properties whereas U, H, G, A are properties.
9. The principle of corresponding states may be stated thus: “At same TR and PR all gases have the
same .”
10. A gaseous phase may be termed a vapour, if it can be condensed by .
11. A refrigerator of capacity 2 tons is working on ammonia at 273 K. The heat of vaporisation of
ammonia is 1260 kJ/kg. The circulation rate of ammonia under this condition is approximately
kg/h.
12. Gibbs–Helmholtz equation relates the change in with changes in .
13. A system from which finite quantities of heat can be removed without affecting its temperature
is called .
14. The maximum velocity attained by a fluid in a pipe of uniform cross-section is equal to the
in the fluid.
15. The maximum velocity attainable in a convergent nozzle is equal to and it is attained
when the equals the critical value.
16. The ratio of the intake volume to the displacement volume in a single-stage compressor is
called the .
17. The ratio of the velocity of flow to the sonic velocity is designated as .
18. The decrease in is a measure of the maximum work obtainable in an isothermal
process.
19. As pressure tends to zero, fugacity of a pure gas becomes equal to its .
20. The ratio of the fugacity to the fugacity in the standard state is called .
21. Isothermal mixing of pure gases always produces a decrease in the . Hence work has to
be done the system for separating a mixture of gases into its components.
(1990)
22. The maximum work obtainable from a closed system under isothermal expansion is given by
; For one mole of an ideal gas expanding isothermally to twice its volume this is equal to
.
(1990)
23. The phase rule is given as .
(1994)
24. Raoult’s law states that the of a component over an ideal solution is directly
proportional to its mole fraction in the solution.
25. In a dilute solution, the obeys Henry’s law and the obeys Raoult’s law.
26. When the Henry’s law constant is equal to , Henry’s law becomes identical to Raoult’s
law.
27. The activity coefficient (gi ) in a solution is related to the chemical potential as = .
28. The phase rule indicates the number of variables needed to specify the intensive state of the
system whereas the indicates those needed to specify the extensive state of the system.
29. A mixture exists as a superheated vapour above its temperature.
30. If the intermolecular forces between unlike molecules are than those between like
molecules, the solution will exhibit negative deviation from ideality.
31. The constant boiling mixtures are called .
32. The vaporisation equilibrium constant Ki is defined as Ki = .
33. Among three liquids A, B and C, the A-B binary is partially miscible whereas A-C and
B-C binaries are totally soluble. On the binodal curve, the A-rich and B-rich phases
in equilibrium become identical in properties at the of the system.
34. The number of degrees of freedom for a system consisting of two miscible non-reacting species
which exists as an azeotrope in vapour–liquid equilibrium is .
35. The equilibrium state for a closed system is the state for which the total is a minimum
at constant temperature and pressure.
36. A binary hydrocarbon liquid mixture of A and B (KA = 1.5) containing 60% (mol) A is flash
vaporised. If 40% of the feed is vaporised, the mole fraction of A in the liquid product is
.
(1990)
37. A system of unit mass at equilibrium consists of two phases a and b of extent x and (1 – x)
respectively. Write down expressions for the pressure and the specific enthalpy of the system as
a whole in terms of the properties Pa, Pb, Ha and Hb of the individual phases: (a) P = ,
(b) H = .
(1990)
38. The heats of formation of CO (g), H2O (g), and CO2 (g) are respectively –110.525 kJ,
–393.509 kJ and –241.818 kJ. The heat of reaction for
CO (g) + H2O (g) CO2 (g) + H2 (g)
is kJ.
39. reactions are favoured by increase in temperature.
40. The following data on heats of combustion at 298 K are given:
Heats of formation of CO2 (g) and H2O (l) are –390 and –280 kJ/kmol respectively.
(a) The heat of formation of gaseous n-heptane at 298 K is .
(b) The heat of formation of gaseous ethyl alcohol at 298 K is .
(1990)
41. The heat absorbed for isothermal reaction
C4H10 (g) C2H4 (g) + C2H6 (g)
at 298 K and 101.3 kPa is . Standard heat of combustion in kJ/kmol are:
C4H10 (g) = – 2873.5, C2H4 (g) = – 1411.9 and C2H6 (g) = – 1561.0
(1991)
42. The heat of formation of a compound is defined as the heat of reaction leading to the formation
of the compound from its .
(1994)
C.3 Say, whether the following statements are TRUE or FALSE. Give correct statements to the false
ones.
1. Internal energy is a state function whereas entropy is a path function.
2. Heat capacity and specific heat are extensive properties whereas volume and temperature are
intensive properties.
3. Heat and work are not properties of a system; they are properties of a process.
7. So long as the process is reversible, the value of is the same for the change of the gas
from any given state to another.
8. Heat involved in any process can be expressed as dQ = T dS.
9. Enthalpy and entropy of an ideal gas are functions of temperature alone.
10. If DS refers to the entropy change between the same initial and final states of the system for two
different processes, one reversible (R) and the other irreversible (I), then DSI = DSR.
11. The heat capacities CP and CV of an ideal gas are independent of temperature.
12. The second law of thermodynamics states that heat cannot be completely converted to work.
13. For any process, the second law of thermodynamics requires that the entropy change of the
system is either zero or positive.
14. The entropy change of a chemical reaction is calculated as the ratio of the heat of reaction to the
temperature of the reaction.
15. When water freezes to form ice, the atoms arrange themselves in a highly ordered manner.
Since the increasing order is associated with the decreasing entropy, we must conclude that
entropy of the universe decreases as a result of this process.
16. Entropy of a rotating flywheel is the same as that of the flywheel at rest.
17. For given operating temperatures all heat engines have the same efficiency regardless of the
nature of the working substance.
18. Real gases behave ideally at high pressures and temperatures.
19. A reversible adiabatic process is essentially isenthalpic.
20. The heat capacity at constant pressure and constant volume of all gases are related as
CP – CV = R.
21. For an ideal gas, the activity and fugacity are numerically equal.
22. For an ideal solution, all property changes of mixing are zero.
23. Raoult’s law is applicable to all ideal liquid solutions.
24. On the P-T diagram of a pure substance, the vaporisation curve and the fusion curve extend up
to infinity.
25. The change in internal energy of an ideal gas is DU = irrespective of the nature of the
process.
26. For gases, the Joule–Thomson coefficient is always positive.
27. Work required for isothermal compression is less than that of adiabatic compression.
28. The clearance has no effect on the work of compression in a single-stage compressor.
29. The reversible work of expansion in a non-flow process under isentropic condition is equal to
–(DU)S.
