In Problem 30.10, the production of ultra pure silicon tetrachloride for the chemical vapor deposition of silicon was described. Hydrogen chloride, HCl, was produced in the elimination of trichlorosilane, SiHCl3, and the HCl was stripped from the liquid silicon tetrachloride into an N2 carrier gas

Question

In Problem 30.10, the production of ultra pure silicon tetrachloride for the chemical vapor deposition of silicon was described. Hydrogen chloride, HCl, was produced in the elimination of trichlorosilane, [latex]\mathrm{SiHCl}_3[/latex], and the HCl was stripped from the liquid silicon tetrachloride into an [latex]\mathrm{N}_2[/latex] carrier gas stream.

Accepted Answer

(a) If the HCl concentration was reduced from 1.8 to 0.2 mol % by its transfer into the 298 K and [latex]1.013 imes 10^5 \mathrm{~Pa}[/latex] nitrogen gas carrier stream, determine the [latex]\left(L_s / G_s ight)_{\min }[/latex] ratio. At the involved low concentration levels, the [latex]\mathrm{HCl}-\mathrm{H}_2[/latex] system obeys Henry’s law with the Henry’s constant, H’, equal to 47.5 atm/mol fraction HCl in the liquid.

(b) Determine the HCl concentration in the exiting [latex]\mathrm{N}_2[/latex] stream when 40 times the minimum [latex]G_s[/latex] is used in the stripper.

The compositions of the known streams, [latex]L_2, L_1[/latex], and [latex]G_1,[/latex] on a solute-free basis are evaluated from the known mole fractions as

[latex]X_{\mathrm{HCl}, 2}=\frac{x_{\mathrm{HCl}, 2}}{1-x_{\mathrm{HCl}, 2}}=\frac{0.018}{0.982}=0.0183[/latex]

[latex]X_{\mathrm{HCl}, 1}=\frac{x_{\mathrm{HCl}, 1}}{1-x_{\mathrm{HCl}, 1}}=\frac{0.002}{0.998}=0.002[/latex]

[latex]Y_{\mathrm{HCl}, 1}=\frac{y_{\mathrm{HCl}, 1}}{1-y_{\mathrm{HCl}, 1}}=0[/latex]

Note that the solute-free compositions are essentially equal to the mole-fraction compositions for low concentrations.

The equilibrium concentration is evaluated with Henry’s law

[latex]p_{\mathrm{HCl}}(\mathrm{atm})=\left(47.5 \frac{\mathrm{atm}}{ ext { mole fraction }} ight)\left(x_{\mathrm{HCl}} ight)[/latex]

At end 2 of the tower

[latex]p_{\mathrm{HCl}, 2}=\left(47.5 \frac{\mathrm{atm}}{ ext { mole fraction }} ight)(0.018)=0.855 \mathrm{~atm}[/latex]

and

[latex]Y_{\mathrm{HCl}, 2}^*=\frac{p_{\mathrm{HCl}, 2}}{P-p_{\mathrm{HCl}, 2}}=\frac{0.855}{0.145}=5.90[/latex]

Figure 31.13 illustrates the operating lines for both [latex]G_{S, ext { min }} ext { and } G_{S, ext { actual }}[/latex] for this example. As one notes, the operating lines for the stripping process are below the equilibrium line.

[latex]\left(\frac{L_S}{G_S} ight)_{\min }=\frac{Y_2^*-Y_1}{X_2-X_1}=\frac{5.90-9}{0.0183-0.002}=362[/latex]

and

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=\frac{L_S}{(G s)(40)}=\frac{362}{40}=9.05[/latex]

[latex]\left(\frac{L_S}{G_S} ight)_{ ext {actual }}=9.05=\frac{Y_2-Y_1}{X_2-X_1}=\frac{Y_2-0}{0.0183-0.002}[/latex]

[latex]Y_2=0.147[/latex]

The mole fraction of the exiting gas stream is

[latex]y_{\mathrm{HCl}, 2}=\frac{Y_2}{1+Y_2}=\frac{0.147}{1.147}=0.128[/latex].

Question 31.4: In Problem 30.10, the production of ultra pure silicon tetra...

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