Production of high-purity silicon, as well as growth of silicon nanowires and other silicon nanostructures, often involves the conversion of crude silicon into silane and chlorosilanes followed by purification by distillation and, finally, decomposition back into solid silicon plus H2, HCl, and

Question

Production of high-purity silicon, as well as growth of silicon nanowires and other silicon nanostructures, often involves the conversion of crude silicon into silane and chlorosilanes followed by purification by distillation and, finally, decomposition back into solid silicon plus [latex]H_2, HCl, and Cl_2 [/latex]. Here, we consider the decomposition of dichlorosilane [latex](SiH_2Cl_2) [/latex] vapor to solid silicon and HCl vapor. Thermodynamic data for these species are available in the NIST Chemistry WebBook. Prepare plots of the fraction ofdichlorosilane converted to solid silicon at equilibrium vs. temperature from 700 to 1300 K for the following conditions:

(a) Starting from pure dichlorosilane at a pressure of 5 bar.

(b) Starting from pure dichlorosilane at a pressure of 1 bar.

(c) Starting with a dichlorosilane mole fraction of 5% in hydrogen at a pressure of 1 bar.

(d) Starting with pure dichlorosilane at a pressure of 0.001 bar.

These conditions are representative of different processes employed in the photovoltaics and microelectronics industries, from growth of bulk silicon in a fluidized bed reactor to deposition of silicon thin films by atmospheric-pressure or low-pressure chemical vapor deposition. Various processes like these may employ silane [latex] (SiH_4) [/latex] or trichlorosilane [latex] (SiHCl_3) rather than SiH_2Cl_2. [/latex]

Accepted Answer

The reaction of interest can be written as:

[latex]SiH_2Cl_2(g) → Si(s) + 2HCl(g)[/latex]

The equilibrium constant as a function of temperature will be the same for all parts of the problem, so we compute it first. Note that the NIST Chemistry WebBook adopts the convention that enthalpies of formation are zero for elements in their standard state at 298.15 K, but uses absolute (third-law) values of entropy. Thus, the Gibbs energy of formation not zero for elements in their standard states at 298.15 K. The polynomial used for the temperature dependence of heat capacity also differs from the one we have been using. Fortunately, the NIST WebBook directly provides values of −(G° − [latex]H^{°}_{f,298})∕T[/latex], which we use along with [latex]H^{°}_{f,298})∕T[/latex] for each species to find G° of each species at several temperatures, as shown in the accompanying table. We then compute [latex] \Delta G^{\circ}=2 G^{\circ}(\mathrm{HCl})+G^{\circ}(\mathrm{Si})-G^{\circ}\left(\mathrm{SiH}_2 \mathrm{Cl}_2 ight) ext { and } K=\exp \left(-\Delta G^{\circ} /(R T) ight) ext {, } [/latex] , as also given in the table.

[latex] \mathrm{SiH}_2 \mathrm{Cl}_2(\mathrm{~g}) [/latex] −320.49 [latex] H_{f, 298}^z [/latex] [latex]kJ·mol{^−1}[/latex]		HCl(g) −92.31 [latex] H_{f, 298}^z [/latex] [latex]kJ·mol{^−1}[/latex]	Si(s) 0 [latex] H_{f, 298}^z [/latex] [latex]kJ·mol{^−1}[/latex]	Reaction
T/K	[latex] -\frac{G^{\circ}-H_{f, 298}^{\circ}}{T} [/latex] G° [latex] \mathbf{J} \cdot \mathrm{mol}^{-1} [/latex] [latex] \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} [/latex]	[latex] -\frac{G^{\circ}-H_{f, 298}^{\circ}}{T} [/latex] G° [latex] \mathbf{J} \cdot \mathrm{mol}^{-1} [/latex] [latex] \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} [/latex]	[latex] -\frac{G^{\circ}-H_{f, 298}^{\circ}}{T} [/latex] G° [latex] \mathbf{J} \cdot \mathrm{mol}^{-1} [/latex] [latex] \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} [/latex]	ΔG° [latex]J·mol^{−1}[/latex] K
700	306.7 −535180	195.1 −228880	25.0 −17486	35.1 −45682 × [latex] 10^{-5} [/latex]
800	313.0 −570890	197.4 −250230	26.9 −21480	48950 6.37 ×[latex] 10^{-4} [/latex]
900	319.1 −607680	199.7 −272040	28.7 −25794	37806 6.40 × [latex] 10^{-3} [/latex]
1000	325.1 −645590	201.9 −294210	30.4 −30390	37806 6.40 × [latex] 10^{-2} [/latex]
1100	330.8 −684370	203.9 −316600	32.0 −35244	15926 1.75 × [latex] 10^{-1} [/latex]
1200	336.3 −724050	205.9 −339390	33.6 −40356	15926 1.75 × [latex] 10^{-1} [/latex]
1300	341.5 −764440	207.7 −362320	35.1 −45682	−5882 1.72

We can now use these values of the equilibrium constant to analyze each of the cases given.

