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Lec-40 Crystallisation

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50 views15 pages

Lec-40 Crystallisation

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© © All Rights Reserved
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1

Crystallization

In this chapter we will deal with crystallization of


dissolved solute from a saturated solution.

Why is crystallization done?


What is the Industrial importance of crystallization?

 A crystal formed from an impure solution is


itself pure.
 Crystallization affords a practical method of
obtaining pure chemical substances in a
satisfactory condition for packaging and storing.

Magma-In industrial crystallization from solution, the


two –phase mixture of mother liquor and crystals of all
sizes, which occupy the crystallizer and is withdrawn as
product, is called a magma.

Purity of the product. A sound well formed crystal itself


is pure, but it retains mother liquor when removed from
the final magma. When retained mother liquor of low
purity is dried on the product, contamination results, the
extent of which depends on the amount and degree of
impurity of the mother liquor retained by the crystals.
2
How to remove the retained mother liquor?

Much of the retained mother liquor is separated from the crystals by filtration or
centrifuging, and the balance is removed by washing with fresh solvent.

Prime objectives in crystallization

 Good yield
 High purity
 Quality- individual crystals should be strong, non aggregated, uniform in size
and non-caking in the package.
 Size range of crystalline product- crystal size distribution (CSD) must be under
control.

Crystal geometry. A crystal is the most highly organized type of non living matter. It
is characterized by the fact that its constituent particles, which may be atoms,
molecules or ions, are arranged in orderly 3-dimensional array called space lattices.
Crystals of the same substance have the same interfacial angles. Crystal forms are
classified on the basis of these angles. The seven classes are cubic, tetragonal,
hexagonal, trigonal, orthorhombic, monoclinic and triclinic. A given material
may crystallize in two or more different classes depending on the conditions of
crystallization.
3
A given material may crystallize in two or more different classes depending on the
conditions of crystallization.

hexagonal form (calcite)

Calcium carbonate
Orthorhombic form
(aragonite)

Invariant crystals. Under ideal conditions, a growing


crystal maintains geometric similarity during growth. Such a
crystal is called invariant.

Each of the polygons in the figure represents the outline of the crystal at a
different time. The rate of growth of any face is measured by the velocity
of translation of the face away from the centre of the crystal in a direction
perpendicular to the face.
In most crystallizers the conditions are far from ideal and growth is
often far from invariant. In extreme cases one face may grow far more
rapidly than any of the other faces, giving rise to long needle like crystals.
Slow growth of one face may give rise to thin plate or disc shaped crystals.

Size of a crystal may be specified by its characteristic length L, where L is


defined as Фs DP.
Where L = (Фs = 6VP/DP. SP )
is the volume and the total surface area of the crystal.

This holds where sphericity is close to 1.0, in practice, but not for needles
or disks with very small values of Фs

L is usually taken as equal to the size determined by screening.


Equilibria and yields
4
Equilibrium in crystallization process is reached when the solution is
saturated.
Most materials follow curves similar to curve 1 for KNO 3; that is their
solubility increases more or less rapidly with temperature. A few
substances follow curves like curve 2 for NaCl, with little change in
solubility with temperature others have what is called an inverted solubility
curve, which means solubility decreases as the temperature is raised.
5

How to attain supersaturation

Many important inorganic substances crystallize with water of


crystallization. In some systems several different hydrates are
formed, depending on the concentration and temperature, and the
phase equilibria in such system can be quite complicated.
6

The equilibrium temperature in degree Farenheit is plotted against the


concentration in mass fraction of anhydrous magnesium sulfate. The entire
area above and to the left of the broken solid line represents under-saturated
solutions of magnesium sulfate in water. The broken line eagfhij represents
complete solidification of the liquid solution to form various solid phases.
The areas pae represents mixtures of ice and saturated solution. Any
solution containing less than 16.5 % MgSO4 precipitates ice when the
temperature reaches line pa. Broken line abcdq is the solubility curve. Any
solution more concentrated than 16.5 % precipitates on cooling a solid
when the temperature reaches this line. The solid formed at point a is called
a eutectic. It consists of an intimate mixture of ice and MgSO 4.12H2O.
Between points a and b the crystals are MgSO 4.12H2O, between points c
and d the crystals are MgSO4.6H2O ; and above point d they are
MgSO4.H2O . In the area cihb, the system at equilibrium consists of
mixtures of saturated solution and crystalline MgSO 4.7H2O. In area dkjc,
the mixture consists of saturated solution and crystals of MgSO 4.6H2O. In
area qdk, the mixture is saturated solution and MgSO4.H2O.
7
Yields

In many industrial crystallization processes, the crystals and the


mother liquor are in contact long enough to reach equilibrium, and
the mother liquor is saturated at the final temperature of the process.
The yield of the process can be calculated from the concentration of
the original solution and the solubility at the final temperature,
allowing for any evaporation by making solvent and solute
balances.

Solvent balance
Initial solvent present = final solvent in mother liquor + water of
crystallization + water evaporated

W1= W2 + y + W1E ---------(1)

W1 = initial mass of solvent in the liquor


W2 = final mass of solvent in the liquor (saturated solution)
Y = yield of crystals
R = ratio of molecular mass of hydrate/ molecular mass of
anhydrous salt.
E = ratio of solvent evaporated / solvent in initial solution.

