DIFFUSION CONTROLLED GROWTH OF A MOVING SPHERE. THE

DIFFUSION CONTROLLED GROWTH OF A MOVING SPHERE. THE KINETICS OF CRYSTAL GROWTH IN POTASSIUM PERCHLORATE PRECIPITATION...
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ARNEE. NIELSEN

46

Vol. 65

DIFFUSION CONTROLLED GROWTH OF A MOVING SPHERE. THE KINETICS OF CRYSTAL GROWTH IN POTASSIUM PERCHLORATE PRECIPITATION BYARNE E. NIELSEN Universitetets fysisk-kemiske institut, Copenhagen, Denmark Received April E#, 1960

The crystals of potassium perchlorate precipitating from aqueous solution were found to grow with a higher velocity than calculated for pure diffusional rate control. This effect is explained by the convection around the falling crystals, which keeps the concentration near the crystals higher than diffusion alone.

matter what is the type of reaction in the surface

Experimental

The supersaturated aqueous solution was prepared by transferring it from the dissolved to the crystalline mixing50ml.of 0.5MKCland 50ml.of 0.5MNaC104. The state. The only possible explanation is that conmixture was stirred continuously during the experiment. vection is also contributing to the transport from The electric conductance was followed by means of a Philips Recording Conductance Bridge. The readings were the bulk of the solution to the crystal. Mathematical.-The expression quoted for K D converted into the degree of advancement a = (a - c)/ (cg - s) (where c = concentration of dissolved KC104, co was derived2 using an approximation, since the = c a t t = 0, s = solubility of KC1o4) by means of the results from measurements on solutions having the composi- concentration gradient at the surface was calculated tions corresponding to different known values of a. The from the stationary solution of Fick’s second law. particle size at the end of the experiment was determined But when the particles grow it must be a little by sedimentation in a mixture of glycerol and water, satu- larger. In order to be sure whether this gives any rated with KC10,. The weight of the sediment was re- appreciable contribution to the increased growth corded and, from the relation between weight and time, the weight distribution curve was obtained by differentiation rate observed we shall first consider the case of according to 0dBn.l In the following the weight average purely diffusional growth. Later on we shall treat radius of the crystals (assumed spherical) is used. Since the combined diffusion-convection-control of the the distribution is rather sharp, the different kinds of aver- rate of particle growth. age differ only by a few per cent.

The crystals have a rather compact shape so that it will not lead to great errors to treat them as spheres in the calculations of the present work.

1. Purely Diffusional Growth of a Sphere.The diffusion process follows Fick’s second law and since the case is spherically symmetric this law can be written, assuming D independent of c

Results I n Fig. 1 is shown the degree of advancement a as function of time t. In a previous work2 it was shown that, in case of pure diffusional rate control the relation between a and t is given by

where c = concentration and r = radius vector. With a = radius of the particle and s = solubility the boundary conditions are

t =

where

KDID

c(T,O) = c,;

c(a,t) = s; c( m , t ) = em

If a and

C m are constant the solution is easily obtained by standard methods (e.g., through the Laplace transformation)

K D = ala/3vD(co - s)

al = radius a t a = 1; v = molar volume of precipitate, D = diffusion coefficient, co = initial con- where 2 2 centration and s = solubility. ID is a function of erf x 3 Jo exp( - f2)d.$ a which has been tabulated.2 We insert al = 4.64 X cm. as found from the sedimentation analy- There is a non-trivial stationary solution (this is sis, v = 55.0 ~ m . ~ / m o lco e , = 2.5 X 10-4mole/ ~ r n . ~ not , true for the analogous one- or two-dimensional s = 1.58 X lov4rnole/~m.~, determined from the cases) obtained for t -+ 03 final value of the conductivity in the kinetic experic = s + ( e , - s)(l - a h ) ment and D = 1.61 X cm.2/sec. (From ionic equivalent conductivities we find D = 1.88 X lob6 The rate of growth must satisfy da/dt = vD(dc/dr),, cm.2/sec.; this is corrected to ionic strength 0.5 by means of the factor 1 0.5864df/(1 t/j)2 and if the stationary value of dc/dr is used, we get = 0.858). This gives K D = 88 see., and the corda/dt = vD(c, - s ) / a is shown on Fig. 1. responding curve, t = 881~(a), :. a = d 2 v D ( c m - s ) t Discussion This is valid at low concentration only. At high It is clear from this that the crystals have grown concentration the growth may be so fast that the faster than possible if the solute can reach the sur- stationary concentration field cannot be established face of the growing crystals only by diffusion-no with sufficient accuracy. We shall therefore derive the exact solution to Fick‘s second law for a growing (1) 5. OdBn, Kolloid-Z., 18, 33 (1916). sphere. (2) A. E. Nieleen, Acto Chem. Scond., 13, 784, 1680 (1959).

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+

DIFFUSION C O N T R O L L E D GROWTH OF A MOVISG SPHERE

Jan., 1961

The radius a is initially zero. Since no geometric parameter is &xed in advance, the function c(r, tr) at any time t1: is derived from ~ ( ~ $ 1 1 )valid a t any other time simply by multiplying all the radii by a common factor. This means that if

47

1

----3,

c(r1,td = c(r2,tId

and c(rr,t~)= c(r&I)

then

0

0

?=?=ar

r2 r d a11 Therefore c = c ( z ) ; z = a/r; a = a(t).