30. For an ideal gas, the fugacity and pressure are equal.
31. The excess volume and the volume change on mixing are the same.
32. For a multicomponent system, equilibrium between two phases is established when the
concentrations in both the phases are uniform.
33. For a solution at a given pressure, the vapour phase can exist in equilibrium with the liquid
phase only at its bubble point.
34. In an ideal binary solution, component A obeys Raoult’s law and component B obeys Henry’s
law.
35. Maximum boiling azeotropes may be formed if the solution exhibits very large positive
deviation from ideality.
36. Azeotropic composition can be shifted by changing the pressure.
37. For a chemically reacting system at equilibrium at constant temperature and pressure, the Gibbs
free energy is maximum.
38. The numerical value of the equilibrium constant depends upon the stoichiometric equation.
39. If there is decrease in the total number of moles during a gas-phase chemical reaction, the
increase in pressure decreases the formation of products.
40. The equilibrium conversion in a gaseous reaction which produces no change in the number of
moles (e.g., the water-gas shift reaction) is not affected by the change in pressure.
Answers
C.1
1. C 2. B 3. C 4. B 5. A 6. C 7. A
8. C 9. C 10. C 11. B 12. B 13. C 14. A
15. D 16. B 17. D 18. C 19. B 20. A 21. C
22. B 23. A 24. B 25. C 26. B 27. B 28. D
29. A 30. C 31. B 32. C 33. C 34. D 35. D
36. C 37. A 38. B 39. A 40. A 41. B 42. D
43. B 44. C 45. C 46. D 47. B 48. B 49. B
50. D 51. C 52. A 53. A 54. C 55. B 56. A
57. B 58. C 59. C 60. B 61. A 62. C 63. C
64. B 65. A 66. D 67. B 68. D 69. B 70. A
71. C 72. B 73. D 74. D 75. D 76. C 77. A
78. C 79. B 80. B 81. C 82. C 83. B 84. A
85. D 86. B 87. B 88. B 89. C 90. B 91. B
92. A 93. D 94. D 95. A 96. D 97. A 98. B
99. C 100. C 101. A 102. C 103. B 104. B 105. D
106. A 107. A 108. B 109. B 110. B 111. D 112. B
113. B 114. A 115. B 116. B 117. B 118. B 119. B
120. C 121. C 122. D 123. C 124. A 125. B 126. C
127. A 128. B 129. B 130. D 131. A 132. B 133. D
134. A 135. D 136. C 137. A 138. C 139. A 140. C
141. D 142. B 143. B 144. B 145. D 146. D 147. A
148. A 149. B 150. C 151. A 152. C 153. D 154. D
155. B 156. D 157. B 158. C 159. B 160. C 161. A
162. D 163. B 164. B 165. A 166. B 167. A 168. D
169. D 170. C 171. C 172. A 173. D 174. D 175. C
176. A 177. A 178. D 179. C 180. B 181. A 182. A
183. C 184. D 185. C 186. B 187. B 188. B 189. C
190. B 191. A 192. C 193. D 194. B 195. B 196. D
197. C 198. C 199. B 200. B 201. A 202. B 203. B
204. D 205. B 206. A 207. C 208. C 209. D 210. B
211. C 212. D 213. C 214. C 215. D 216. B 217. B
218. C 219. B 220. D 221. C 222. D 223. C 224. D
225. C 226. A 227. B 228. D 229. B 230. C 231. A
232. A 233. A 234. D 235. A 236. D
C.2
1. Intensive
2. Mass and energy
3. Temperature
4. G = H – TS
5. Enthalpy, entropy
6. 70%
7. Irreversible
8. Reference, energy
9. Z (Compressibility factor)
10. Compression at constant temperature
11. 20 kg/h
12. G/T with T
13. Heat reservoir
14. Sonic velocity
15. Sonic velocity, Pressure ratio
16. Theoretical volumetric efficiency
17. Mach number
18. Helmholtz free energy
19. Pressure
20. Fugacity coefficient
21. Gibbs free energy, on
22. P dV, RT ln 2
23. F = C – p + 2
24. Partial pressure (fugacity)
25. Solute, solvent
26. Vapour pressure
27. RT ln gi
28. Duhem’s theorem
29. Dew point
30. Stronger (greater)
31. Azeotropes
32. yi/xi
33. Plait point
34. One
35. Gibbs free energy
36. 0.46
37. (a) Pa = Pb (b) xHa + (1 – x)Hb
38. 262.22
39. Endothermic
40. (a) – 120 kJ/kmol (b) – 210 kJ/kmol
41. 99.4 kJ
42. Constituent elements
C.3
1. False. Both are state functions.
2. False. Heat capacity and volume are extensive properties whereas specific heat and temperature
are intensive properties.
3. True.
4. True.
5. False. Entropy can have absolute values.
6. True.
7. False. The value of P dV is dependent on the path followed.
8. False. Heat involved in a reversible process can be expressed as dQ = T dS.
9. False. Enthalpy of an ideal gas is a function of temperature only. Entropy depends on pressure as
well.
10. True.
11. False. The heat capacities CP and CV of an ideal gas are dependent on temperature only.
12. False. The second law of thermodynamics states that heat cannot be completely converted into
work continuously (or in a cyclic process).
13. False. For any process, the second law of thermodynamics requires that the entropy change of
the system and the surroundings together is either zero or positive.
14. False. The entropy change of a chemical reaction is to be computed as the sum of the absolute
entropies of the products minus the sum of the absolute entropies of the reactants.
15. False. When water freezes to form ice, greater disorder may result in the surroundings due to
transfer of heat with a consequent increase in the total entropy.
16. True.
17. False. For given operating temperatures all reversible heat engines (Carnot engines) have the
same efficiency regardless of the nature of the working substance.
18. False. Real gases behave ideally at low pressures and or high temperatures.
19. False. A reversible adiabatic process is essentially isentropic.
20. False. The heat capacity at constant pressure and constant volume of ideal gases are related as
CP – CV = R.
21. True.
22. False. For an ideal solution, property changes of mixing are not zero for entropy and entropy-
related functions such as the free energy.
23. False. Raoult’s law is applicable to all ideal liquid solutions provided the vapour phase is an
ideal gas.
24. False. On the P-T diagram of a pure substance, the vaporisation curve lies between the triple
point and the critical point whereas the fusion curve extends up to infinity.
