1
MME6701: Lecture 5
Solution ThermodynamicsPart 1
A. K. M. B. RashidProfessor, Department of MMEBUET, Dhaka
Composition of solutions Partial molar properties The ideal and non-ideal solutions Dilute solutions Multi-component systems Free energy of mixing Problem solving
2
Introduction
During the formation of solutions , also known as the multi-component, homogeneous non-reacting systems, chemical composition changes due to the transfer of atoms or molecules across the boundary
Materials of practical applications are either:
mixtures of metals (alloys), or
mixture of compounds (slags, mattes, glasses, ceramics)
When mixed, materials can exist either as separate mixtures of several components, or as solutions.
A solution is a homogeneous phase composed of two or more chemical substances, whose concentration may be varied without the precipitation of a new phase.
the mixing is entirely physical where only intermolecular interactions take place in such a way that the molecules maintain their individual identity.
The thermodynamic properties of a component in solution are significantly different from the properties of that component when it is pure.
Number of moles of component k (nk) and mole fraction (Xk)
Molality of solute k in solution, mk
Molarity of solute k in solution, ck
Weight percentage of component k (wt.% k)
ck =nk
V
mk = =nk
w0x 1000
1000 nk
M0n0
wt.% k =wk
wTx 100
Xk =nk
nT
nk = gram moles of solute k w0 = gram of solvent n0 = gram moles of solvent M0 = molecular weight of solvent
V volume of solution in litres
3
How can we express the molar properties of a component in solution?
For single component systems, volume occupies by two moles of oxygen at 298 K and 1 atm is exactly twice that occupies by one mole under the same conditions, i.e., V = V’/ n.
What is the molar volume of each of the components present in solution?
We can measure this by adding a very small quantity of a component in solution without significantly changing the overall composition of the solution and then converting the measured change in volume to one mole of the component being added.
For a multi-component system, V is a function not only of T and P (as in one-component system) but also of the number of moles of each component in the system (n1, n2, ….., nC).
V’ = V’ (T, P, n1, n2, …., nC)
For an arbitrary infinitesimal change in state
dV’ = dT + dPV’T P, nk
V’P T, nk
V’n1 P, T, n2, n3, …., nC
+ dn1 + dn2 + …..V’n2 P, T, n1, n3, …., nC
V’nC P, T, n1, n2, …., nC-1
+ dnC
4
dV’ = dT + dPV’T P, nk
V’P T, nk
V’nk P, T, nj≠nk
+ dnk∑C
K=1
V’nk P, T, nj≠nk
Vk (k = 1, 2, 3, …. C)
The coefficient of each of the changes in number of moles
These quantities are defined to be the partial molar volumes for each component in the system.
It represents the rate of change of volume with respect to addition of substance k and is equal to the increase in volume resulting the addition of one mole of k to an infinite amount of all the components in the system without significantly altering the overall composition of the system.
dZ’ = MdT + NdP + Zk dnk∑C
K=1
Z’nk P, T, nj≠nk
Zk (k = 1, 2, 3, …. C)
For any extensive variable Z’,
where the partial molar property of Z of component k
Self-Assessment Question 6.2Write an expression for the partial molar Gibbs free energy of component A in the A-B binary solution.
dV’ = dT + dPV’T P, nk
V’P T, nk
V’nk P, T, nj≠nk
+ dnk∑C
K=1
5
General Properties of Partial Molar Properties
Z’ = n1Z1 + n2Z2 + ….. + nCZC = nk Zk∑C
K=1
Z = X1Z1 + X2Z2 + ….. + XCZC = Xk Zk∑C
K=1
Z = Z’ / nT
Xk = nk / nT
Differentiating completely
dZ = (Xk dZk + Zk dXk)∑C
K=1
dZ’ = MdT + NdP + Zk dnk∑C
K=1
dZ = (Xk dZk + Zk dXk)∑C
K=1
Since
Xk dZk = 0∑C
K=1
(At constant T and P) Gibbs-Duhem Equation
At constant T and P
dZ’ = Zk dnk∑C
K=1dZ = Zk dXk∑
C
K=1
and
6
Each of the four categories of relationships developed earlier for the total properties of system between the various state functions (the laws, definitions, coefficient relations, and Maxwell relations) are applicable also to the partial molar properties of the component of a system.
