## Step-by-Step Solution We are given concentration vs. time data for a reaction \( A \rightarrow B + C \). We need to determine: 1. **The order of the reaction** (with respect to \( A \)). 2. **The value of the rate constant \( k \)**. ### **Step 1: Organize the Data** | Time, \( t \) (min) | Concentration, \( [A] \) (mg/L) | |---------------------|----------------------------------| | | 90 | | 10 | 72 | | 20 | 57 | | 40 | 36 | | 60 | 23 | --- ### **Step 2: Test Possible Reaction Orders** #### **Zero-Order** - **Equation:** \( [A] = [A]_ - kt \) - Plot: \( [A] \) vs \( t \) should be a straight line if zero order. #### **First-Order** - **Equation:** \( \ln[A] = \ln[A]_ - kt \) - Plot: \( \ln[A] \) vs \( t \) should be a straight line if first order. #### **Second-Order** - **Equation:** \( \frac{1}{[A]} = \frac{1}{[A]_} + kt \) - Plot: \( 1/[A] \) vs \( t \) should be a straight line if second order. Let's calculate values for each. --- ### **Step 3: Calculate Values for Each Order** #### **First-Order Test: Compute \(\ln[A]\)** | \( t \) (min) | \( [A] \) (mg/L) | \( \ln[A] \) | |---------------|------------------|--------------| | | 90 | 4.50 | | 10 | 72 | 4.28 | | 20 | 57 | 4.04 | | 40 | 36 | 3.58 | | 60 | 23 | 3.14 | #### **Second-Order Test: Compute \(1/[A]\)** | \( t \) (min) | \( [A] \) (mg/L) | \( 1/[A] \) (L/mg) | |---------------|------------------|--------------------| | | 90 | .0111 | | 10 | 72 | .0139 | | 20 | 57 | .0175 | | 40 | 36 | .0278 | | 60 | 23 | .0435 | --- ### **Step 4: Analyze Which is Linear** Look for approximately constant differences (for zero order), or constant increments (for first or second order, based on the transformed values). - **First-Order:** Differences of \( \ln[A] \) per 10 min: - –10: 4.50 – 4.28 = .22 - 10–20: 4.28 – 4.04 = .24 - 20–40: 4.04 – 3.58 = .46 (over 20 min → .23 per 10 min) - 40–60: 3.58 – 3.14 = .44 (over 20 min → .22 per 10 min) This is quite consistent, indicating **first-order** kinetics. --- ### **Step 5: Calculate the Rate Constant \( k \)** First-order: \( \ln[A] = \ln[A]_ - kt \) Rearrange: \( k = \frac{\ln[A]_ - \ln[A]}{t} \) Let's use the first and last data points: - \( \ln[A]_ = \ln(90) = 4.50 \) - \( \ln[A]_{t=60} = \ln(23) = 3.14 \) - \( t = 60 \) min \( k = \frac{4.50 - 3.14}{60} = \frac{1.36}{60} = .0227 \, \text{min}^{-1} \) --- ## **Final Answers** ### **1. Order of Reaction:** **First Order** ### **2. Rate Constant:** **\( k = .023 \, \text{min}^{-1} \) (rounded to 2 significant digits)** --- ### **Summary Table** | Step | Result | |---------------------|--------------------------| | Reaction Order | First Order | | Rate Constant, \(k\)| \(.023 \, \text{min}^{-1}\) | --- **Let me know if you want to see the plots or need an explanation for another reaction order!**