25. True.
26. False. For gases the Joule–Thomson coefficient may be positive, zero or negative.
27. True.
28. True.
29. True.
30. True.
31. True.
32. False. For a multicomponent system in equilibrium, the chemical potentials in both phases are
uniform.
33. False. For a solution at a given pressure, the vapour phase can exist in equilibrium for a range of
temperatures lying between the bubble point and the dew point.
34. False. In an ideal binary solution both components obey Raoult’s law.
35. False. Minimum boiling azeotropes may be formed if the solution exhibits very large positive
deviation from ideality.
36. True.
37. False. The Gibbs free energy is minimum.
38. True.
39. False. If there is decrease in the total number of moles during a gas-phase chemical reaction, the
increase in pressure favours the formation of products.
40. False. If the compressibility of the components are affected by the change in pressure, the
equilibrium conversion also will be affected.
References
1. Abrams, D. and Prausnitz, J.M., AIChE J., 21, 116, 1975.
2. Atkins, P.W., Physical Chemistry, 4th ed., ELBS-Oxford, Oxford University Press, 1990.
3. Benedict, M., Webb, G. and Rubin, L., J. Chem. Phys., 8, 334, 1940.
4. Callen, H.B., Thermodynamics, John Wiley, New York, 1960.
5. Daubert, T.E., Chemical Engineering Thermodynamics, McGraw-Hill, New York, 1985.
6. Denbigh, K., The Principles of Chemical Equilibrium, 4th ed., Cambridge, New York, 1981.
7. Dodge, B.F., Chemical Engineering Thermodynamics, McGraw-Hill, New York, 1944.
8. Fredenslund, Aa., Jones, R.L. and Prausnitz, J.M., AIChE J., 21, 1086, 1975.
9. Glasstone, S., Thermodynamics for Chemists, Van Nostrand, New York, 1958.
10. Hill, L., Statistical Mechanics, McGraw-Hill, New York, 1956.
11. Hougen, O.A., Watson, K.M. and Ragatz, R.A., Chemical Process Principles, Part II,
2nd ed., John Wiley, New York, 1960.
12. Karapetyants, M.Kh., Chemical Thermodynamics, Mir Publishers, Moscow, 1978.
13. Kirkwood, I.J. and Oppenheim, I.,Chemical Thermodynamics, McGraw-Hill, New York,
1961.
14. Kyle, B.G., Chemical and Process Thermodynamics, 2nd ed., Prentice-Hall of India,
New Delhi, 1994.
15. Lewis, G.N., Randall, M., Pitzer, K.S. and Brewer, L., Thermodynamics, McGraw-Hill,
New York, 1981.
16. Modell, M. and Reid, R.C.,Thermodynamics and Its Applications, Prentice Hall, New Jersey,
1974.
17. Peng, D.Y. and Robinson, D.B., Ind. Eng. Chem. Fundam., 15, 59, 1976.
18. Perry, J.H. and Chilton, C.H. (Eds.),Chemical Engineers’ Handbook, 5th ed., McGraw-Hill,
Tokyo, 1973.
19. Prausnitz, J.M., Lichtenthaler, R.N. and Azevedo, E.G.,Molecular Thermodynamics of Fluid
Phase Equilibria, 2nd ed., Prentice Hall, New Jersey, 1986.
20. Rao, Y.V.C., An Introduction to Thermodynamics, Wiley Eastern, New Delhi, 1993.
21. Redlich, O. and. Kwong, J.N.S., Chem. Rev., 44, 233, 1949.
22. Renon, H. and Prausnitz, J.M., AIChE J., 14, 135, 1968.
23. Reynolds, W.C. and Perkins, H.C., Engineering Thermodynamics, McGraw-Hill, Tokyo,
1977.
24. Saad, M.A., Thermodynamics for Engineers, Prentice-Hall of India, New Delhi, 1969.
25. Sandler, S.I., Chemical and Engineering Thermodynamics, John Wiley, New York, 1977.
26. Smith, J.M. and Van Ness, H.C.,Introduction to Chemical Engineering Thermodynamics, 4th
ed., McGraw-Hill, New York, 1987.
27. Smith, N.O., Chemical Thermodynamics: A Problems Approach, Reinhold, New York, 1967.
28. Soave, G., Chem. Engg. Sci., 27, 1197, 1972.
29. Sonntag, R.E. and Van Wylen, G.J., Introduction to Thermodynamics, John Wiley,
New York, 1971.
30. Treybal, R.E., Mass-Transfer Operations, 3rd ed., McGraw-Hill, New York, 1981.
31. Wark, K., Thermodynamics, 2nd ed., McGraw-Hill, New York, 1971.
32. Weber, H.E. and Meissner, H.C., Thermodynamics for Chemical Engineers, 2nd ed.,
John Wiley, New York, 1957.
33. Wilson, E.D. and Ries, H.C., Principles of Chemical Engineering Thermodynamics,
McGraw-Hill, New York, 1956.
