Crystallization

Crystallization Principle

Primary nucleation: Can be either homogeneous or heterogeneous

The rate of primary nucleation modeled by power law expression:

No foreign particles present

Foreign particles present

B = no. of nuclei formed per unit volume per unit timeN = no. of nuclei per unit volumekn = rate constantc = instantaneous solute conc.c* = solute conc. at saturation(c – c*) = supersaturationn = exponent typically 3 to 4.

Crystallization Principle

Secondary nucleation (usually predominates):

The rate of secondary nucleation:

Shear nucleation

Contact nucleation

K1 = rate constantMT = suspension densityb = exponent typically 2.j= 1 (most probable value)

Occurs as a result of fluid shear on

growing crystal faces

Occurs because of crystals colliding with each other and with impeller and other vessel internal surfaces

Crystallization Principle- As in precipitation, the solution must be supersaturated in order

for particles to form crystal (c – c*) in both Eqs.- Supersaturation must be above a certain value before nucleation

will begin.

Crystallization Principle

1• In Metastable region, the

supersaturation is so low that nucleation will not start.

2• The supersaturation is raised to Labile

region, nucleation can begin. • At this point, crystals begin to grow, and

the supersaturation decreases.• Point • is possible way of carrying out a

crystallization.

3• If the supersaturation becomes too

high, the nucleation rate will be too great, and an amorphous precipitate will result.

Crystallization Principle3) Crystal Growth- Crystal growth is the post-nucleation process in which the

molecules in solution are added to the surface of existing crystals.

- For designing a crystallizer, the useful relationship describing the rate of mass deposition, R during crystal growth is:

W = mass of crystals per volume of solvent

A = surface area of crystals per volume of solvent kG = overall mass transfer coefficient (depend on T, crystal size, hydrodynamic conditions, and presence of impurities.) g = the order usually between 0 – 2.5 (near unity is most common)

Crystallization Principle3) Crystal Growth- Also, to express overall linear growth rate as (delta L law):

L = characteristic single dimension of the crystal, such as length

It is shown that geometrically similar crystals of the same material grow at the rate described by Eq 9.2.4.

Crystallization Principle3) Crystal Growth- Crystal growth is actually a process that consists of two steps

in series.

(1) Solute molecules must reach the crystal surface by means of diffusion.

(2) At the surface, the solute must be integrated into the crystal lattice.

Crystallization Principle3) Crystal Growth

Where ci = concentration at the interface between liquid and solid phase. kd & kr = mass transfer coefficients.

- When the exponents are unity, combining Eqs (9.2.3), (9.2.5) & (9.2.6) gives:

Thus, if surface integration is very fast compared with bulk diffusion, then, and

Crystallization Principle4) Crystallization Kinetics from Batch Experiment

Process Crystallization of ProteinWhy use crystallization for purification of protein?

1) One or more expensive chromatography steps possibly can be eliminated.

2) Relatively inexpensive to carry out, since costly adsorbents are not required (as in chromatography)

3) Proteins crystals often can be stored for long periods at low T without being degraded or denatured after the addition of stabilizing agents such as ammonium sulfate, glycerol, or sucrose.

Process Crystallization of ProteinHow to crystallize protein?- Adjustment of pH to isoelectric point (isoelectric precipitation)- Addition of organic solvents- Addition of salts (salting-out)- Addition of non-ionic polymers.

Another way to crystallize protein?- Reduction of ionic strength by dialysis or diafiltration, which

relies on the limited solubility of many proteins at low ionic strength (reverse of “salting-in” effect).

- Key strategy: move the system very slowly to a state of min. solubility of the desired protein until a limited degree of supersaturation is reached.

Process Crystallization of ProteinStudy case:

1) Alcohol oxidase enzyme - from Pichia pastoris yeast

- grown in a 100-L pilot plant fermentor- crystallized by lowering ionic strength by diafiltration with deionized water- crystallization was preceded by lysis of yeast cells, diafiltration with microfilter to obtain alcohol oxidase in the permeate, and concentration and then diafiltration with an ultrafilter.

Process Crystallization of ProteinStudy case:

2) Ovalbumin - crystallized in the presence of conalbumin and lysozyme by

addition of 2.5 µm seed crystals to a solution (600 mL) that had been made supersaturated by slowly adding ammonium sulfate solution. - the supersaturation was kept in the metastable region to avoid nucleation.- the crytals growth rate is a second-order dependence on ovalbumin supersaturation, and the presence of the other 2 proteins did not affect the growth rate constant.

Crystallizer Scale-up & DesignImpurity is a problem in crystallization, can be in the

following locations in crystal sample:

1) Deposited on crystal surfaces due to incomplete removal of impure mother liquor.

2) Trapped within voids between separate crystals in materials that agglomerate.

3) Contained in inclusions of mother liquor within individual crystals.

4) Distributed throughout the crystals by molecular substitution at the lattice sites.

Crystallizer Scale-up & DesignIt is recommended that geometric similarity and

constant power per volume be used in scale-up of crystallizers.

For turbulent flow in vessels (Reynolds number >

10,000), constant power per volume:

For an agitated tank, the Reynolds number is:

Ni = impeller rotation ratedi = impeller diameter

Ni = rev. per minρ = fluid densityµ = fluid viscosity

Crystallizer Scale-up & Design2 additional strategies also used in Scale-up:

1) Maintaining constant impeller tip speed;

2) Scale-up at the minimum speed required for particle suspension;

Practice 1: Solubility in Crystallization

Given solubility data for Phthalic acid as follows:

18 g / 100 mL water at 100 ºC

0.54 g / 100 mL water at 15 ºC

A student obtained 3.2 g of crude phthalic acid. After recrystallization and drying, 2.5 g of pure acid was obtained. Calculate:

a) The percent recovery

2.5g/3.2gx100 =78%

b) Amount of water needed for recrystallization

100mL/18gx3.2g = 17.7mL

c) Amount of product lost to the filtrate.

0.54g/100mLx17.7mL=0.1g