Hk = Uk + PVk
Gk = Hk – TSk
dGk = –SkdT + VkdP
k
n T,
k V P
G
k
k
n P,
k S - T
G
k
Evaluation of Partial Molar Properties
1. Measurements of the partial molar properties from the total properties of the solution.
Z Zk
2. Measurements of the partial molar properties of one component knowing the partial molar properties of another.
Z1 Z2
Two methods:
7
Partial molar properties from the total molar properties
Z = Zk Xk = Z1 X1 +Z2 X2
Z = Z1 X1 + X2 Z1 + dZdX2
Z = Z1 + X2dZdX2
Z = Z1 X1 + X2 + X2dZdX2
Z1 = Z – X2dZdX2
= Z + (1 – X1)dZdX1
Z2 = Z – X1dZdX1
= Z + (1 – X2)dZdX2
dZ = Z1 dX1 +Z2 dX2
dZ = (Z2 – Z1 ) dX2
X1 + X2 = 1
dX1 = - dX2
Z2 = Z1 + dZdX2
∑C
K=1
Z1 andZ2 can be found from knowledge of Z either analytically or graphically
Analytical Solution
Example 6.3Derive expressions for the partial molar volumes of each components as functions of composition if the volume change in a binary solution obeys the relation V = 2.7 X1 X2
2 cc/mol.
dV/dX1 = 2.7 ( X22 + X1 . 2X2 . dX2/dX1 ) = 2.7 (X2
2 – 2X1X2)
dV/dX2 = – dV/dX1 = –2.7 (X22 – 2X1X2)
V1 = V – X2 (dV/dX2) = 2.7 X1 X22 – X2 –2.7 (X2
2 – 2X1X2)
V1 = 2.7 ( X23 – X1 X2
2 )
V2 = V – X1 (dV/dX1) = 2.7 X1 X22 – X1 2.7 (X2
2 – 2X1X2)
V2 = 5.4 X12X2
8
Graphical Solution
E
P
F
DB
C
G A
Pure 1 Pure 2Mole Fraction, X2
Molar Property, Z
Z
Z1
Z2
Z20
Z10
Schematic representation of molar property, Z, versus composition, X2, diagram of a binary solution at constant temperature and pressure
X20
dZ / dX2 = BC / PB
Slope of curve at point P
Z2 = Z + (1 – X2) (dZ/dX2)
= AB + PB (BC/PB)
= AC
To determine partial molar properties for a solution of a given composition, construct a tangent to the Z curve at that composition and read the intercepts on the two sides of the graph.
Z1 = GD
Z2 = Z + (1 – X2) (dZ/dX2)
Z2 = Z – X1 (dZ/dX1)
Partial molar properties of one component from partial molar properties of another
X1 dZ1 + X2 dZ2 = 0
dZ1 = – dZ2X2
X1
Using Gibbs-Duhem equation for a binary solution
Integrating this equation from X1=1 to X1=X1
Z1 – Z1X1 = X1X1 = 1 = – dZ2
X2
X1X1 = 1
X1 = X1
9
Z1 – Z1X1 = X1X1 = 1 = – dZ2
X2
X1X1 = 1
X1 = X1
For pure component, partial molar property does not exist.
So at X1=1, Z1 = 0
Z1 = – dZ2
X2
X1X1 = 1
X1 = X1
Thus, knowing the value ofZ2, the partial molar properties of one component, and a reference state, the partial molar property of the other component can easily be calculated.
