Question 24.7: It is desired to separate a mixture of two industrial enzyme...
It is desired to separate a mixture of two industrial enzymes, lysozyme and catalase, in a dilute, aqueous solution by a gel filtration membrane. A mesoporous membrane with cylindrical pores of 30 nm diameter is available (Figure 24.5). The following separation factor (a) for the process is proposed
\alpha=\frac{D_{A e}}{D_{B e}}Determine the separation factor for this process. The properties of each enzyme as reported by \text { Tanford }^{27} are given below.
{}^{27} C. Tanford, Physical Chemistry of Macromolecules, John Wiley & Sons, New York, 1961.
The transport of large enzyme molecules through pores filled with liquid water represents a hindered diffusion process. The reduced pore diameters for lysozyme and catalase are
\varphi_{A}=\frac{d_{s, \mathrm{~A}}}{d_{\text {pore }}}=\frac{4.12 \mathrm{~nm}}{30.0 \mathrm{~nm}}=0.137 \quad \text { and } \quad \varphi_{B}=\frac{d_{s, \mathrm{~B}}}{d_{\text {pore }}}=\frac{10.44 \mathrm{~nm}}{30.0 \mathrm{~nm}}=0.348For lysozyme, F_{1}\left(\varphi_{A}\right) by equation (24-64) and F_{2}\left(\varphi_{A}\right) by the Renkin equation (24-65) are
F_{1}(\varphi)=\frac{\text { flux area available to solute }}{\text { total flux area }}=\frac{\pi\left(d_{\text {pore }}-d_{s}\right)^{2}}{\pi d_{\text {pore }}^{2}}=(1-\varphi)^{2} (24-64)
F_{2}(\varphi)=1-2.104 \varphi+2.09 \varphi^{3}-0.95 \varphi^{5} (24-65)
F_{1}\left(\varphi_{A}\right)=\left(1-\varphi_{A}\right)^{2}=(1-0.137)^{2}=0.744
F_{2}\left(\varphi_{A}\right)=1-2.104 \varphi_{A}+2.09 \varphi_{A}^{3}-0.95 \varphi_{A}^{5}
=1-2.104(0.137)+2.09(0.137)^{3}-0.95(0.137)^{5}=0.716
The effective diffusivity of lysozyme in the pore, D_{A e} is estimated by equation (24-62)
D_{A e}=D_{A B}^{\circ} F_{1}(\varphi) F_{2}(\varphi) (24-62)
D_{A e}=D_{A-\mathrm{H}_{2} \mathrm{O}}^{\circ} F_{1}\left(\varphi_{A}\right) F_{2}\left(\varphi_{A}\right)=1.04 \times 10^{-6} \frac{\mathrm{cm}^{2}}{\mathrm{~s}}(0.744)(0.716)=5.54 \times 10^{-7} \frac{\mathrm{cm}^{2}}{\mathrm{~s}}
Likewise, for catalase F_{1}\left(\varphi_{B}\right)=0.425, F_{2}\left(\varphi_{B}\right)=0.351, and D_{B e}=6.12 \times 10^{-8} \mathrm{~cm}^{2} / \mathrm{s}. Finally, the separation factor is
\alpha=\frac{D_{A e}}{D_{B e}}=\frac{5.54 \times 10^{-7} \mathrm{~cm}^{2} / \mathrm{s}}{6.12 \times 10^{-8} \mathrm{~cm}^{2} / \mathrm{s}}=9.06It is interesting to compare the value above with \alpha^{\prime}, the ratio of molecular diffusivities at infinite dilution
\alpha^{\prime}=\frac{D_{A-\mathrm{H}_{2} \mathrm{O}}^{\circ}}{D_{B-\mathrm{H}_{2} \mathrm{O}}^{\circ}}=\frac{1.04 \times 10^{-6} \mathrm{~cm}^{2} / \mathrm{s}}{4.1 \times 10^{-7} \mathrm{~cm}^{2} / \mathrm{s}}=1.75The small pore diameter enhances the value for a because the diffusion of the large catalase molecule is significantly hindered inside the pore relative to the smaller lysozyme molecule.