(a) On a basis of 1 mole of [latex]SiH_2Cl_2[/latex], the gas-phase mole fractions in this case are:

[latex] y_{\mathrm{SiH}_2 \mathrm{Cl}_2}=\frac{1-\varepsilon}{1+\varepsilon} \quad ext { and } \quad y_{\mathrm{HCl}}=\frac{2 \varepsilon}{1+\varepsilon} [/latex]

Note that the solid silicon does not play a role in the gas-phase mole fractions. Moreover, in the general equilibrium expression of Eq. (14.10),

[latex] \prod_i\left(\hat{f}_i / f_i^0 ight)^{ u_i}=K [/latex] (14.10)

the ratio [latex] \hat{f}_i / f_i^0 [/latex] for pure solid silicon can be taken to be 1.0. The effect of pressure on the fugacity of a solid is negligible, and the standard state for silicon is the pure solid. Thus, it simply does not appear in the equilibrium expression, which assuming that dichlorosilane and HCl are in their ideal-gas state, is simply:

[latex] \frac{y_{\mathrm{HCl}}^2}{y_{\mathrm{SiH}_2 \mathrm{Cl}_2}}=K\left(\frac{P^{\circ}}{P} ight) [/latex]

Using the expressions for the mole fractions:

[latex] \frac{4 \varepsilon^2}{(1-\varepsilon)(1+\varepsilon)}=K\left(\frac{P^{\circ}}{P} ight) [/latex]

From which:

[latex] \left(4+K\left(\frac{P^{\circ}}{P} ight) ight) \varepsilon^2=K\left(\frac{P^{\circ}}{P} ight) [/latex]

And finally:

[latex] \varepsilon=\sqrt{^{K\left(\frac{P^{\circ}}{P} ight)} /\left(4+K\left(\frac{P^{\circ}}{P} ight) ight)}=\sqrt{K /(20+K)} [/latex]

Evaluating this for temperatures from 700 to 1300 K using the corresponding values of equilibrium constant gives the values shown in the following table and in Fig. 14.5.

(b) This is the same as part (a) except with P°/P = 1 instead of P°/P = 1/5, so:

[latex] \varepsilon=\sqrt{^{K }/(4+K)} [/latex]

Results of this expression are shown later in tabular form and plotted in Fig. 14.5.

(c) Here, if we again take 1 mol SiH2Cl2 as our basis, so the extent of reaction is equal to the fraction of dichlorosilane converted to silicon, we also have 19 mol [latex]H_2[/latex] and the expressions for the mole fractions become:

[latex] y_{\mathrm{SiH}_2 \mathrm{Cl}_2}=\frac{1-\varepsilon}{20+\varepsilon} \quad ext { and } \quad y_{\mathrm{HCl}}=\frac{2 \varepsilon}{20+\varepsilon} [/latex]

Using these expressions for the mole fractions at a pressure of 1 bar:

[latex] \frac{4 \varepsilon^2}{(1-\varepsilon)(20+\varepsilon)}=K [/latex]

And finally:

[latex] E=\frac{-19 K+\sqrt{361 K^2+80 K(4+K)}}{2(4+K)} [/latex]

We have picked the positive sign in the quadratic formula to obtain a positive value of ε. Again, results are shown in the table and in Fig. 14.5.

(d ) This is the same as part (a) except with P°/P = 1000 instead of P°/P = 1/5, so:

[latex] \varepsilon=\sqrt{^{1000 K} /(4+1000 K)} [/latex]

Again, results are presented in the table below and in Fig. 14.5.

This problem illustrates the effect of reduced pressure and dilution on increasing the conversion of this type of reaction at a given temperature. In reality, formation of additional products— including [latex] \mathrm{H}_2, \mathrm{SiHCl}_3 ext {, and } \mathrm{SiCl}_4 [/latex] should be considered under many conditions of relevance to silicon refining. Indeed, [latex]SiCl_4[/latex] is a major byproduct of solar-grade silicon production.

ε, Fractional Conversion of [latex]SiH_2Cl_2[/latex] to Si in Ex. 14.7

T/K	(a)	(b)	(c)	(d)
700	0.0013	0.0029	0.0129	0.0914
800	0.0056	0.0126	0.0549	0.3706
900	0.0179	0.0400	0.1642	0.7844
1000	0.0446	0.0994	0.3605	0.9534
1100	0.0932	0.2049	0.6005	0.9888
1200	0.1722	0.3640	0.7991	0.9967
1300	0.2816	0.5487	0.9084	0.9988

Question 14.7: Production of high-purity silicon, as well as growth of sili...

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