Crystal-Growth Kinetics After Nucleation

From Supersaturation to Crystal Growth

After a viable crystal nucleus exists, dissolved drug can deposit on its surface. The local driving force is supersaturation: if the dissolved concentration exceeds the equilibrium solubility, deposition is thermodynamically favored. The concentration excess alone does not specify how quickly a crystal grows. A growth law also represents the rate of molecular attachment at the surface and may depend on the crystal form, medium, temperature, and hydrodynamic conditions.

Let $C$ be the dissolved drug concentration and $C_s$ its equilibrium solubility under the conditions being modeled. Define the supersaturation ratio $S=C/C_s$. Thus $S=1$ is saturation and $S>1$ is supersaturation. The supersaturation ratio is dimensionless; the radial growth velocity $G$ has units of length per time.

Definition: A one-sided power-law crystal-growth model is $$ G=\frac{dr}{dt}=k_g\,[\max(S-1,0)]^\gamma , $$ where $r$ is the crystal radius, $k_g$ is a growth-rate coefficient, and $\gamma>0$ is an empirical exponent. With this convention, $G$ describes deposition during supersaturation and is set to zero at or below saturation. It does not describe redissolution.

For example, if $\gamma=1$, radial growth velocity is proportional to the supersaturation excess $S-1$. If $\gamma>1$, the modeled velocity increases more sharply as supersaturation rises. Because $S-1$ is dimensionless, $k_g$ has the same units as $G$, such as micrometres per minute. If a law is instead written in terms of $C-C_s$, its coefficient has different units; the chosen driving-force definition must therefore be reported with the fitted coefficient.

This law is a phenomenological description of growth over a specified range of conditions. It assumes that the chosen concentration or supersaturation represents the driving force at the crystal surface, and that $k_g$ and $\gamma$ remain applicable over the interval considered. It does not determine when nuclei first appear, how many appear, or how the growth law changes as crystals interact.

Worked Example: Radius Increase at Fixed Supersaturation

Suppose a hypothetical spherical crystal has initial radius $2.00\ \mu\text{m}$. During a short interval, the surrounding solution remains at $S=1.50$. Let $k_g=0.020\ \mu\text{m}/\text{min}$ and $\gamma=1$. Calculate the radial growth velocity and the radius after 30.0 minutes.

The model gives $$ G=k_g(S-1)^\gamma =(0.020\ \mu\text{m}/\text{min})(1.50-1)^{1} =0.0100\ \mu\text{m}/\text{min}. $$

Because $G$ and $S$ are held constant, integrate $dr/dt=G$: the radius increase is $Gt=(0.0100\ \mu\text{m}/\text{min})(30.0\ \text{min})=0.300\ \mu\text{m}$. The final radius is $2.00+0.300=2.30\ \mu\text{m}$.

This calculation assumes constant supersaturation, a constant growth coefficient, and no change in crystal shape. If the dissolved concentration fell appreciably as drug deposited, the assumed constant $S$ would no longer be justified.

Converting Radial Growth into Solid-Mass Accumulation

A radius increase does not by itself state how much drug has precipitated. The conversion requires crystal geometry, density, and the number of crystals. For a spherical crystal of density $\rho_s$, its solid mass is its density multiplied by its volume. Differentiating that mass with respect to time gives the instantaneous deposition rate.

m_s=\rho_s\frac{4\pi}{3}r^3 \qquad\Longrightarrow\qquad \frac{dm_s}{dt} =\rho_s\,4\pi r^2\frac{dr}{dt} =\rho_s A G , \quad A=4\pi r^2 .

The result has a direct physical interpretation: mass accumulates at a rate equal to solid density times surface area times radial growth velocity. For $N_c$ identical crystals growing at the same rate, the total rate is $N_c\rho_s A G$. For crystals with different radii, sum the individual contributions; the total surface area, not simply the mean radius, determines the instantaneous mass rate.

If the number of crystals is fixed and each spherical crystal has a constant radial velocity $G$, then $r(t)=r_0+Gt$. The exact mass added over an interval $t$ is

\Delta M_s =N_c\rho_s\frac{4\pi}{3} \left[(r_0+Gt)^3-r_0^3\right].