Answers to Exercises
CHAPTER 1
1.1 89.55 kg; 878.51 N
1.2 102.14 N
1.3 698 mm
1.4 1.6783 bar
1.5 3.02 bar
1.6 1.5453 105 N/m2
1.7 17.845 m, 25 kJ
1.8 1.667 102 kJ
1.9 7.355 kJ; 61.29 W
1.10 16.8 W
1.11 20.39 m
1.12 (a) 4.164 103 N
(b) 1.3254 105 N/m2
(c) 2.082 103 J
(d) 490.5 J
1.13 2.156 105 J
1.14 17.15 m/s
1.15 1.146 104 kJ
1.16 6.073 104 J
CHAPTER 2
2.1 40.75 kJ
2.2 400 K, 500 kPa
2.3 (a) 183.94 kJ
(b) 1839.4 kJ
(c) 1655.46 kJ
(d) 38.36 m/s
(e) 1839.4 kJ
2.4 (a) 981 J
(b) 981 J
(c) 0.234 K
2.5 (a) 14 m/s
(b) 0.78 K
2.6 18,467 kJ
2.7 4466 kJ
2.8 195,814 kJ
2.9 2423.9 kJ/kg; 2214 kJ/kg
2.10 (a) 3.572 kJ
(b) 10.97 min
2.11 Q = 914.86 kJ, W = 139.35 kJ
2.12 227 V
2.13 Q = 1.246 106 kJ
2.14 Q = 58.4 104 J
2.15 435.5 HP
2.16 35 kJ
2.17 70 kJ
2.18 2453 kJ/kg
2.19 389.9 K
2.20 (a) 78.9 m/s
(b) 72.7 kPa
2.21 870.9 kJ/kg, 5.74 kg/s
2.22 55.5 kW
2.23 3.68 MW
2.24 75%
2.25 141.3 MW
2.26 1.722 K
2.27 100 kJ
2.28 (a) 22,480 kJ
(b) 8730 kJ
2.29 166.67 kJ
2.30 (a) V = 4.65584 m3, P = 1 bar, T = 56 K
DU = 4656 kJ, DH = 6518 kJ,
W = 0, Q = 4656 kJ
(b) V = 23.2792 m3, P = 5 bar, T = 1400 K, DU = 22,380.3 kJ, DH = 32,592.0 kJ, W = 0, Q =
22,380.3 kJ
2.31 DU = 2168.6 kJ, DH = 3000 kJ, W = – 125.32 kJ, Q = 2043.28 kJ
CHAPTER 3
3.1 (a) 38.301 103 J
(b) 0
(c) 4.31 bar
(d) – 47.817 103 J
3.2 (a) 24.0685 kJ
(b) – 17.695 kJ
(c) 6.374 kJ
3.3 (a) 2282 kJ
(b) 2206 kJ
(c) 0
3.4 (a) 472.5 kJ
(b) – 4863 kJ
(c) 0
(d) 4863 kJ
(e) 5981 kJ
3.5 (a) DU = 997.68 kJ/kmol;
DH = 1662.8 kJ/kmol;
W = – 997.68 kJ/kmol; Q = 0
(b) DU = – 997.68 kJ/kmol;
DH = – 1662.8 kJ/kmol;
W = – 665.12 kJ/kmol;
Q = – 1662.8 kJ/kmol
(c) DU = 0; DH = 0;
W = 1490 kJ/kmol;
Q = 1490 kJ/kmol
(d) DU = 0; DH = 0;
W = – 172.8 kJ/kmol;
Q = – 172.8 kJ/kmol
3.6 DU = – 2329 kJ/kmol;
DH = – 3252 kJ/kmol;
W = 3395 kJ/kmol;
Q = 1066 kJ/kmol
3.7 (a) 0.373 kg
(b) 0.304 kg
3.8 418.5 K, 7.65 bar
3.9 (a) 4605 kJ
(b) 132.9 K, 2376 kJ
(c) 900 kJ
3.10 3.66 10–4 m3/mol
3.11 (a) 3 10–3 m3/mol
(b) 2.98 10–3 m3/mol
3.12 23.84 bar
3.13 (a) 65.54 bar
(b) 57.23 bar
(c) 57.87 bar
3.14 1.8 10–4 m3/mol
3.15 7.134 10–5 m3/mol (liquid),
1.712 10–3 m3/mol (vapour)
3.16 3.485 10–3 m3/mol, 0.8862
3.17 (a) 4.157 10–5 m3/mol
(b) 6.44 10–5 m3/mol
(c) 5.3 10–5 m3/mol
3.19 – 110.6 kJ/mol
3.20 – 1207.69 kJ/mol
3.21 – 1655.07 kJ/mol
3.22 48.70 kJ/mol
3.23 – 42.62 kJ
3.24 – 395.2 kJ/mol
3.25 – 799.3 kJ/mol
3.26 = – 3.8235 104 – 31.82 T + 1.776 10–2T2 – 3.108 10–6 T3
3.27 – 224.673 kJ
3.28 – 207.2 103 kJ
3.29 32,528.5 kJ
3.30 1360 K
3.31 1216 K
3.32 – 311.627 kJ/mol
3.33 – 3283.5 kJ
3.34 2.7989 104 kJ
3.35 2089.5 K
3.36 – 114.408 kJ/mol
3.38 (a) – 103.2 kJ/mol
(b) = – 75,964 – 62.71T + 4.496 10–2T2 – 9.561 10–6T3 + 11.224 104(1/T)
3.39 2055.8 K
3.40 3.766 106 kJ
3.41 – 3535.50 kJ
3.42 2141.9 K
CHAPTER 4
4.1 329.84 kJ
4.2 1.448 109 J/s
4.3 (a) 2.1502 106 kJ/h
(b) 18.1%
4.4 (a) 2.9736 kJ
(b) 3.9736 kJ
4.5 Unacceptable
4.6 Unacceptable
4.7 Unrealistic
4.8 (a) –19.1437 J/mol K
(b) 30.63 J/K
(c) 11.49 J/K
4.9 79.91 J/K
4.10 (a) 6.66 kJ/K, – 4.6204 kJ/K, 2.04 kJ/K
(b) 6.66 kJ/K, – 4.925 kJ/K, 1.735 kJ/K
(c) 6.66 kJ/K, – 4.89 kJ/K, 1.77 kJ/K
(d) 6.66 kJ/K, – 6.66 kJ/K, 0
4.11 18.531 kJ/kmol K, Possible
4.12 4579.6 kJ/kmol
4.13 7.1 105 kJ
4.14 0.0508 kJ
4.16 31.28 kJ/kmol K, 111.9 kJ/kmol K
4.17 – 2.46 kJ/kmol K
4.18 (a) – 1.99 kJ/K
(b) 3.44 kJ/K
(c) 1.45 kJ/K
4.19 (a) 0.37
(b) 0.56
4.20 (a) 0
(b) 19.5 kJ
4.21 5966 kJ/kmol
4.22 Yes
4.23 10.65 J/K
4.24 1.07 kJ/K
4.25 (a) 5.46 kJ/kmol K, – 4.48 kJ/kmol K, 0.98 kJ/kmol K
(b) 8.6 kJ/kmol K, – 7.9 kJ/kmol K, 0.70 kJ/kmol K
4.26 348.5 K
4.27 64.6%
4.28 21.995 103 kJ/kmol