If a relationship between Z2 and X2 is given, the above relation can be modified as
Z1 = – X2
X1X1 = 1
X1 = X1
dZ2
dX2dZ2 = dX2
dZ2
dX2dX2
If, on the other hand, the input relationship is between Z2 and X1, then
Z1 = – X2
X1X1 = 1
X1 = X1
dZ2
dX1dX1
Z1 = – dZ2X2
X1X1 = 1
X1 = X1
10
Example 6.4The partial molar enthalpy of component 2 in a binary solution is given by the equation DH2 = aX1
2. Compute the partial molar enthalpy for component 1 and
the molar enthalpy for the solution.
DH1 = – X2
X1X1 = 1
X1 = X1
d(DH2)dX1
dX1
d(DH2)dX1
= 2aX1
DH1 = – X2
X1X1 = 1
X1 = X1
. 2aX1 . dX1 = – 2a X2 dX1X1 = 1
X1 = X1
= + 2a X2 dX2X2 = 0
X2 = X2
DH1 = aX22
DH = X1DH1 + X2DH2 = X1 . aX22 + X2 . aX1
2
DH = aX1 X2 (X1 + X2) = aX1 X2
Every pure substance in liquid or solid state has a definite vapour pressureat any given temperature
Pure solid or liquid, when placed in a closed, initially evacuated chamber will evaporate spontaneously until the vapour phase and the solid or the liquid phase comes to a dynamic equilibrium.
At any temperature, every substance in liquid or solid state has been in equilibrium with its vapour phase and has a definite vapour pressure.
Pure Metal
Approx. Vapour Pressureat 1600 C (atm)
Mg > 20
Mn 0.07
Al 0.006
Fe 0.0007
Ni 0.0001
11
Evaporation
Acquire sufficient kinetic energy and escape into the vapour phase
Condensation
Molecules collide with condensed phase and trapped through loss of kinetic energy
Condensed Phase
Gas Phase
Kinetic View
At equilibrium,
Rate of evaporation, re = Rate of condensation, rc
The condensation rate, rc (A)
Proportional to the number of A atoms in the vapour phase which strike the liquid surface in unit time
This, for a fixed temperature, is proportional to the pressure of the vapour, that is, rc (A) = k pA
In case of a pure condensed phase A
The evaporation rate, re (A)
Depends upon the magnitude of the mutual attraction of A particles
re (A) = rc (A) = k . p0A
At Equilibrium
p0A = standard vapour
pressure of A
12
When a small quantity of B is added to A
Conditions:
Particles of A and B were inert to each another Diameters of A and B are comparable
Some fraction of the surface area (XA) of liquid A will be occupied by B
The rate of evaporation of liquid A will be reduced by a factor XA.
The new evaporation rate will be re(A) . XA
The equilibrium vapour pressure exerted by A is decreased from p0A to pA
At equilibrium, for A containing small quantity of B
re(A). XA = k.p0A . XA = k . pA
re(A) . XA = k.p0A . XA = k . pA
pA = p0A XA
pB = p0B XB
A solution that obeys Raoult’s Law is called an ideal solution, and such behaviour of a solution is called the Raoultian behaviour.
These equations are known as the Raoult’s Law:
The vapour pressure exerted by a component k in a solution is directly proportional to the mole fraction of k in the solution and the proportionality constant is the saturated vapour pressure of component k at the temperature of the solution
13
XB
VapourPressure
p0A
p0B
p = pA+pB
Pure A Pure B
Addition of solute
the solute itself also exerts its own vapour pressure.
lowers the vapour pressure of solvent
The vapour pressure of the solution is the sum of the separate vapour pressures of solvent and solute.
pA = p0A XA
The relative lowering of vapour pressure of solvent due to addition of solute is equal to the mole fraction of solute in the solution.
Alternate Formulation of Raoult’s Law
pA = p0A . XA = p0
A . (1 - XB)
p0A – pA
p0A
= XB
In order to do that, the following three conditions must be satisfied:
1. Atoms of A and B of the solution must be of same size,
2. Atoms of A and B must be inert to each other, and
3. The vapours of A and B should behave like ideal gas.
The Raoult’s law presupposes that
The intrinsic evaporation rates of A and B be independent of the composition of the solution.
