Introduction to Chemical Engineering Thermodynamics
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: McGraw-Hill Education
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### Reactor Operating at Steady State - Example Problem

**Balanced Reaction Equation:**
\[1A + 5B \rightarrow 2M + 1N\]

**Problem Description:**

A reactor is operating at a steady state. The feed to the reactor contains only reactants A and B, with B being fed in excess by 14% of what is required to react with all of A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream.

All of the A fed to the reactor is consumed by the reaction.

**Objective:**

Calculate the mole fraction of M in the outlet from the reactor.

**Steps to Solve:**

1. Determine the moles of B fed to the reactor compared to A.
2. Use stoichiometry to find the amount of M produced.
3. Calculate the total moles in the outlet stream.
4. Calculate the mole fraction of M.

**Hint:**
- Remember, B is in excess by 14%, so for every mole of A, the moles of B are given by \(5 \times 1.14 = 5.70\) moles of B.

### Calculation Method:

1. Let's assume 1 mole of A is fed.
2. Moles of B fed = \(1 \times 5.7 = 5.7\) moles.
3. From the reaction, 1 mole of A produces 2 moles of M.
4. Since all A is consumed, moles of M produced = 2 moles.
5. Moles of N produced = 1 mole.
6. Total moles in the outlet stream:
   \[ \text{Total moles} = \text{moles of M} + \text{moles of N} + \text{excess B} \]

Given data:
- Moles of M = 2
- Moles of N = 1
- Excess B = \( 5.7 - 5 \) moles = 0.7 moles

Thus,
Total moles = \(2 + 1 + 0.7 = 3.7\) moles

Mole fraction of M in the outlet stream:
\[ \text{Mole fraction of M} = \frac{\text{moles of M}}{\text{total moles}} = \frac{2}{
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Transcribed Image Text:### Reactor Operating at Steady State - Example Problem **Balanced Reaction Equation:** \[1A + 5B \rightarrow 2M + 1N\] **Problem Description:** A reactor is operating at a steady state. The feed to the reactor contains only reactants A and B, with B being fed in excess by 14% of what is required to react with all of A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream. All of the A fed to the reactor is consumed by the reaction. **Objective:** Calculate the mole fraction of M in the outlet from the reactor. **Steps to Solve:** 1. Determine the moles of B fed to the reactor compared to A. 2. Use stoichiometry to find the amount of M produced. 3. Calculate the total moles in the outlet stream. 4. Calculate the mole fraction of M. **Hint:** - Remember, B is in excess by 14%, so for every mole of A, the moles of B are given by \(5 \times 1.14 = 5.70\) moles of B. ### Calculation Method: 1. Let's assume 1 mole of A is fed. 2. Moles of B fed = \(1 \times 5.7 = 5.7\) moles. 3. From the reaction, 1 mole of A produces 2 moles of M. 4. Since all A is consumed, moles of M produced = 2 moles. 5. Moles of N produced = 1 mole. 6. Total moles in the outlet stream: \[ \text{Total moles} = \text{moles of M} + \text{moles of N} + \text{excess B} \] Given data: - Moles of M = 2 - Moles of N = 1 - Excess B = \( 5.7 - 5 \) moles = 0.7 moles Thus, Total moles = \(2 + 1 + 0.7 = 3.7\) moles Mole fraction of M in the outlet stream: \[ \text{Mole fraction of M} = \frac{\text{moles of M}}{\text{total moles}} = \frac{2}{
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