Reaction is Irreversible in Gas Phase in Tubular Reactor. T = Constant & Ignore Pressure Drops. Feed Volumetric Flowrate = 2400 h Concentration of Y₁ = 5.0 gmol L gmol Concentration of Y₂ = 7.0% L Concentration of Inerts = 3.0 gmol L 3Y₁ +4Y₂ → 3Y3 +4Y4 Q1) Determine the limiting reactant for the given system & reaction. Q2) Fill out table Species Y₁ Y₂ Y3 Y₁ Symbol Initial Moles gmol Fjo hour 12,000 16,800 0 0 Change Final F; Concentration C; A B C D Inert | 0 Q3) Calc the concentration of Y3 & the concentration of Y4 @ at point in reactor where conversion of the limiting reactant is 60%

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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**Chemical Reaction in Tubular Reactor**

**Reaction:**
The reaction is irreversible and occurs in the gas phase within a tubular reactor.

\[ T = \text{Constant} \quad \& \quad \text{Ignore Pressure Drops.} \]

\[ 3Y_1 + 4Y_2 \rightarrow 3Y_3 + 4Y_4 \]

**Feed Conditions:**

- **Volumetric Flowrate:** \( 2400 \, \frac{L}{h} \)
- **Concentration of \( Y_1 \):** \( 5.0 \, \frac{g \, mol}{L} \)
- **Concentration of \( Y_2 \):** \( 7.0 \, \frac{g \, mol}{L} \)
- **Concentration of Inerts:** \( 3.0 \, \frac{g \, mol}{L} \)

**Questions:**

**Q1)** Determine the limiting reactant for the given system & reaction.

**Q2)** Fill out the table:

| Species | Symbol | Initial Moles \( F_{j0} \, \frac{g \, mol}{hour} \) | Change | Final \( F_j \) | Concentration \( C_j \) |
|---------|--------|------------------------------------------|--------|---------|-------------------|
| \( Y_1 \) | A      | 12,000                                   |        |         |                   |
| \( Y_2 \) | B      | 16,800                                   |        |         |                   |
| \( Y_3 \) | C      | 0                                        |        |         |                   |
| \( Y_4 \) | D      | 0                                        |        |         |                   |
| Inert   | I      | 0                                        | 0      | 0       |                   |

**Q3)** Calculate the concentration of \( Y_3 \) and the concentration of \( Y_4 \) at the point in the reactor where the conversion of the limiting reactant is 60%.
Transcribed Image Text:**Chemical Reaction in Tubular Reactor** **Reaction:** The reaction is irreversible and occurs in the gas phase within a tubular reactor. \[ T = \text{Constant} \quad \& \quad \text{Ignore Pressure Drops.} \] \[ 3Y_1 + 4Y_2 \rightarrow 3Y_3 + 4Y_4 \] **Feed Conditions:** - **Volumetric Flowrate:** \( 2400 \, \frac{L}{h} \) - **Concentration of \( Y_1 \):** \( 5.0 \, \frac{g \, mol}{L} \) - **Concentration of \( Y_2 \):** \( 7.0 \, \frac{g \, mol}{L} \) - **Concentration of Inerts:** \( 3.0 \, \frac{g \, mol}{L} \) **Questions:** **Q1)** Determine the limiting reactant for the given system & reaction. **Q2)** Fill out the table: | Species | Symbol | Initial Moles \( F_{j0} \, \frac{g \, mol}{hour} \) | Change | Final \( F_j \) | Concentration \( C_j \) | |---------|--------|------------------------------------------|--------|---------|-------------------| | \( Y_1 \) | A | 12,000 | | | | | \( Y_2 \) | B | 16,800 | | | | | \( Y_3 \) | C | 0 | | | | | \( Y_4 \) | D | 0 | | | | | Inert | I | 0 | 0 | 0 | | **Q3)** Calculate the concentration of \( Y_3 \) and the concentration of \( Y_4 \) at the point in the reactor where the conversion of the limiting reactant is 60%.
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