4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient?
4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient?
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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Can you please show me how to do part b step by step answer is shown but i would like to see the steps
![**Problem Statement:**
A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hours.
**Question:**
a. If the treatment reaction is zero-order, what is the decay rate?
b. If the treatment reaction is first-order, what is the decay coefficient?
**Solution:**
a. **Zero-order reaction:**
The formula for a zero-order reaction is:
\[ C = C_0 - kt \]
Given:
- \( C_0 = 100 \, \text{mg/L} \)
- \( C = 30 \, \text{mg/L} \)
- \( t = 5 \, \text{hr} \)
Substitute the values:
\[ 30 \, \text{mg/L} = 100 \, \text{mg/L} - k \times (5 \, \text{hr}) \]
Solve for \( k \):
\[ k = 14 \, \text{mg/(L}\cdot \text{hr)} \]
b. **First-order reaction:**
The formula for a first-order reaction is:
\[ C = C_0 e^{-kt} \]
Given:
- \( C_0 = 100 \, \text{mg/L} \)
- \( C = 30 \, \text{mg/L} \)
- \( t = 5 \, \text{hr} \)
Substitute the values:
\[ 30 \, \text{mg/L} = 100 \, \text{mg/L} \times e^{-k \times (5 \, \text{hr})} \]
Solve for \( k \):
\[ k = 0.24 \, \text{hr}^{-1} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1919734-3b26-4a2f-b5e0-ed6740cf625e%2F4168cdee-40f0-4745-b17e-b0cfdb9c6588%2Fluxxnlm_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hours.
**Question:**
a. If the treatment reaction is zero-order, what is the decay rate?
b. If the treatment reaction is first-order, what is the decay coefficient?
**Solution:**
a. **Zero-order reaction:**
The formula for a zero-order reaction is:
\[ C = C_0 - kt \]
Given:
- \( C_0 = 100 \, \text{mg/L} \)
- \( C = 30 \, \text{mg/L} \)
- \( t = 5 \, \text{hr} \)
Substitute the values:
\[ 30 \, \text{mg/L} = 100 \, \text{mg/L} - k \times (5 \, \text{hr}) \]
Solve for \( k \):
\[ k = 14 \, \text{mg/(L}\cdot \text{hr)} \]
b. **First-order reaction:**
The formula for a first-order reaction is:
\[ C = C_0 e^{-kt} \]
Given:
- \( C_0 = 100 \, \text{mg/L} \)
- \( C = 30 \, \text{mg/L} \)
- \( t = 5 \, \text{hr} \)
Substitute the values:
\[ 30 \, \text{mg/L} = 100 \, \text{mg/L} \times e^{-k \times (5 \, \text{hr})} \]
Solve for \( k \):
\[ k = 0.24 \, \text{hr}^{-1} \]
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