4.29 13.38 kJ/kmol K
4.30 76.1%
4.31 (a) – 1375.35 kJ/K
(b) 326.45 kJ/K
(c) 1.16887 105 kJ
4.32 (a) 397.7 kW
(b) 329.2 kW
4.33 8.62 J/g K
CHAPTER 5
5.1 13.54 m/s
5.2 0.9948 kg/s
5.3 350 K
5.4 722.51 kW
5.5 6.877 105 kJ
5.6 144.1 kJ/h
5.7 567 K
5.8 6558 kg
5.9 759 K
5.10 0.9159 kg
5.11 229 K, 3.82 bar
5.12 420 K
5.13 (a) 865.8 K
(b) 18.4 kg
5.14 (a) 45.61 kg
(b) –104594 kJ
5.15 0.0744, 9018 kJ
5.16 343.3 K
5.17 109.3 HP
5.18 1481.3 W, 2.95 105 N/m2
5.19 4.316 kW, 0.8662 105 N/m2,
9.66 105 N/m2
5.20 (a) 402.64 m/s
(b) 5.751 10–4 m2
5.21 1.14 kg/s
5.22 2.19
5.23 61.04 K
5.24 813.8 m/s
5.25 11.33
5.26 (a) 0.5457, 542.96 m/s
(b) 28.323 kPa
5.27 429.4 K, 3.65P2
5.29 541 kW, 3060.97 kJ/kg, 7.3544 kJ/kg K
5.30 (a) 114.91 kW
(b) 0.49 m3/s
5.31 31.84 kW, 0.2963 m3/s, 393 K
5.32 71.8 kJ/kg, 339 K
5.33 14
5.34 1.11 kJ/s, 2.5 kJ/s
5.35 2.525 kg/s, 280 kJ/s, 4
5.36 419.32 kg/h
5.37 (a) 0.5079 kg/min
(b) 0.311 kW
(c) 2.857
5.38 (a) 3.42 HP
(b) 63.3 kg/h
(c) 1731.1 kg/h
(d) 6.9
5.39 11.39 kg/h, 5.03
5.40 (a) 162.2 kg/h
(b) 3.5%
(c) 131.55 kg/h
(d) 4.11, 5.0, 3.76
5.41 4.78 kW, 8.9214 103 kg/h
5.42 (a) 3
(b) 4 kJ/s
(c) 231 K
5.43 6976.7 kg/h, 57.67 ton
5.44 9.73 kW
5.45 0.084 (winter), 0.048 (summer)
5.46 (a) 2% liquefied
(b) 23% liquefied
5.47 (a) 0.048, 4.92 kg/h
(b) 0.006, 0.65 kg/h
5.48 (a) 0.054
(b) 230 K
5.49 (a) 13.3 kg/h
(b) 58.87 kW
5.50 40%
5.51 (a) 2.9245 kJ/kg
(b) 578.17 kJ/kg
(c) 55.4 kJ/kg
(d) 2867.4 kJ/kg
(e) 2211.6 kJ/kg
5.52 (a) 38.83%
(b) 37.22%
5.53 38.42%
5.54 54.14%
5.55 (a) 34.95%
(b) 29.68%
(c) 4232 kg/h
5.56 34.25%
5.57 34.81%
5.58 (a) 29.72%
(b) 4000 kg/h
5.59 (a) 39.66%
(b) 29.61%
(c) 3.0505 105 kg/h
(d) 2.7016 105 kW, 1.9016 105 kW
5.60 39.12%
5.61 25.45%, 0.941 (at 700 kPa), 33.34%,
0.8627 (at 3500 kPa), 36.16%,
0.771 (at 7000 kPa), 38.22%,
0.6944 (at 14,000 kPa)
5.62 32.42%, 0.7835 (at 573 K), 32.99%,
0.8128 (at 623 K), 36.69%,
0.920 (at 873 K), 37.64%,
0.9361 (at 923 K)
5.63 (a) 510.34 kJ/kg
(b) 222.15 kJ/kg
(c) 56.47%
5.64 1129.4 kJ/kg
5.65 (a) 7.233
(b) 54.68%
(c) 909.24 kJ/kg
5.66 6.13
5.67 (a) 13.94
(b) 1.578
(c) 17.874 kJ
(d) 61.49%
5.68 (a) 80 kPa, 310 K (step 1),
2009.5 kPa, 778.69 K (step 2),
2368.9 kPa, 917.96 K (step 3),
94.3 kPa, 365.45 K (step 4)
(b) 300.93 kJ
(c) 60.2%
5.69 43.2%
5.70 (a) 9
(b) 2167.4 kPa
(c) 49.68%
(d) 78.618 kW
5.71 (a) 13.94
(b) 1.64
(c) 366.81 kJ/kg
(d) 61.13%
5.72 Compression ratio = 2.69
Expansion ratio = 2.408
5.73 (a) 3101.3 K
(b) 3.3
(c) 2172.5 kJ/kg
(d) 55.74%
5.74 19.16
5.75 (a) 40.62%
(b) 543.27 kJ/kg
(c) 1.22 kJ/kg K
5.76 (a) 1828 kJ/kg
(b) 1040.6 kJ/kg
(c) 56.93%
5.77 (a) 1203.26 kJ/kg
(b) 1976.9 kJ/kg
5.78 (a) 220 kJ/kg
(b) 372.2 kJ
(c) 22.2%
5.79 (a) 100 kPa
(b) 244.6 kJ
(c) 585.2 kJ
(d) 44.79%
5.80 (a) 40.06%
(b) 24.7%
5.81 (a) 1416 kW
(b) 3616 kW
(c) 32.72%
(d) 421 kPa
5.82 (a) 24,714.4 kJ/s
(b) 0.3973
(c) 66.54 kg/s
5.83 (a) 300 K, 100 kPa (point 1), 445.8 K, 400 kPa (point 2), 900 K, 390 kPa(point 3), 618.6 K,
105 kPa (point 4)
(b) 29.85%
CHAPTER 6
6.11 m = – b/CP, Temperature increases
6.12 z = (1 + 2aP)/(1 + aP)
6.13 RT ln f = PV – RT – RT ln
CHAPTER 7
7.12 103 (m3/kmol) are given in brackets against x : 0.0667 (10.332), 0.16(13.86), 0.30
(19.78), 0.3634 (22.16), 0.84 (36.83)
7.14 0.8866 10–3 m3/kg
7.15 = – 1.3873 10–6 m3/mol (salt)
= 18.023 10–6 m3/mol (solvent)
7.16 = 0.0389 10–6 m3/mol (methanol)
= 0.0175 10–6 m3/mol (water)
7.17 – 22 kJ/kmol (HCl),
– 9.1 kJ/kmol (water)
7.18 0.02903 m3/mol
7.19 5.056 10–4 mol/kg (water)
7.20 533 bar
7.21 – 9.74 kJ, – 9.46 kJ
7.22 gA and are shown in brackets against mole fraction x.
