# Step-by-Step Solution ## **Given Data** - Reaction: \( A \rightarrow B \) - Rate Law: \(-r_A = k C_A\) - Rate Constant: \(k = .95\, \text{h}^{-1}\) - Single-pass conversion: \(X_{sp} = .40\) - Flash separator: - \(k_A = 3.5\) - \(k_B = .5\) - Recycle ratio (\(R\)): 4 ## **Find** 1. **Overall conversion** 2. **Reactor feed concentration** --- ## **Step 1: Material Balances and Definitions** Let: - \(F_\): Fresh feed (moles/hr) - \(F_r\): Recycle feed (moles/hr) - \(F\): Total feed to reactor (\(F = F_ + F_r\)) - Recycle ratio \(R = \frac{F_r}{F_}\) Given \(R = 4\), so \(F = F_ + 4F_ = 5F_\). ### **Single Pass Conversion** Single-pass conversion: \(X_{sp} = .40\) So, **fraction of A converted in one pass**: - A fed to reactor: \(F \cdot C_{A}\) - A unreacted after reactor: \(F \cdot C_{A} \cdot (1 - X_{sp})\) --- ## **Step 2: Flash Separator** A and B are separated after reactor. The separator splits both A and B according to partition coefficients \(k_A, k_B\): Let: - \(y_A, y_B\): Fraction of A and B in vapor phase after separator - \(x_A, x_B\): Fraction of A and B in liquid phase after separator Given partition coefficients: \[ k_A = \frac{y_A}{x_A} = 3.5, \quad k_B = \frac{y_B}{x_B} = .5 \] Let’s **assume total moles in separator as 1 for simplicity** (can be scaled later). Let \(L\) = moles in liquid, \(V\) = moles in vapor. For a separator, the material balance and partition coefficient relate \(L\) and \(V\). But since overall conversion is mainly determined by recycle, let's focus on the effect of recycle. --- ## **Step 3: Overall Conversion with Recycle** Overall conversion with recycle is given by: \[ X_{overall} = \frac{\text{moles A reacted}}{\text{moles A in fresh feed}} \] The fraction of fresh feed that is converted is: \[ X_{overall} = \frac{X_{sp}}{1 - (1-X_{sp})R} \] But this is for simple recycle with no separator. With the separator, only a fraction of unreacted A is recycled. Let \(\alpha\) be the fraction of unreacted A that is recycled. **In this problem, let's presume that all liquid phase goes to recycle (usual for flash + recycle systems), and all vapor phase is product.** The split depends on the partition coefficients, but the main calculation for recycle ratio remains. --- ## **Step 4: Reactor Feed Concentration** The **reactor feed concentration** can be found from the total flow to the reactor (\(F\)) and the fresh feed (\(F_\)), using the recycle ratio. \[ F = F_ + F_r = (1 + R)F_ \] Given \(R = 4\), \(F = 5F_\). The **reactor feed concentration**: \[ C_{A,reactor~feed} = \frac{F_ C_{A} + F_r C_{A,recycle}}{F} \] If the recycle stream is pure A (often assumed for ideal problems), then \(C_{A,recycle} = C_{A}\). So, \[ C_{A,reactor~feed} = \frac{F_ C_{A} + 4F_ C_{A}}{5F_} = C_{A} \] But if some A is lost in the separator, the concentration may be slightly less. --- ## **Step 5: Final Calculations** ### **1. Overall Conversion** With recycle, the overall conversion is: \[ X_{overall} = \frac{X_{sp}}{1 - (1-X_{sp}) \cdot \text{fraction~of~unreacted~A~that~is~recycled}} \] Given that recycle ratio is 4, and if all unreacted A is recycled, \[ \text{fraction recycled} = \frac{R}{1+R} = \frac{4}{5} = .8 \] So, \[ X_{overall} = \frac{.40}{1 - (1-.40) \times .8} = \frac{.40}{1 - .60 \times .8} = \frac{.40}{1 - .48} = \frac{.40}{.52} = .769 \] ### **2. Reactor Feed Concentration** Assuming no loss in separator (ideal case): \[ C_{A,reactor~feed} = C_{A} \] If the recycle is slightly diluted, you can use: \[ C_{A,reactor~feed} = \frac{F_ C_{A} + F_r C_{A,recycle}}{F} \] But with all unreacted A recycled and same concentration, it's just \(C_{A}\). --- ## **Final Answers** \[ \boxed{ \begin{align*} 1. &~\text{Overall Conversion} = .769~(\text{or } 76.9\%) \\ 2. &~\text{Reactor Feed Concentration} = C_{A} \end{align*} } \] If you want to consider the flash separator's effect in more detail (depending on what fraction of A is recycled), please provide more details about separator operation. But with the given data, above is the standard approach.