This requires that the magnitudes of A-A, B-B, and A-B interactions be identical (i.e., no preferential mixing or bond formation).
If a solution violates one or more of the above conditions, Raoult’s is not obeyed, and the solution is termed as the non-ideal solution.
14
Case I: Existence of attraction/repulsion between components
vapours behave as ideal gas molecules are of similar size not inert to each other (A-B bond A-A / B-B bond)
A-B attraction >> A-A / B-B attractions
Negative deviation from Raoult’s Law
Example: Fe – 50 wt%Ni solution at 1600 °C(pNi is roughly three-quarter of its ideal value)
A BMole Fraction
1
23Vap
ou
r P
ress
ure
p0A p0
B
A-B attraction << A-A / B-B attractions
Positive deviation from Raoult’s Law
Example: Fe – 1 wt.% Cu solution at 1550 °C(pCu is about 10 times higher than its ideal value)
A BMole Fraction
1
23
Vap
ou
rP
ress
ure
p0A
p0B
Fe – Mn solution is virtually an ideal solutionat 1600 °C.
Case III: Vapour phases do not behave ideally
At high pressures or low temperatures (approaching the critical temperature of the gas), the pressure exerted by gases are not equal to their vapour pressure, p.
In such cases the ideal gas law pV = nRT is no longer obeyed by the gas phase.
Case II: Components are of different sizes
A difference in size in atoms of the components of a solution will affect the distance between the centres of adjacent particles, and consequently, there exists a mutual attraction or repulsion between them.
15
This equation is valid for all ideal solutions with imperfect gases.
At higher pressures or low temperatures, the gases no longer obey the ideal gas law, PV = nRT.
Also, the linear relationship between Gibbs free energy, G, and the vapour pressure, p, becomes complicated.
For such conditions, G. N. Lewis introduced the quantity fugacity, F, so that, under all conditions, the equation of state remains the familiar form, and ensure the linear relation between G and F in any state for all gas.
Using fugacity the Raoult’s law can then be restated as: Fk = F0k Xk
FV = RT
If the vapour above the solution is ideal, then F = p, and
Real solutions do not obey Fk = F0k Xk
Thus the fugacity of a component is not proportional to its mole fraction.
For real solutions, another thermodynamic quantity, activity, is defined
ak = Fk / F0k or, Fk = F0
k ak
This is an alternate and more commonly used expression of Raoult’s law.
ak = pk / p0k = Xk
16
Activity of a component is the fraction of molar concentration that is available for reaction.
In real solutions, the activity of a component can be considered as the active mass of the component in the solution, ie. a measure of its availability for reaction, perhaps with another phase.
For ideal solutions, interactions between components do not exist and, therefore, activity equals mole fraction, i.e., ak = Xk
Example: Removal of Si in steelmaking process
Si + O2 = SiO2
The rate of reaction is not proportional to XSi.pO2 but to aSi.pO2
(assuming oxygen to behave as an ideal gas at 1600 °C)
Reason:
Silicon in solution does not behave as if it were pure silicon dissolved in ideal solution in iron. This is because the Fe-Si bond is very strong.
Thus, the activity of silicon is less than its mole fraction (a –ve deviation).
In fact, the activity of silicon in a 1 wt.% solution in iron at 1600 °C is about 1/60 of its mole fraction.
There are may standard states used to refer activity.
For liquid and solid solutions, one standard state commonly used is that of pure substance at the temperature and pressure of the system.
For pure substance, gk =1 and Xk = 1, so that ak =1.
The Standard State
The numerical value of activity depends on the concentration, temperature and pressure of the solution as well as on the choice of the standard state.
The activity is meaningless unless it is accompanied by a detailed description of the standard state to which it refers.