This finite-interval formula matters because surface area increases as crystals grow. Multiplying the initial mass-deposition rate by the whole interval is only an approximation when the radius changes little. In a well-mixed batch of constant volume $V$, if deposition is the only process changing dissolved drug, the corresponding concentration change is $\Delta C=-\Delta M_s/V$, with compatible mass and volume units.

Worked Example: Mass Added to a Fixed Crystal Population

Consider a hypothetical suspension containing $N_c=2.00\times10^6$ identical spherical crystals. Each has initial radius $r_0=1.00\ \mu\text{m}$, and each grows at constant radial velocity $G=0.0100\ \mu\text{m}/\text{min}$ for 20.0 minutes. Let the solid density be $\rho_s=1.20\ \text{g}/\text{cm}^3$. Calculate the total mass added.

First, the final radius is $r_0+Gt=1.00+(0.0100)(20.0)=1.20\ \mu\text{m}$. Convert radii to centimetres: $1.00\ \mu\text{m}=1.00\times10^{-4}\ \text{cm}$ and $1.20\ \mu\text{m}=1.20\times10^{-4}\ \text{cm}$. Apply the finite-interval formula:

\Delta M_s =(2.00\times10^6)(1.20\ \text{g}/\text{cm}^3) \frac{4\pi}{3} \left[(1.20\times10^{-4}\ \text{cm})^3 -(1.00\times10^{-4}\ \text{cm})^3\right].

The difference of the cubed radii is $(1.728-1.000)\times10^{-12}=7.28\times10^{-13}\ \text{cm}^3$. Thus each crystal adds $(4\pi/3)(7.28\times10^{-13})=3.049\times10^{-12}\ \text{cm}^3$, and the whole population adds $6.098\times10^{-6}\ \text{cm}^3$. Multiplying by density gives $(1.20)(6.098\times10^{-6})=7.318\times10^{-6}\ \text{g}$, or **7.32 $\mu\text{g}$** of solid drug.

The calculation assumes fixed crystal number, spherical crystals, constant density, and constant radial growth velocity. It accounts for the increasing surface area through the change in radius rather than assuming a constant mass-deposition rate.

Growth-Limited and Nucleation-Limited Precipitation

Nucleation and crystal growth are distinct potential bottlenecks. Nucleation-limited precipitation occurs when formation of viable crystals is slow relative to the relevant observation period: little solid may form because few growth surfaces are available. Growth-limited precipitation occurs when crystals are already present but their growth is too slow to account for the observed rate of drug removal from solution. The terms describe the rate-controlling process in a particular setting, not permanent properties of a drug.

A long delay before detectable solid appears does not, by itself, prove that growth is slow. There may be no viable crystals during that delay. Conversely, observing crystals does not prove that growth is fast enough to explain the measured precipitation. To interpret a concentration-time profile, distinguish evidence about the appearance of crystals from evidence about their subsequent size or mass change. A growth law applied after nucleation describes the latter and does not predict the former.

Worked Example: Assessing the Effect of Supersaturation on Growth

In a hypothetical system, assume $k_g=0.00400\ \mu\text{m}/\text{min}$ and $\gamma=2$. Compare the radius increase over 60.0 minutes at $S=1.25$ and $S=2.00$, assuming existing crystals and constant supersaturation in each case.

At $S=1.25$, the velocity is $G=(0.00400)(1.25-1)^2=0.000250\ \mu\text{m}/\text{min}$. The radius increase is $(0.000250)(60.0)=0.0150\ \mu\text{m}$.

At $S=2.00$, the velocity is $G=(0.00400)(2.00-1)^2=0.00400\ \mu\text{m}/\text{min}$. The radius increase is $(0.00400)(60.0)=0.240\ \mu\text{m}$. Under this assumed power law, the second condition produces a 16-fold larger radial increase over the same interval. This comparison concerns growth of existing crystals; it does not establish how many crystals form or whether the system is nucleation-limited.

Choosing and Checking a Growth Description

A radial-velocity law is convenient when measurements or a mechanistic hypothesis describe crystal-face motion. An alternative is to specify a mass flux per unit surface area, such as $j=k_m(C-C_s)$ for $C>C_s$, where $j$ has units of mass per area per time. For a spherical crystal, that flux implies $dm_s/dt=A j$. The two forms can be related through $j=\rho_sG$ when the deposited solid has density $\rho_s$ and the surface area is defined consistently. They are alternative parameterizations of deposition, not interchangeable coefficients with the same units.

