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

Question

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

[latex]\alpha=\frac{D_{A e}}{D_{B e}}[/latex]

Determine the separation factor for this process. The properties of each enzyme as reported by [latex] ext { Tanford }^{27}[/latex] are given below.

[latex]{}^{27}[/latex] C. Tanford, Physical Chemistry of Macromolecules, John Wiley & Sons, New York, 1961.

Accepted Answer

[latex]\begin{array}{ll} ext { Lysozyme (species } A ext { ) } & ext { Catalase (species } B ext { ) } \ M_{A}=14100 \mathrm{~g} / \mathrm{g} \mathrm{mol} & M_{B}=250000 \mathrm{~g} / \mathrm{g} \mathrm{mol} \ d_{s, \mathrm{~A}}=4.12 \mathrm{~nm} & d_{s, \mathrm{~B}}=10.44 \mathrm{~nm} \ D_{A-\mathrm{H}_{2}\mathrm{O}}^{\mathrm{o}}=1.04 imes 10^{-6} \mathrm{~cm}^{2} / \mathrm{s} & D_{B-\mathrm{H}_{2} \mathrm{O}}^{\mathrm{o}}=4.10 imes 10^{-7} \mathrm{~cm}^{2} / \mathrm{s}\end{array}[/latex]

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

[latex]\varphi_{A}=\frac{d_{s, \mathrm{~A}}}{d_{ ext {pore }}}=\frac{4.12 \mathrm{~nm}}{30.0 \mathrm{~nm}}=0.137 \quad ext { and } \quad \varphi_{B}=\frac{d_{s, \mathrm{~B}}}{d_{ ext {pore }}}=\frac{10.44 \mathrm{~nm}}{30.0 \mathrm{~nm}}=0.348[/latex]

For lysozyme, [latex]F_{1}\left(\varphi_{A} ight)[/latex] by equation (24-64) and [latex]F_{2}\left(\varphi_{A} ight)[/latex] by the Renkin equation (24-65) are

[latex]F_{1}(\varphi)=\frac{ ext { flux area available to solute }}{ ext { total flux area }}=\frac{\pi\left(d_{ ext {pore }}-d_{s} ight)^{2}}{\pi d_{ ext {pore }}^{2}}=(1-\varphi)^{2}[/latex] (24-64)

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

[latex]F_{1}\left(\varphi_{A} ight)=\left(1-\varphi_{A} ight)^{2}=(1-0.137)^{2}=0.744[/latex]

[latex]F_{2}\left(\varphi_{A} ight)=1-2.104 \varphi_{A}+2.09 \varphi_{A}^{3}-0.95 \varphi_{A}^{5}[/latex]

[latex]=1-2.104(0.137)+2.09(0.137)^{3}-0.95(0.137)^{5}=0.716[/latex]

The effective diffusivity of lysozyme in the pore, [latex]D_{A e}[/latex] is estimated by equation (24-62)

[latex]D_{A e}=D_{A B}^{\circ} F_{1}(\varphi) F_{2}(\varphi)[/latex] (24-62)

[latex]D_{A e}=D_{A-\mathrm{H}_{2} \mathrm{O}}^{\circ} F_{1}\left(\varphi_{A} ight) F_{2}\left(\varphi_{A} ight)=1.04 imes 10^{-6} \frac{\mathrm{cm}^{2}}{\mathrm{~s}}(0.744)(0.716)=5.54 imes 10^{-7} \frac{\mathrm{cm}^{2}}{\mathrm{~s}}[/latex]

Likewise, for catalase [latex]F_{1}\left(\varphi_{B} ight)=0.425, F_{2}\left(\varphi_{B} ight)=0.351[/latex], and [latex]D_{B e}=6.12 imes 10^{-8} \mathrm{~cm}^{2} / \mathrm{s}[/latex]. Finally, the separation factor is

[latex]\alpha=\frac{D_{A e}}{D_{B e}}=\frac{5.54 imes 10^{-7} \mathrm{~cm}^{2} / \mathrm{s}}{6.12 imes 10^{-8} \mathrm{~cm}^{2} / \mathrm{s}}=9.06[/latex]

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

[latex]\alpha^{\prime}=\frac{D_{A-\mathrm{H}_{2} \mathrm{O}}^{\circ}}{D_{B-\mathrm{H}_{2} \mathrm{O}}^{\circ}}=\frac{1.04 imes 10^{-6} \mathrm{~cm}^{2} / \mathrm{s}}{4.1 imes 10^{-7} \mathrm{~cm}^{2} / \mathrm{s}}=1.75[/latex]

The 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.

Question 24.7: It is desired to separate a mixture of two industrial enzyme...

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