Chemical processes. Use the atomic species balances to calculate n3 and n4. This has two chemical reactions. ## Example 4.6.3### Reaction:\[ \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_4 + \text{H}_2 \]\[ \text{C}_4\text{H}_{10} \rightarrow 2\text{C}_2\text{H}_4 \]### Diagram:A diagram illustrating a reactor is shown with arrows indicating inflow and outflow:- **Inputs:** - 100 mol of $\text{C}_4\text{H}_{10}$ and other reactants.- **Outputs:** - Mixture including $\text{C}_2\text{H}_{6}$, $\text{C}_2\text{H}_{4}$, etc.### Problem Statement:Determine the fractional yield of ethylene ($\text{C}_2\text{H}_4$).### Calculations:**Answers:**- $ n_1 = 423 $ mol $\text{C}_2\text{H}_4$- $ n_2 = 400 $ mol $\text{C}_2\text{H}_6$- $ n_3 = 162 $ mol of another substance### Notes:- Use the atomic species balance for N2 and N4.- Ensure precise calculations for optimal yield assessment.

Two chemical reactions are occurring. Given, total number of moles entering the reactorThere is…

Answered: ค Example 9.6-3 C₂H6C₂lly + H₂ сенье Иг…

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

Section: Chapter Questions

Problem 1.1P

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Chemical processes. Use the atomic species balances to calculate n3 and n4. This has two chemical reactions.

## Example 4.6.3

### Reaction:
\[ \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_4 + \text{H}_2 \]
\[ \text{C}_4\text{H}_{10} \rightarrow 2\text{C}_2\text{H}_4 \]

### Diagram:
A diagram illustrating a reactor is shown with arrows indicating inflow and outflow:
- **Inputs:**
- 100 mol of $\text{C}_4\text{H}_{10}$ and other reactants.
- **Outputs:**
- Mixture including $\text{C}_2\text{H}_{6}$, $\text{C}_2\text{H}_{4}$, etc.

### Problem Statement:
Determine the fractional yield of ethylene ($\text{C}_2\text{H}_4$).

### Calculations:

**Answers:**
- $ n_1 = 423 $ mol $\text{C}_2\text{H}_4$
- $ n_2 = 400 $ mol $\text{C}_2\text{H}_6$
- $ n_3 = 162 $ mol of another substance

### Notes:
- Use the atomic species balance for N2 and N4.
- Ensure precise calculations for optimal yield assessment.

Expert Solution

Step 1: Moles of hydrogen and methane at reactor exit

Two chemical reactions are occurring.

$C subscript 2 H subscript 6 rightwards arrow for blank of C subscript 2 H subscript 4 plus H subscript 2 C subscript 2 H subscript 6 space plus space H subscript 2 rightwards arrow for blank of 2 C H subscript 4$

Given, total number of moles entering the reactor

$F equals 100 space m o l$

There is 0.850 mol ethane per mole feed.

$F subscript C subscript 2 H subscript 6 end subscript space equals 0.85 cross times space 100 space equals space 85 space m o l$

There is n₁, n₂, n₃ & n₄ moles of ethane, ethylene, hydrogen and methane respectively at the reactor exit.

Now we will write atomic species balance for carbon (C)

$2 F subscript C subscript 2 H subscript 6 end subscript space equals space 2 n subscript 1 plus 2 n subscript 2 plus n subscript 4 space space 2 n subscript 1 plus 2 n subscript 2 plus n subscript 4 space equals space 2 cross times 85 space equals space 170 space space space space space space...... left parenthesis 1 right parenthesis$

Similarly, atomic species balance for hydrogen (H)

$6 F subscript C subscript 2 H subscript 6 end subscript space equals space 6 n subscript 1 plus 4 n subscript 2 plus 2 n subscript 3 plus space 4 n subscript 4 space space 6 n subscript 1 plus 4 n subscript 2 plus 2 n subscript 3 space end subscript plus 4 n subscript 4 space equals space 6 cross times 85 space equals space 510 space space space space space 3 n subscript 1 plus 2 n subscript 2 plus n subscript 3 plus 2 n subscript 4 space equals space 255 space space space space space space...... left parenthesis 2 right parenthesis$