Deriving Dissolution Rate from Interfacial Mass Transfer

From an Interfacial Gradient to a Dissolution Rate

A solid can dissolve only as dissolved molecules move away from its surface into the surrounding liquid. The concentration immediately at the solid–liquid interface may be close to a saturation value, while the well-mixed bulk liquid has a lower concentration. This spatial difference creates a concentration gradient and, under a diffusion-layer model, a flux away from the surface.

The Noyes–Whitney equation expresses this flux as a rate of increase in bulk concentration. The Nernst–Brunner form makes the mass-transfer coefficient explicit as diffusivity divided by diffusion-layer thickness. Both equations follow from the same physical argument; neither is a separate law that removes the need to state the assumptions behind the flux and mass balance.

Definition: Let \(A\) be the wetted, exposed solid surface area, \(D\) the solute diffusivity in the liquid, and \(h\) the thickness of an idealized diffusion layer adjoining the surface. Let \(C_s\) be the dissolved concentration at the surface and \(C_b\) the concentration in the well-mixed bulk. The bulk-to-surface concentration difference is \(C_s-C_b\). The outward molar or mass flux \(J\) is the amount of dissolved solute crossing unit area per unit time, using a consistent concentration basis.

For example, if concentration is expressed as mass per volume, \(J\) has units of mass per area per time. The total dissolution rate \(R_{\mathrm{diss}}\) is the amount of solid mass entering solution per unit time, so its units are mass per time. With spatially uniform flux over the exposed surface, \(R_{\mathrm{diss}}=AJ\).

Deriving the Interfacial Flux

Use a coordinate \(x\) that begins at the solid surface (\(x=0\)) and increases outward through the diffusion layer to the bulk edge (\(x=h\)). Assume one-dimensional transport normal to a planar surface, constant diffusivity, no reaction or loss within the layer, and a quasi-steady concentration profile. The boundary concentrations are \(C(0)=C_s\) and \(C(h)=C_b\). Fick’s first law states that flux is proportional to the negative concentration gradient:

J=-D\frac{dC}{dx}

Under the quasi-steady assumption, there is no net accumulation within the layer. With constant \(D\) and no internal source or sink, the flux is constant across it. Therefore \(dJ/dx=0\), and Fick’s law implies \(d^2C/dx^2=0\). The concentration profile is linear between its boundary values. Integrating the gradient across the layer gives:

\int_0^h \frac{dC}{dx}\,dx=C_b-C_s \qquad\Longrightarrow\qquad \frac{dC}{dx}=\frac{C_b-C_s}{h}

Because \(C_s>C_b\) during dissolution, the concentration decreases outward, so this derivative is negative. Substituting it into Fick’s law gives a positive outward flux:

J=D\frac{C_s-C_b}{h}

This is the interfacial mass-transfer result. The assumptions matter: the linear profile and constant flux follow from the idealized steady layer, not merely from the existence of a concentration gradient. The rate leaving the solid surface can differ from the rate reaching the bulk if solute accumulates within the layer.

Connecting Flux to Bulk Concentration

Let \(M_s\) denote the mass remaining in the solid. The positive dissolution rate is \(R_{\mathrm{diss}}=-dM_s/dt\). For uniform flux across area \(A\), the rate leaving the solid is \(AJ\). To equate that rate with the rate entering the bulk, additionally assume that accumulation in the diffusion layer is negligible, as in the quasi-steady diffusion-layer model. Also assume that the bulk is well mixed, its volume \(V\) is constant, and no dissolved drug is removed from it. Under these conditions:

R_{\mathrm{diss}}=AJ \qquad\text{and}\qquad \frac{dM_b}{dt}=R_{\mathrm{diss}}

Here \(M_b=VC_b\) is the dissolved mass in the bulk. Since \(V\) is constant, differentiating \(M_b=VC_b\) yields \(dM_b/dt=V\,dC_b/dt\). Combining this balance with the flux equation gives the Noyes–Whitney form and its Nernst–Brunner form:

V\frac{dC_b}{dt} =R_{\mathrm{diss}} =AJ =\frac{DA}{h}(C_s-C_b)

\frac{dC_b}{dt} =\frac{DA}{Vh}(C_s-C_b) \qquad\text{or, equivalently,}\qquad R_{\mathrm{diss}}=\frac{DA}{h}(C_s-C_b)

The factor \(D/h\) is the diffusion-layer mass-transfer coefficient in this idealization; it has units of length per time. In the Nernst–Brunner notation one often writes \(k=D/h\), giving \(R_{\mathrm{diss}}=kA(C_s-C_b)\). This shorthand does not make \(k\) a universal material constant: its value depends on the layer model and conditions represented by \(h\).

Assumptions for the bulk-concentration equation: The interfacial concentration is specified (often taken as the saturation concentration under the local conditions); diffusion across a planar layer is one-dimensional, quasi-steady, and governed by constant \(D\); the bulk is well mixed; layer accumulation is negligible; \(A\), \(h\), and \(V\) are defined consistently; and no dissolved drug is removed from the bulk. If bulk removal occurs, its rate must be included in the bulk balance. If the layer accumulates solute, the surface dissolution rate cannot be set equal to the bulk influx without an additional layer balance.