A linear law in $S-1$ is simpler and may be adequate over a restricted range. A power law with an estimated exponent can represent curvature in growth-rate data, but additional flexibility does not guarantee a more reliable model. More detailed surface-kinetic models may represent attachment steps or face-specific behavior; they require information that may not be available in a precipitation experiment. The appropriate choice depends on the measurements and intended use.

Useful checks include verifying that $G$ is zero at $S=1$ under the one-sided convention, that predicted velocities have length-per-time units, and that estimated growth produces a plausible radius or mass change over the observed interval. A fitted velocity that predicts crystals larger than observed, or a mass rate incompatible with the measured loss of dissolved drug, signals a problem with the parameter values, assumptions, or model structure. Check whether the concentration used to compute $S$ is the free dissolved concentration relevant to the crystal surface, and whether the solubility used corresponds to the same conditions.

Specify the growth variable.
State whether the model predicts radial velocity $G$ or surface mass flux $j$, and give its units.

State the driving force and range.
Define $S$ or $C-C_s$, identify the conditions for which the law applies, and say how the model treats saturation and undersaturation.

Translate rate into mass.
Use the appropriate surface area, geometry, density, and crystal number; integrate over time if area changes materially.

Check the interpretation.
Separate evidence for crystal appearance from evidence for growth, and compare the predicted solid accumulation with measurements.

Assumptions, Failure Modes, and Limits

The fixed-population mass calculation assumes that no new crystals appear, existing crystals do not disappear, and their shape and density remain as specified. These assumptions are useful for isolating growth after nuclei are present, but they are not a complete precipitation model. In particular, if the number of crystals changes, total surface area can change independently of radial growth. Aggregation, breakage, changing crystal form, or non-spherical geometry can also invalidate a calculation based on identical spheres.

A common mistake is to use the initial surface area to calculate a constant mass rate for a long interval, then treat that rate as exact. The reasoning fails because a growing sphere has $A=4\pi r^2$, so its area increases with radius. Use the exact finite-interval volume difference when radius changes appreciably, or integrate the changing area over time. Another mistake is to interpret a slow rise in solid mass as direct evidence of a small $k_g$; a small number of available crystals can also limit total deposition. Independent information about crystal number or size helps distinguish these explanations.

The one-sided law is also not a model of what happens below saturation. Setting $G=0$ there is an explicit growth-only convention, not a claim that crystals cannot dissolve. Redissolution requires a law for the reverse process and consistent accounting of dissolved and solid drug. A growth model can therefore be informative while still being only one component of a larger precipitation description.

Key takeaway: A growth law predicts how fast existing crystal surfaces advance under a specified driving force. Converting that velocity to accumulated mass requires crystal geometry, density, surface area, and population size; it does not by itself predict when or how many nuclei form.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

If $S=1.40$, $k_g=0.0150\ \mu\text{m}/\text{min}$, and $\gamma=1$, what is $G$ under the one-sided power law?
Why can two suspensions with the same radial growth velocity have different total mass-deposition rates?
A spherical crystal has $r_0=3.00\ \mu\text{m}$ and grows at $0.0200\ \mu\text{m}/\text{min}$ for 10.0 minutes. What is its final radius?
Does observing a long delay before measurable precipitation establish that growth is the rate-limiting step? Explain.
Name two assumptions needed to use the fixed-population spherical mass formula.

Solutions. 1. $G=(0.0150)(1.40-1)^1=0.00600\ \mu\text{m}/\text{min}$. 2. Total mass rate depends on total crystal surface area and density as well as radial velocity; different crystal numbers or radii can produce different areas. 3. The radius increase is $(0.0200)(10.0)=0.200\ \mu\text{m}$, so the final radius is $3.20\ \mu\text{m}$. 4. No. The delay may reflect slow formation of viable nuclei, and the observation alone does not separate nucleation from growth. 5. Examples include fixed crystal number, spherical shape, constant density, and a specified radial growth rate; any two are sufficient.