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Crystallisation Flashcards

1
Q

Crystal Habit

A

Describes general external shape of crystal

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2
Q

Gibbs crystal habit theory

A

Crystal growth attempts to minimise total free energy of the surface
High surface energy = low morphological importance

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3
Q

Wulff crystal habit theory

A

Distance of facet from its centre is proportional to surface energy

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4
Q

BFDH crystal habit theory

A

facet growth is inversely proportional to the interplanar spacing, d_hkl

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5
Q

Hartman & Perdock crystal habit theory

A

Growth rate is proportional to the attachment energies of crystal facets

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6
Q

Miller Indices

A

Integers that are inversely proportional to the intercepts of the crystal face

h = a/x, k = b/y, l = c/z

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7
Q

Miller Indices methodology

A
  • Determine the intercepts of the face along the crystallographic axes
  • Take reciprocals
  • Clear fractions
  • Reduce to the lowest terms
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8
Q

Polymorphs

A

Molecularly identical crystals with different lattice structure

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9
Q

Dihedral angle

A

Angle between 2 facets

consistent irrespective of particle size

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10
Q

Supersaturation

A

Solution containing more solute than thermodynamically stable in solutions (and able to overcome energy barrier to form crystal surface)

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11
Q

Saturation

A

Crystal is at thermodynamic equilibrium with solution (maximum concentration that is thermodynamically stable)

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12
Q

Degree of Supersaturation

A

Delta_C = C - C*

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13
Q

Supersaturation Ratio

A

S = C/C*

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14
Q

Relative Supersaturation

A

Phi = Delta_C/C* = S - 1

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15
Q

Moyers & Rousseau (van’t Hoff Relationship)

A

ln(x) = Hf/RT * (T/Tm - 1)

Hf = Latent heat of fusion
Tm = melting point
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16
Q

Predicition of solubility for very small particles

A

ln (cr/c) = 2Msigma/(niRTro_sr)

17
Q

Supersaturation Equation (Sigma)

A

sigma = Delta(mu)/RT

18
Q

Primary Nucleation

A

In the absence of seeded crystals

19
Q

Homogeneous nucleation

A

formation of crystals from supersaturation only

20
Q

Heterogeneous nucleation

A

formation of crystals from the presence of insoluble material

21
Q

Secondary nucleation

A

Occurs in the presence of seeded crystals

22
Q

Empirical equation for secondary nucleation

A

dN/dt = k_n(c - c)^i = B

23
Q

Classical Nucleation Theory

A

Surface energy of small crystals provides a thermodynamic barrier to forming new surface

Delta Gr = - 4/3 pir^3 * Delta Gv + 4pir^2 *sigma

(Assumes Nucleus to be spherical and interacial tension to isotropic - non directionally dependent)

24
Q

Critical Radius

A

Minimum size of crystal for it to be more energy beneficial to grow than redissolve

25
Q

Method of solving for r_crit and Delta Gr_crit

A 
Set d(delta Gr)/dr = 0
Solve for r_crit
put equation for r_crit into delta Gr equation to solve for delta Gr_crit
26
Q

Homogeneuous rate of nucleation

A

J = Fexp((-16pi*sigma^3 vm^2)/(3k^3 T^3(lnS)^2)

27
Q

Heterogeneous rate of nucleation

A

J = K_n * delta(c)^n

28
Q

Derivation of homogeneous nucleation rate

A
  • Start with Arrhenius:
    J = F*exp(-delta G/kT)
  • Insert delta G_crit
  • Rearrange ln(cr/c*) for r and insert
  • Use molar volume to remove terms and insert some constants in pre-exponential factors
29
Q

Define induction period

A

Delay between supersaturation & first crystal formation

ti is proportional to 1/J

30
Q

2 Stage process of crystallisation

A
  • Birth = Nucleation

- Growth (Volume Diffusion & Surface diffusion)

31
Q

3 Crystal growth regimes

A
  • Continuous Growth (lots of kink sites)
  • Surface Nucleation (several nuclei on suface)
  • Spiral growth (single nucleus on surface)
32
Q

Crystal habit modification

A
  • growth rates are not constant along facets

- Specific adsorption of species on particular facets modifies crystal growth

33
Q

Crystal Yield Equation

A

y = Rw1 (c1 - c2(1 - E))/(1 - c2(R - 1))

R = ratio of hydrate: anhydrous MW
E = ratio of solvent evaporated to initial solvent
34
Q

Crystal yield derivation

A
  • Solute balance
    w1c1 = w2c2 + y/R
  • Solvent Balance
    w1 = w2 + y(R - 1)/R + w1E
  • Rearrange solvent balance for w2
  • Insert w2 into solute balance
  • rearrange for y
35
Q

Derivation of mass of seeds required equation:

A
  • Take the mass of crystal assuming spherical
    m_s = ro*pi/6 * d_s^3
  • Take equation for yield based on volume difference:
    y = ropi/6(d_p^3 - d_s^3)
  • Write out m_s/y
  • Then multiply both sides by y
36
Q

Optimum cooling curve

A
  • Growth rate will increase as crystal grows
  • dC/dT of solubility curve decreases as C decreases, therefore need to increase (-dT/dt) to accommodate
  • Need to cool faster to remain in metastable region
  • Slow initial cooling but increasing over time
37
Q

Practice sketching crystallisation curves

A

slides 50 and 51

Particles (6 decks)

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