0 (–, 1.00), 0.2 (0.5361, 1.0652), 0.4 (0.7330, 1.4565), 0.6 (0.8862, 1.7609), 0.8 (0.9710,
1.9273), 1.0 (1.0, 1.987)
7.23 Yes
7.24
7.25 (a) 1.27, 0.652
(b) 0.4174, 0.2145
7.26 1.331
7.27 (a) 0.0834
(b) 0.36 bar
7.28 0.953
7.29 0.8939
7.30 34.2589 J/K
7.31 – 4675.3 kJ/kmol K h
7.32
7.39
7.40
7.41
+
and ME/x1x2 are respectively,
A – B + C, 0, A – B + C (for x1 0) and
0, A + B + C, A + B + C (for x1 1)
7.42
7.43 GE/RT = x1x2 [A + B(x1 – x2)]
7.44 = 0.1275 m3/kmol
= 0.1625 m3/kmol
CHAPTER 8
8.2 (a) 2, overdefined (b) 1 (c) 2 (d) 0 (e) 1 (f) 2 (g) 1
8.3 0.0855
8.4 x and y are given in brackets against T
353.1 (1, 1), 358 (0.78, 0.90),
363 (0.581, 0.777), 368 (0.411, 0.632), 373 (0.258, 0.456), 378 (0.130, 0.261), 383 (0.017,
0.039), 383.6 (0, 0)
y = 2.45 x/(1 + 1.45 x)
8.5 8.5239 kPa, 76.3% benzene
8.6 120.3 kPa
8.7 131.24 kPa, 46.2% A
8.8 N2: 67.4% (liquid), 90.34% (vapour)
8.9 (a) x1 and y1 are given in brackets against P in kPa:
33.79 (0, 0), 43.426 (0.2, 0.3775),
53.062 (0.4, 0.6179), 62.698 (0.6, 0.7844), 72.334 (0.8, 0.9066), 81.79 (1, 1)
(b) x1 and y1 are given in brackets against T in K:
311.45 (1, 1), 315 (0.787, 0.902),
319 (0.581, 0.773), 323 (0.405, 0.623), 327 (0.254, 0.449), 331 (0.123, 0.250),
335.33 (0, 0)
8.10 (a) x1 and y1 are given in brackets against P in kPa:
34.02(0, 0), 63.03(0.2, 0.568),
92.04(0.4, 0.778), 121.04(0.6, 0.888),
150.05(0.8, 0.995), 179.06(1, 1)
(b) x1 and y1 are given in brackets against T in K:
353.3(1, 1), 363.3(0.686, 0.925),
373.3(0.458, 0.816), 383.3(0.287, 0.666), 393.3(0.156, 0.464), 403.3(0.053, 0.198), 409.4(0,
0)
8.11 (a) 77.3% benzene (b) 57.5% benzene
37.9% benzene
8.12 (a) x1 = 0.3138, y1 = 0.7730
(b) 334.4 K, 79.6% pentane
(c) 110.25 kPa, 85.7% pentane
8.15 x1 and y1 are given in brackets against T in K:
334.4 (0, 0), 336.5 (0.2, 0.165), 337.7 (0.334, 0.334), 337 (0.4, 0.4277), 335 (0.6, 0.6858),
331.5 (0.8, 0.8750), 329.5 (1, 1)
8.17 (a) A = 1.0624, B = 1.0217
(b) ln g1 and ln g2 are shown in brackets against x1:
0 (1.0624, 0), 0.2 (0.6299, 0.0435), 0.4 (0.3706, 0.1713), 0.6 (0.1621, 0.3793), 0.8 (0.0399,
0.6640), 1.0 (0, 1.0217)
8.18 (a) 7.19% acetone
(b) 108.35 kPa
8.19 (a) A = 1.6625, B = 2.7475
(b) 67.81% hexane
(c) 101.56 kPa
8.20 (a) A = 0.8940, B = 0.8426
(b) 60.3% acetate, 54.72 kPa
(c) 48.64% acetate
8.21 A = 0.9376, B = 3.0119
8.22 A = 0.1365, B = 0.1122
8.23 A = 1.7492, B = 1.4446
8.24 Yes
8.25 (a) 335.53 K, 80.9% acetone
(b) 351.96 K, 4.07% acetone
8.26 A = 3.8297, B = 2.3540
8.27 83.16% alcohol
8.28 y1 is shown in brackets against x1:
0 (0), 0.2 (0.1355), 0.4 (0.3299),
0.6 (0.6123), 0.8 (0.9212), 1.0 (1.0)
8.29 g1 = 1.3551, g2 = 1.682
8.30 108.4 kPa, 43.46% acetone
8.31 (a) 91.50 kPa, 53.77% propanol
(b) 96.75 kPa, 43.91% propanol
(c) 353.84 K, 81.5% propanol
(d) 360.615 K, 6.38% propanol
8.32 329.7 K, 356.9 K
8.33 (a) 330 K
(b) 340.6 K
(c) Mole % in the liquid and vapour are given in brackets: Ethane (0.48, 0.09), Propane (36.3,
17.45), Isobutane (18.18, 18.74), n-Butane (44.98, 63.35), Isopentane (0.13, 0.38)
8.34 (a) 930.3 kPa
(b) 337.4 K, Composition of condensate: Ethane (1.8%), Propane (6.2%), Isobutane (17.3%),n-
Butane (67.1%), Isopentane (7.50%) 330.2 K, Composition of the liquid and vapour: Ethane
(3.22%, 16.81%), Propane (9.31%, 18.7%), Isobutane (19.55%, 18.46%), n-Butane (63.38%,
44.57%), Isopentane (4.54%, 1.46%)
8.35 2205 kPa, Methane: 41.91, Ethane: 20.25, Propane: 21.96, Isobutane: 8.57, n-Butane: 7.31
706 kPa, Methane: 0.2, Ethane: 1.91, Propane: 16.14, Isobutane: 30.88, n-Butane: 50.88
8.36 861.4 kPa, 2446 kPa
8.37 758 kPa
8.38 (a) 717 kPa
(b) Propane: 63.93%, n-Butane: 36.07%
(c) Propane: 36.07%, n-Butane: 63.93%
8.39 Consistent
8.40 A = 0.685, B = 0.785, consistent
8.41 Inconsistent
8.42 Consistent
8.43 Consistent
8.44 (a) g1 given in brackets against x1: 1.0 (1.00), 0.87 (1.016), 0.50 (1.2074), 0.30 (1.3778)
(b) (kPa) in brackets against x1: 1.0 (0), 0.87 (7.45), 0.50 (20.05), 0.30 (31.63), 0 (37.72)
8.45 Inconsistent
8.46 g2 in brackets against x1: 0 (1.0), 0.0033 (0.9999), 0.0168 (0.9982), 0.0486 (0.9854), 0.0986
(0.9746), 0.168 (0.9313), 0.2701 (0.8535), 0.424 (0.7688)
8.47 1.0493
8.48 (kPa) in brackets against x1: 0.065 (24.20), 0.14 (47.90), 0.211 (66.67), 0.293 (84.56),
0.383 (100.31), 0.483 (114.08), 0.587 (125.48), 0.713 (137.19), 0.854 (150.67)
8.49 A = 1.7405, B = 1.4012. g1 and g2 are in brackets against x1: 0 (5.7, 0), 0.04 (4.8231,
1.0034), 0.11 (3.6990, 1.0251), 0.28 (2.2064, 1.1603), 0.43 (1.5902, 1.3880), 0.61(1.2226,
1.8417), 0.80 (1.0561, 2.6406), 0.89 (1.0144, 3.1869), 0.94 (1.004, 3.5522), 1.00 (1.0, 4.06)