17
For an ideal solution, gk =1.
gk > for positive deviation
gk < for negative deviation
For non-ideal/actual/real solution, activities of the components are not equal to their mole fractions, i.e., ak ≠ Xk
The Raoultian activity coefficient of a component in a solution is defined as the ratio of the activity of the component to its mole fraction
gk = =ak
Xk
Fk
F0k Xk
Xk dZk = 0∑C
K=1
Gibbs-Duhem Equation
For A-B binary solution
XA dZA + XB dZB = 0
Using partial molar Gibbs free energy change for components in solution,
XA d(DGA) + XB d(DGB) = 0 (6.33)
18
For a system of fixed composition, dG = – dSdT + VdP
At constant temperature, dGT = VdPT
And for 1 mole of an ideal gas at constant temperature, using V = RT/P,
dG = (RT/P) dP = RT (dP/P) = RT d lnP
Hence the partial molar free energy of component k in solution will be
dGk = RT d ln pk
For real solutions, replacing partial pressure with activity, we have
dGk = RT d ln (p0k . ak ) = RT d ln ak
p0k is constant
dG = RT d ln a (6.38)
Integrating at constant temperature from the standard state to any other state
G – G0 = DG = RT ln a – RT ln a0
By definition, the activity of a substance in its standard state is unity
G – G0 = DG = RT ln a
In terms of partial properties, similar integration of will give
Gk – G0k = DGk = RT ln ak
dG = RT d ln a
d (DGk) = RT d ln ak (6.40)
19
XA d(DGA) + XB d(DGB) = 0 d (DGk) = RT d ln ak
XA (RT d ln aA) + XB (RT d ln aB) = 0
XA d ln aA + XB d ln aB = 0
d ln aA = – d ln aBXB
XA
XA = 1
XA = XA
d ln aA = – d ln aBXB
XA
XA = 1
XA = XA
ln aA = – d ln aBXB
XA
XA = 1
XA = XA
(6.42)
(6.41)
The integration is done by graphical method using the plot (XB/XA) vs. ln aB
XB=XB
- ln aB
Shaded area = ln aA
1. For XB1, XB/XA and –ln aB=0.Curve exhibits a tail to plus infinity.
2. For XB0, XB/XA0 and – ln aB – Curve exhibits a tail to minus infinity.
XB
XA
Point 2 is important since the area under investigation includes a tails that extends to minus infinity.
This includes an uncertainty into the calculation.
20
Use of activity coefficient to measure activity
dXA + dXB = 0
dXA
XA
dXB
XBXA + XB = 0
XA d ln XA + XB d ln XB = 0
XA d ln aA + XB d ln aB = 0
Subtracting eq.(6.43) from eq.(6.41) gives
XA d ln gA + XB d ln gB = 0
For a binary A-B solution, XA + XB = 1
(6.41)
(6.43)
d ln gA = – d ln gB
XB
XA
XA = 1
XA = XA
d ln gA = – d ln gB
XB
XA
XA = 1
XA = XA
ln gA = – d ln gB
XB
XA
XA = 1
XA = XA
XA d ln gA + XB d ln gB = 0
(6.45)
21
XB=XB
ln gB
Shaded area = ln gA
XB
XA
1. For XB1, XB/XA and lngB=0.Curve exhibits a tail to infinity.
2. Since the value of gB is always finite, ln aB
will always has a finite value when XB/XA0.
The sign of ln gB depends on positive or negative deviation of solution
Example 6.5
The following data have been obtained for Cr-Ti solutions at 1250 °C.
XCr 0.09 0.19 0.27 0.37 0.47 0.67 0.78 0.89aCr 0.302 0.532 0.660 0.788 0.820 0.863 0.863 0.906
Calculate the activity of titanium in a Cr-Ti solution containing 60 atom% Ti.
Integrating Gibbs-Duhem equation between the limits (1.0, 0.6) of titanium,
ln gTi = – d ln gCr
XCr
XTi
XTi = 1
XTi = 0.6
Answer
The integration will be done using graphical methods by using the plot (XCr/XTi) vs. ln gCr.
