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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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 \( k \):

\[ k = 0.24 \, \text{hr}^{-1} \]
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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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