Let's solve each question **step-by-step**. --- # **Question 1: Determining Reaction Order and Rate Constant** ## **Step 1: Test for Zero, First, and Second Order** ### **Zero Order:** For a zero-order reaction: \[ C = C_ - kt \] Plot **C vs t**. If linear, zero order. ### **First Order:** For a first-order reaction: \[ \ln C = \ln C_ - kt \] Plot **ln(C) vs t**. If linear, first order. ### **Second Order:** For a second-order reaction: \[ \frac{1}{C} = \frac{1}{C_} + kt \] Plot **1/C vs t**. If linear, second order. ## **Step 2: Prepare Data Table** | t (min) | C (mg/L) | ln(C) | 1/C | |---------|----------|----------|-------| | | 20. | 2.996 | .050 | | 5 | 13.3 | 2.587 | .075 | | 10 | 10. | 2.303 | .100 | | 15 | 8. | 2.079 | .125 | | 20 | 6.7 | 1.902 | .149 | | 25 | 5.7 | 1.740 | .175 | | 30 | 5. | 1.609 | .200 | ## **Step 3: Check Linearity** Calculate differences (quick check): - **ln(C) drops nearly linearly with t** (difference per 5 min: about -.20 to -.17). - **1/C increases nearly linearly with t** (difference per 5 min: about +.025). But **ln(C) vs t** is slightly more linear. ## **Step 4: Determine Order** **First-order fits best.** ## **Step 5: Find Rate Constant \( k \)** Use first-order equation with two points: \[ k = \frac{\ln(C_) - \ln(C)}{t} \] Using \( t = , C_ = 20 \), \( t = 30, C = 5 \): \[ k = \frac{\ln(20) - \ln(5)}{30} = \frac{2.996 - 1.609}{30} = \frac{1.387}{30} = .0462\ \text{min}^{-1} \] --- ### **Final Answers (Q1):** - **Order:** First-order - **Rate constant:** \( k = .046\ \text{min}^{-1} \) --- # **Question 2: TCE Decay in Batch Reactor** Given: - \( C_ = 180\ \text{mg/L} \) - \( k = .21\ \text{day}^{-1} \) ## **a) Mass Balance Equation** No inflow/outflow: \[ \frac{dC}{dt} = -kC \] --- ## **b) Integrated Expression** First-order decay: \[ \frac{dC}{dt} = -kC \implies \ln C = \ln C_ - kt \] Or: \[ C = C_ e^{-kt} \] --- ## **c) Concentration After 6 Days** \[ C = 180\, e^{-.21 \times 6} \] \[ C = 180\, e^{-1.26} \approx 180 \times .284 = 51.1\ \text{mg/L} \] --- ## **d) Time to Reach 25 mg/L** \[ 25 = 180\, e^{-.21 t} \] \[ \frac{25}{180} = e^{-.21 t} \] \[ \ln \left(\frac{25}{180}\right) = -.21 t \] \[ \ln(.139) = -.21 t \] \[ -1.971 = -.21 t \] \[ t = \frac{1.971}{.21} = 9.39\ \text{days} \] --- ### **Final Answers (Q2):** a) \(\frac{dC}{dt} = -kC\) b) \(C = C_ e^{-kt}\) c) \(C = 51.1\ \text{mg/L}\) after 6 days d) \(t = 9.4\) days to drop below 25 mg/L --- # **Question 3: Mixed Tank with Two Streams and Decay** Given: - \( V = 10\ \text{m}^3 \) - \( Q_1 = 2\ \text{m}^3/\text{hr},\ C_1 = 60\ \text{mg/L} \) - \( Q_2 = 3\ \text{m}^3/\text{hr},\ C_2 = 20\ \text{mg/L} \) - \( k = .25\ \text{hr}^{-1} \) - Steady state --- ## **a) Steady-State Mass Balance** General steady-state: \[ \text{Input} = \text{Output} + \text{Decay} \] \[ Q_1 C_1 + Q_2 C_2 = (Q_1 + Q_2) C + k V C \] --- ## **b) Solve for \( C \) (Tank/Effluent Concentration)** \[ Q_1 C_1 + Q_2 C_2 = (Q_1 + Q_2 + kV) C \] \[ C = \frac{Q_1 C_1 + Q_2 C_2}{Q_1 + Q_2 + kV} \] Plug in numbers: - \( Q_1 C_1 = 2 \times 60 = 120 \) - \( Q_2 C_2 = 3 \times 20 = 60 \) - \( Q_1 + Q_2 = 5 \) - \( kV = .25 \times 10 = 2.5 \) \[ C = \frac{120 + 60}{5 + 2.5} = \frac{180}{7.5} = 24.\ \text{mg/L} \] --- ## **c) Percent Reduction vs Flow-Weighted Influent** Flow-weighted influent (no decay): \[ C_{mix} = \frac{Q_1 C_1 + Q_2 C_2}{Q_1 + Q_2} = \frac{180}{5} = 36\ \text{mg/L} \] Percent reduction: \[ \%\ \text{reduction} = \frac{36 - 24}{36} \times 100\% = \frac{12}{36} \times 100\% = 33.3\% \] --- ### **Final Answers (Q3):** a) **Mass balance:** \( Q_1 C_1 + Q_2 C_2 = (Q_1 + Q_2 + kV) C \) b) **Tank concentration:** \( C = 24.\ \text{mg/L} \) c) **Percent reduction:** \( 33.3\% \) --- **Let me know if you need any calculations explained in more detail or want to see any plots!**