= + d ln gCr
XCr
XTi
XTi = 0.6
XTi = 1
22
XCr XTi XCr/XTi aCr gCr ln gCr
0.09 0.91 0.0989 0.302 3.355 1.210452
0.19 0.81 0.2346 0.532 2.8 1.029619
0.27 0.73 0.3699 0.66 2.444 0.893636
0.37 0.63 0.5873 0.778 2.103 0.743365
0.47 0.53 0.8868 0.82 1.745 0.556755
0.67 0.33 2.0303 0.863 1.288 0.253091
0.78 0.22 3.5454 0.863 1.106 0.10075
0.89 0.11 8.0909 0.906 1.018 0.01784
Data Sheet
XCr
XTi
ln gCr
0
1
2
3
4
5
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
The tail of the curve is extrapolated to zero value of (XCr/XTi).
The area bounded by the curve and the axis for (ln gCr) between the values of ln gCr
corresponding to XTi=0.6 and XTi=1.0 is calculated by the trapezoidal rule.
From the plot,a = 0.64b = 1.36
Let , n = 6Cord width, h = (b-a)/n = 0.12
f(a) = 0.67 f(x1) = 0.58 f(x2) = 0.38 f(x3) = 0.28
F(x4) = 0.18 f(x5) = 0.08 f(b) = 0
f(a)
f(b)
f(x5)f(x4)f(x3)f(x2)f(x1)
XTi = 0.6
XTi = 1.0
23
The area under the curve
S = h + f(x1) + f(x) + …… + f(xn-1)f(a) + f(b)
2
= (0.12) ( 0.67/2 + 0.58 + 0.38 + 0.28 + 0.18 + 0.08 )
= 0.22
Then at XTi = 0.6,
ln gTi = 0.22 or, gTi = 1.25
Hence, the activity of Ti at XTi = 0.6,
aTi = gTi . XTi = 1.25 x 0.6 = 0.75
Difficulties in measuring activity for solutions having a small amount of solutes
Darken and Gurry introduce a function to determine activity when a solution contains small amount of solute XA :
ak =ln gk
(1 – Xk)2
The a function is always finite and well behaved by virtue of the fact that, as Xk1, gk1.
For a binary A-B solution,
aA =ln gA
XB2
aB =ln gB
XA2
and
24
aB =ln gB
XA2
ln gB = aB XA2
d ln gB = d ( aB XA2 ) = 2 aB XA dXA + XA
2 daB
ln gA = – d ln gB
XB
XA
XA = 1
XA = XA
ln gA = – (2aB XB dXA) – (XA XB daB)1
XA
1
XA
d (xy) = x dy + ydx
d (XA XB aB) = XA XB daB + aB d (XA XB)
ln gA = – (2aB XB dXA) – (XA XB daB)1
XA
1
XA
XA XB daB = d (XA XB aB) – aB d (XA XB)
XA XB daB = d (XA XB aB) – aB XA dXB – aB XB dXA
25
ln gA = – (2aB XB dXA) – (XA XB daB)1
XA
1
XA
XA XB daB = d (XA XB aB) – aB XA dXB – aB XB dXA
ln gA = – 2aB XB dXA –1
XA
1
XA
d (XA XB aB)
+ aB XA dXB + aB XB dXA1
XA
1
XA
ln gA = – 2aB XB dXA –1
XA
1
XA
d (XA XB aB)
– aB XA dXA + aB XB dXA1
XA
1
XA
ln gA = – 2aB XB dXA –1
XA
1
XA
d (XA XB aB)
– aB XA dXA + aB XB dXA1
XA
1
XA
ln gA = – d (XA XB aB ) – aB (XA + XB) dXA1
XA
1
XA
ln gA = – XA XB aB – aB dXA1
XA
When aB is plotted against XA, the area under the curve from XA=XA to XA=1 equals the right hand side of the above equation.
Then, ln gA at XA=XA is obtained as (-aBXAXB) minus the area under the curve.
As aB is everywhere finite, this integration does not involve atail to infinity.
26
The variation of aCu with composition in iron-copper solution