Worked Rates and Concentration Changes

Worked Example: Calculating an Instantaneous Interfacial Rate

Suppose a hypothetical dissolving surface has \(D=5.00\times10^{-10}\ \mathrm{m^2/s}\), \(A=2.00\times10^{-4}\ \mathrm{m^2}\), and \(h=1.00\times10^{-4}\ \mathrm{m}\). At the instant of interest, let \(C_s=0.80\ \mathrm{kg/m^3}\) and \(C_b=0.20\ \mathrm{kg/m^3}\). Assume the idealized layer and uniform-flux conditions above.

The concentration difference is \(C_s-C_b=0.60\ \mathrm{kg/m^3}\). The gradient magnitude is therefore:

\frac{C_s-C_b}{h} =\frac{0.60\ \mathrm{kg/m^3}}{1.00\times10^{-4}\ \mathrm{m}} =6.00\times10^3\ \mathrm{kg/m^4}

The outward flux and total instantaneous rate are:

J=(5.00\times10^{-10})(6.00\times10^3) =3.00\times10^{-6}\ \mathrm{kg/(m^2\,s)}

R_{\mathrm{diss}}=AJ =(2.00\times10^{-4})(3.00\times10^{-6}) =6.00\times10^{-10}\ \mathrm{kg/s} =0.600\ \mathrm{\mu g/s}

The positive sign indicates outward transfer from solid to liquid. This is an instantaneous rate: if the bulk concentration rises while \(C_s\), \(A\), and \(h\) remain fixed, the driving difference and rate decrease.

Worked Example: Integrating the Bulk Concentration Equation

Consider a hypothetical well-mixed vessel with \(V=0.500\ \mathrm{L}=5.00\times10^{-4}\ \mathrm{m^3}\), \(D=4.00\times10^{-10}\ \mathrm{m^2/s}\), \(A=1.00\times10^{-4}\ \mathrm{m^2}\), and \(h=2.00\times10^{-4}\ \mathrm{m}\). Take \(C_s=0.500\ \mathrm{kg/m^3}\) and \(C_b(0)=0.100\ \mathrm{kg/m^3}\). Assume all coefficients and \(C_s\) stay constant for one hour, with no bulk removal and negligible layer accumulation.

Define \(\lambda=DA/(Vh)\). Its units are inverse time, and its value is:

\lambda= \frac{(4.00\times10^{-10})(1.00\times10^{-4})} {(5.00\times10^{-4})(2.00\times10^{-4})} =4.00\times10^{-7}\ \mathrm{s^{-1}}

The governing equation is \(dC_b/dt=\lambda(C_s-C_b)\). Rearranging gives \(d(C_s-C_b)/(C_s-C_b)=-\lambda\,dt\). Integrating from time zero to \(t\) and applying the initial concentration yields:

C_b(t)=C_s-\bigl(C_s-C_b(0)\bigr)e^{-\lambda t}

For \(t=3600\ \mathrm{s}\), the exponent is \(-\lambda t=-0.00144\). Thus:

C_b(3600) =0.500-(0.500-0.100)e^{-0.00144} \approx0.100576\ \mathrm{kg/m^3}

The concentration increase is approximately \(5.76\times10^{-4}\ \mathrm{kg/m^3}\). The corresponding bulk mass increase is:

V\Delta C_b =(5.00\times10^{-4})(5.76\times10^{-4}) =2.88\times10^{-7}\ \mathrm{kg} =0.288\ \mathrm{mg}

This small change is consistent with the low initial rate: \(R_{\mathrm{diss}}(0)=V\lambda(C_s-C_b(0))=8.00\times10^{-11}\ \mathrm{kg/s}\). The rate falls as the bulk approaches \(C_s\); the integrated result does not imply a constant rate over the hour.

Worked Example: How the Bulk Concentration Changes the Driving Force

For a hypothetical surface, let \(D=6.00\times10^{-10}\ \mathrm{m^2/s}\), \(h=1.50\times10^{-4}\ \mathrm{m}\), \(A=1.50\times10^{-4}\ \mathrm{m^2}\), and \(C_s=0.400\ \mathrm{kg/m^3}\). Compare two instants with \(C_b=0.0400\ \mathrm{kg/m^3}\) and \(C_b=0.320\ \mathrm{kg/m^3}\). Assume the same quasi-steady, uniform-flux model at both instants.

