Question 4.9: Solar-grade silicon can be manufactured by thermal decomposi...
Solar-grade silicon can be manufactured by thermal decomposition of silane at moderate pressure in a fluidized-bed reactor, in which the overall reaction is:
SiH_{4}(g) → Si(s) + 2 H_{2} (g)
When pure silane is preheated to 300°C, and heat is added to the reactor to promote a reasonable reaction rate, 80% of the silane is converted to silicon and the products leave the reactor at 750°C. How much heat must be added to the reactor for each kilogram of silicon produced?
For a continuous-flow process with no shaft work and negligible changes in kinetic and potential energy, the energy balance is simply Q = ΔH, and the heat added is the enthalpy change from reactant at 300°C to products at 750°C. A convenient path for calculation of the enthalpy change is to (1) cool the reactant to 298.15 K, (2) carry out the reaction at 298.15 K, and (3) heat the products to 750°C.
On the basis of 1 mol SiH_{4}, the products consist of 0.2 mol SiH_{4}, 0.8 mo l Si, and 1.6 mol H_{2}. Thus, for the three steps we have:
\Delta H_{1} = \int_{573.15K}^{298.15K}{C^{o}_{P}(SiH_{4} )dT }
\Delta H_{2}= 0.8 \times \Delta H^{o}_{298}
\Delta H_{3} = \int_{298.15K}^{1023.15K}{\left[0.2 \times C^{o}_{P} (SiH_{4}) +0.8 \times C^{0}_{P} (Si) + 1.6\times C^{0}_{P}(H_{2} ) \right]dt}
Data needed for this example are not included in App. C, but are readily obtained from the NIST Chemistry WebBook (http://webbook.nist.gov). The reaction here is the reverse of the formation reaction for silane, and its standard heat of reaction at 298.15 K is \Delta H^{o}_{298} = − 34,310 J. Thus, the reaction is mildly exothermic.
Heat capacity in the NIST Chemistry WebBook is expressed by the Shomate equation, a polynomial of different form from that used in this text. It includes a T³ term, and is written in terms of T/1000, with T in K:
C_P^{\circ}=A+B\left(\frac{T}{1000}\right)+C\left(\frac{T}{1000}\right)^2+D\left(\frac{T}{1000}\right)^3+E\left(\frac{T}{1000}\right)^{-2}
Formal integration of this equation gives the enthalpy change:
\Delta H=\int_{T_0}^T C_P^{\circ} d T
\Delta H=1000\left[A\left(\frac{T}{1000}\right)+\frac{B}{2}\left(\frac{T}{1000}\right)^2+\frac{C}{3}\left(\frac{T}{1000}\right)^3+\frac{D}{4}\left(\frac{T}{1000}\right)^4-E\left(\frac{T}{1000}\right)^{-1}\right]_{T_0}^T
The first three rows in the accompanying table give parameters, on a molar basis, for SiH_4 , crystalline silicon, and hydrogen. The final entry is for the collective products, represented for example by:
A(products) = (0.2) (6.060) + (0.8)(22.817) + (1.6)(33.066) = 72.3712
with corresponding equations for B, C, D, and E.
| Species | A | B | C | D | E |
| SiH_4(g) | 6.060 | 139.96 | −77.88 | 16.241 | 0.1355 |
| Si(s) | 22.817 | 3.8995 | −0.08289 | 0.04211 | −0.3541 |
| H_2(g) | 33.066 | −11.363 | −0.08289 | −2.773 | −0.1586 |
| Products | 72.3712 | 12.9308 | 2.6505 | −1.1549 | −0.5099 |
For these parameters, and with T in kelvins, the equation for ΔH yields values in joules. For the three steps making up the solution to this problem, the following results are obtained:
1. Substitution of the parameters for 1 mol of SiH_4 into the equation for ΔH leads upon evaluation to: [/latex]ΔH_1 = −14,860 J[/latex].
2. Here, ΔH_2 = (0.8)(− 34,310) = − 27,450 J .
3. Substitution of the parameters for the total product stream into the equation for ΔH leads upon evaluation to: ΔH_3 = 58,060 J.
For the three steps the sum is:
ΔH = − 14,860 − 27,450 + 58,060 = 15,750 J
This enthalpy change equals the heat input per mole of SiH_4 fed to the reactor. A kilogram of silicon, with a molar mass of 28.09, is 35.60 mol. Producing a kilogram of silicon therefore requires a feed of 35.60/0.8 or 44.50 mol of SiH_4. The heat requirement per kilogram of silicon produced is therefore (15,750)(44.5) = 700,900 J.