Let's break down and solve the problem step by step. --- ## **Given Data** - **Reaction:** R-CN + 2 H₂O → NH₃ + R-COOH - **Order:** First order in R-CN - **Activation energy (Ea):** 46 kJ/mol - **Rate constant at 150°C (k₁):** 4.5 h⁻¹ - **Initial concentration (C₀):** 200 mg/L - **Target concentration (C):** <5 mg/L - **Pipe flow rate (Q):** 30 L/min - **Pipe diameter (d):** 15.2 cm = .152 m - **Pipe length (L):** 50 m - **Temperatures:** 150°C and 200°C --- ## **(a) Is 50 m sufficient at 150°C? What length is required at 200°C?** ### **Step 1: Calculate rate constant at 200°C using Arrhenius Equation** #### **Arrhenius Equation:** \[ k_2 = k_1 \exp\left[\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right] \] Where: - \(k_1 = 4.5\, \text{h}^{-1}\) at \(T_1 = 150^\circ \text{C} = 423\, \text{K}\) - \(E_a = 46,000\, \text{J/mol}\) - \(R = 8.314\, \text{J/mol·K}\) - \(T_2 = 200^\circ \text{C} = 473\, \text{K}\) \[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{423} - \frac{1}{473} = \frac{473-423}{423 \times 473} = \frac{50}{200079} \approx .00025 \] \[ k_2 = 4.5 \exp\left[ \frac{46000}{8.314} \times .00025 \right] \] \[ = 4.5 \exp\left[ 5532.5 \times .00025 \right] \] \[ = 4.5 \exp\left[ 1.383 \right] \] \[ = 4.5 \times 3.987 = 17.94\, \text{h}^{-1} \] --- ### **Step 2: Calculate Residence Time in Pipe** #### **Pipe Volume:** \[ \text{Volume} = \pi r^2 L \] Where \( r = \frac{.152}{2} = .076\, \text{m} \), \(L = 50\, \text{m}\) \[ \text{Volume} = \pi \times (.076)^2 \times 50 \] \[ = 3.1416 \times .005776 \times 50 \] \[ = 3.1416 \times .2888 \approx .907\, \text{m}^3 = 907\, \text{L} \] #### **Residence Time (\(t\)):** \[ t = \frac{\text{Volume}}{\text{Flow rate}} = \frac{907\, \text{L}}{30\, \text{L/min}} = 30.23\, \text{min} \] Convert to hours: \[ t = \frac{30.23}{60} = .504\, \text{h} \] --- ### **Step 3: Plug into First-Order Plug Flow Reactor (PFR) Equation** \[ \frac{C}{C_} = \exp(-kt) \] We want \(C < 5\, \text{mg/L}\): \[ \frac{5}{200} = \exp(-kt) \] \[ .0025 = \exp(-kt) \] \[ \ln(.0025) = -kt \] \[ -5.991 = -kt \] \[ kt = 5.991 \] #### **At 150°C (\(k = 4.5\, \text{h}^{-1}\)):** \[ t = \frac{5.991}{4.5} = 1.33\, \text{h} \] Available residence time is **.504 h**, which is **not sufficient**. #### **At 200°C (\(k = 17.94\, \text{h}^{-1}\)):** \[ t = \frac{5.991}{17.94} = .334\, \text{h} \] Required residence time is **.334 h** (20.04 min). --- #### **Required Pipe Length at 200°C:** \[ \text{Required Volume} = Q \times t = 30\, \text{L/min} \times 20.04\, \text{min} = 601.2\, \text{L} \] \[ \text{Length} = \frac{\text{Volume}}{\pi r^2} = \frac{601.2}{.005776 \times 100} = \frac{601.2}{5.776} \approx 104.1\, \text{m} \] Oops! Check units: \[ \text{Pipe cross-section (m}^2\text{)} = \pi \times (.076)^2 = .0181\, \text{m}^2 \] \[ \text{Length} = \frac{.6012\, \text{m}^3}{.0181\, \text{m}^2} = 33.2\, \text{m} \] --- ### **Final Answers:** #### **At 150°C:** - Required residence time: **1.33 h (~80 min)** - Pipe only provides **.504 h (~30 min)** (with 50 m), **not sufficient**. #### **At 200°C:** - Required residence time: **.334 h (20 min)** - Required length: **33.2 m** (so a 50 m pipe is sufficient) --- ## **(b) Trade-Offs** - **Larger pipe (length):** Higher **capital cost**, more material, space, installation. - **Higher temperature:** Higher **operating cost** (energy to heat), but allows for **shorter pipe**. - **Trade-off:** **Optimize** between capital (pipe size/length) and operating (energy for heating). --- ## **(c) Concentration Profile at 200°C** \[ C = C_ \exp(-kt) \] Assume \(C_ = 200\, \text{mg/L}\), \(k = 17.94\, \text{h}^{-1}\), \(t\) in hours. Sample calculation for a range of times: | Time (min) | Time (h) | C (mg/L) | |------------|----------|------------------------| | | | 200 | | 5 | .083 | \(200 \exp(-17.94 \times .083) = 200 \exp(-1.49) = 449\) | | 10 | .167 | \(200 \exp(-3.) = 99\) | | 15 | .25 | \(200 \exp(-4.49) = 22\) | | 20 | .334 | \(200 \exp(-6) = 4.98\) | ### **Plot** ```plaintext Time (min) | Concentration (mg/L) | 200 5 | 449 10 | 99 15 | 22 20 | 5 ``` --- ## **Summary Table** | Temp (°C) | k (h⁻¹) | t required (h) | Pipe length required (m) | Is 50 m sufficient? | |-----------|---------|----------------|-------------------------|---------------------| | 150 | 4.5 | 1.33 | 132 | No | | 200 | 17.94 | .334 | 33.2 | Yes | --- ## **Final Answers** ### **(a)** - **At 150°C, 50 m is not sufficient.** - **At 200°C, required length is 33.2 m, so 50 m is more than enough.** ### **(b)** - **Trade-off:** Longer pipe = higher capital cost; higher temp = higher operating cost. Must balance to find optimal design. ### **(c)** - **Concentration profile at 200°C** (see table above; plot is exponential decay). Initial 200 mg/L drops to <5 mg/L in about 20 min. --- **If you need the plot or have further questions, let me know!**