8.50 (a) 1.7951, 1.4679
(b) 1.7951, 1.4679
(c) 65.21% water, 65.82 kPa
8.51 x1 = 0.20, g1 = 1.0720, g2 = 1.0059, P = 100.15 kPa, y1 = 0.2126
x1 = 0.90, g1 = 1.0007, g2 = 1.0815,
P = 100.03 kPa, y1 = 0.894
8.52 g1 = 1.0288, g2 = 2.6100
8.53 20.22 kPa, 53.57% ethanol
8.54 93.3 kPa, 57.1% A
8.55 (a) 128.6 kPa, 34.25% A
(b) No change
8.56 123.21 kPa, 38.54% A
8.57 Azeotrope exists at
CHAPTER 9
9.5 6.09 105
9.6 – 57,350 J/mol
9.7 5.7498 10–4
9.8 Yes
9.9 (a) – 24,800 J/mol, feasible
(b) 2.067
(c) DG0 = – 7.53305 104 + 6.12 104(T)–1 + 63.710T ln T – 181.11T – 44.958 10–
3T2 + 4.7805 10–6T3
(d) 1.7297
9.10 14.3692
9.11 (a) Above 812.4 K, quite favourable
Below 550 K, unfavourable
(b) 75.42%, 45.69%
(c) 81.94%, 55%
9.12 1.0506 10–3
9.13 DG0 = – 4.35946 104 + 13.003 T lnT
– 1.8564 10–2T2 + 2.43835 10–6T3
+ 54.972T
Efficiency
Carnot, 96–100, 172
compressor, 144, 148–149
ejector, 143
gas liquefaction, 169
gas turbine, 190
of internal combustion engine, 182, 184–185, 188
of steam power plant, 171–172, 175, 178
theoretical volumetric, 147–149
Endothermic reactions, 70, 437–438
Energy, 6, 8, 24, 34–36
availability of, 90, 93–94
balance, 35, 127–129
conservation of, 24, 127
degradation of, 90, 91, 93–95
equation, 127–129
internal, 24–27, 34–35, 41, 51, 93
kinetic, 8, 24–28, 35–36
potential, 8, 24–27, 35–36
properties, 206
Engines
Carnot, 171–172
heat, 14, 180
internal combustion, 180–187
Enthalpy, 31–33
calculation of, 32–34, 220–226
change of chemical reaction, 69–70
change of combustion, 70
change of formation, 70–71
change of mixing, 309–312, 351
change of phase change, 213
change, constant pressure process, 32, 41
change, constant volume process, 32
change, steady–state flow process, 36
departure, 257–259, 263
differential equation for, 210, 217–218
effect of T and P on, 220–222
ideal gas, 55, 218–219
partial molar, 281–284, 300
residual (see Enthalpy departure)
Enthalpy–entropy diagram, 260–261
Entrance work, 35
Entropy, 84–86, 92–97, 99–109
absolute, 119
calculation of, 103–107, 216, 220–226
change as criterion of equilibrium, 330–331
change of mixing, 105–106, 310–311
departure, 257–259, 263–264
differential equation for, 216–217
effect of T and P on, 220–222
and heat, 93–94
heat capacity relations, 215–216
for ideal gas, 103–104, 106–107
and irreversibility, 115–116
of isolated systems, 111–112
and nature of processes, 94–95
partial molar, 286, 310
principle of increase of, 112
and probability, 118–119
residual (see Entropy departure)
statistical explanation of, 118–119
and temperature, 94
Equality of chemical potentials, 339
Equations of state, 51, 60–65, 250
approach for VLE, 345, 386
Benedict–Webb–Rubin, 64
limiting conditions for, 61
Peng–Robinson, 64, 387
Redlich–Kwong, 63, 387
Redlich–Kwong–Soave, 64, 387
van der Waals, 61–62
Virial, 65
Equilibrium, 1, 3, 329–332
chemical reaction, 425–479
constant for vaporisation, 363
constant, 431–436
criteria of, 330–332, 429–431
liquid–liquid, 408–410
mechanical, 10, 330
phase, 329–410
state, 1, 3, 329, 331–332, 436
thermal, 10–12, 330
vapour–liquid, 344–406
Equilibrium constant, 431–436
and activity, 431–433
effect of pressure on, 446
effect of reaction stoichiometry on, 432–433
effect of temperature on, 436–440
and free energy change, 433–434
evaluation of, 440
for multiple reactions, 466–467
Equilibrium conversion, 1, 2, 425–426, 428
effect of excess reactants on, 454
effect of inerts on, 451
effect of pressure on, 446–448
effect of products on, 457
in multiple reactions, 465–467
Equilibrium yield (see Equilibrium conversion)
Exact differential equation, 209–210, 237
Excess properties, 317–319, 333
Exothermic reactions, 69, 436–437
Expansion engine vapour compression, 156–157
Extensive properties, 4, 206–207, 273, 276
Extent of reaction, 426–428, 430, 467–468
Extract, 409
Extraction, 409
Gas
ideal, 12, 51–57
liquefaction, 166–169
real, 60–65
solubility, Henry’s law for, 294
turbines, 188–190
Gas turbine cycle, 188–190
Generalised charts
compressibility factor, 68–69, 258
enthalpy departure, 259
entropy departure, 259
fugacity coefficient, 248–249
Generalised correlation for
enthalpy departure, 258–259
entropy departure, 258–259
Giauque functions, 440–442
Gibbs–Duhem equation, 302–304, 334, 368, 395–396, 398–399
Gibbs free energy, 206–209
at equilibrium, 331–332, 335–336, 428–430
change as criteria of equilibrium, 330–331, 429–431
change of mixing, 308, 311, 315, 332
change on reaction, 429–431, 433–435
differential equation for, 210
effect of temperature on, 235–236
excess, 318–319, 333, 369, 398
of formation, 440
