Unformatted text preview: lving (3) for a range of crystal size,
Dp, microns
0.01
0.10
1.00
10.00 Dp , m
0.00000001
0.0000001
0.000001
0.00001 c/cs
1.417
1.0355
1.0035
1.00035 c, g/100 g water
300
219.5
212.7
212.1 The results show that the effect on solubility is greatest for the smallest crystals. Exercise 17.17
Subject: Required supersaturation ratio.
Given: Crystals of sucrose of 0.5 microns diameter. Properties of sucrose: MW = 342; crystal
density = 1,590 kg/m3; and interfacial tension, σs,L = 0.01 J/m2
Assumptions: Spherical crystals. Temperature of 25oC.
Find: Supersaturation ratio, S = c/cs required for the crystals to grow.
Analysis: Use (1716) with SI units: ln S = ln 4v σ
c
= s s,L
cs
υRTD p Substituting into (1) gives,
342
(0.01)
1,590
S = exp
0.5
1( 8315 )( 298 )
106
4 = 1.007 (1) Exercise 17.18
Subject: Explanation for the experimental maximum in the crystal solubility curve with crystal
size.
Given: Surface energy of crystals may depend on surface electrical charge as well as on
interfacial tension. Relationship for electrical charge effect.
Assumptions: Spherical crystals.
Find: A modification of (1716) that predicts the maximum.
Analysis: From (1716), ln S = ln 4v σ
c
= s s,L
cs
υRTD p (1) To predict a maximum, modify (1) by incorporating the given electrical charge effect, as
follows:
4vs σ s , L
2q 2 vs
c
(2)
ln S = ln
=
−
4
cs
υRTD p πKRTD p
Equation (2) is of the form:
ln v
c
A
B
=s
−4
cs
RT D p D p (3) Differentiate (3) with respect to Dp to find min/max:
v
d
c
A 4B
ln
= s − 2 + 5 =0
dD p
cs
RT
Dp Dp
Solving (4) for Dp,
4B
Dp =
A 1/ 3 2q 2 υ
=
πK σs,L 1/ 3 , which can be shown to be a maximum (4) Exercise 17.19
Subject: Effect of supersaturation ratio on rate of primary homogeneous nucleation.
Given: Aqueous solutions of AgNO3, NaNO3, and KNO3 at 25oC, together with values for
crystal density and interfacial tension.
Assumptions: Spherical crystals.
Find: Rates of primary homogeneous nucleation over the range of supersaturation ratio,
S = c/cs , of 1.005 to 1.02.
Analysis: Use (1718) with SI units in the exponential term:
B o , number of nuclei formed/cm 3 s = 1030 exp
The properties are:
Component
AgNO3
NaNO3
KNO3 υ
2
2
2 MW
169.6
85.0
101.1 ρ,
ρ, kg/m3
4,350
2,260
2,110 −16πvs σ3 N a
s,L 3υ2 ( RT ) ( ln S )
3 (1) 2 σs,L, J/m2
0.0025
0.0015
0.0030 vs = MW/ρ
0.0390
0.0376
0.0479 Na = Avagadro’s number = 6.022 x 1026 molecules/kmol
R = 8315 J/kmolK
T = 298 K
Use of a spreadsheet gives the following results from (1): Rate of primary nucleation, Bo , nuclei/scm3
Component
AgNO3
NaNO3
KNO3 S = 1.02
4.37 x 1025
1.33 x 1029
4.20 x 1018 S = 1.01
5.43 x 1012
3.39 x 1026
8.72 x 1016 S = 1.005
1.91 x 1039
1.54 x 1016
4.50 x 10150 Exercise 17.20
Subject: Primary homogeneous nucleation of BaSO4.
Given: Crystal density of 4.50 g/cm3 and interfacial tension of 0.12 J/m2.
Assumptions: Spherical crystals.
Find: Effect of relative supersaturation on primary nucleation from an aqueous solution at 25oC.
Analysis: Since from Table 17.6, BaSO4 is sparingly soluble in w...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details