Absolutely! Let's focus on **part (b): Calculate the recycle ratio \( R/V \)**. I'll walk you through step by step, tutor-style. --- ### **Step 1: Understand the Problem** - **Reaction:** \( 2\text{C}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{C}_2\text{H}_4\text{O} \) - **Single-pass conversion:** 45% - **Stoichiometric feed:** Ethylene and oxygen in stoichiometric ratio. - **Some output is recycled, rest is withdrawn.** - **Asked:** Find \( \frac{R}{V} \), where - \( R \) = moles recycled - \( V \) = moles fresh feed --- ### **Step 2: Basis of Calculation** Let's **assume a basis**: Let **fresh feed (\( V \)) = 100 mol** (easy number). --- ### **Step 3: Moles Entering Reactor** Each pass, the reactor feed is \[ \text{Total feed to reactor} = V + R \] --- ### **Step 4: Amount Converted per Pass** - **Single-pass conversion: 45%** - So, **45% of material entering reactor is converted.** - Amount **reacted** = \( 0.45(V + R) \) - Amount **unreacted** = \( 0.55(V + R) \) (This is what's left after reaction) --- ### **Step 5: Recycle and Product Streams** - **The unreacted gas** splits: - **Recycle (\( R \))** - **Product (purge, \( P \))** So: \[ R + P = \text{Unreacted} = 0.55(V + R) \] --- ### **Step 6: Overall Conversion** **Fresh feed (\( V \))** is all that we care about (the rest is recycled). - **Total product formed** comes only from the converted amount of fresh feed. - All product (converted) must exit as net product/withdrawn. **Fraction of V actually converted overall?** - Since all unreacted is recycled except what is purged, in the long run, all the fresh feed is converted to product (if recycle continues forever). But, with recycle, **overall conversion is 100%** for the fresh feed, because eventually, all of it will react. --- ### **Step 7: Relate Streams** From above: - **Product withdrawn = amount reacted per pass** (since only fresh feed can contribute to net product in steady-state) - But strictly, **Product withdrawn (\( P \))** = \( 0.45(V + R) \) (amount reacted per pass) But **some unreacted is also withdrawn, balancing the system.** **But, key equation:** \[ R = 0.55(V + R) - P \] But at steady state, **amount of reactant in net product stream = V**. So, let's make a material balance on the limiting reactant (say ethylene): - **Input of ethylene to system:** V - **Output in net product stream:** All gets converted (eventually) to product. - So, **overall conversion = 100%** But let's do standard recycle ratio calculation. --- ### **Step 8: Standard Recycle Ratio Formula** For a **single-pass conversion, \( X \),** and **overall conversion = 100%**, the **recycle ratio** is: \[ \frac{R}{V} = \frac{1 - X}{X} \] --- ### **Step 9: Plug in the Numbers** Given \( X = 0.45 \): \[ \frac{R}{V} = \frac{1 - 0.45}{0.45} = \frac{0.55}{0.45} = \boxed{1.22} \] --- ### **Final Answer** **Recycle ratio \( \frac{R}{V} = 1.22 \) (rounded to two decimal places).** --- **Summary Table:** | Step | Description | Value | |------|-------------|-------| | 1 | Single-pass conversion | 45% | | 2 | Recycle ratio formula | \( \frac{R}{V} = \frac{1 - X}{X} \) | | 3 | Calculation | \( \frac{0.55}{0.45} = 1.22 \) | --- If you want to try with different conversion, just plug into the formula! Let me know if you want a worked example with actual mole numbers!

Certainly! Here's an explanation for the calculation of the streams based on the given data: Since only 45% of the ethylene reacts in a single pass, it means that 55% of the incoming gases remain unreacted after the reaction. To achieve a steady state, the unreacted gases are partially recycled back into the reactor, while the rest are withdrawn as the final product. In terms of streams: - The fresh feed of ethylene and oxygen enters the reactor. - A portion of the gases leaving the reactor, which includes unreacted ethylene and oxygen, is recycled back to the inlet. - The remaining gases, which include the reacted products and unreacted gases that are not recycled, are withdrawn as the product stream. The key point here is that the recycle stream helps increase the overall conversion of ethylene beyond the 45% achieved in a single pass. By recycling the unreacted gases, the process allows more ethylene to react over multiple passes until a desired conversion or steady state is reached. This setup ensures efficient utilization of reactants and minimizes waste, which is crucial in industrial chemical processes.