Question 24.6: It is desired to separate a mixture of two industrial enzyme......

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, A}}{d_{\mathrm{pore}}}}={\frac{4.12\,\mathrm{nm}}{30.0\,\mathrm{nm}}}=0.137\,\mathrm{and}\,\varphi_{B}={\frac{d_{s, B}}{d_{\mathrm{pore}}}}={\frac{10.44\,\mathrm{nm}}{30.0\,\mathrm{nm}}}=0.348

For lysozyme, F_{1}(\varphi_{A}), by equation (24-64), and F_{2}(\varphi_{A}), by the Renkin equation (24-65), are

F_{2}(\varphi)=1-2.104\varphi+2.09\varphi^{3}-0.95\varphi^{5}          (24-65)

F_{1}(\varphi_{A})=(1-\varphi_{A})^{2}=(1-0.137)^{2}=0.744\\

F_{2}(\varphi_{A})\,=\,1-2.104\varphi_{A}+2.09\varphi_{A}^{3}-0.95\varphi_{A}^{5}\\

=\;1-2.104(0.137)+2.090(0.137)^{3}-0.95(0.137)^{5}=0.716

The effective diffusivity of lysozyme in the pore, D_{Ae} , is estimated by equation (24-62)

\frac{D_{A e}}{D_{A B}^{o}}=F_{1}(\varphi)F_{2}(\varphi)        (24-62)

D_{A e}=D_{A-\mathrm{H}_{2}O}F_{1}(\varphi_{A})F_{2}(\varphi_{A})=1.04\times10^{-6}\frac{\mathrm{cm}^{2}}{\mathrm{s}}(0.744)(0.716)=5.54\times10^{-7}\frac{\mathrm{cm}^{2}}{\mathrm{s}}

Likewise, for catalase F_{1}(\varphi_{B})=0.425,F_{2}(\varphi_{B})=0.351,\;\mathrm{and}\;D_{B e}=6.12\times10^{-8}\;\mathrm{cm}^{2}/s. Finally, the separation factor is

\alpha={\frac{D_{A e}}{D_{B e}}}={\frac{5.54\times10^{-7}\,\mathrm{cm}^{2}/s}{6.12\times10^{-8}\,\mathrm{cm}^{2}/s}}=9.06

It is interesting to compare the value above with \alpha^{\prime}, the ratio of molecular diffusivities at infinite dilution:

\alpha^{\prime}={\frac{D_{A-H_{2}O}}{D_{B-\mathrm{HzO}}}}={\frac{1.04\times10^{-6}\mathrm{cm}^{2}/{\mathrm{s}}}{4.1\times10^{-7}\mathrm{cm}^{2}/{\mathrm{s}}}}=1.75

The small pore diameter enhances the value for \alpha, because the diffusion of the large catalase molecule is significantly hindered inside the pore relative to the smaller lysozyme molecule.