Chapter 10: Catalysis and Catalytic Reactors
For the homogeneous reaction:
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\( \text{SiH}_4 \leftrightarrow \text{SiH}_2 + \text{H}_2 \) \( \text{SiH}_2 \rightarrow \text{Si} + \text{H}_2 \) |
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We found that |
\( r_{\text{Dep}} = \frac{k P_{\text{SiH}_2}}{1 + K_{\text{SiH}_2} P_{\text{SiH}_2}} \) |
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The first reaction is in equilibrium |
\( \frac{-r''_{\text{SiH}_4}}{k_g} \approx 0 \) |
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\( K_P = \frac{P_{\text{SiH}_2} P_{\text{H}_2}}{P_{\text{SiH}_4}} \) |
Case A: Small Consumption of SIH2 |
| In many cases a very thin film of the material is deposited (a few molecular layers) and as a result very little SiH2 is consumed. For example if SiH4 dissociates to form 1 mole SiH2 and 1 mole of H2 at equilibrium and then 0.01 mole of SiH2 is consumed in the deposition, then for all practical purposes \( P_{\text{SiH}_2} \approx P_{\text{H}_2} \) (i.e., 0.99 » 1.0). Consequently, |
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\( P_{\text{SiH}_2} = \sqrt{K_P P_{\text{SiH}_4}} \) \( r_{\text{Dep}} = \frac{k \, P_{\text{SiH}_4}^{1/2}}{1 + K \, P_{\text{SiH}_4}^{1/2}} \) |
| Case B: Significant Consumption of SIH2 |
| One the other hand if significant amounts of SiH2 are consumed, we can still obtain \( P_{\text{SiH}_2} \) from the equilibrium relationship. However \( P_{\text{SiH}_2} \ne P_{\text{H}_2} \) and we must include \( P_{\text{H}_2} \) in the rate law |
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\( P_{\text{SiH}_2} = \frac{P_{\text{SiH}_4} K_P}{P_{\text{H}_2}} \) \( r_{\text{Dep}} = \frac{k \, P_{\text{SiH}_2}}{1 + K'_A \, P_{\text{SiH}_2}} = \frac{k \, K_P \, P_{\text{SiH}_4}}{P_{\text{H}_2} + K_P \, K'_A \, P_{\text{SiH}_4}} \) |
| which is the same as Eqn. below (S10-12) on p666. |
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\( r_{\text{Dep}} = \frac{k \, P_{\text{SiH}_4}}{P_{\text{H}_2} + K \, P_{\text{SiH}_4}} \) |