Question

Semiconductor microchip processing often involves with chemical vapor deposition (CVD). In this process a stream of Silane gas (SiH4) enters a reactor chamber, reacts with oxygen (O2) which comes to the chamber from another stream; this reaction results in production of silicon dioxide (SiO2) and hydrogen gas (H2). Silicon dioxide leaves the chamber in the form of solid thin film, but the rest of materials leave the chamber in gas phase. The chamber has rigid non insulated walls and it is fixed to the floor of the cleanroom.

In a cleanroom CVD reactor, we use the following procedure to fabricate thin films; a stream of Silane gas (SiH4) at 1.00 atm, 17.0 °C and volumetric flow rate of 12.57 gal/min enters the reactor. 25 % excess oxygen enters the system from another stream at 27.0 °C, 1.00 atm. Also, argon gas (Ar is an inert gas) is entering the system at 27.0 °C, 1.00 atm and its molar flow rate ratio to Silane’s molar flow rate is (5:1). If the temperature of the thin film is 850 °C, and all the gases that are leaving the chamber has the temperature of 397 °C. How much heat is transferred to/from our reactor in KW?

Reaction: SiH₄ (g) + O₂ (g) ---> SiO₂ (g) + 2H₂​ (g)

Material	Heat of formation at 25°C, 1atm (KJ/mol)	Heat capacity (J/mol*K)	Boiling Temperature (°C)
SiH4	34.31	42.8	-112

I know that the heat transfered from/to the reaction is equal to the change in enthalpy (Q= ΔH). The formula I would be using is ΔH= m ∫ Cp ΔT for each material in the equation. I just really need help on how to get the moles of each component (SiH₄, O₂, SiO₂, H₂, Ar) going in the reaction and coming out.

Answer 1

So for moles going in/coming out of the reactor start by visualising the process and draw a basic flow diagram like this:

Now, you have been given that the inlet flow rate of SiH4 gas is 12.57 gal/min. COnvert these to SI units of m^3/s as follows:

12.57 * 0.00454609 / 60 m^3/s = 0.0009524 m^3/s {1 gal = 0.00454609 m^3}

Now, as per the reaction, for one mole (or m^3) of SiH4, you also need 1 mole (or m^3) of O2 stoichiometrically.

However, you are providing 25% excess of O2. This means that for each mole (or m^3) of SiH4, you are feeding 1.25 moles (or m^3) of O2.

You have been given that the molar flow ratio (or volumetric flow rate ratio) for Ar and SiH4 is 5:1. this means that for each mole (or m^3) of SiH4, you are feeding 5 moles (or m^3) of Ar.

Therefore, you are feeding 0.0009524*1.25 (or 0.00119) m^3/s of O2 and 0.0009524*5 (or  0.004762) m^3/s of Ar.

Assuming 100 % conversion of SiH4 to SiO2 (as it's a combustion reaction and hence the reaction would almost go towards completion).

SiH4 is the limiting reactant (since O2 is taken in excess).

For each mole (or m^3) of SiH4 reacting, you would produce 1 mole (or m^3) of SiO2 and 2 moles of H₂.

Hence, 0.0009524 m^3/s of SiH4 gas would produce 0.0009524 m^3/s of SiO2 and 0.0019048 m^3/s of H2.

The excess O2 entering as feed would come out unreacted as there is no SiH4 left to react. Therefore the amount of O2 flowing out is 0.25*0.0009524 m^3/s (or 0.0002381 m^3/s)

Ar would not participate in the reaction and would come out as it is. Therefore flow rate of Ar coming out is 0.004762 m^3/s.

Things going in		Things coming out
Species	Flow rate (m^3/s)	Species	Flow rate (m^3/s)
SiH4(g)	0.0009524	SiH4(g)	0
O2 (g)	0.00119	O2(g)	0.0002381
Ar (g)	0.004762	Ar (g)	0.004762
SiO2 (g)	0	SiO2 (g)	0.0009524
H2(g)	0	H2(g)	0.0019048

Hope this helps....