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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:CVEN5331: Biological Wastewater Treatment Homework 3 1-7: ‘The following data was obtained for a reactant for the reaction 4 — B + C. Determine the order of the reaction for the reactant and the value of the reaction rate constant k. Time, t Concentration Min mg/L 0 90 10 72 20 57 40 36 60 23

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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:Uploaded ImageCVEN5331: Biological Wastewater Treatment Homework 3 1-7: ‘The following data was obtained for a reactant for the reaction 4 — B + C. Determine the order of the reaction for the reactant and the value of the reaction rate constant k. Time, t Concentration Min mg/L 0 90 10 72 20 57 40 36 60 23

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Created at: 2025-10-02 00:20:52

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# Step-by-Step Solution ## **Given Data** | Time, t (min) | Concentration, C (mg/L) | |:-------------:|:-----------------------:| | | 90 | | 10 | 72 | | 20 | 57 | | 40 | 36 | | 60 | 23 | Reaction: \[ A \rightarrow B + C \] We are to **determine:** 1. The order of the reaction with respect to A. 2. The reaction rate constant, \( k \). --- ## **Step 1: Write Integrated Rate Laws** Let's recall the integrated rate laws for different reaction orders: ### **Zero Order** \[ C = C_ - kt \] ### **First Order** \[ \ln C = \ln C_ - kt \] ### **Second Order** \[ \frac{1}{C} = \frac{1}{C_} + kt \] --- ## **Step 2: Test Reaction Order** We'll check which integrated rate law gives a straight line with the data. ### **Prepare Data Table** | t (min) | C (mg/L) | ln(C) | 1/C | |---------|----------|--------|--------| | | 90 | 4.4998 | .0111 | | 10 | 72 | 4.2767 | .0139 | | 20 | 57 | 4.0431 | .0175 | | 40 | 36 | 3.5835 | .0278 | | 60 | 23 | 3.1355 | .0435 | --- ### **Check First Order (Plot: \( \ln C \) vs \( t \))** Calculate differences: - \( \ln(90) = 4.4998 \) - \( \ln(72) = 4.2767 \) - \( \ln(57) = 4.0431 \) - \( \ln(36) = 3.5835 \) - \( \ln(23) = 3.1355 \) Check if the decrease in \( \ln C \) is linear with time: - From to 10 min: \( 4.4998 - 4.2767 = .2231 \) - From 10 to 20 min: \( 4.2767 - 4.0431 = .2336 \) - From 20 to 40 min: \( 4.0431 - 3.5835 = .4596 \) (for 20 min) - From 40 to 60 min: \( 3.5835 - 3.1355 = .448 \) (for 20 min) The changes are fairly consistent, suggesting **first-order kinetics**. --- ## **Step 3: Calculate the Rate Constant, \( k \)** For first order: \[ \ln C = \ln C_ - kt \implies k = \frac{\ln C_ - \ln C}{t} \] ### Use two points ( and 60 min): \[ k = \frac{4.4998 - 3.1355}{60} = \frac{1.3643}{60} = .0227\, \text{min}^{-1} \] ### Check with other intervals for consistency: - Between and 10 min: \[ k = \frac{4.4998-4.2767}{10} = .0223 \] - Between 10 and 20 min: \[ k = \frac{4.2767-4.0431}{10} = .0234 \] - Between 20 and 40 min: \[ k = \frac{4.0431-3.5835}{20} = .02298 \] Values are very close to each other! So, **first order** is correct and \( k \approx .023 \) min\(^{-1}\). --- ## **Final Answers** ### **1. Order of reaction:** **First order** with respect to reactant A. ### **2. Value of rate constant \( k \):** \[ k \approx .023 \ \text{min}^{-1} \] --- ### **Summary Table** | Order | Plot to Check | Linear? | Rate Constant (\( k \)) | |-------|------------------|---------|--------------------------------------| | | \( C \) vs \( t \) | No | — | | 1 | \( \ln C \) vs \( t \) | Yes | \( k = .023 \ \text{min}^{-1} \) | | 2 | \( 1/C \) vs \( t \) | No | — | If you want, you can plot \( \ln C \) vs \( t \) as a straight line to visualize!

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