functions (see Giauque functions)
partial molar, 284 (see also Chemical potential)
Gibbs–Helmholtz equation, 235–236, 437
Gibbs paradox, 107
Gross heating value, 70
Heat, 4, 7–8, 25, 35, 40–41, 89–90, 93
of combustion, 70–71
and entropy, 93–94
of formation, 70–71
latent, 213, 337
of mixing, 300, 311–312
quality of, 89–90
of reaction, 69–74
reservoir, 14, 94, 96
sign convention for, 25
of solution, 311–312
specific, 40
of vaporisation, 214
Heat capacity, 4, 40–41, 53, 215
constant pressure, 40–41, 53, 239
constant volume, 40–41, 53, 239
effect of pressure and volume on, 229–230
entropy change and, 215
of ideal gas, 53–54, 231
mean, 315–316
ratio of, 229
relationship between, 226–227, 243
Heat engines, 14, 91, 180
efficiency, 14, 91
Heat pump, 15, 91, 165
COP of, 15, 165
Heat of reaction, effect of T on, 72–74
Helmholtz free energy, 206–208
as criterion of equilibrium, 331
differential equations for, 210
Henry’s law, 293–294, 295–296, 298
for activity coefficients, 298
and gas solubility, 294
and Lewis–Randall rule, 293–294, 305
for standard state, 298–301, 460–461
Hess’s law, 71
HS diagram, 260–261
HT diagram, 260, 262–263
construction of, 262–263
Jacobians, 236–238
thermodynamic relations using, 238–244
Jet pumps, 142
Joules experiment, 24
Joule–Thomson
coefficient, 52, 143, 166–167, 233–235, 244
expansion, 143
inversion curve, 233–234
inversion temperature, 143
liquefaction, 166–168
Partial
heat of mixing, 300
pressure, 288, 290
Partially miscible systems, 403–405
Partial molar property, 273–281
determination of, 279–281
physical meaning of, 274–275
and properties of solutions, 276–277
Path functions, 4, 102
Peng–Robinson equation, 64, 387
Perpetual motion machine, 24
Phase, 3
change and entropy change, 103
diagrams, 346–351, 353–355, 362–366, 404–406, 408–410
equilibrium, 329–410
rule, 11, 341–343, 469
Phase equilibrium, 329–410
criterion of, 292, 330–332
in multicomponent systems, 338–340
in single component systems, 335–336
PH diagrams, 155, 259–260
Plait point, 409
Polytropic process, 57
Positive deviation from ideality, 318, 363–365
Potential energy, 8, 25–27, 35–36
Power–plant cycles, 170–179
Poynting correction, 346, 388
Practical efficiency, 169
Pressure, 5–6
critical, 50
of decomposition, 462
drop, 135
osmotic, 340–341
partial, 288, 290, 353–354, 362–364, 432, 462
reduced, 68–69
Pressure ratio
critical, 139
turbine, 190
Principle of corresponding states, 68–69
Probability and entropy, 118–119
Properties, 4
critical, 50, 62
derived, 206–207
energy, 206
excess, 317–319
extensive, 4
intensive, 4
partial molar, 273–281
path, 4, 102
reduced, 68–69
reference, 206
residual, 257–259
Property changes on mixing, 296, 307–311
and excess property changes, 317
for ideal solutions, 310–311
PT diagram, 50–51, 61
PV diagram for IC engines, 181, 184, 187
PV isotherm, 49–50, 61, 65
PVT behaviour of fluids, 49–51
Raffinate, 409
Rankine cycle, 171–172
Raoult’s law, 292–293, 295–296, 298, 305, 351–355, 362–363
Reaction coordinate, 426
Redlich–Kister consistency test, 398
Redlich–Kwong equation, 63, 387
Redlich–Kwong–Soave equation, 64, 387
Reduced properties, 68–69
Reference properties, 206
Refrigerant, choice of, 159–160
Refrigeration cycle, 151–165
absorption, 163–165
air, 161–162
capacity of, 152
Carnot cycle for, 152–153
COP of, 152, 153, 156, 162
vapour compression cycle, 154–156
Regenerative cycle, 177–178
Reheat cycle, 174–175
Relative volatility, 354–355
Residual properties (see Departure functions)
Reverse osmosis, 340
Reversible
process, 16–19, 95, 103, 108–110, 207–209, 213, 330
work, 17–19, 207–209
Saturated phases, 50
Saturation
pressure, 50
temperature, 50
Second law of thermodynamics, 90–92, 111–112
Shaft work, 34–36, 122–129, 144–145
Solubility of gas in liquid, Henry’s law, 294
Sonic velocity, 137
Specific heat (see also Heat capacity), 4, 40–41, 215
Stability criteria, 332–333
Standard
Gibbs free energy change, 433–437, 441
heat of combustion, 70–71
heat of formation, 70–71, 442
states, 69–70, 245, 255, 296–301, 308
vapour power cycle (see Rankine cycle)
Standard heat of reaction, 69–71, 438–439
effect of temperature on, 72–74
State, 3
equilibrium, 3, 329, 331, 332, 436
functions, 4, 25–26, 101–102, 207
steady, 10, 34, 329
Steady–state flow process, 34–36
Steam–power plants, 170–179
Rankine cycle for, 171–173
regenerative cycle for, 177–179
reheat cycle for, 174–176
Steam tables, 483–490
Stoichiometric
numbers, 72, 426
reaction, 426
Sublimation, 51
Surroundings, 3, 26, 112
System, 2
adiabatic, 112
closed, 3, 26, 112
heterogeneous, 3
homogeneous, 3
isolated, 3, 25, 84, 112–113, 330
open, 3