{ eg MgSO4 = 120.4, MgSO4. 7H2O = 246.5 }

R = =
Solute balance
8

W1 C1 = W2 C2 + -------(2)

C1 = initial concentration of the solution ( mass of anhydrous salt/


mass of solvent)

C2 = final concentration of the solution ( mass of anhydrous salt/ mass of solvent in satu soln)
W1= W2 + y + W1E ---------(1)
W1 C 1 = W 2 C 2 + -------(2)

From equation (1) W2 = W 1 - y - W1E

W2 = W1 (1- E) - y
Substitute W2 in equation (2)

W1 C1 = [W1 (1- E) - y ]C2 +

y = RW1 { }

where : C1 = initial concentration of the solution ( mass of anhydrous salt/ mass of solvent)
C2 = final concentration of the solution ( mass of anhydrous salt/ mass of solvent in satu soln)
R = ratio of molecular mass of hydrate/ molecular mass of anhydrous salt.
W1 = initial mass of solvent in the liquor

Supersaturation
Supersaturation may be generated by 3 methods
 A saturated solution becomes supersaturated by simple cooling
ie temperature reduction.
 Supersaturation may be generated by evaporating a portion of
the solvent.
9
 By adding a third component. The third component may act
physically by forming with the original solvent a mixed solvent
in which the solubility is sharply reduced. This process is
called salting.
If a nearly complete precipitation is required, a new solute may
be created chemically by adding a third component, that will
react with the original solute and form an insoluble substance.
This process is called precipitation.

Units for supersaturation. Supersaturation is the concentration difference


between that of the supersaturated solution in which the crystal is growing
and that of a solution in equilibrium with the crystal. Concentrations may
be defined either in mole fraction of the solute denoted by y, or in moles of
solute in unit volume of solution, denoted by c.

y = y - ys
c = c - cs

y supersaturation, mole fraction of solute


y = mole fraction of solute in solution
ys = mole fraction of solute in saturated solution
c = molar supersaturation, moles per unit volume
c = molar concentration of solute in solution
cs = molar concentration of solute in saturated solution

Caking of crystals

Crystalline materials frequently cake or cement together on


storage.
In general caking is caused by
10
 a dampening of the crystal surfaces in storage because of
inefficient drying
 an increase in atmospheric humidity.
 Presence of a hygroscopic trace impurity in the crystals.
 Moisture may be released from inclusions if crystal
fractures under storage conditions.

Thus, Crystal size, shape, moisture content and


storage conditions can all contribute to caking
tendency.

Caking can be minimized by


 efficient drying
 packaging in airtight containers
 avoiding compaction on storage.
 Crystals may be coated with an inert dust that acts as
moisture barrier.
Small crystals are more prone to cake than large crystals-greater number
of contact points.

Narrower the size distribution and more granular the shape, the lower is
the tendency of caking.

Nucleation and growth


The transformation can be divided into two discrete steps that occur one
after another.
 The formation of tiny stable particles of , called nucleation.
 The increase in size of these stable particles, called growth.
11
Nucleation
Nucleation refers to the birth of very small bodies of a new phase within a
supersaturated homogeneous existing phase.

Homogeneous nucleation –is restricted to the formation of new particles


within a phase uninfluenced in any way by solids of any sort, including the
walls of the container.

Heterogeneous nucleation-when solid particles of foreign substances do


influence the nucleation process by catalyzing an increase of nucleation
rate at a given supersaturation.
Homogeneous nucleation almost never happens.

Crystal nuclei may form from various kinds of particles: molecules, atoms,
or ions. Because of their random motion, in any small volume several of
these particles may associate to form what is called a cluster- a rather loose
aggregation which usually disappears quickly. Occasionally however,
enough particles associate into what is called an embryo, usually embryos
have short lives and revert to clusters, but if the supersaturation is large
enough , an embryo may grow to such a size that it is in thermodynamic
equilibrium with the solution. It is then called a nucleus, which is the
smallest assemblage of particles that will not redissolve and can therefore
grow to form a crystal. The sequence of stages in the evolution of a crystal
is, then,

clusterembryonucleuscrystal
12

The nucleation kinetics


If f is the free energy change accompanying the
formation of a spherical new phase particle we can write

……….(1)
Where,
r is the radius of the particle,
is the Gibbs free energy change per unit volume
is the surface energy per unit area of the interface separating
the parent and the product phase.

The surface energy term is always positive.


If is negative, the function f passes through a maximum.
13

Initially as the new phase particle starts to form, the energy of the
system increases, as the surface term is dominant. There after the variation
in the volume term becomes dominant and as this term is negative, there is
a continuous decrease in energy of the system. By setting d(f)/dr = 0, the
values corresponding to the maximum, called the critical values and
denoted by the superscript * are obtained.

d (f )/ dr = =0

r* = - …………..(2)
14

Substituting the value of r* = - in equation (1) we


get
…….(1)
f* =

f* =

The critical value of free energy of nucleation f* and of


the particle radius r* .
Particles which are smaller than the critical size are
called embryos. Those that are larger than the critical
size are called nucleus.
As becomes more negative with a lowering of the
temperature, the critical condition occurs at smaller
values of f and r.

For the L   transformation, the following usually holds good

= =

Where is the enthalpy change (the heat of reaction) per unit volume of
the product.
Tm is the equilibrium melting point at which = 0.

T is the transformation temperature, is the degree of supercooling.


15
Substituting the value of in the equation
f* =
we have
f* =

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