First, \(DA/h=(6.00\times10^{-10})(1.50\times10^{-4})/(1.50\times10^{-4})=6.00\times10^{-10}\ \mathrm{m^3/s}\). At the lower bulk concentration:

R_{\mathrm{diss}} =(6.00\times10^{-10})(0.400-0.0400) =2.16\times10^{-10}\ \mathrm{kg/s} =0.216\ \mathrm{\mu g/s}

At the higher bulk concentration:

R_{\mathrm{diss}} =(6.00\times10^{-10})(0.400-0.320) =4.80\times10^{-11}\ \mathrm{kg/s} =0.0480\ \mathrm{\mu g/s}

The second rate is \(0.0480/0.216=0.222\) of the first because the concentration driving force is smaller. A sink approximation that sets \(C_b\) near zero would instead predict \(2.40\times10^{-10}\ \mathrm{kg/s}\); that is close to the first rate but five times the second. The approximation is therefore not equally appropriate at both concentrations.

Interpreting the Equation and Recognizing Failure

The equation predicts proportional changes in instantaneous rate when one factor changes and the others are held fixed: doubling \(A\) doubles the rate, and doubling \(C_s-C_b\) doubles it. Increasing \(D\) increases the predicted flux; increasing \(h\) reduces it. These are model-based comparisons, not guarantees that changing a formulation or apparatus will alter only one factor.

The integrated exponential solution applies only while \(D\), \(A\), \(h\), \(V\), and \(C_s\) can be treated as constant. If the exposed area changes as solid dissolves, or the interfacial concentration changes, the differential equation can still describe an instantaneous rate if its quantities are updated, but the simple exponential solution no longer follows. A separate particle-size or geometry model is needed to specify how area changes; that geometry is not derived by the interfacial flux equation itself.

Useful checks include verifying that \(C_s\geq C_b\) for net outward dissolution, that \(D A(C_s-C_b)/h\) has units of mass per time, and that the predicted bulk mass change is compatible with the amount of solid lost. A negative predicted rate when \(C_b>C_s\) describes inward transfer under the same gradient convention, not positive dissolution. A calculated bulk concentration exceeding \(C_s\) under constant-\(C_s\), no-source assumptions signals an inconsistent calculation or model setup.

The main alternative is an empirical first-order concentration model, \(dC_b/dt=k_{\mathrm{obs}}(C_s-C_b)\). Its mathematical form matches the Noyes–Whitney concentration equation, but \(k_{\mathrm{obs}}\) may simply be fitted from data. The mechanistic interpretation \(k_{\mathrm{obs}}=DA/(Vh)\) requires the stated interfacial assumptions and known or defensible values for its components. A sink approximation is another simplification: when \(C_b\ll C_s\), \(C_s-C_b\approx C_s\), so the rate is approximately constant if the other factors are also constant. Neither simplification should be used when its conditions are not plausible.

Common Pitfall: Skipping the Layer Balance

A tempting but incomplete argument is: “The solid loses mass at rate \(AJ\), so the bulk concentration rises at \(AJ/V\).” That step is not justified merely by saying there is no removal from the bulk. If dissolved material is accumulating in the diffusion layer, the rate leaving the solid and the rate entering the bulk differ.

The correction is to state the quasi-steady assumption explicitly: accumulation in the layer is negligible, so interfacial dissolution rate equals the rate entering the bulk. With no bulk removal and constant volume, \(dM_b/dt=R_{\mathrm{diss}}\) and \(M_b=VC_b\), giving \(V\,dC_b/dt=R_{\mathrm{diss}}\). If layer accumulation is not negligible, a layer mass balance is needed before writing the bulk equation.

Key takeaway: The Noyes–Whitney and Nernst–Brunner equations combine a diffusion-layer flux, \(J=D(C_s-C_b)/h\), with a bulk mass balance. The gradient determines the flux; the quasi-steady layer and bulk assumptions justify converting that flux into a bulk concentration rate.

Check Your Understanding

Answer each question, then compare with the solutions.

With concentration in \(\mathrm{kg/m^3}\), what are the units of \(J\), \(D/h\), and \(R_{\mathrm{diss}}\)?
For \(C_s=0.30\ \mathrm{kg/m^3}\), \(C_b=0.10\ \mathrm{kg/m^3}\), and \(DA/h=2.0\times10^{-10}\ \mathrm{m^3/s}\), calculate the instantaneous dissolution rate.
Why does the surface dissolution rate not automatically equal the rate of increase of bulk dissolved mass?
Under constant coefficients, what happens to the rate as \(C_b\) approaches \(C_s\), and what happens to it at \(C_b=C_s\)?

Solutions. (1) \(J\) is \(\mathrm{kg/(m^2\,s)}\), \(D/h\) is \(\mathrm{m/s}\), and \(R_{\mathrm{diss}}\) is \(\mathrm{kg/s}\). (2) \(R_{\mathrm{diss}}=(2.0\times10^{-10})(0.30-0.10)=4.0\times10^{-11}\ \mathrm{kg/s}\). (3) Solute may accumulate within the diffusion layer, so the flux leaving the solid need not equal the flux reaching the bulk; negligible layer accumulation is needed to equate them. (4) The driving difference, and thus the predicted rate, decreases toward zero; at \(C_b=C_s\), the model predicts zero net flux.