Inserting this in Fick’s second law we get dc da dz dt

DaZ d2c r3 dz2

or a da Dai

Since dt

a da = - _-

D dt

dc/dz

(2)

da -- = V D

me have

= i3d2c/dc2

%a

-a

(&)

and

=

I= 1

K , a constant

>0

The general solution of this differential equation is c = K1 [zexp

(--&) + 4 2 e r f (i d:)] + K 2

The arbitrary constants Kl and K2 are eliminated by means of the two boundary conditions for z = 1 c = S, dc/dz = - K / v

which gives

From the third boundary condition c = c m for r =

m

follows

In the limit K

+

0 we find

which reproduces the formula (2 =

d2Dv(c,

- s)t

found by means of the stationary solution to Fick’s second law. If we write the exact solution a =

150

200

Fig. 1.-The degree of reaction as function of time during the precipitation of potassium perchlorate from supersaturated aqueous solution. “881~,”theoretical curve, assuming that only diffusion is controlling the rate of growth; “88I~220,”theoretical curve, assuming that both diffusion and convection are controlling the rate of growth; circles, experimental points. The coefficients in the theoretical curves were calculated from the particle size, measured independently. Kq = V(C, - S)

u(cm

(z)s=~

-’

100

t , sec.

Table I gives numerical examples of this relationship.

V D dc

=

50

d2Dv(cm - s ) t / q

defining the quantity q by this equation, we may determine the relation between q and v(c, - s) by means of the equations

- a)

0.00001 .00002 .OM05 ,0001 .0002 .0005 .OOl .002 ,005 .Ol .02

TABLEI Q

0.996 .994 * 991 .987 .982 .972 .960 .944 .912 * 877 ,827

v(cm

- s)

0.05 .1 .2 .3 .4 .5 .6 .7

.8 .9

1.o

4

0.730 .626 .485 .382 .300 .232 ,172 .120 .073 .034 .000

We see that t’heapproximation made when using the stationary concentration gradient is good when v(c, -8) is small. This entity is the ratio between the supersaturation cm -s and the concentration of the matter in the precipitated state, l / v . It is also the volume of precipitate originating from a unit volume of the solution. When v(cm - 8 ) = 1 the transition from the dissolved to the precipitated state does not require any transport through space, and (neglecting latent heat of phase change, etc.) may proceed infinitely fast. In the experiments reported above v = 55.0 ml./ mole, co = 2.5 X and s = 1.6 X mole/ml., so that in the beginning of the experiment v(cm -s) = 0.005, p = 0.904. The spheres will thus reach a certain size in 90.4% of the time they would require if the concentration field were of the ‘(stationary” type. Since it will complicate later calculations to take into account the change of with concentration we shall in the following always assume q = 1. The error will not invalidate the conclusions concerning the cause of the great velocity of growth of the particles. 2. Convection.-According to StokesYthe velocity field in the space around a sphere of radius a falling through the liquid with the velocity U satisfies (3) G . G. Stokes, Trans. Cambridge Phil. Soc., 9 , 8 (1850),reprinted in “Mathematical and Physical Papers,” by G. G. Stokes, VoI. 3, 1 , Cambridge Univ. Press, 1901, see pp. 55 ff.

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Vol. 65

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+ cii

= -Dgradc

The rate of change of the concentration is given by dc/dt = -div

\

9

= div[D grad c

- ca]

Neglecting the compressibility and the partial volume of the solute we may use div ii = 0

With a constant D we therefore derive .

- ii grad c

dc/dt = D div grad c

Introducing the Stokes velocity field we b a l l y get

c = C(T, e, t ) ; C ( T , 0, 0 ) = c,; C(U, e, 1) = S; C(,, e, t ) = cm. Since we need only the stationary solu-

tion we let the left-hand side be zero. Introduction of the dimensionless constant Fig. 2.-Sections of surfaces of equal concentration (z = 0, 0.2, 0.4, 0.6 and 0.8) at difierent velocities ( A = 0, 1 10 and 100) around a sphere growing with a rate controlled by diffusion and convection whilst falling through the liquid.

A

= aU/D

the dimensionless variables p

= ln(y/a),

r

and the two functions of Ql = Qz

uq =

ep

= (c

- s)/cm

p

-3

+ 1 e-%

3

1 - 2~

=ep--,--,e

0

where u,, U,9 and u, are the components in the coordinate directions of the velocity relative to the sphere, T being radius vector (from the center of the sphere), 0 the colatitude and cp the longitude; 6 = 0 when T points vertically upwards. These equations describe the velocity fields for Reynold’s numbers R

= 2aUd/q

transforms the equation into

ar = 0 - AQ2 sin e) ae

(cot e

where { = f ( p , e ) ; 0 S p 5 0 2 0 2 T ; (0, e) = 0 ; f ( a ,0) = 1. A is closely related to Reynolds number 0);

A = Rq/2Ddo up to about 8. For < 8.15 eddies do not occur.4 For R